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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
|---|
签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案
6 f$ W5 j- J2 A1.5请参照本章例题,编写一个C程序,输出以下信息:
0 Z+ F6 u$ }; k1 W _ G& X, x+ cmain()6 y) j0 R2 E4 [! K3 i" b$ C
{
4 U6 @7 u. }% D6 cprintf(" ************ \n");9 r! P: \7 I- R$ U5 K- `
printf("\n");2 |' S2 ?& l6 Z. c U* p, X5 R
printf(" Very Good! \n");7 O- Z! D8 r: I# _
printf("\n");0 e' l2 `+ b/ i* \
printf(" ************\n");" P6 f. `1 E9 H2 S9 p0 J( \
}$ m) Z7 x" U0 f( V
1.6编写一个程序,输入a b c三个值,输出其中最大者。' p: g5 D5 x( x3 H0 g, W' d
解:main()
* D/ X( @3 }& `/ L{int a,b,c,max;# i% [8 v s* z
printf("请输入三个数a,b,c:\n");
/ u4 h {0 K0 T/ L3 I. Y2 w xscanf("%d,%d,%d",&a,&b,&c);& h2 k/ m; V V. _' N' A1 ^( ~5 }
max=a;) S9 I9 M1 y2 i' s( s
if(maxmax=b;. h! q! x7 Z# T3 h0 g8 d' A
if(maxmax=c;; S" C1 w' \$ X# ~2 Z
printf("最大数为:%d",max);2 T+ v& u$ ~/ B1 h5 \- I
}0 ]3 g% W" C% D u* W/ N
第三章
8 ` j6 t6 m( t/ ]/ N3.3 请将下面各数用八进制数和十六进制数表示:: F4 q# W6 S; m! Q
(1)10 (2)32 (3)75 (4)-617# i% F' U1 u' ]' W; Z+ y
(5)-111 (6)2483 (7)-28654 (8)21003
; t R4 n) Q5 z解:十 八 十六
/ n; j5 S; R" K$ ~+ s0 A; D (10)=(12)=(a)) x6 t; m% r s/ W# Z2 i) s
(32)=(40)=20
4 h4 k4 [5 U/ L" r6 O' j; W L" B5 I (75)=(113)=4b
' n! A2 u" R2 s (-617)=(176627)=fd97; h3 L# k! S8 T
-111=177621=ff91
# R4 J# ]8 w+ M2 F 2483=4663=963
7 c, t, E5 W# M% u/ t -28654=110022=9012
3 b5 r! @" U$ @ 21003=51013=520b
1 n8 S/ x! W& T; v, {3.5字符常量与字符串常量有什么区别?
; y% p* `9 Q4 X8 O解:字符常量是一个字符,用单引号括起来。字符串常量是由0个或若干个字符
& g/ [3 l! _( o5 U7 P x) B而成,用双引号把它们括起来,存储时自动在字符串最后加一个结束符号'\0'.
/ U, N! k [! _3.6写出以下程序的运行结果:8 A. j! |9 e7 g) e
#include/ c3 a: J# P" h3 ^5 \0 R/ }& L. V
void main()/ m- H2 d3 E' |7 {
{
7 r% C+ m; N( Z4 I$ N% _, `( R, W! jchar c1='a',c2='b',c3='c',c4='\101',c5='\116';
" }- L _' z" U3 W' oprintf("a%c b%c\tc%c\tabc\n",c1,c2,c3);
4 t$ U- a g+ [printf("\t\b%c %c\n",c4,c5);2 i2 ]7 E) {' Q. |4 X( {# {' Q
解:程序的运行结果为:1 ^3 z/ X" q/ t+ [8 S& ]
aabb cc abc
6 L- A+ m% F* f" M. l, N% s A N1 ?1 N! c# R) N) p Q9 T! @
3.7将"China"译成密码.密码规律:用原来的字母后面第4个字母代替原来的字母,2 q# M. c s# o/ D' j0 v
例如,字母"A"后面第4个字母是"E",用"E"代替"A".因此,"China"应译为"Glmre".% }: ?2 q2 O) y& G) L; V' {, V
请编一程序,用赋初值的议程使c1,c2,c3,c4,c5分别变成'G','1','m','r','e',并
4 N$ M+ ^/ v3 s$ I输出.
, v$ |1 v+ Y: t9 G+ `- emain(). y F" L% D+ Y- ^, @: I2 u! |: }
{char c1="C",c2="h",c3="i",c4='n',c5='a';) p# _5 X" v7 }9 y/ [; _
c1+=4;
/ J7 D m4 L, Zc2+=4;
B9 a3 ?6 J: lc3+=4;* u! N: n) [7 j) u! ?: E* g
c4+=4;: ^, f P7 `' H. n9 \# U+ k
c5+=4;" M' p. l, l+ }+ ~6 E
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);
7 ]- O# N4 m+ h" j. H1 m7 u, M6 E}) P& ]2 ~3 x0 J4 A! N3 i, [1 R: b- F! G
3.8例3.6能否改成如下:2 o6 a2 I# Y0 r
#include
8 r! J. W3 `- I4 u( K {: zvoid main()
" B+ c$ x- d9 L' ]' e{
, w+ F2 Y8 ^( q- B0 S+ i- Tint c1,c2;(原为 char c1,c2)
( j0 Z) w1 w8 j( ?; ]c1=97;1 f0 ^4 Z& V. x/ W8 X0 D* F: |
c2=98;. q% l! d3 K8 j3 i* }
printf("%c%c\n",c1,c2);
: P9 M" m2 f# {4 p) ] j0 h! hprintf("%d%d\n",c1,c2);% R) i+ q2 v9 r! a5 ]
}
8 d- \% Z0 m+ T! ?1 ]0 @0 r解:可以.因为在可输出的字符范围内,用整型和字符型作用相同.7 C) O8 V5 X8 z3 y! `( m
3.9求下面算术表达式的值., v( v: Z# e2 G
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
+ @* O, o" I, h5 n, Y2 P(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
/ J* O7 ^; r5 u- F& c! U3.10写出下面程序的运行结果:! r6 b, t& q$ K; `( k
#include- }$ ]! x; u: [# L V( s
void main()
2 q ?: w5 o+ Y( t6 @{4 W2 t$ D2 i0 g5 s
int i,j,m,n;' N" z# S( j( f% h( y: O7 U6 f# B) N
i=8;* h8 I; U1 M4 Y2 ^5 f
j=10;+ y9 ^" ]5 c5 H
m=++i;/ [% j: N; p9 |( F' C, f. ]
n=j++;; ~( O# B/ c% f# X5 B/ N, I8 g
printf("%d,%d,%d,%d\n",i,j,m,n);
. [( c! O' n6 {$ J+ p l4 H}
# [5 L% {3 H) s9 I9 A' p解:结果: 9,11,9,10
. v4 A. H; U E: v2 { D第4章" N/ S( U5 l" } }4 D
4.4.a=3,b=4,c=5,x=1.2,y=2.4,z=-3.6,u=51274,n=128765,c1='a',c2='b'.想得
! {" H7 ^/ P8 u& A; o. a& R* e# @到以下的输出格式和结果,请写出程序要求输出的结果如下:
- _4 ?, T! ]9 Y) I1 C1 ea= 3 b= 4 c= 5, e( C1 f8 E3 N, R2 |
x=1.200000,y=2.400000,z=-3.600000
5 O$ N5 t3 {2 ^x+y= 3.60 y+z=-1.20 z+x=-2.40
2 ?" ?, o3 H1 x2 |7 M vu= 51274 n= 128765/ U; K2 L4 `; U* C2 b7 y: M
c1='a' or 97(ASCII)
' G; Z/ P6 O# o) @8 n/ c$ X0 |c2='B' or 98(ASCII)+ b# a/ ^9 e, Z7 E+ I7 v$ ^
解:
* o) B# `6 X9 i" G4 o8 j# omain(); g, g! ]5 @& L- v
{- x% S9 k; W" |
int a,b,c;
) w1 {4 `+ W. Olong int u,n;
1 K9 i- C: I$ N- Rfloat x,y,z;" X" q$ X0 H* a. S$ ?& I! v
char c1,c2;7 g: z+ J% J/ n, x* K7 e, Z5 d
a=3;b=4;c=5;
% {: S2 M2 u. a& ?. E' \% Qx=1.2;y=2.4;z=-3.6;; m! J+ x& P3 C5 D' {' S
u=51274;n=128765;! n9 Q" p% ?5 D
c1='a';c2='b';
9 ?2 O2 M* _4 qprintf("\n"); ]7 m: ?( o$ `, }1 w/ [# V) d+ U) p) L
printf("a=%2d b=%2d c=%2d\n",a,b,c);; p. {3 h1 ?4 ]7 e
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
& ]/ J9 p; {' h8 L5 I, mprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);
; g2 j; [5 d8 ~printf("u=%6ld n=%9ld\n",u,n);! }; a0 [+ z* p9 z- i5 v5 W) {- i
printf("c1='%c' or %d(ASCII)\n",c1,c2);$ Y+ s2 |% J) m, P* P$ f
printf("c2='%c' or %d(ASCII)\n",c2,c2);
( ~3 c& I1 O! r2 U2 o. X; `}: {, K3 e" w" [) W J0 W# J; b
4.5请写出下面程序的输出结果.7 W( L' ]- b* G' d3 f. z6 ~5 @) [6 d
结果:: l9 }" i" p% [& Z4 h+ k, J: T8 V
57' E$ |/ W9 ^5 M' m5 J* Y
5 7& x8 l' w* m8 Q8 |, U; V
67.856400,-789.123962, g3 ~& u# ]) W
67.856400 ,-789.123962
3 q; s: Z& S% N5 u5 @- `0 n 67.86,-789.12,67.856400,-789.123962,67.856400,-789.1239624 Q% k9 Q; i- O% Z" R
6.785640e+001,-7.89e+002- |; N: V2 l9 b" Y8 D
A,65,101,41
& H1 w+ [4 C: a. [7 ?1234567,4553207,d687% r9 a# Q# g: B) X$ `$ ?2 {$ s- S
65535,17777,ffff,-1
6 `$ a8 I1 n; n, r1 C5 L) s( } X1 X& nCOMPUTER, COM1 B7 M; k3 z* o' n0 c. l7 K
4.6用下面的scanf函数输入数据,使a=3,b=7,x=8.5,y=71.82,c1='A',c2='a',
2 |, N6 g9 a0 o# |) ^0 c( I _) M2 z; ~问在键盘上如何输入?3 _. O1 a. r. R7 F) _
main()8 n, K" p; k/ J
{0 z' L$ t: W" u$ z
int a,b;
: K5 {. s1 S( h1 C; N) r5 c: h, \# d4 Rfloat x,y;
/ Q0 J/ R$ V8 S2 |char c1,c2;. G" G- V- N" M8 @- `' D* E
scanf("a=%d b=%d,&a,&b);
" h$ H* f8 c5 H5 J) z6 D* J% Hscanf(" x=%f y=%e",&x,&y);
$ w7 o1 J5 ^6 w/ H. k1 {2 mscanf(" c1=%c c2=%c",&c1,&c2);% N2 T& m! u0 Y) @3 X) @; S9 `
}
7 a) w6 X! r- g, M, X* S# G% a9 o解:可按如下方式在键盘上输入:( L$ t4 t! S. k& v
a=3 b=75 E" @) x5 Y- a: X, `
x=8.5 y=71.82. ~7 o' R# a, m2 J+ Y- S
c1=A c2=a' }8 t1 O1 @3 ~& x( p& g' n
说明:在边疆使用一个或多个scnaf函数时,第一个输入行末尾输入的"回车"被第二3 b8 e5 D. R& ~3 C& h
个scanf函数吸收,因此在第二\三个scanf函数的双引号后设一个空格以抵消上行
. @: x) I0 e; r) B) ?入的"回车".如果没有这个空格,按上面输入数据会出错,读者目前对此只留有一
; u) w4 L L& t8 [初步概念即可,以后再进一步深入理解.9 j/ q6 E S, J- w! L! X' g
4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-& h$ @& b* N' P+ V ^5 R
3.75,z=57.8,请问/ U' P( M7 K/ G9 o; n
在键盘上如何输入数据?
V" ]5 \6 R. n) \3 V* A8 f, Xscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);7 {* N% v$ @3 A, f; H
解:
8 i `" A1 E a0 j, xmain()
+ ], Y+ o( e6 m8 m" E X{7 d, k% ^0 E+ `$ H' [
int a,b;! {5 K" n5 @! R! Y0 b1 M1 P
float x,y,z;
' `. y* ]' }* G K1 }4 y6 Uchar c1,c2; _1 q, d5 S! x- ]
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);5 g( { S3 C5 n+ b
}/ h+ t4 ^9 B" ]* z# r0 A$ p
运行时输入:2 Z. _- E0 p; j
10 20Aa1.5 -3.75 +1.5,67.8
" ]* [+ L* u9 |7 v, l u+ y注解:按%5d格式的要求输入a与b时,要先键入三个空格,而后再打入10与20。%*f
2 j/ f( i# \% b5 q是用来禁止赋值的。在输入时,对应于%*f的地方,随意打入了一个数1.5,该值不
1 `7 ?* }9 r Z+ c9 S会赋给任何变量。' {6 B( s! Z x+ _9 t
4.8设圆半径r=1.5,圆柱高h=3,求圆周长,圆面积,圆球表面积,圆球体积,圆柱体积,
7 m8 |/ A) d5 C9 k3 C6 E用scanf输入数据,输出计算结果,输出时要求有文字说明,取小数点后两位数字.请编( a( r' m! k( e6 s
程.
9 ]& L! l2 W6 p+ ?+ A& D8 c# H# K解:main()' X8 `% o3 a$ v E' `4 v
{
1 G0 c2 G; x- B6 P. K6 v3 Dfloat pi,h,r,l,s,sq,vq,vz;0 T) |1 y- G) V% a T8 g
pi=3.1415926;
8 s+ M3 O( L( i2 `9 }( Yprintf("请输入圆半径r圆柱高h:\n");$ V; e; }' d9 q4 G9 j
scanf("%f,%f",&r,&h);1 P5 |! Y6 P! A# W% A' d$ m N3 B
l=2*pi*r;
5 L; i* j7 j# B& q. i2 Vs=r*r*pi;
& w9 i) k& h' [& hsq=4*pi*r*r;, @9 x0 U; ]# q# ~4 n
vq=4.0/3.0*pi*r*r*r;: ]; N# w; `" T U3 L8 S$ q
vz=pi*r*r*h;
! z5 t) j) v; _/ qprintf("圆周长为: =%6.2f\n",l);8 U1 G" V0 r8 y
printf("圆面积为: =%6.2f\n",s);3 x: E$ k' d" q
printf("圆球表面积为: =%6.2f\n",sq);' i0 C( p1 K+ T* U
printf("圆球体积为: =%6.2f\n",vz);7 _6 D1 A" n* g4 V: `& [
}% y" @; m \" h) z( f$ y2 F! v5 J
4.9输入一个华氏温度,要求输出摄氏温度,公式为C=5/9(F-32),输出要有文字说明,4 ~: o9 z/ ~2 u3 W) j
取两位小数.; ~; i0 h& P" P2 `9 q/ i
解: main()
6 Z- Y: r" y, B6 [8 H6 y0 q( U8 T{
0 D$ b: X0 S5 ^: d! Afloat c,f;3 K! r& [1 m! Q2 e1 W: `
printf("请输入一个华氏温度:\n");
( F/ ]" r, E2 t: c+ g0 gscanf("%f",&f);
8 d9 x7 F1 k& Y% v: } Rc=(5.0/9.0)*(f-32);
O! k( ]. b( ?, I+ b0 S! Cprintf("摄氏温度为:%5.2f\n",c);* \, c( x2 R9 @4 K( o
}. G1 C3 q. G( l4 [
第五章 逻辑运算和判断选取结构7 H- Y8 F& D$ c" Y
5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
7 u( |+ v. u5 \main()
; @, @# N& }; q! P{
! D9 l% f. ^% V4 h" c$ m+ i% xint a,b,c;
" p4 X& ^# d% E' Kprintf("请输入三个数:");6 [, J R, C3 K8 s; {: d
scanf("%d,%d,%d",&a,&b,&c);) Y% ~: E( V1 r
if(a if(b printf("max=%d\n",c);& f/ r8 R7 ~3 [+ }, G
else
1 ]3 ?1 |! n9 u4 K8 w Z8 L printf("max=%d\n",b);
) E. s+ |& [0 P3 m1 d+ o6 o# helse if(a printf("max=%d\n",c);
% ~( {6 V$ U U3 x- A* K y( @ else
& q$ D) {7 v) Y a3 J8 m printf("max-%d\n",a);/ `' ?5 |* z# n ~0 A: i# J
}
& S3 u6 |; w. O: z* L方法2:使用条件表达式.* {0 y d! y% N: S
main()
# J0 x: O1 W" ]9 u. M8 O{int a,b,c,termp,max;( x" y* u; i G2 v' L# K
printf(" 请输入 A,B,C: ");
( o+ R. r: V3 @! U% C% N scanf("%d,%d,%d",&a,&b,&c);
3 F8 ?; C6 G: C+ y, l* z8 s printf("A=%d,B=%d,C=%d\n",a,b,c);
4 ~, `. Z1 O2 C9 \! O temp=(a>b)?a:b;
8 j0 y5 L! P$ |9 A( f$ I max=(temp>c)? temp:c;3 J" B; u a; v+ D- R' M) |1 [
printf(" A,B,C中最大数是%d,",max);
4 K. x. J3 L; k2 E" C5 |, H}# Q& @% P/ T) I$ C* o# d0 M' p
5.5 main()
; B; h+ ]0 u+ o& @2 X" h{int x,y;
% o! ?3 G( _& ]4 o& x7 N, D/ D, Mprintf("输入x:");# N H" ^4 d( _3 E' j0 G
scanf("%d",&x);
" o' p6 {% |2 Q* K8 H8 I. vif(x<1)+ E& O. \- a% j+ ^( @$ y
{y=x;; B, T7 P/ }1 S+ o2 U2 ]/ ~
printf("X-%d,Y=X=%d \n",x,y);) E. Z* r- r6 f" A
}
5 A6 _, I6 T f# U4 x" T1 ?else if(x<10)
' a; Z+ D. t. A- N {y=2*x-1;6 Q8 K* d; Q- C7 f' y8 D
printf(" X=%d, Y=2*X-1=%d\n",x,y);
1 u _1 q/ h$ I: @ }
: l- ^( W1 I4 A8 I0 gelse5 S5 s% R5 H2 V9 k, E3 {( f, l
{y=3*x-11;2 V" K5 p: g3 A
printf("X=5d, Y=3*x-11=%d \n",x,y);% d( |+ c5 H6 |" U0 Z. ^
}3 d6 S5 \: p/ M1 f2 {1 b7 N; q
}5 A$ p8 ?. J% U. s7 f- C9 n
(习题5-6:)自己写的已经运行成功!不同的人有不同的算法,这些答案仅供参考!
' ~7 b# T3 w* u5 }+ n7 vvoid main()
; y* C/ G, x+ G/ c( l4 \( j5 T{
g! W# l$ q; ~8 f% l# H6 \float s,i;
5 z7 f5 m& b0 T: g4 uchar a;" M3 |+ j) W) n+ P% m+ B1 R
scanf("%f",&s);
; t. N. Y9 D2 X- B# Q+ c; |( ~while(s>100||s<0)
2 Q% a h, G' w& p* r{
5 w: C+ C- ]+ W. y! Hprintf("输入错误!error!");7 L' d6 y% K$ ~ G4 k3 y
scanf("%f",&s);! K1 T/ A; F! ]' X
}- {& y. F; t, z2 h( H p
i=s/10;
7 V9 b% Z# I" Pswitch((int)i)9 \% P+ C$ @8 K0 B! w3 o1 c
{
4 K5 K1 y% o$ i9 o1 ecase 10:* n' x3 M% K& Z$ x" `& w( [: ^; ~
case 9: a='A';break;
6 N% l1 P; x6 bcase 8: a='B';break;
. k9 I. o s( S. Ccase 7: a='C';break;
+ C' C: V3 {3 o) L: x% dcase 6: a='D';break;
, j3 W4 c6 D! L) k3 y+ ~/ `- p* D& Ncase 5:
2 B5 j: N; ^0 U4 m, u# w& G* Lcase 4:% M- R; x O+ F2 m7 ]& T9 F: f
case 2:: o: P- [7 o1 }* w) _( A) g
case 1:
5 `7 R8 h- J1 Y" L' Ucase 0: a='E';2 r* r% G# V; |9 u% D0 O
}: A {. d/ e1 m1 x
printf("%c",a);! j" I* F8 J- w4 t; m G! k& j
}
+ k( [( b9 S' L! w5.7给一个不多于5位的正整数,要求:1.求它是几位数2.分别打印出每一位数字3.
& Y9 a! O) {7 |% U按逆序打印出各位数字.例如原数为321,应输出123.
- t4 W9 z( h3 U2 x `& F* w9 dmain()
1 h$ ~+ {7 F; s9 m& c0 S {
& Y6 o. C4 m0 H# c/ H' E2 t long int num;8 x* A: e u9 a0 T# l, m% e7 A
int indiv,ten,hundred,housand,tenthousand,place;
, f& X) Y8 \6 Y% C8 F6 Q printf("请输入一个整数(0-99999):");
$ A& k0 H% m6 M& H) H1 [6 [9 o scanf("%ld",&num);
3 E; q2 u1 b5 K1 W if(num>9999)/ G; U$ I0 X* p1 c' E
place=5;
; g! D) g1 m! s7 velse if(num>999)
& F2 j/ [' G5 R1 E& r6 J% P place=4;5 O8 w" J4 M( l2 l9 | k
else if(num>99)
9 b# y5 Y( A, m( S+ g6 Q* j- k place=3;& }: Q& m! }, s; k2 J
else if(num>9)& \& U7 b/ V$ A
place=2;
5 x* g0 A, Q: H$ I7 M' k' _+ eelse place=1;. t( S" q% |* r- f' [- G9 G
printf("place=%d\n",place);
6 {1 b! W G n1 M9 ~1 s6 `% Cprintf("每位数字为:");' W+ W+ v' y6 q" k* H; \
ten_thousand=num/10000;
" J. |$ m. A/ W" Vthousand=(num-tenthousand*10000)/1000;
; p7 M* Z. D; a) `hundred=(num-tenthousand*10000-thousand*1000)/100;6 P' S3 f1 t3 J/ d/ n% G3 D- @! V
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;. a: ~' Y6 w# J; T" ^2 F
indiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;
! D0 r$ ?. z" [& u. p$ x6 r) pswitch(place)
7 o! R( |- \4 F{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
+ w' n7 X" E+ g* c5 p T( n6 k: O printf("\n反序数字为:");
8 Q" U0 T8 f$ A2 ~+ v1 s printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);1 [+ u8 H& S0 \! z/ n
break;4 G9 [% D: c; `9 S9 I
case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
9 E" U# x/ i8 `7 F printf("\n反序数字为:");- G: J5 V# g( \# z) A1 s
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);
! z5 w5 v1 j+ S+ d' A* V break;, ?& {: i7 [$ Z3 U* K C
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);! M. q- [7 x7 F3 G( W& W' W. c
printf("\n反序数字为:");
9 K5 S1 r# }+ T printf("%d%d%d\n",indiv,ten,hundred);+ R! R2 A0 \& k8 s* t3 ?- \. ?( S* N
case 2:printf("%d,%d\n",ten,indiv);# A6 a; t4 k# T( J+ H6 p- x( H, ^
printf("\n反序数字为:");5 H; W) L; J# I" Z
printf("%d%d\n",indiv,ten);3 a/ K6 i3 _! V% i9 B( x
case 1:printf("%d\n",indiv);2 {2 n! d- r+ C# H4 d& ~
printf("\n反序数字为:");2 G; b5 S/ a6 B5 t5 _: {# x# X
printf("%d\n",indiv);+ i: X8 C6 g+ U1 v! n, H! |
}) d, a9 n/ B% V" O: d: H9 J! |
}
% c1 |6 `" E" f( p6 X) _5.8
$ W* x- ^7 A3 V" ~$ F m9 ?1.if语句
5 }+ p! G, ^% _. d* q1 N! @ X! zmain()- h, s, e3 N* W/ p5 e+ ^* f
{long i;% j3 @( e" E* G+ Q3 H
float bonus,bon1,bon2,bon4,bon6,bon10;+ |# e* r/ U2 B3 a+ @! N
bon1=100000*0.1;& i- {0 M) c8 N& l! D
bon2=bon1+100000*0.075;
0 K4 T; @* o' [2 u( T, [7 ^; n bon4=bon2+200000*0.05;# r( b$ N( m% \- o8 E) i! a
bon6=bon4+200000*0.03;- y* R4 u! ~5 W
bon10=bon6+400000*0.015;
, T0 `3 [' Q1 y/ E/ Y scanf("%ld",&i);
% w6 n- |8 J( C if(i<=1e5)bonus=i*0.1;
' H/ x3 L7 E7 c, U1 S9 q3 B/ | r+ {- i else if(i<=2e5)bonus=bon1+(i-100000)*0.075;) k! q7 K& j8 ~8 A- h
else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
' \& ^8 E, H0 C4 C. @ else if(i<=6e5)bonus=bon4+(i-400000)*0.03;1 p2 \* l& b8 b+ t) v
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;& C# H$ U0 k% _( ?
else bonus=bon10+(i-1000000)*0.01;/ y" N; l" V& \ V! H4 T9 |
printf("bonus=%10.2f",bonus);
: x9 u2 W. F3 B) O}) Y6 [6 [- t7 Y" e
用switch语句编程序! A; q; j# ^5 |# @! \
main()
E1 S8 Z, v- s" }) f, z{long i;$ Z q) D, p; ?4 e$ H
float bonus,bon1,bon2,bon4,bon6,bon10;2 x. C c* m+ y1 U9 A
int branch;& G$ I) ^- i: `5 m
bon1=100000*0.1;4 V9 {# K3 V$ l" `
bon2=bon1+100000*0.075;7 V/ f7 s& j: l. r" U% c
bon4=bon2+200000*0.05;
6 y) e& P9 C ~" s d1 X bon6=bon4+200000*0.03;
5 r6 _4 T E4 S% i3 a: l bon10=bon6+400000*0.015;# f6 _) u: K0 a, i [" i
scanf("%ld",&i);: U& V+ }) b' D- a
branch=i/100000;, p( i1 ~- d; _) c- H: {9 \+ p1 d$ K
if(branch>10)branch=10;0 a2 `& y) \; L# n; y
switch(branch)& c }! E5 E+ Z
{case 0:bonus=i*0.1;break;5 U W; w! v7 r% }
case 1:bonus=bon1+(i-100000)*0.075;break;# j" \( \1 H+ a7 o5 n, w
case 2:/ |; R* z+ o- [7 ^& I9 L7 j
case 3:bonus=bon2+(i-200000)*0.05;break;# F9 Y' k/ X# L* Z6 |! X: W; h
case 4:
D& W- k# y# L: g8 A( J2 R case 5:bonus=bon4+(i-400000)*0.03;break;5 V8 i& e; x1 ]+ O% y* r0 J3 ]
case 6: N9 k' C H m( ~) \
case 77 L/ q0 w: q2 ]
case 8:" H* q0 e% C7 d, i- { H) A
case 9:bonus=bon6+(i-600000)*0.015;break;! t* L! j" P, o! p& }) p8 e
case 10:bonus=bon10+(i-1000000)*0.01;% e& m9 N/ k' Y( ]6 H" H; E
}% Y% z3 \1 g- ^
printf("bonus=%10.2f",bonus);; y" W' W, x1 ^6 F
}
) @0 R3 d3 H6 u, `9 U9 C( R5.9 输入四个整数,按大小顺序输出.& p) C. t: Q0 Y! i
main()
! E% |# S& s; p' q4 W {int t,a,b,c,d;
) `% N# G6 x2 H5 I+ b+ U printf("请输入四个数:");/ L# s; K8 }( m; e3 ?1 q6 P7 Q
scanf("%d,%d,%d,%d",&a,&b,&c,&d);
7 P& S- y6 P& n* q B6 K' @ printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
2 A- c5 P9 b O; m% M if(a>b)+ ~, E- D1 Y# S: v; q
{t=a;a=b;b=t;}
6 X; W6 b2 A/ u! Q1 v) @7 G if(a>c). s! @( u' u+ G: C# E; I
{t=a;a=c;c=t;}
' c6 M# {8 L' Q( V( c+ a/ j if(a>d), u& G5 u. t# J7 P
{t=a;a=d;d=t;} p- G' Y* c( n- u/ o/ g" Y0 V
if(b>c)
, m& M0 |5 u( e2 }! x* H$ c {t=b;b=c;c=t;}$ ?! s& z A3 x2 N6 {
if(b>d)
' @5 U, Z* A! O( ^0 Y. P7 z% G {t=b;b=d;d=t;}7 X, U2 t" m; w* d+ N+ x
if(c>d)3 [- H2 H; m! Q, h6 z& x
{t=c;c=d;d=t;}8 R8 [3 P* E6 J: ^ K0 V
printf("\n 排序结果如下: \n");
" W5 P4 |4 T! U& T: T. o' ]7 y& [printf(" %d %d %d %d \n",a,b,c,d);
2 H* H2 W1 A( @}
! H" y7 i/ v3 v) k- `3 V4 l5.10塔; @: D( ^; v2 I# |+ x b4 k
main(): ~4 s4 O$ \. h9 H
{/ k9 v2 |) f! Q0 C7 A4 I& t, v
int h=10;
2 a# Y5 n7 n; [) _float x,y,x0=2,y0=2,d1,d2,d3,d4;/ y K ]+ N3 w x
printf("请输入一个点(x,y):");% O/ t9 h4 P: B* x G+ _
scanf("%f,%f",&x,&y);
* S, C/ o$ }: A! c" m" M$ Md1=(x-x0)*(x-x0)+(y-y0)(y-y0);- V2 T7 r' l" V0 m7 f: [
d2=(x-x0)*(x-x0)+(y+y0)(y+y0);& f& X, F) K/ z' g% _! o" m
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
" z" D+ J9 V( wd4=(x+x0)*(x+x0)+(y+y0)*(y+y0);7 E: C9 S! p- |! P# D2 L
if(d1>1 && d2>1 && d3>1 && d4>1)( [7 `, I% R: a
h=0;
4 E$ \2 e" Z9 t- B5 wprintf("该点高度为%d",h);
) n7 R% ?- M0 H# y* ]2 T, g}
3 D. r( \) s' {! k0 e第六章 循环语句
, Q6 x) ]: U' G; |! `6.1输入两个正数,求最大公约数最小公倍数.- A: n7 d3 `7 c5 M! a
main(); T& W6 m8 I4 k1 r# N; t7 i$ w
{* V5 _4 C! s t7 F1 v
int a,b,num1,num2,temp;- d7 i" {* N `7 z
printf("请输入两个正整数:\n");
1 p8 F. q' Y& g& a: C: gscanf("%d,%d",&num1,&num2);
, P4 k7 v; C. T4 j& l" ?, Qif(num1{
/ X# \# q; V* j" Ktemp=num1;
; ]# h6 M: D" t2 V6 Z. Q3 |, n/ O9 ynum1=num2;
; ~3 W. W4 ?8 V: w% C4 Qnum2=temp;# S" P$ H+ s U" z @
}" I) Y T7 U" X1 @! i0 J7 ?3 k
a=num1,b=num2;& l! z3 ] j0 |" ]. r
while(b!=0)
3 R: ~& G& j) [% p4 I {# [: Z8 d$ ]* P3 q' B
temp=a%b;7 s1 r M2 P8 D, J
a=b;1 F `- e1 I( c) B$ W9 W
b=temp;
0 Y! D( c8 K2 ~: `% z+ y }8 E8 h0 K6 U) ]
printf("它们的最大公约数为:%d\n",a);, U" \( f( w' q4 O. G1 B9 j
printf("它们的最小公倍数为:%d\n",num1*num2/2);
& p; F+ @* K7 k4 W}( }) y) }$ b; c0 P7 n) B
6.2输入一行字符,分别统计出其中英文字母,空格,数字和其它字符的个数.
" k& f; `& H% f# z解:
: d& t- S& p3 m& d6 ~5 Q#include < >
1 Z% P3 p# i0 z) [5 K& Vmain()3 P/ n. o, ?1 Z$ e$ |+ s+ T
{
3 t( A! T0 d( Z, n. r- N/ {char c;! J+ S1 ~3 Y( W2 Y% g. Q* n. c
int letters=0,space=0,degit=0,other=0;9 J3 y. p. H; U- n4 A# H- e! c
printf("请输入一行字符:\n");0 s3 \; l8 B( j: q8 E) ?% C: t
scanf("%c",&c);' B. _- V% T- V: [1 _, t6 ^1 ^+ M
while((c=getchar())!='\n')4 P/ t2 t6 E4 d. R1 Y
{
7 `' K h. J2 V! [5 Xif(c>='a'&&c<='z'||c>'A'&&c<='Z')3 }! {, O- D1 l5 Y- t! k
letters++;
; V$ |. \" v* ^4 Z7 Qelse if(c==' ')
! {6 f- @) m$ N6 V) M& dspace++;$ k* D. V- W& `
else if(c>='0'&&c<='9')! w7 Z- X! W# _
digit++;# N/ l# B! Z! B1 u* I* a
else
. G6 b1 f) M: D2 R8 P! P& L1 Lother++;
' g! ^+ B* Y* L) x5 r3 b: o}( |5 @/ l5 z" \- I; s0 }4 ^3 F+ [
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%0 t6 e D- K0 X, k' S
d\n",letters,space,' K, X6 m* `7 I! i% L- c# ^
digit,other);
- @1 [$ h3 M `2 t# x4 D' t}2 e! I1 W! E$ E3 v8 ]" B9 o7 k. `3 Z! ^
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.
" w8 Z& W6 T$ x _解:9 ?9 i0 P" ~9 ^2 r
main()" q3 j$ ~( I/ \5 g7 S5 S& d
{
' C3 u2 S/ i) O1 Uint a,n,count=1,sn=0,tn=0;7 E$ C, O# _2 P. ?# y
printf("请输入a和n的值:\n");! R1 z7 G* Z u- Z0 m" i
scanf("%d,%d",&a,&n);
) ?+ \, o$ L- s; q8 s- bprintf("a=%d n=%d \n",a,n);
0 y0 Y$ B+ Q! f. `* E* ^while(count<=n)# J7 `$ T8 m& v) T2 U( z0 M1 E
{
9 w" a4 ]- |8 Wtn=tn+a;
/ f0 K) g4 \+ w& y! s2 h! M, nsn=sn+tn;
8 x$ T1 j7 ~( ~4 p' A, xa=a*10;6 i8 f4 H4 N' f f2 @
++count;- D% _! }0 }; u) i) d3 y/ R. h
}# m3 N! F5 X& a$ ~- S9 N0 T
printf("a+aa+aaa+…=%d\n",sn);2 |# R) T% Q3 Q, x0 y7 [. e1 U8 q+ r
}+ X- o, B2 ~& a) \
6.4 求1+2!+3!+4!+…+20!.
$ [9 v8 t# {2 Y3 `: |' ]main(); `) ^/ p: @$ J/ }
{
8 B: A) R) J( G! O2 u! ^float n,s=0,t=1;
) ?9 P4 G: \, ?1 R4 H- Ofor(n=1;n<=20;n++)
' o1 D! E5 \% }, t4 X{
# L5 _0 I2 k! b" ?t=t*n;2 |- \( `1 o( R4 J4 P& Z }, Q i
s=s+t;0 B# _( B x) s& R0 `
} N- k! I6 s. L! x' G k
printf("1!+2!+…+20!=%e\n",s);
0 V" `1 ~) D: i: M! F1 z8 z$ K}8 g' c: m* Y9 M9 r/ O7 D
6.5 main()
1 G$ Y( y K1 u* V{
$ Q1 B7 E$ C/ f% a7 M) U8 f1 ]. x' Cint N1=100,N2=50,N3=10;. J$ `6 G' F1 U3 `0 t1 |# e8 e' h
float k;! T* j! O g2 m) a* L, v
float s1=0,s2=0,s3=0;
# |1 b8 a6 Q0 ?. Y. g- bfor(k=1;k<=N1;k++)/ H' B/ u! l, q% [ @
{
! [1 Z8 K5 M9 m( i( H# Ks1=s1+k;
8 k2 Z8 E$ c3 v: s+ ^/ _}
2 Z# D7 R' G1 g# U% Lfor(k=1;k<=N2;k++)6 ]' ~& m6 j8 B- t' \+ l' E
{$ x5 r. p5 s! o' H0 a5 O/ \5 j- K
s2=s2+k*k;4 I" }0 N( X8 q7 H
}0 s/ k9 ^, h' I& A- I3 Y/ t
for(k=1;k<=N3;k++)
( j" }& T( l0 Y{
" h' Q" ]1 H6 J, Q2 V/ ` h) ts3=s3+1/k;
' r4 I, m% R- P* D0 Y, G) R8 p( g}$ _, a. q7 A! i; d& d
printf("总和=%8.2f\n",s1+s2+s3);+ v l+ l3 a, e+ z3 F. {: i( l
}- j. I2 t$ E: c- `
6.6水仙开花
% K& r( E# \* Q4 ?main()% G# o; H! E9 m3 X
{8 i5 a2 [7 k; g) }
int i,j,k,n;) V3 ~1 J) l- I, W- c. w- `/ Q
printf(" '水仙花'数是:"); @( ~& ]& G8 K1 G' G( ^# h
for(n=100;n<1000;n++)
+ W S- R, ]$ h4 {4 ]8 ^( {) W( `{; k' _6 ?6 e8 a Z I6 y: f
i=n/100;
; Q5 Q0 n3 m& M: u0 Tj=n/10-i*10; `+ c6 M, z8 k/ S
k=n%10;
. ~0 h/ C' v8 o; _if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
4 n: C! @) ?/ N{! x7 W; @2 p4 ^ Z8 F' O; d
printf("%d",n);
9 R, ~8 L6 f$ g5 Y}- h. l; O$ G. r2 A6 u" e, F
}
( \' @. j1 u$ Cprintf("\n");
* l+ V. b6 Y; b- h! D1 D}3 P C" l# o+ ?
6.7完数2 ]: `) k+ c. b+ j2 o; t
main()
5 T* o! n; B; h5 J% D! l#include M 10008 I6 ]/ ?! e- q w5 v3 M
main(). n- p' i q9 d
{) a0 J: V& X3 P& ^! ~8 x$ [
int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;
8 q0 [6 a* H6 b0 p4 f7 |3 w( Sint i,j,n,s;
) N2 k* u; g6 J& pfor(j=2;j<=M;j++)
" B7 Q% n6 `- I4 T{
/ e. x; J' z5 ~n=0;+ }- h! J& Y8 g( T3 ~! Y5 f4 o- ?
s=j;4 ]* q I* n2 |* i% s, X
for(i=1;i {
. }! z/ F+ z: zif((j%i)==0)1 b$ t' Y/ _3 x) `! S
{) o0 K5 b, q/ x! V2 r
if((j%i)==0)
. _+ O7 O8 \/ h U {
$ O$ Y- M8 j: ^- ] n++;5 S1 l) T7 v. M( k, P* P
s=s-i;
/ m, G- {5 k, b1 p switch(n)
% v, z- ?5 Z- Y4 U4 l2 T {
- S0 \8 U. r) D3 J case 1: u( P- h ~) J1 b) M: |$ t
k0=i;
3 Y( c0 p B% I+ z) ?# N break;
' }, N# D% `. Z# W case 2:9 F! ~8 I s+ u4 _" X
k1=i;9 H# O5 W" R$ K q6 ~( q E
break;. R- c8 l" O. N% w \0 a
case 3:
$ g5 j" Y A7 _3 T# v; A, v/ g' f k2=i;
8 ?6 U+ { \& m1 n break;; l( c! {9 `+ ]1 D. S) H2 M; y
case 4:
% r5 S9 }1 Z" C; X% g k3=i;1 D" @) j9 O# a% c
break;
/ G% J3 O, w6 S4 x8 a6 N case 5:
" c9 _3 T1 Q. [ k4=i;" v+ o, v0 @8 W( p( _) Q1 K
break;% W* {! h! {! x' @' M% y7 y: f2 L
case 6:
+ \" ]& d N$ w. R: w k5=i;' ~4 r1 D) |# p9 W. O
break;* t, w- ~( @6 D( |" A4 {
case 7:
% Q i5 K% m' d# n1 r, X+ \7 t k6=i;
* t; R! n$ J" _: e break;, G$ H( W! s) M I
case 8:
/ ?1 F0 Z+ p# S; J: \( x k7=i;
/ g9 \3 ?) t8 ]4 i% j1 z D break;4 |! @9 a9 U$ F" P7 x
case 9:5 B% {. [$ O/ f: B' c; r7 Z
k8=i;1 M* ^- H1 @- b* o, O
break;
! d$ n* M( Q1 n case 10:) z2 i2 ?3 `* @; h7 ^0 O4 X% L- G+ S
k9=i;
2 v b0 E8 z2 G9 P! e break;
4 r7 |& v J9 V5 E4 v) B }! D+ a1 f! S3 z) ^( N5 y# A
}
( z, g8 j6 \& r6 K }
* T: `# ~/ W( c( {! Tif(s==0)
1 G; k/ y' _9 Y2 K- c {. J0 c% Y! ?1 R
printf("%d是一个‘完数’,它的因子是",j);! D2 d- q& O ]* l8 f3 S
if(n>1)- q+ Z7 y; R p+ b/ O3 D0 `0 ~
printf("%d,%d",k0,k1); K$ u& w! S5 E/ O7 o9 h' ?
if(n>2)
0 m6 s: {& n) v& Y printf(",%d",k2);8 `' \) A- A( i5 B- G: x3 c6 j9 b
if(n>3)+ p' W; O* v) @+ w" n6 ^& B
printf(",%d",k3);
9 d4 Q$ W4 m# T1 K6 Xif(n>4)
% S* M' \4 ^! g( F. p' e F% x printf(",%d",k4);# e+ R# ~( x0 `# A8 M
if(n>5)
" j9 {" d4 t3 L5 j$ W printf(",%d",k5);
. G* v& _6 N/ Lif(n>6)0 [4 w' y( B# {9 k4 u1 v
printf(",%d",k6);
) p: l6 T) }+ |4 w; G/ Qif(n>7)2 V8 A; Y2 X# R/ F7 M1 [/ u+ Z# y' I
printf(",%d",k7);% Q1 g4 Y; l9 b
if(n>8)/ _9 R. p% X: B6 G( [3 F
printf(",%d",k8);
/ s; n. M; N; b1 s1 X6 uif(n>9)
4 G y, _/ R+ X; s& |+ k+ s3 a printf(",%d",k9);
9 V! G. B8 W7 d& p+ e& K' bprintf("\n");
0 [. P, V# r) @5 a4 p }* Y* N2 ^0 E& M- q& X2 h; a8 U
}
( Y9 X* a" e) v2 H. i方法二:此题用数组方法更为简单.
9 U6 B6 X; W/ X$ y5 tmain() i4 G/ W) @) t9 ?4 L
{
$ F, Z( U8 X4 s0 Z( J' x$ Lstatic int k[10];9 _- M1 X$ S' C* K7 i
int i,j,n,s;3 {* V" ]$ k. E# K0 t p. P1 ]1 I0 H$ v
for(j=2;j<=1000;j++)4 g9 ~% u/ H; k1 }- ]$ i" u0 R
{
+ \' t) _. F7 I! ]" Qn=-1;
( s+ U. ~: B' p4 h* Rs=j;' W e8 g% N9 R+ a* G
for(i=1;i{# D- z' l0 {, W! a4 S6 h* i
if((j%i)==0)8 b5 J) n* |6 M+ P9 x4 {& z
{
+ I3 j8 H, i8 ]5 z9 dn++;
+ ]* B" w# z: w( ]: Os=s-i;
5 G1 N, b$ o' x8 C; `3 xk[n]=i;
) i- A0 k9 ~ x6 j3 h% L8 l+ m }& v! p8 q7 t" J% n+ O1 `$ w$ U
}
2 x5 L( S- @8 p `1 L8 L& B4 Xif(s==0)
6 }0 s7 p+ U9 I{
0 c! k. N e7 a. Lprintf("%d是一个完数,它的因子是:",j);% M; O: N3 P7 T5 w
for(i=0;iprintf("%d,",k[i]);
5 R( y, @1 ]4 Lprintf("%d\n",k[n]);# N [2 Q3 V" p) w6 @
}
2 N6 B- p) a8 C8 c}
0 z" r4 d4 ~. k# }. }# y" N6.8 有一个分数序列:2/1,3/2,5/3,8/5……求出这个数列的前20项之和.
% R0 V( D n9 I解: main()
2 O Z- P7 R" ?+ Q+ ]/ S2 g{% R+ I& K3 a$ l+ Z+ h9 m! U* w/ G+ j
int n,t,number=20;
" P Y+ A; a: d3 N |7 D2 Cfloat a=2,b=1,s=0;" ~% w# r, D( @) {5 x- `/ e: Y
for(n=1;n<=number;n++); u9 v) S4 V) ?2 r
{4 @- F% o2 v' b6 m( q& F5 V# S6 w
s=s+a/b;0 s7 H0 l, b/ t* y: o+ P; T" M+ a
t=a,a=a+b,b=t;- M* v7 r) i- ]7 u# d
}/ [. g) F: d& i
printf("总和=%9.6f\n",s);
- w8 `7 F8 z( R. f0 }- g2 Z3 W}
& g2 A% ?1 \ Y* o+ O6.9球反弹问题1 n P, R8 N$ d- l& Q
main(): ?* M# \1 }. {
{2 f+ O( f) b( r! R2 n
float sn=100.0,hn=sn/2;
7 G+ i0 j. u' dint n;
4 x- t* h0 l8 s! r# ifor(n=2;n<=10;n++)
$ d9 e6 p! E7 ]( B4 \0 f& Y! u: w{
1 Q- @/ v+ q4 n2 D& msn=sn+2*hn;. U$ A4 ~* U, j! h1 d2 F% W
hn=hn/2; C1 d0 l+ h% m: G& u+ m
}
+ r$ Y- b7 o( ?" U. j" Iprintf("第10次落地时共经过%f米 \n",sn);1 \1 K" a# x: L% p# |0 `) `" ^
printf("第10次反弹%f米.\n",hn);( V1 h4 y% F$ c7 _
}0 ^7 p+ G$ _9 H1 o1 L, k
6.10猴子吃桃( M( F$ ^; E* Y
main(); Q) q ]3 N/ {# W5 u) _& w& D
{8 W# D, O' {" n) @* F! P* J9 c' |
int day,x1,x2;
8 Y' g; J. K" ^ Rday=9;
* M, }1 y6 A. {; z9 n) Fx2=1; P$ }( y/ E/ c2 S2 R
while(day>0)
3 n3 |- `. M6 T{
6 m0 |' w. |1 ~# Y; Gx1=(x2+1)*2;
% l- q. ?7 Y- A, h0 P0 Wx2=x1;
; ~6 Q5 ]6 L. ~" Z" ~2 P: F" Xday--; S. r4 v7 o" C, N2 Q
}3 K: s3 \) C. d: Z) e) ^# B
printf("桃子总数=%d\n",x1);6 e- O9 U1 [# B! u2 N* |0 \) @
}/ i: g, { o. l! q' n' o9 d
. s9 {' Q- T, r# U9 U
6.12) ~" _! w a) @- s8 J' L
#include"math.h"
3 [2 m# U+ E5 i3 p' i$ }main()
/ R9 B) ^) {1 Q, ~* T1 |1 A8 J$ D{float x,x0,f,f1;
$ t, L! S4 K6 y3 h1 X( O x=1.5;
6 \2 u4 @- _4 z7 r, A+ @3 }. ~5 p do9 r/ _9 J9 p. L; F* ?
{x0=x;& C6 o' j5 s# _6 Q- P- A* x* U/ {/ N: [
f=((2*x0-4)*x0+3)*x0-6;4 A3 L+ ?" K9 Y( O# F. w5 i
f1=(6*x0-8)*x0+3;
" W) Z& {. [- g' B* ^: e x=x0-f/f1;
p4 ^$ B9 d6 @ }: E- _. K# `/ p. m f
while(fabs(x-x0)>=1e-5);5 K( \5 x8 @3 S% V" _
printf("x=%6.2f\n",x);# A o `; r& V" I
}8 n3 `+ i6 H+ L/ l
K! n+ F4 H: g7 q$ z6.13
% V n. @, @, G8 G( _3 b3 d' `#include"math.h"% ~5 l$ ~; ^9 m- w6 t
main()
' S. L9 H+ O) N% e9 U{float x0,x1,x2,fx0,fx1,fx2;3 ~& V$ \9 s5 @" p" {/ L
do9 x/ M3 P' m! v( v! S' C
{scanf("%f,%f",&x1,&x2);* Q7 \) `" q/ {0 b+ I1 Y6 p
fx1=x1*((2*x1-4)*x1+3)-6;
0 P6 w' M( ~* \& {; f9 ` fx2=x2*((2*x2-4)*x2+3)-6;! W2 K% ]9 G0 t/ y7 Z& L* u% A
}4 G4 H- d3 N6 y" X/ n
while(fx1*fx2>0);
. `" `6 X( P1 W6 `; R do
( V- ]; y$ F' o3 }& ^; g" A. K {x0=(x1+x2)/2;% j1 W l0 d! C
fx0=x0*((2*x0-4)*x0+3)-6;
) |. T) H- H3 ~' {' s; t$ N if((fx0*fx1)<0)
2 @" C( H( F |+ I2 K. s# p {x2=x0;- a$ w9 y6 j% \
fx2=fx0;) w" z P- p1 n' X
}, x0 _- ^4 z5 }0 Z
else, {% U) i4 j" x0 @3 ^
{x1=x0;
+ F7 M+ ~2 X! P. D$ k5 B fx1=fx0;
4 R& \# J* d: Z. u; y# [ }( }5 a5 i$ Q+ j1 E1 u. n
}% a/ ^, p# e" `! X1 n8 y
while(fabs(fx0)>=1e-5);
: ]3 W# B- }, j: ?3 f1 j printf("x0=%6.2f\n",x0);* z* Y% T5 {8 i4 S; u
}2 z+ a. R5 K, N+ Q; e" e
6.14打印图案
I! o0 R! y. G6 x- f* Emain()9 E1 N' `) i8 c/ t% B
{int i,j,k;
4 {8 U4 X0 W u5 N1 }2 h: t A3 o t for(i=0;i<=3;i++)
) A: Y- `' E* W {for(j=0;j<=2-i;j++)9 y C# E) y& o! i4 {8 V6 }
printf(" ");
8 U1 m. H: K0 W7 G for(k=0;k<=2*i;k++)
- D- K" W+ ~6 ^/ S c printf("*");
3 r/ e1 M9 O/ g4 Q printf("\n");0 p3 R5 P0 U+ A$ u% |
}) r9 [: @3 p' ~, C1 z4 \
for(i=0;i<=2;i++)2 J9 P9 G, z5 \% L! V! G4 ]
{for(j=0;j<=i;j++)
, V9 L" S" k9 h# Y: `6 } printf(" ");
% y& j3 Q: n" E( K% O& O, x for(k=0;k<=4-2*i;k++)
t7 s0 L) G* V5 e8 [+ o printf("*");% G, ?$ F: }' }/ D$ o* b5 r
printf("\n");
$ a; o, I: g# d5 e9 Z2 L' l- k* s }& O) x5 d: q2 i- s% `1 G0 c, ^# r
}7 V" `3 A( j; i0 \
6.15乒乓比赛# S$ Y7 Z" C4 r% `; s& B _. t" c
main()9 a# d5 b' Y" F
{8 D" _, t$ m5 }0 O% A% _. W
char i,j,k; n& z8 |2 n: D% y
for(i='x';i<='z';i++); D/ _5 r" C" j( I" \% N* r
for(j='x';j<='z';j++)$ w4 Y" Z) }" t1 [/ D1 G
{
& H* P! A, O2 R% j: Pif(i!=j)
& x4 e) |: S9 |: K1 }for(k='x';k<='z';k++)
( w+ c1 b% f1 E8 f# }/ ?4 W {& W- \2 a7 n0 u# h
if(i!=k&&j!=k)
0 [% o% X8 P% [6 { {if(i!='x' && k!='x' && k! ='z')" _% L( D, L9 t& K2 L
printf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);! n9 b; I, s3 b4 H7 j* _
}
9 n& u. U( u6 C w/ K }# R/ B7 }) ~! ~" i* k& @
}
8 y, o# Z, z, g5 m8 |3 S( G3 w! ~}
* R0 Z2 J- ~/ u9 [& }' e7 s: cC语言设计谭浩强第三版的课后习题答案) Z1 c; V6 j- Z
7.1用筛选法求100之内的素数.6 O4 u- v5 k' E
#include
+ [$ O: |5 ^+ n* R+ g! L8 h#define N 101
+ @, Q5 p Q) J3 Rmain()
. L& S7 w6 q- ^; y* c# |" t q{int i,j,line,a[N];
. q m, R9 G7 ]( X4 Efor(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)/ M( C3 @( P6 \- N( h- U
if(a[j]%a[i]==0)' R# G! [: J: `, P
a[j]=0;
% v: [6 v, Q) m7 ~) Qprintf("\n");. ~4 A: V( o/ i7 [& K% m$ d G
for(i=2,line=0;i{ if(a[i]!=0)4 A) f* O) a: f* H
{printf("%5d",a[i]);6 m3 ]1 K6 Y: a% o* f# S
line++;
! V- |) _/ m1 T# K4 { if(line==10)
9 z1 ?% L) k( u) D7 I" N {printf("\n");8 [9 v8 G9 C7 g9 [: z
line=0;}
" D: {# p( L n9 L2 u: x9 g }& s [& E6 }) |1 J- ]6 o+ C
}' d9 F* O# o( M2 V8 N: E
7.2用选择法对10个数排序.3 Y8 u, D( w5 R, X$ F
#define N 10
. T2 K$ S1 }- imain()
/ }# z. a/ H' V; c{ int i,j,min,temp,a[N];* N% r5 {+ |& u: Z, c3 c
printf("请输入十个数:\n");" X; A; |. D- R# r. p" n
for (i=0;i{ printf("a[%d]=",i);
2 t) l. a: e8 C; b$ K- ]0 H scanf("%d",&a[i]);
# W* V3 f5 \. e$ ?1 i, X; q! W}
& B v: H" q9 Aprintf("\n");& x2 t2 Y7 k3 ~7 ~2 h5 e9 e) X
for(i=0;i printf("%5d",a[i]);( Q- {$ B4 \9 |: ]) U$ Y+ w% o
printf("\n");- g( U1 R5 h3 @$ I
for (i=0;i{ min=i;7 g! _& |- D h8 K
for(j=i+1;j if(a[min]>a[j]) min=j;! @; \) f5 D5 |- G* s& a! R
temp=a[i];
|7 ?/ k) {# p5 G" w+ S$ F a[i]=a[min];
F1 }1 i- S ?( ~. e& |: G a[min]=temp;
: {5 N0 z. H, f3 L0 n* m" p}; @1 w0 D5 H" v. F% d4 z
printf("\n排序结果如下:\n");) k' W& \6 V: F. w% h1 a
for(i=0;iprintf("%5d",a[i]);
v0 c4 J/ V+ o' t- n3 ~7 P}
, R5 _. U* A" t* t# t c7.3对角线和:
- |1 e* ]* e6 V! H* z4 ?main()
5 E" {" u. I+ w" H) Q{7 z8 L( s3 f% t$ J0 k1 Y% E
float a[3][3],sum=0;8 U/ T. Q7 D5 J
int i,j;
- V% W# O, v9 b3 o, Jprintf("请输入矩阵元素:\n");7 |6 D7 M1 M4 s }1 k
for(i=0;i<3;i++)
, e" n3 d9 @: L/ n" x7 P9 a" X2 N for(j=0;j<3;j++)# D ^. ]1 f8 `6 }: Q( M8 h+ y4 _
scanf("%f",&a[i][j]);+ A, S+ y- G' [8 d" V
for(i=0;i<3;i++)
$ o6 \# ]; R( L$ I7 ? sum=sum+a[i][i];. p! {/ l6 p3 |
printf("对角元素之和=6.2f",sum);7 V, n% z2 K* d$ R( ~0 k
}
& Y/ N9 ?: f. h; s: B7.4插入数据到数组
2 ~( W' h, |8 S/ Rmain()7 I) q; M# k. j' `$ \, |' E
{int a[11]={1,4,6,9,13,16,19,28,40,100};% Q: H: U2 ?) b7 q: i+ a7 u, E
int temp1,temp2,number,end,i,j;
8 x& ]* {& |& X0 O, r. U# {printf("初始数组如下:");
5 b* o1 J I8 m7 Y; g2 wfor (i=0;i<10;i++)5 C% b; w, i; C
printf("%5d",a[i]);: M2 O9 ]! q. b# O7 j
printf("\n"); w6 C" t v% Z6 a3 d6 t* u( t& o
printf("输入插入数据:");
4 _" r! J3 u- Y* ~scanf("%d",&number);7 T, y( d7 [/ n& R7 X
end=a[9];5 V" I" S- C" a2 g* i Q
if(number>end); y! e8 r; `) A7 F6 i5 ~5 G+ b2 R! a1 q
a[10]=number;% m% X1 _5 O6 w4 y( i
else
L, V7 x9 E- f0 u3 o9 ^" q8 u9 O {for(i=0;i<10;i++); `$ }6 Y: x5 E! S- r+ q0 F$ T7 q
{ if(a[i]>number)
& f! y4 q! S0 W0 X" j" L. K9 T% ~ {temp1=a[i];
) E/ c4 l! V' T- M a[i]=number;+ h# U# Q0 T" z
for(j=i+1;j<11;j++)9 n% \7 l% i5 e+ o
{temp2=a[j];
- \0 V) c4 n$ ]6 V a[j]=temp1;
0 ^3 Q4 w( k+ `4 e! o2 o8 R9 N- j temp1=temp2;
4 n: T5 o3 q& m' B! v" h }% H w: [% B4 E9 q8 }! w
break;
+ ^) R m5 S' Y( }- I) ^ }
6 H6 e. {, e. x9 _" n0 n$ F: {! B }
* g; x) |4 {, f3 P }
7 J. Z w. _) Y0 W! T; R! } for(i=0;j<11;i++)5 ]4 G. G' V; a( n9 V* [
printf("a%6d",a[i]);
% ]' [# D9 E7 [# ?# H! B" Y* D& E}
- @/ ]' `3 O: X$ `' B7.5将一个数组逆序存放。, T3 k6 M8 g0 p
#define N 5
6 t/ \. F$ M, Q1 z/ w+ Xmain()
: K# S* a3 `6 R+ b# ^9 B8 w n1 ^{ int a[N]={8,6,5,4,1},i,temp;& ]7 }% K, ?' K; K
printf("\n 初始数组:\n");0 s$ G! D; X0 @; K v
for(i=0;iprintf("%4d",a[i]);
2 C( U- j/ o2 S5 A+ j7 `for(i=0;i{ temp=a[i];+ D8 t/ o6 @ V w
a[i]=a[N-i-1];; p) h7 p% e# M' z" B: f, q
a[N-i-1]=temp;$ E/ l0 @- M" C, g0 z( T
}
9 B4 _3 f( D" T+ p" v7 D* N' Aprintf("\n 交换后的数组:\n");
& u1 \7 ?$ Y& q, wfor(i=0;i printf("%4d",a[i]);2 [1 u, M u2 x# q9 e) k% ~/ n
}- Y% H1 r- s4 W9 I
7.6杨辉三角
# g3 Y$ o7 ]8 O; t4 l @& I/ U#define N 11
0 A# r1 `& j5 }+ nmain(), t1 U5 l& ?3 @; b
{ int i,j,a[N][N];
# d/ c% s; F# B# R" A for(i=1;i {a[i][i]=1;( Z) }" B% C% y- I# @5 |
a[i][1]=1;+ I) @/ E3 C7 O6 O& s
}
7 H/ p3 I2 T+ P+ {0 `3 |( C2 H for(i=3;i for(j=2;j<=i-1;j++)5 |; L9 |# i9 G. B% J
a[i][j]=a[i01][j-1]+a[i-1][j];9 r! l1 k1 t+ r, H
for(i=1;i { for(j=1;j<=i;j++)% F2 w0 T I/ b- K& O
printf("%6d",a[i][j];2 w! e0 p1 J( N, \" v
printf("\n");3 o# ?0 X2 U: E) y
}4 Z% N& x6 l& _! c( @5 E: e
printf("\n");
F) V' M& L' f9 J |! w}
' X# O9 O8 B1 L9 i& E; }/ ~7.8鞍点9 z# [& M( P3 T @
#define N 10. D1 z1 z* A0 i8 \
#define M 10
) b" O6 F' Y8 I% O. Imain()
1 P: s: o$ Z$ o0 J{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;, j7 `1 ]# ]. y4 M/ c, {
printf("\n输入行数n:");; P7 P6 p' R( U4 w' u( B
scanf("%d",&n);
6 d) F5 X" x* g3 @# l" [$ o# x printf("\n输入列数m:");% F8 w% B3 ~1 H3 @3 V5 z9 ?, f5 N. A
scanf("%d",&m);6 c$ P8 i6 s; V3 d T
+ \+ i2 \% J+ v) a+ m- l
for(i=0;i { printf("第%d行?\n",i);: x9 b+ u' m! t" p6 }
for(j=0;j scanf("%d",&a[i][j];
' A' Y" J8 _- E9 L) _! a$ w } V' t6 x, M+ m9 p4 q
for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
- v! ]' ~7 A, B pritf("\n");. v, d2 L( a- o2 ?! v! C
}* P1 \& J$ w* h# X, |
flag2=0;4 i# |# c8 {( p$ X* Z
for(i=0;i { max=a[i][0];% q5 p1 K3 q- C4 P
for(j=0;j if(a[i][j]>max)% _+ H% _% L2 y! L: o& ]0 o# U$ z* W
{ max=a[i][j];
4 s& t; H7 |- S$ r maxj=j;
! f+ d. ]+ K% H' H) I }9 a. y( g2 B" X
for (k=0,flag1=1;k if(max>a[k][max])
6 R* Z- h5 V7 o, \ flag1=0;" m1 {/ R0 ~9 ]$ M- b: P6 B
if(flag1)
7 O5 a4 N2 z. @/ { { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);4 L3 O1 K% \3 g# h- v) }/ t
flag2=1;
. D# t) _- L z) f$ h }. c7 ^" h; o5 H0 l3 ~
}: m/ E1 ]* p3 u
if(!flag2)1 i! T' i. d) e9 O7 v1 C& Y$ L8 X
printf("\n 矩阵中无鞍点! \n");
; S% a z5 I8 ?0 p0 v}2 r# S! {1 Y( f! J9 Y
% Y4 c! n% u/ V" e8 W: t
7.9变量说明:top,bott:查找区间两端点的下标;loca:查找成功与否的开关变量.# I: g& A/ k7 e7 W$ S
#include- B2 ~1 b. ~6 O. W" n1 K
#define N 15: I* Z2 P b7 p" K
main()
8 f9 Y+ ]! u# @, Y8 W# k% ^{ int i,j,number,top,bott,min,loca,a[N],flag;
6 K! \7 m- e) ^. v! J3 L( \( u2 j char c;) u6 w- s% ~+ C
printf("输入15个数(a[i]>[i-1])\n);
1 R/ c+ ^ e1 E' m) ^ scanf("%d",&a[0]);3 n, J' ~$ Z* A0 n7 U9 k, \/ T. @7 h
i=1;" t4 W; a7 W$ K( i! G" Z' t
while(i { scanf("%d",&a[i]);, T/ M7 l- g$ B5 F. |" Y
if(a[i]>=a[i-1])3 Y2 p8 Z# D, {' S& [( H8 b
i++;; s0 z+ l- a6 U( b; k7 ~+ D. S) T
esle3 N" W8 D0 D( B# y
{printf("请重输入a[i]");0 V5 h, G& [; Q# `" I
printf("必须大于%d\n",a[i-1]);" _" J. T5 Z& b8 r2 s
}
/ V0 h- |; b5 u6 j2 ~3 k }
' G& L* e: \; n% w0 ]% Y+ O& ?" j" M printf("\n");
% W& W' H) f; e% B for(i=0;i printf("%4d",a[i]);' [ \. a" b7 p5 ~, D/ F0 b) q+ n
printf("\n");
* C/ n7 `9 M; } C6 h3 H
; v% t, C1 a8 G! k+ I flag=1;7 ~# ^" c( a) g! |. }5 P! r
while(flag)) O7 G$ y2 Q2 s1 @+ j# G$ j3 y
{5 Q; V* Q% D" U2 x+ f8 {
printf("请输入查找数据:");# z7 R1 r* q T. _) u4 g" C! e, T7 A
scanf("%d",&number);
7 A2 G+ |3 b. r% @' @; Q& @: E loca=0;
$ E) v4 O! w& @% i7 c5 W top=0;
6 F* g E6 v0 i( h4 R bott=N-1;. ]+ D; N* N) o
if((numbera[N-1]))
. T' y! ]- k' m- G loca=-1;& j$ c% g+ {' ?8 p3 B3 Q: `3 H' g
while((loca==0)&&(top<=bott))& k4 f- l3 z$ R% o3 n$ V; E m
{ min=(bott+top)/2;
3 A: [* H1 F- M' u' x# m if(number==a[min])1 n o0 W7 e# B5 `) Y1 h/ ]
{ loca=min;
$ H4 t+ o c: ]9 v4 m2 G& [9 N printf("%d位于表中第%d个数\n",number,loca+1);& G! q; X8 Q' X! d0 C$ F
}
7 @8 ~5 C! L/ M: w, _$ G! O else if(number bott=min-1;
4 x- s. o6 G2 T1 t7 ? else4 V- v( O" [0 F" a
top=min+1;. {/ D) J, m( X( x0 E
}1 Z9 h! t) G2 |5 C
if(loca==0||loca==-1)( s2 |: X; z" s, z% u& [! n
printf("%d不在表中\n",number);2 e3 _ O2 q5 V! o; g9 C. l
printf("是否继续查找?Y/N!\n");
* n/ q9 [- b7 R. t! J c=getchar();
' Y! G( X! k, {+ m1 p2 B if(c=='N'||c=='n'), L+ \$ {: t7 ]! j
flag=0;' b% C; y4 t& h9 {, }3 O
}
; u0 W- U1 b+ n. l3 i0 \}& _ [1 S9 k2 r( H+ M
5 ], r6 L: M* N4 l! j7.10
- O$ j l' I2 T ^1 {main()7 r' L' p, \1 c) M' I0 n
{ int i,j,uppn,lown,dign,span,othn;
- @$ c6 H8 J4 F7 D7 X/ n char text[3][80];- ]8 T; {2 @$ z6 A' q+ h
uppn=lown=dign=span=othn=0;
4 a2 K2 n# C2 W5 ]" B1 D for(i=0;i<3;i++)% B& ~+ T7 ] w0 _! X
{ printf("\n请输入第%d行:\n",i);
( K! F! z- I$ |0 p6 Z o% n gets(text[i]);
: E3 Q) A. I3 G5 B for(j=0;j<80 && text[i][j]!='\0';j++)6 B! t8 l& V8 t* p
{if(text[i][j]>='A' && text[i][j]<='Z')
3 `7 `% L# o! X, d% [ uppn+=1;
% I. W. N' V, _ else if(text[i][j]>='a' && text[i][j]<='z')
* T2 S( [" }/ m4 p8 W lown+=1;& F& f& S' l( X' ^! G0 J- n
else if(text[i][j]>='1' && text[i][j]<='9')9 g9 N1 D* |& H
dign+=1;& w2 Y+ ?9 P9 D* u
else if(text[i][j]=' ')6 ]4 H. I; g( o% } T2 V; X4 H) t
span+=1;
1 R6 b+ d% V% L% i7 Y% x/ l) D else
! ~2 S6 N* ?0 g+ t% H+ o othn+=1;
( x" @8 @* x Z# m }
# ^9 a, K. k' ]( l8 Q }
/ g e" ?% T8 X2 { g/ B8 V for(i=0;i<3;i++)' n' V0 M. `# d4 [9 U
printf("%s=n",text[i]);
' U) ]$ v/ F* c/ F printf("大写字母数:%d\n",uppn);! k* e' b% D8 `
printf("小写字母数:%d\n",lown);
* |3 f- r$ }& }2 \" S printf("数字个数:%d\n",dign);" y0 _: |/ o, ^- _
printf("空格个数:%d\n",span);6 I! K' D% K+ W( _' ]6 U4 F/ c# p
printf("其它字符:%d\n",othn);7 d# g8 `, b! K/ k+ J7 Q" r. H, y4 _
}3 b. o3 l+ b; ]: j4 m1 `# B1 `7 `
2 U" R% q7 @( p" j3 C3 K
1 T5 W8 A% S2 k' E! K5 ]7 V4 z
7.11/ N. \1 e( K/ j8 h$ z) L- Q
main()* V3 R2 n9 E) n2 W- }% U3 N# r0 B
{static char a[5]={'*','*','*','*','*'};
# F# I, \. s8 l8 J B+ j( H# O7 l int i,j,k;
) S; H+ p2 R: r1 C( B# `( L char space=' ';
. n; ^6 t! X% h& h3 f/ c0 z for(i=0;i<=5;i++)% e5 y% E6 [) }9 P
{printf("\n");
% |1 W% F. e! s* W; y! [ for(j=1;j<=3*i;j++)7 v) Q2 _1 m( e) x b0 i) Q T
printf("%lc",space);& p$ x0 k1 c$ g$ A
for(k=0;k<=5;k++)
3 {4 s- h- ~% L; B printf("%3c",a[k];' V' i- u+ u+ J
}% z" E4 Q6 r1 Y3 c8 j' O" Q- n& F$ |
}
% F* E( q9 _( l9 l7.12! Y0 v, U t/ z5 j( I7 k/ ~6 g- O, O
#include
' b6 l: S# N% K9 U- l! Tmain()4 b6 M$ m6 {: J0 e3 X
{int i,n;5 G5 M3 S& X; j; P4 R5 h
char ch[80],tran[80];. S- f2 _% e: ^: o9 Q7 X
printf("请输入字符:");
+ `" a# H& f( ~' \2 M gets(ch);; Q' q+ A" m' m! _7 n: Q& S
printf("\n密码是%c",ch);3 X& _, B7 e2 u1 y& J
i=0;
, N7 z( m L' ?9 H; r' m3 `( Cwhile(ch[i]!='\0')
0 S8 ?1 P8 W, D' ]$ s( J. J( |% l{if((ch[i]>='A')&&(ch[i]<='Z')); Z+ S2 S9 F c
tran[i]=26+64-ch[i]+1+64;
8 w `" f' X/ F4 |, G5 qelse if((ch[i]>='a')&&(ch[i]<='z'))' O7 p2 B0 K* l
tran[i]=26+96-ch[i]+1+96;; s+ T) s& h) S+ l0 N
else
4 g; Q5 q' a, }9 }7 l) d/ Z tran[i]=ch[i];
' L) o4 C( B2 T6 R i++;
( u2 L. {" n. d4 v% _}. O$ d( I" T8 g- h+ c$ c+ n. r% o
n=i;
5 |7 K. d; _( u7 M: f* oprintf("\n原文是:");; N% ~3 y" Z' P3 H; N; {
for(i=0;iputchar(tran[i]);, ]8 Y; j! m; {7 l" k
}+ s) h# I* A2 l! K
7.13
& H4 E i! e% I- i! C0 s# d) Y1 Tmain()# h3 {) [5 U- a6 d. O% r
{4 V3 M7 h; \4 P5 S8 v
char s1[80],s2[40];
. X0 v" f4 `: L- }1 { int i=0,j=0;
9 c2 z; w: \! x/ L3 F" z$ { printf("\n请输入字符串1:"); d7 h7 P% v. G+ T5 q- a9 C
scanf("%s",s1);1 j% z& G2 l, K1 r+ }, k' G! |
printf("\n请输入字符串2:");5 Y) x2 E( ~+ @3 L+ _3 H
scanf("%s",s2);4 V7 y+ }, i) a! M! G3 B
while(s1[i]!='\0')3 @, H8 h: h: w
i++;
# s. i) f& Z: y. m4 owhile(s2[j]!='\0')
. z n; `# l! b b; q. ~3 m s1[i++]=s2[j++];$ K- P0 R2 {( p
s1[i]='\0';# b1 b L: n# ~& w$ {+ b& I. o) c/ \
printf("\n连接后字符串为:%s",s1);
8 }6 h' [4 E6 Q. W: `. I) G }" Q0 @2 K8 M O; G
- p* w. S M- R2 s5 ~# x( |3 Y0 C3 g
7.14* b$ G6 R/ ^- U9 j! i
#include
! M8 y3 e: p# gmain()
& }4 m5 \5 e0 V0 x- g{int i,resu;
) A6 A4 i* v4 L# N char s1[100],s2[100];. }+ B4 I: C- M7 s0 S( t4 L
printf("请输入字符串1:\n");, N1 \7 Y6 q S# |8 F
gets(s1);0 b0 c: F, N$ O* K" b# p, {/ j
printf("\n 请输入字符串2:\n");
: u3 s* q4 G2 y; i! T& j* d( z# x gets(s2);; w+ i5 H' H5 \
i=0;
! N; g* P! d5 q& ]8 J while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;- J; Z$ V: j5 ^( x2 G' n; I
if(s1[i]=='\0' && s2[i]=='\0')resu=0;
* p6 k2 g" j7 W4 C { else
5 L1 G# D) Y z9 ^: \ resu=s1[i]-s2[i];) B6 o2 x+ \! F! i* x3 z
printf(" %s与%s比较结果是%d",s1,s2,resu);
) q# c! o; R8 V5 _}
, L2 T: l5 ^$ M7.15
2 o% d# Z: s& s#include; B! T1 \' b6 U& v
main()& C* h$ {8 ^7 ~5 M3 g, H+ q2 y
{
: z/ f3 v" C. e- d! F" }) t) s H char from[80],to[80];/ A+ s. g0 _1 M$ R
int i;
* q1 y- V8 Y. i, T0 C$ U" V- E printf("请输入字符串");( ?& j. \2 m$ g7 ~, O* i
scanf("%s",from);
5 o" Y- i% R7 S for(i=0;i<=strlen(from);i++)/ H6 ?0 _7 { W+ K! p# I# S
to[i]=from[i];
9 e, P" f. ?" l printf("复制字符串为:%s\n",to);
; C8 T/ C: O9 V* v: D$ U A- Q1 a }" j4 T9 R- P7 w( y e) x
: @0 V. Y2 H$ T9 E
$ P# \2 F7 j, Y+ b! p$ s第八章 函数
$ R2 J2 l0 E8 X5 I3 [! [8.1(最小公倍数=u*v/最大公约数.)
7 l4 m$ T8 G* f( F( Vhcf(u,v)
' X0 V% D+ W! G3 s0 ?* h6 f& ^int u,v;
, S# g: T- m+ O3 a0 h(int a,b,t,r;. c# |1 o# Q' k, } G4 h
if(u>v)' T3 `, j( x J; _( p
{t=u;u=v;v=t;}
, m1 k* @% R& P' C a=u;b=v;5 p/ u% a3 |4 L) X
while((r=b%a)!=0)
) C' ^! e8 D8 \9 @1 M {b=a;a=r;}
1 \ q( f" l9 _! T: B return(a);8 M4 x- K# y% W; n/ N
}
+ ~4 k/ M! _9 ? lcd(u,v,h)
1 U/ e% X/ J. y. c1 o | int u,v,h;
0 c! s" {& g6 I {int u,v,h,l;
J# r% P$ Q: Q scanf("%d,%d",&u,&v);
- ^8 U# e' L+ b# h6 J h=hcf(u,v);
0 E" g7 U1 {0 f/ G: w printf("H.C.F=%d\n",h);/ b! g' j# H7 f3 g% I! L
l=lcd(u,v,h);
: P6 z% M% t" i- R9 x printf("L.C.d=%d\n",l);
* Q8 D6 |3 j9 O: T* D }
5 }: P! E3 a K' r) a4 G; C {return(u*v/h);}" K% n1 ]. A- j4 [+ w
main()
) X8 Z+ [6 }% R {int u,v,h,l;
5 A$ [' _- H' u' p! P scanf("%d,%d",&u,&v);
; h6 G5 D& F% s: h) y' Z h=hcf(u,v);. t2 E U: q& T, @3 i3 u
printf("H.C.F=%d\n",h); ~6 U0 V3 l3 ?" T2 ]3 K
l=lcd(u,v,h);
* w5 w1 F4 r' f; o1 E) } printf("L.C.D=%d\n",l);/ N( e0 c) d4 y" d( v5 {4 k
}: p$ I* h. ]4 ]" q( T6 H
5 X( [* Q2 P0 K4 B2 i/ s' e" `. T% Z2 V- Z8 m5 ~3 o0 _3 q
6 M5 g; d/ w; w; \" L& b8.2求方程根
, b- H- r. @2 S& H& U) t#include
3 i7 e) [, S. t' r5 b z, Xfloat x1,x2,disc,p,q;
6 ~5 J3 G, Z! Z. p/ V" u( ~/ ?( e9 Egreater_than_zero(a,b)
, J" M# O( L% v; e: B% f1 x( Pfloat a,b;
, _3 N+ t2 q4 u; F1 p- ^{2 E( c/ f9 A+ p4 ?7 `" ]% A
x1=(-b+sqrt(disc))/(2*a); H" C$ q8 j1 i% Z" b$ ~
x2=(-b-sqrt(disc))/(2*a);, d, t( Y- W2 N; w' \' d% P. o
}) `) U5 S* F" ~1 J: \: v; W
equal_to_zero(a,b)
7 a# b" h* `; L4 T% }1 @float a,b;4 K. M( G8 w& Z, @
{x1=x2=(-b)/(2*a);}# l7 {5 b6 x4 }+ |
smaller_than_zero(a,b)' I3 C2 e: c) {+ y8 i- t5 v( H
float a,b;
+ W0 `% \& w Q+ m( j{p=-b/(2*a);: {2 b: r' u! M8 z! A! \
q=sqrt(disc)/(2*a);
1 s Q/ V1 K7 l}
0 z H/ p: r+ `' n a/ f0 omain()
5 C* M0 K8 Q) c" @4 V& C! b3 f{3 I) ?0 ?5 ~* V7 l) U
float a,b,c;
3 h3 c( }: B( Z S+ x$ R; Xprintf("\n输入方程的系数a,b,c:\n");
* H0 l8 }. @1 fscanf("%f,%f,%f",&a,&b,&c);
/ w0 S. I% e; h* L, d$ d2 q" A0 @printf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);6 g4 }! p, s/ l2 r4 a* x; Q3 _
disc=b*b-4*a*c;
1 s* X5 r8 X$ c. F. z; b3 P7 L. yprintf("方程的解是:\n");3 S0 a p! x4 d6 [+ x
if(disc>0)" h+ i& M& o0 S: G3 U. H* |0 Y
{great_than_zero(a,b);
" T$ p2 ^0 p( J W2 E' kprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);* F- W* N- ~ `0 i5 K+ v* O
}
5 N7 }5 _% i, B) zelse if(disc==0)
$ I0 x- R8 _: c, p. V {
0 G8 h2 G+ t7 a ^' }zero(a,b);* P! m: c/ g8 I a2 O K% i
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);4 T* h y" a: F' }& ^
}" d0 N; b; K7 w- R7 K% V6 E
else1 j l. K/ v1 m% L+ W
{
$ ?) c! Z' [( R* i# N) h small_than_zero(a,b,c);! O( q4 s/ a7 a7 o4 r
printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);
9 C% |3 X& W1 ]) y }7 `& g# u- J2 N3 x) U- a
}+ A+ f! `' k3 `/ p
8.3素数1 x3 m4 m. \: c5 _# B
#include"math.h"
/ a4 q9 b; r) zmain()
5 F# y0 ^% t6 Y8 u+ Y7 L, ]{int number;# G# Q4 j4 v; k
scanf("%d",&number);! K' I# v' S5 A7 X4 b o! w3 m% N
if(prime(number))' c; c' Q4 _7 B; |5 B
printf("yes");
# d( E i0 G6 z" h( G else
% y2 V; y: D0 {' _. |6 ]6 r printf("no");
8 h* h' G: s6 m& N}3 W$ G7 q, N: t: B6 v6 Y; F
int prime(number)
: H) C" Z. z X |int number;
) O8 [7 \3 X+ W{int flag=1,n;
0 J7 k' j. O& I for(n=2;n if(number%n==0)/ L6 C9 z: S4 V8 O0 S: `
flag=0;
( m" r( r6 K$ c return(flag);
- s- J2 `5 z7 C}' ^5 v4 S0 k/ [
! V3 Q# p1 ^. L4 |4 s/ B: s
! ?) G" ~0 Y! E) N
9 O4 n% i- t! s8.4
8 M) U; G, l9 F4 `& D, Z# m L#define N 3
$ n4 i7 y( ^6 q) k) ?int array[N][N];
. [: l' E' \1 o6 L% u5 Cconvert(array)% q4 p: ^- k/ F( q7 P+ X
int array[3][3];
M8 `5 M2 C! e { int i,j,t;
5 V- r' }0 ]6 T+ W for(i=0;i for(j=i+1;j { t=array[i][j];! _2 G8 v/ }) l
array[i][j]=array[j][i];' ~4 Q$ `- |% G1 r7 n2 W
array[j][i]=t;6 i, f5 H2 A; D: z, v/ g
}$ n& n1 t: j+ u9 A& B, |
}& o" C8 Y/ L7 p5 [8 }; O
main()
, L7 X( S' O8 l; [( \4 R* A{
. ^, G L; m+ M4 n+ ~" m int i,j;0 G" A2 S; h+ r
printf("输入数组元素:\n");
6 m& O9 G8 i" X% I for(i=0;i for(j=0;j scanf("%d",&array[i][j]; X4 N2 W# T& Z9 ]$ A3 Q, j
printf("\n数组是:\n");
4 Z0 V8 G0 E1 o6 ufor(i=0;i { for(j=0;j printf("%5d",array[i][j]);
4 X$ y# E9 y8 o: V* d- _/ l O) M printf("\n");: C5 y! J% t6 y; E C
}
6 v4 G% U2 e' [ [) t% p, d convert(array);
9 O+ C7 a! [% }6 j' l. f$ u/ K printf("转置数组是:\n");
9 x. \2 H! w: R5 J y for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
: G" p" V5 d! O8 H printf("\n");6 K3 v6 q% [6 u. H' w
}
( |" f0 S9 Q" A8 X}1 l% b3 {$ [/ u+ l! G9 h) l0 f
3 A2 D' x2 d* d; c& v- b9 K
# `$ L, \7 }; D5 P6 F4 M- @3 J3 i" Y- L+ r( U
8.5. w; w, T( u( l3 p P! {6 W* Y2 h
main()7 T$ y& g0 ]6 T! M7 m2 y! o9 A, E
{1 m" J( c+ P/ G7 N; D% r+ o/ \7 ^
char str[100];
5 P/ c, i% D, p1 Q0 t+ o printf("输入字符串:\n");
N' ?6 ?; a m* c: L% b7 l scanf("%s",str);! w4 ?# e6 S! p- `5 J; G" z. k
inverse(str);
. Y Z' [; X# q1 h6 O' j2 V printf("转换后的字符串是: %s\n",str);) S6 j3 t6 o4 c; m
}
5 V* T5 J( I! ]) F/ Vinverse(str)
& L0 ~& I4 C3 y. Tchar str[];
. [# a" g: G3 A4 {{
. ?8 _5 }! a, E' `' J+ c char t;
1 r) H: a3 C8 j: [5 f! W int i,j;
9 _8 m& j6 k8 N9 ~* Y; o5 O for(i=0,j=strlen(str);i {
: N( ]) D- }( n; y t=str[i];; F; L8 M/ x, N( X" N1 s
str[i]=str[i-1];
$ ]* _4 H% s$ z7 [ str[i-1]=t;' D, `/ f6 O2 W( b( {/ W0 F
}3 u. p& R4 x, C4 M4 V* u4 U
}3 t% o' ~3 W* |- ]$ @
6 V: t$ i5 q$ ~% A$ }
9 d! Z$ p/ k. v& [% ]8 \; w* m- N) L1 x* L; L% a+ f' J
8.6
" g- \5 {0 O" b. U/ Z! l U4 Gchar concatenate(string1,string2,string);
& a; W1 H2 Y. \+ c+ ]char string1[],string2[],string[]; r6 U6 e% R- e2 M! W0 W
{
7 S: V; x, @' W+ jint i,j;5 I2 @3 u7 l3 U* Y5 _
for(i=0;string1[i]!='\0';i++)
# Y$ V0 ~1 H1 y7 M9 J% c) X3 P& x string[i]=string1[i];6 Q) L* Z, n; p6 o/ r% D1 z/ s& [ K
for(j=0;string2[j]!='\0';j++)
+ o& ~% \% @3 [ string[i+j]=string2[j];1 _7 p# V$ M8 d
string[i+j]='\0';, f" k% J/ r, q+ j4 }; j9 Z
}
+ n" c. R* p, O" N: i) Amain()
: s$ i/ {6 A" p0 l! N* B. H; k/ a{
7 j3 {, @/ S$ q3 m char s1[100],s2[100],s[100];) Q; f" I0 [! s0 u1 s7 p/ A
printf("\n输入字符串1:\n");5 K. v. v' i% ?0 T- b
scanf("%s",s1);0 @6 c4 G Y' F H0 ~
printf("输入字符串2:\n");% a5 |! S9 Y. F5 Y- [ Z' N7 U3 \4 J$ c
scanf("%s",s2);, a( T( y9 y. X! m; L! J
concatenate(s1,s2,s);
; ~$ ^ Z& J5 t4 B printf("连接后的字符串:%s\n",s);
: S3 N( a4 [* G& l- m}" S( _+ G1 n7 ?1 E3 P- u" F, x
. F+ S1 p ? u p3 X0 n) b
- @ H) Z7 V( G' E' f* o
8.8
' B% Z0 S6 i* ?5 i0 i" ~main()
! L3 w `% A6 I/ O" p6 D{! }. G% _$ _9 @
char str[80];5 p# s" \& R. m8 t1 ?4 A7 u T* P
printf("请输入含有四个数字的字符串:\n");) {" ]. ~; q6 a
scanf("%s",str);) I1 N: i# b& H7 F! s3 w0 u
insert(str);! Z/ `) s& O9 B$ Q5 E
}6 G' P! {6 S; p. `# ^0 e* m7 O! m) p
insert(str)4 H& O; t q7 Y: F1 Y& y* E
char str[];
% B% N4 @( q6 U( ~! Y{) U0 c6 Q3 F! x; ^3 D
int i; k! a. g- H7 R( N
for(i=strlen(str);i>0;i--)
0 Y' Z7 {4 A5 B0 g3 k { str[2*i]=str[i];2 K% M9 ?( c( A; X0 G
str[2*i-1]=' ';
/ ]1 G' t+ M8 X3 s% Z# }% P/ q# w }
5 n0 r/ \9 b2 Q8 x printf("\n 结果是:\n %s",str);8 p t9 C8 l0 \# x0 B/ |
}: _" y5 R* C3 ^! S
2 z) ]7 A+ p( \6 e1 T, U* O/ Q7 p
+ S& ]$ w) h$ M1 B2 v: b# V& h( N: o: a( o! \2 v. V
8.9
- v- f, j: w- j+ M. X#include"math.h"* x& i* {9 J/ Z$ x
int alph,digit,space,others;6 M8 _- ^/ s) n7 Z# X: X' I
main()# C+ A) ]$ w+ o
{char text[80];0 K& S! _0 g3 Q
gets(text);. d5 n: j9 A! m
alph=0,digit=0,space=0,others=0;
1 D* L! Z- G0 L3 L6 V) t count(text);
7 n% h/ g9 R, }" } printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);
; t+ B2 w3 p# a4 |' @! A8 H}
9 |1 q0 w5 r0 n" N" E+ r& Ycount(str)
/ E+ R+ b, t7 |7 Q2 S1 q5 gchar str[]; P# Q7 v7 x, c$ Q
{int i;$ z, _! O% F; f: I, U; S+ r
for(i=0;str[i]!='\0';i++), v/ Y2 Q8 V( h9 [# R) `- n
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
n- i6 B0 W: p alph++;
. W% i2 j; E2 h' J- n& `: @ else if(str[i]>='0'&&str[i]<='9'): Q3 G4 C. S$ d/ T# q/ \9 f5 u4 A
digit++;( S9 u3 r6 T9 R4 I
else if(strcmp(str[i],' ')==0)
* T) u" W8 F1 w7 ^: ~( f space++;
" R& V6 b+ @( T% z else
8 h3 K5 }7 n7 B others++;
2 O. S% P+ d" {* u1 m/ l; P}) [5 I# ]0 K. q' J
2 U; c4 F7 i) D5 e! j% C
2 O9 L. _+ y! s& Q8.10& k) j8 v- c, R v) B/ ]
int alphabetic(c);
: w2 ~! w" b9 V7 D4 N" Qchar c;
# N0 p) M2 Z+ {. S( z{& _# _2 n w! {( G' a3 K: Z
if((c>='a' && c<='z'||(c>='A' && c<='Z'))$ L" D1 s0 |& Y
return(1);* E& U7 W$ W: ]$ N
else
) c! w8 [ V+ Z% f0 w- d: |% | return(0);9 ~4 n% o* g; d1 V
}* j. \; B% q7 R- W. C! A
7 {, J8 }( N7 }! Y# O: G0 c' ` |int longest (string) a% j7 V3 F" L* o9 _+ \
char string[];0 y+ E" X- T7 T# s5 p0 L/ a
{
. T) L8 C6 }0 U& ^* c int len=0,i,length=0,flag=1,place,point;% l3 d8 n: s( t' k9 r
for(i=0;i<=strlen(string);i++)
" n7 N: O+ r; Q) l1 @, S if(alphabctic(string[i]))5 v, X; N Q9 Y, E- p# V I+ X1 W
if(flag)
( j# o8 g: T4 ?& z {' E" @8 b. Q9 X; t: [- n
point=i;3 U- s0 _5 n4 j. u+ Q3 h8 c$ Y
flag=0;
7 A3 L9 n: p& J3 j) B: f }
7 m. Y8 h/ ^* \/ Q1 J2 T- y else( x0 v9 `! R% o& ?/ A5 k! k# k$ U. j
len++;6 k' g, d) x9 r
else% t+ X8 r& @ Z: \% ^0 X
{ flag=1;
\% {2 q! |) m+ V6 g3 Y if len>length)
( ?+ Y3 n( Q* G/ x# k {length=len;8 M& p' c/ k& |: U: V. d. }- n# w/ v
place=point;! p: E3 ^9 G( n# e$ m$ H/ m
len=0;
, T7 |) Z, ~- s: u }( R' g, C; K2 B/ X9 {2 Q
}. I3 Q3 O; V! j, |3 z
return(place);! R- ^/ ~) z4 ]9 S
}
; u. T, p: l! z* _2 ]5 r/ H8 gmain()
# e# I7 }! R8 \# z{' _2 m$ G2 _1 ]9 R- }! U3 u; J9 P, ^& P
int i;) H5 J$ f9 N; K: N, J
char line[100];
- T: U+ Y" u" j; ]6 Nprintf("输入一行文本\n");
+ v6 d. g' J1 M& L {gets(line);. j8 ^. v: z, p& Y
printf("\n最长的单词是:");5 h( C- z, e. P, \' Z8 Z- L, g
for(i=longest(line);alphabctic(line[i]);i++)
' j9 I+ b4 o, H8 f/ W printf("%c",line[i];* e% E, {. \; W5 B* Y; v
printf("\n");
! ?$ s9 U7 \! i1 q" C9 \}( f# p! Z9 Y' A7 K. |' `
0 \- l6 Q @5 W+ I9 T9 B( a+ k8 {. m
8 ^3 W/ @$ B r; S7 g
4 R4 _& u. v0 f7 K4 x+ ^8.11
: a/ l& A6 E9 q1 j#include
; ~* t: z. {6 o' U! L0 L
9 J5 L8 Q' e( i8 @#define N 103 H. \; Q5 ^2 |
char str[N];. g4 N% g) ~0 r5 v# A" G0 o9 W
main()1 I2 I6 o4 L& [) L+ X
{& b4 ^; n1 @& C8 r3 s' }
int i,flag;
' o7 d l$ B2 O6 P+ `. g# l4 |% N) efor(flag=1;flag==1;)
" d) b) n' C4 f0 n8 D+ }2 q, K1 j2 \7 {" f{$ |; W% w# L( j5 x
printf("\n输入字符串,长度为10:\n");! Q- p# [$ D* T1 |+ I& R$ |
scanf("%s",&str);* G9 ?9 W: B; ?+ b D6 L4 ]
if(strlen(str)>N)
* @. @" ~3 {: |8 U, _/ R" `$ [4 l0 V printf("超过长度,请重输!");4 x6 a6 `; Z9 C3 q" O# D0 m9 t
else
8 L- f; A9 C# i flag=0;& }0 Q( r \3 ~
}
8 v1 [" s: Y% L q; D1 psort(str);" L4 U5 ^* \; r$ e3 f! b; y
printf("\n 排序结果:");& c. H# A3 G1 b- c( e8 `. E ^+ L2 l
for(i=0;i printf("%c",str[i]);
2 T, L. q5 W9 G% t" u}
! x' y6 e$ \ \$ \sort(str)
2 X: I! X2 ^( m% _. dchar str[N];
* S/ f j. x2 d1 e6 S/ ?; `{
2 ? l1 x7 l% O! P+ @: H4 r$ Wint i,j;* g( e1 T3 ?$ J2 O# J) E+ G
char t;
& ]1 h( l' M; vfor(j=1;j for(i=0;(i if(str[i]>str[i+1])
- F" [9 U5 s& L7 \! Q { t=str[i];/ ]3 z" G3 H, L8 |' I- x1 }
str[i]=str[i+1];7 l! B# A" h* F" @8 i7 l
str[i+1]=t;& h j# ?+ H/ U; n2 Y5 ~) H
}
" R3 m, o, d* l) `% m _! _# K}
' H/ S/ d" u1 X/ L8 a8.12
P* D1 c0 n- J, b#include
- Z; y1 }% I) u. k#include
/ ]3 w7 S2 a) L' gfloat solut(a,b,c,d)
, ]- W- d# H K9 m! M0 C5 \float a,b,c,d;
) a9 m0 s2 \3 V{float x=1,x0,f,f1;2 I: Z* {; f5 i6 l' Y9 [; g% e
do9 T9 G8 A1 O' m/ c# d9 b. i- `
{x0=x;) M$ v% k j e8 p
f=((a*x0+b)*x0+c)*x0+d;; X8 O2 q" m+ }! y \5 g: \
f1=(3*a*x0+2*b)*x0+c;
. S# V$ m% ]' N) y# ?- m x=x0-f/f1;
! [' J. k" }+ |: _! h1 i }
3 H( r" G" O! D1 l& W while(fabs(x-x0)>=1e-5);+ T; T% i( F8 G8 c
return(x);
8 t5 W3 n* J& L& b1 c- @: P1 l& S, ]}
1 u8 _) i2 A& |* S2 Fmain()
8 \' b' H+ s: I{float a,b,c,d;
! M8 a9 f- [# ~4 h5 B) i scanf("%f,%f,%f,%f",&a,&b,&c,&d);9 v" ?' G$ h( o! Q' F i% }' F' W- k
printf("x=%10.7f\n",solut(a,b,c,d));
@, e, x+ ~: e% i# S. j2 O/ x}
) B+ e+ N; I/ ?5 o$ y* x6 m8.13$ {8 @# L {" D) ?
#include2 z- [0 F/ S) E0 z) }
main()
% w" Y- C' E4 f8 t" }' f5 d{int x,n;+ [- t6 J' `# X2 j3 K
float p();
1 U, R: Y4 H- ? scanf("%d,%d",&n,&x);
+ }0 Q- M; v: T2 q printf("P%d(%d)=%10.2f\n",n,x,p(n,x));3 ]8 \& X1 x# J2 F' [$ h; ?+ I% h: ?6 r
}
# @7 W; d- ~' p! Ufloat p(tn,tx)
6 P% Y% D9 ^# A$ ]; Gint tn,tx;
4 h5 U5 |* \5 s7 V{if(tn==0)7 {4 G" x* _2 y; t9 ^$ k
return(1);. @' i2 I7 q6 @ E7 |% r6 x
else if(tn==1)
0 z9 r. P7 y, J& e return(tx);; G2 W7 C/ U/ D# [9 n2 Q
else
7 ]& y/ t5 d+ y, U5 Z. @ return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
, A- b# c; L3 i, i}
6 t0 R j! m3 {& R/ }8 r8.14
5 {7 e4 L( n% Z* d7 j# F3 a#include "stdio.h"
" a7 c! C# l6 a4 ^5 s$ i#define N 10( m$ k* u. |9 y7 m
#define M 5
+ W3 N$ f/ Q I, U! n! mfloat score[N][M];
3 M k1 g. `. p s0 r+ c. gfloat a_stu[N],a_cor[M];
8 n. K% R6 L: h) T2 f tmain()
! ~, d: U& Q- p5 }{int i,j,r,c;1 M9 J1 m8 j# O. f9 ^1 I- d: {
float h;+ a f( N( B: {0 K: @7 z2 k& L
float s_diff(); j* u- Y- S. q7 |' I& h' ~+ \# ]% E
float highest();* O6 W" `- C! z6 i# C
r=0;
2 C6 I, m3 s2 u+ B3 L7 M, R* X, a c=1;
, o" G8 q) z0 x. N input_stu();8 i. u7 ^$ W% ]0 v- x
avr_stu();
. w) r* ?. r+ v avr_cor();
/ C' z! l8 E E9 q k; x* a printf("\n number class 1 2 3 4 5 avr");
# M F7 W. `# Y( Y: Q for(i=0;i {printf("\nNO%2d",i+1);& W% t1 j4 `! c
for(j=0;j printf("%8.2f",score[i][j]);
5 L) A4 F N% g% U/ ?5 ]1 o printf("%8.2f",a_stu[i]);
1 L0 `$ D) |( I! ^0 w; O" D Y+ t }
% q' I( u7 B- _0 x$ E7 j printf("\nclassavr");: Z+ d+ h. X" I
for(j=0;j printf("%8.2f",a_cor[j]);
3 G, y' q! N2 Z* g) j h=highest(&r,&c);# m5 S. e7 F( @3 h
printf("\n\n%8.2f %d %d\n",h,r,c);
! t) J1 s I; V& u# I printf("\n %8.2f\n",s_diff());
- @' N& h. c. h. G}
( Q) E2 J4 d! ?input_stu()
5 O! a- i1 d. V8 r{int i,j;
/ J+ U+ r; l5 f' m* ^; f2 x float x;; _. ?9 {: B3 _ I, \5 ~* G
for(i=0;i {for(j=0;j {scanf("%f",&x);0 r$ X8 X+ f, c2 M
score[i][j]=x;3 A* P& ]5 X# o- s$ Q, P6 P% S
}. n% |$ k, ]: V
}
3 c) K7 P Q! H, t' @}: l4 y1 g7 W8 g) t! y, ~
avr_stu()
# j, W, ^* _) S) ~; T2 S/ g( `{int i,j;
. H, a- m( |; F$ ]! \. @& \- T3 U float s;
1 N" T& z5 g, M5 x% o" x/ l for(i=0;i {for(j=0,s=0;j s+=score[i][j];
' m5 c) m g6 `& M: @# q a_stu[i]=s/5.0;
8 ~( G1 ~! T6 B* j$ H& q0 }/ y }
( q4 a2 m8 `# S}
$ Y0 `. {* ]# o2 H w, K2 Uavr_cor()% g5 d$ s) a! H, Y f: ^& Q/ X3 w
{int i,j;
; q7 I: U. h8 p4 u8 F float s;
, h% `6 Q1 ?1 V# r# u for(j=0;j {for(i=0,s=0;i s+=score[i][j];- |. K0 M! r, n8 p9 g
a_cor[j]=s/(float)N;
1 k' ~4 M& C7 s/ a0 C }
% ^& N5 U4 i) K8 {& Z/ l e3 }}
1 q* U8 Z) H8 ^2 E; p& Pfloat highest(r,c)
0 ]4 \, y' ]7 {int *r,*c;
# \5 | h! P" B0 B{float high; z2 q4 D. P! H) {7 `
int i,j;( d) N# S7 Z3 _. q# ] k
high=score[0][0];
" B. s% K, ~' {. e1 @5 e. | c for(i=0;i for(j=0;j if(score[i][j]>high). x- G3 L* J& F) e) k9 U( r
{high=score[i][j];
* n2 Z# g) |/ ~ b' \ *r=i+1;
5 S! u ?# M+ b *c=j+1;2 Z: ?- ?* @2 T, A; D
}. {- v; F9 Z; L! w: o7 W
return(high);
* z& B8 Y. U c" F}% o, L" ]( z% Q
float s_diff()6 t! p% u1 B9 y9 b2 A. p; H4 t
{int i,j;
; r5 \) U. ?1 m: S) G float sumx=0.0,sumxn=0.0;
B% i' h3 C' { for(i=0;i {sumx+=a_stu[i]*a_stu[i];
1 p$ J3 |: ^) X$ q sumxn+=a_stu[i];
1 }; b: V! ?" r. Q }! J: b& C v7 S" l7 Y6 ?/ j
return(sumx/N-(sumxn/N)*(sumxn/N));
2 o5 ?5 J# G$ ]$ k}# u1 [9 _ Q3 n U" f" r
8.15, C2 y( Y7 t7 B, r w- z
#include
& l& ?' S) a2 f6 I" m* \#define N 10
- S3 u4 ?$ ?9 G+ V2 t, A' e Q2 Lvoid input_e(num,name)5 E$ h: ~& V- O) B9 W- l
int num[];
; b/ p1 C2 B) v7 D" Vchar name[N][8];
3 w( o( }2 W f# y{int i; n5 H" F; Q* D! E6 I8 X; p
for(i=0;i {scanf("%d",&num[i]);
' b- D8 G& E! v, Y6 m gets(name[i]);
: b7 `3 @) t' |1 ? ?' K5 R, l0 @ }% ?& O9 Y m1 X
}
, H# k$ Y3 h& ?, [; B. J% Zvoid sort(num,name)4 `# v9 d! R7 f
int num[];: C! t% F2 J% u7 S* K; ^
char name[N][8];! V3 a3 m: u9 l( Y; H8 G- G
{int i,j,min,temp1;% [% E, l8 p& ]1 e! d' [! Q' z
char temp2[8];
9 W# I8 s8 d% P/ v. m, T, Q for(i=0;i {min=i;4 _- x3 d! @8 ]9 K
for(j=i;j if(num[min]>num[j])min=j; Z0 y" ^) _+ d- V, A
temp1=num[i];; q# _8 r+ v& d- L
num[i]=num[min];; h/ i( H+ t3 f8 u# h& y( Z
num[min]=temp1;. r' F7 A0 G1 V6 `: M2 @' @
strcpy(temp2,name[i]);! ^7 J3 y7 l0 D6 {# c2 ~" }+ P
strcpy(name[i],name[min]);" }; z5 a" I4 @* o2 R( T
strcpy(name[min],temp2);
9 B4 Q7 ^/ c. Z2 O7 U }3 v% h1 w5 Y P* G! k+ p( V
for(i=0;i printf("\n%5d%10s",num[i],name[i]);. E4 s0 H; t. P. T/ S
}& @6 Y. v4 d1 d
void search(n,num,name)
" ?, V! x* J( Q& v4 u9 |; Oint n,num[];
4 ~2 W2 S3 }0 [4 J5 z5 e2 `9 h& qchar name[N][8];" X: L! O7 O0 C8 }. ^# B% D
{int top,bott,min,loca;
6 |7 o# @4 s' e) i$ ] loca=0;
2 \/ k! ~' i+ D0 d8 @ top=0;
, f8 t8 V8 p7 B% ] bott=N-1;7 U/ _; W1 O' N y) L9 P' C/ Z
if((nnum[N-1]))+ P" ?* ?; U" C; C; Y+ R3 X$ Q
loca=-1;
. z- Y, g& @8 \" D while((loca==0)&&(top<=bott))
/ B9 d8 w3 y6 E1 { {min=(bott+top)/2;
* C) A& z: Q2 {) l# J ] if(n==num[min])* Y: A3 q6 |" ~1 ]
{loca=min;2 Q! }+ y U$ J2 S3 ~6 |
printf("number=%d,name=%s\n",n,name[loca]);
# D9 h8 D2 ]- x4 e! j7 e }+ y# K4 g' z- K" ^+ G
else if(n bott=min-1;
( D5 ]0 u$ |6 w9 M8 R% E else T" W& G3 O. p. N2 _
top=min+1;, D4 V6 }. N7 R/ y! u+ `/ W, r/ `' H
}! o. r. O+ X9 ?* y; o
if(loca==0||loca==-1)3 S2 i' B$ b A, a9 a
printf("number=%d is not in table\n",n);
& o4 L9 P9 y) [. u) k5 w. \}
~% N/ D$ c( [! Y6 K9 W2 h' H( qmain()
3 ]: x z7 n1 Z& J: E/ N{int num[N],number,flag,c,n;
$ {: Z+ [: z7 g" S7 ~. m0 c char name[N][8];2 Y$ [$ {! s0 S \! U: A
input_e(num,name);
3 ~: A. o& n8 n/ n2 i" y$ e% K sort(num,name); D9 s, {6 M8 V- X! M' j5 d9 g
for(flag=1;flag;)7 t1 _3 T$ T V& J1 T1 _/ x1 Z
{scanf("%d",&number);
2 o2 G- J+ X) B0 \) l$ S& R# @9 K search(number,num,name);
; k2 R/ E5 o2 M* j printf("continue?Y/N!");
$ L4 U% f1 a1 F c=getchar();
: C( W) [ L9 _9 g5 n if(c=='N'||c=='n')
* D1 h; D, E- C ?8 Q2 ]- \" D flag=0;
$ ]+ I$ m/ Y. u' ~) A }) ]. Q- d4 C/ j, r( q9 r
}
* L8 I i' ]& J% ?) H4 F+ u8 q8 g
8.16
9 B7 @% K" j" L1 u) |0 R t" F6 c9 Y#include
2 p" ?6 i4 f* u1 L4 r1 U, o6 U#define MAX 1000+ x V7 e3 U2 Y* r. v& |
main()
Z. ?1 N; r% ]; D. w) @{ int c,i,flag,flag1;9 s5 N/ c6 |7 G9 k {
char t[MAX];5 j; a" q2 E; k3 ?1 c
i=0;
' s2 C# U2 t7 a8 d7 n flag=0;0 F) i7 u2 a, f h$ y8 T4 z
flag1=1;: W/ U) e- T0 I# R
printf("\n输入十六进制数:");! r: O5 ~6 n2 n' \ B
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')0 H$ B6 W. g: W: O% _+ F0 q8 j* X
{flag=1;+ ]- E" ^$ G8 H5 u% Q
t[i++]=c;, C1 v$ o( }3 C- u) |! L
}
2 @; w1 W/ N7 U% ~0 A& ~( q else if(flag)5 e5 i8 |' K6 h, O, `/ r
{6 |- e% T4 Q0 s7 X# X* x+ j
t[i]='\0';2 J. \2 a8 @- t( I) l2 Q6 }% I
printf("\n 十进制数%d\n",htoi(t));) c$ a, I; g* h( J, b
printf("继续吗?");
- A# h5 E4 f! m: W c=getchar();1 h7 W' @- x1 {0 l3 q" ?7 P
if(c=='N'||c=='n')+ @& E% g0 m3 }0 x1 a) a- M1 T
flag1=0;& O$ t( N! l6 N0 O& w+ j$ }
else, i, k, I: C. g5 D
{flag=0;# e% b' h4 M/ { ^. e
i=0;! |# y5 f, a$ a, z. j! v0 w# |
printf("\n 输入十六进制数:");# y* L7 L1 P3 }* R9 ~# }
}
$ ?5 ^4 T4 N- o# E! g" O}( U# U8 g5 M) c S& j
}' F4 v- @/ O0 l1 g* Q; b8 W( L( Y
}; e. Z& c4 J0 J
htoi(s)
5 g% l3 O/ O0 N( `/ U3 Gchar s[];0 s) C! C7 T3 t" M
{ int i,n;$ V0 O. c; I" R; K: \
n=0;- u6 Z) T. t2 G9 G/ T
for(i=0;s[i]!='\0';i++)4 C, V0 S7 X$ A) X4 V8 G& f
{if(s[i]>='0'&&s[i]<='9')
+ Z9 T: l0 s; M8 u2 S g- x n=n*16+s[i]-'0';" b Q9 T1 p& }* P
if(s[i]>='a'&&s[i]<='f')2 S$ {( i+ t' q" p6 }
n=n*16+s[i]-'a'+10;
6 Q5 o O2 [# |7 H if(s[i]>='A'&&s[i]<='F')( G7 l- Q q; n. L5 v
n=n*16+s[i]-'A'+10;, [1 m c; E% B
}# Y9 E5 [9 U2 I: d: E: S
return(n);
4 s: W( C+ h% U% l$ {) O! Q3 @! J}
1 X+ y8 D. W+ j- [
* x& |0 k* y$ E7 k! K9 g/ E& J0 J" `1 k3 m% \- i1 T- ?
& T) i! v1 m7 ]" N: t, Z$ K8.17
* G) E4 F; q+ ~8 C. w#include
) _& S/ y& ^* K4 uvoid counvert(n) Z# v+ d. F# V, Y: n' ?6 [; L
int n;6 p* [+ z t R0 \' E4 u0 n
{ int i;
( G, b0 J" s) z4 `' X' _ f if((i=n/10)!=0)
) J: a+ I& ?( c9 h) s3 e( A convert(i);' J, ]$ t" Q& e
putchar(n%10+'0');* B7 P( W2 |0 C' G6 O
}
7 k- H. q5 b6 k h! l f( p" tmain()1 p6 C7 r8 N9 D/ x; f
{ int number;& o9 b, s) t: M8 F ^
printf("\n 输入整数:");, R+ ^! m! ]8 o7 M& `
scanf("%d",&number);5 f. W. e% K# q" l+ t- [ m! {6 B# ^
printf("\n 输出是: ");
9 w) [4 \0 K3 P I+ k/ m. P if(number<0)
: o% \6 y* }: c* Z+ _ { putchar('-');$ z0 w# R$ K& b; U4 g
number=-number;
1 H, c; k G# F# a1 V; o+ C& h* \ }
" D0 G* l q- Bconvert(number);2 [; ^( t% V5 Z) q% K
}
: i8 C3 y5 O" b5 I" c" ^) N9 C4 }8 d" E8 h9 K9 X
# @! z$ N9 b" |) Z/ @1 A2 M3 g
2 c I, R) B2 q, L# M
8.18
& V$ J+ j. X) w# g# Y& w5 O( rmain()
+ N* z( `8 k" `3 {; T! @{5 ^7 A; `0 O% f8 g5 w) K
int year,month,day;* j) f5 m) y% D1 r& b
int days;8 Q, n/ V! X7 |1 V O
printf("\n 请输入日期(年,月,日)\n");1 A6 a( {0 b# J$ k; u
scanf("%d,%d,%d",&year,&month,&day);
. f4 N/ M- M/ n/ } printf("\n %d年%d月%d日",year,month,day);. p# c+ \& |7 {# @. b1 W8 x
days=sum_day(month,day);! N9 `9 t# ^+ T" m* h4 ~; {
if(leap(year)&&month>=3)
% Z9 Y; p* Q6 s3 K/ j: t$ |: F( N days=days+1;$ v- ?* K6 |) C- e: C( G! I* h
printf("是该年的%d天.\n",days);
" P( x0 y8 [$ ^' a6 g }) I& Y! t- t6 L% ?+ y8 ~& G
static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
* B6 z+ g1 H# v- y1 n& B# t7 T. t- [ int(sum_day(month,day)
3 T7 P7 w/ D7 o int month,day;; {7 T2 i" [! I
{
/ G1 @; k! W# U2 r int i;
' G* O* f5 A9 r5 M for(i=1;i day+=day_tab[i];
9 h, r4 C; @* q7 g I6 A p return(day);
* ?. _' h* g1 a2 d, D }
/ n9 N2 I! ?/ c' s; Z int leap(year)
+ h) v# }) `4 ?9 G% @. g5 ` int year;! ~' u5 U% T2 ?6 \
{
: m v5 f" O! ~* F; O int leap;
% r+ x9 z/ v3 t( M7 h leap=year%4==0&&year%100!=0||year%400==0;
5 E+ b: l7 p" a0 J" @5 P) W return(leap);
6 C6 P5 ]5 i+ X4 q; b1 z' Z }
3 b- X! k4 v9 q. |5 [第九章 编译预处理 s. _ s6 w! q7 n
9.1. _$ L3 e" u; P+ ]
#define SWAP(a,b) t=b;b=a;a=t: F8 T7 E w1 j6 G: l0 a
main()
8 f: F n4 ~4 c3 b/ T{
8 O- J7 M+ H" t* l+ uint a,b,t;4 V( B* m3 F) A+ a( n
printf("请输入两个整数 a,b:");
+ p5 k0 r& s4 l. ^- n& l& Jscanf("%d,%d",&a,&b);
" Q6 V7 {4 C2 j6 l& A' ISWAP(a,b);
( ]( q$ M5 h0 x8 Y) _" D. y" xprintf("交换结果为:a=%d,b=%d\n",a,b);% ^2 }. z! `1 m" b* u+ l: l$ C
}
2 `* h9 G" g- n9 q6 _+ _- U
4 l: R: h M6 o/ q& t8 F8 G& F( D5 U( g7 O% p& g3 L7 K$ p. y( \
9.2
9 v- E- k7 v3 E4 y& g8 w#define SURPLUS(a,b) ((a)%(b))
X t6 g+ |- r1 h$ t" H, S+ N1 G1 dmain(): i! Y, D& ?( }3 S/ S1 ]- U! @
{7 p% h" K0 K( U% A& F5 P; t; t
int a,b;
! p3 d0 r1 o" {& @" X* r: C( _. s printf(" 请输入两个整数 a,b:");
# w2 y9 m' P' _6 K$ i scanf("%d,%d",&a,&b);
& V7 N' i3 Z. T$ h5 T' p& Nprintf("a,b相除的余数为:%d\n",SURPLUS(a,b));4 \4 w) E2 V, m; v
}
F" O. x. q' S4 z4 r* ^7 v
# n" k! K1 d! `1 [- s
# }0 s# i3 n5 [- e# v2 j# Q' n9.3
% N, O, {" e" Y! s% t#include; _+ v6 [6 K+ w8 q( T' l& u# k: O. x
#defin S(a,b,c) ((a+b+c)/2)
1 M( Q: f- y+ F7 {#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-
* C X, t. @* Y/ [c)))( @" B0 k( c$ H6 D b& b5 O$ @
main()
" V, e+ Z0 Q" R! h) D; _! b/ H/ i {
+ q. r: L, M& k& g3 w$ Q float a,b,c;% L( k3 v9 z* V' k
printf("请输入三角形的三条边:");" @& G3 j8 D; x9 ^" \
scanf("%f,%f,%f",&a,&b,&c);
. }: v' H7 w' y& P; @4 y3 `( l if(a+b>c && a+c>b && b+c>a)
0 s1 d* I. z4 {- I" M8 Y printf("其面积为:%8.2f.\n",AREA(a,b,c));8 l- A2 W9 ]7 k9 C
else2 j! w* T7 V& P9 i: i2 o, x; j
printf("不能构成三角形!");
! U" m2 x) h7 g- x B }8 ~4 j3 R" v3 n# X0 A6 E
' |0 `* \7 o; L% I: z1 O/ m
# R1 L0 y" L3 K9 j: E# G
3 _' [* \) A' M$ D$ w+ }. s! q9.4, S% ~' `3 z! S, V" K/ z1 |
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)/ C6 G; U5 U- R; l! ]) ]& s
main()
% J" q% P; N& ~, ^* f {9 S5 e' d, d. ?) b
int year;
. t2 U) e8 r# W0 V printf("\n请输入某一年:");7 `% U" {0 w7 J5 Q7 h: ~+ y# S
scanf("%d",&year);
! W5 O0 h7 V3 b. F+ }9 J7 v. O+ u if(LEAP_YEAR(year))% M- f/ r2 B! ~- `$ u; i- i" M2 ]
printf("%d 是闰年.\n",year);3 c/ o) A8 \% W9 E. V
else9 `; p; _6 Y5 {+ Y
printf("%d 不是闰年.\n",year);1 A9 v: C* F; q9 E- V$ K
}3 O$ @/ U) T( q$ {! @$ s
: i3 }/ \ `, \' ?8 m: W8 T) _. R5 \
3 c6 [) z- k6 I
6 u1 _) _' ^# _7 ^9.5解:展开后:+ M: R. S7 Y# a5 q
printf("&#118alue=%format\t",x);
% e6 q, h" y, uprintf("&#118alue=%format\t",x);putchar('\n');5 {( I2 \5 j: o$ a8 g1 I
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');
1 @+ U Y( h5 G输出结果:- T( Z# u; J' k
&#118alue=5.000000ormat &#118alue=5.000000ormat
$ C+ O( u3 X# j&#118alue=3.000000ormat &#118alue=8.000000ormat+ T6 w; Y0 S8 W
& O' H1 v# U' B/ U2 N
- e @* Z, X, M- O+ S+ B9 j% S- q9.89 c. u, }1 L5 b* Q; @: F
main()
9 V4 b% S" i, M$ s' y/ s, v {! Q% S* \; M. q' x7 d" w! M* R6 Z
int a,b,c;
% q) F, b0 M. ?8 }* F j" F+ d printf("请输入三个整数:");3 r+ k2 d5 G( n
scanf("%d,%d,%d",&a,&b,&c);7 [ {7 F, l( j( B7 R) K% Y2 Z
printf("三个之中最大值为:%d\n",max(a,b,c));2 I1 D Q+ W1 l
}* L9 j8 N% O; j
max(x,y,z)3 a! Z- G5 Y3 ]5 a( `+ V
int x,y,z;' A, q9 h+ t* L9 |1 r
{/ x, O: C" x* }5 M; i% f2 q5 m- H
int t;5 n7 z! G+ c8 N5 d. G6 F
t=(x>y? x:y);
- V* T2 r# f$ S0 A return(t>z? t:z);
! h9 i/ {6 }3 ?& R* ~* M }& w: D0 \0 z4 E7 U4 ~
7 M8 F0 M* G# y3 d5 _$ @
1 N. k. o' x; e* {! k9 m
7 h! p: R) g9 x+ h2 r9 K' O! Q/ o9 K9.10
. D5 x6 B/ ~- e+ M$ I5 {#include( ~8 t! ^5 S' }' E7 T
#define MAX 80, H- ?% i3 D' |# L
#define CHANGE 1
: H" J _6 }, \4 c6 j3 o1 W( I2 Amain()+ d1 _7 i# J6 ?7 Q5 u
{$ u" ~8 s p: T' o1 f: x
char str[MAX];
2 i1 K9 q, r# ? int i;( ^! ^( ^: }! O( W% e J% B* V0 X7 G
printf("请输入文本行:\n");
$ i v6 ]& l% R5 A' t% y4 J scanf("%s",str);; ~7 \# {# v$ G! @0 L
#if(CHANGE)& P% V& k7 @/ v/ R+ h
{4 m$ m/ A# K, v, Q/ Y/ S
for (i=0;i {9 U1 l, o, s5 S0 U% I& O" P+ E
if(str[i]!='\0'
/ h' W, F6 ~0 D J( Z. ~ if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
( k4 m4 C; [8 V4 G- @ str[i]+=1;8 Z- o( P }( _5 h8 N$ c
else if(str[i]=='z' || str[i]=='Z')5 O% Q0 ]: D h9 m
str[i]-=25;
$ \) x: W9 e. Q& T1 i/ d }
/ A) I2 w! {& Y8 W( @3 h}
6 }; t2 `6 V3 i) w. ^#endif( y2 \1 o6 F; W) p; f
printf("输出电码为:\n%s",str);2 p( O" g% `: J* B5 D) h0 h
}0 a+ J4 I0 M# ~+ @" q+ c4 l
第十章 指针( `9 _! |9 w# H& I/ q
10.1
! {5 O- ?2 I' J( k) Bmain()7 M6 [' ?8 p: ]1 m; j% K
{int n1,n2,n3;$ J0 {& R* ?0 P& u0 P. s' |
int *p1,*p2,*p3;
) J, i6 d! m: h: D: u scanf("%d,%d,%d",&n1,&n2,&n3);
) X. t/ w$ Q; ?, k: e p1=&n1;( [& J; @% `- O/ C8 z d g$ w
p2=&n2;0 \, W$ O f+ J2 \( b. S
p3=&n3;/ z% R, [, N' V3 ]. S
if(n1>n2)swap(p1,p2);& G/ I& P8 {, K4 O2 [
if(n1>n3)swap(p1,p3);
; M7 E$ Z/ K- F3 B$ q1 n if(n2>n3)swap(p2,p3);
, |9 t7 _1 V* ]- h; }, u printf("%d,%d,%d\n",n1,n2,n3);3 `7 f; b8 P1 g7 v
}2 K i" m7 T$ ?& G
swap(p1,p2)
; G$ x L% D6 @int *p1,*p2;% R- u1 ~% I& g
{int p;9 g+ R, @; }$ s2 L
p=*p1;*p1=*p2;*p2=p;
9 K) {: }9 E* G- p3 s}. M) S- g; m' E
10.2, V/ L0 c) R& o9 x
main()
* s+ C! ?! z+ `! R; Y+ o{char *str1[20],*str2[20],*str3[20];! N3 C) |: G5 Y" Y# u8 q
char swap();9 W3 f6 l* C) \# `% T8 _
scanf("%s",str1);3 @" ` r1 S5 y4 _8 M; }) p
scanf("%s",str2);& U) W) k6 Y3 c9 n
scanf("%s",str3);
6 z0 @1 S9 R1 T9 N2 a+ }$ e if(strcmp(str1,str2)>0)swap(str1,str2);* U. W- A( l5 P& }3 j$ P
if(strcmp(str1,str3)>0)swap(str1,str3);
7 _% y) ?. T8 ~, Z if(strcmp(str2,str3)>0)swap(str2,str3);. v m f' P, x, l& {
printf("%s\n%s\n%s\n",str1,str2,str3);, c5 Q" H: C+ q- S! J
}
; L" p! R+ u3 j9 J4 R2 {) Jchar swap(p1,p2)
; I; f& r' y& L( r% cchar *p1,*p2;0 w( R+ R2 S" G6 u" D9 S$ j* p' c7 y
{char *p[20];5 E& m4 r, H" D' P+ j
strcpy(p,p1);
' A/ }$ a. M" T9 G; A0 ]7 e4 q; F strcpy(p1,p2);. J7 | g* g7 {- ?* Q0 a4 r- Z
strcpy(p2,p);) N; B' b+ r* p6 f+ u
}, I: e2 F$ y8 e, Y4 E# h
10.3
( n( A" x' s3 z9 b# T R8 L+ a* wmain()
) p& S2 y. f" O; _{int number[10];' _0 R9 P/ r u$ B0 s: H
input(number);0 Y% _6 i) }' x& y. w4 h
max_min_&#118alue(number);
: f; b- Z" i4 l* F! \ output(number);$ U4 w8 f2 X2 I1 Z2 A
}) g/ D8 w( a u' r1 u- s
input(number)6 w/ L! ^- Y- T# I& D
int number[10];
4 A9 F6 O1 y! D5 X* z4 |{int i;
6 K0 A. c8 L! C for(i=0;i<10;i++)
. Y9 _! E: |; U6 p: q4 f) w* M0 o scanf("%d",&number[i]);
& j/ R0 ?5 ^. ~5 F7 N5 }}. E3 r5 S% c* y3 e4 H
max_min_&#118alue(number)" g$ W0 Y7 i! R/ R7 ?% v
int number[10];
' Q# q a% Q3 q+ d! g# N{int *max,*min;5 b' _9 q" m! ~* d
int *p,*end;3 c2 M" U! s* o7 `! E* F, w
end=number+10;, s. |) s7 F Z/ G' s$ N" x* P4 b( u
max=min=number;
2 r* a( _9 m4 u) L. @$ p for(p=number+1;p if(*p>*max)max=p;
' [' C! {8 ?5 f+ \" C else if(*p<*min)min=p;
7 h" }% y: _( A9 V3 G4 q *p=number[0];
+ B& V8 ]4 C, c$ F+ H6 V8 ` number[0]=*min;7 }6 U5 r% G* _: F4 m
*min=*p;- D. y# A* X7 [ F
*p=number[9];+ i9 E L0 X$ ^
number[9]=*max;1 T9 Y" ?; g H9 A
*max=*p;
# e; Q) ^$ T% V6 {* S3 b' S9 h4 H return;
2 |) B" o, R8 F7 \) g5 G. p}! ~7 [3 B; X# S; D
output(number)9 ^( }; _0 s2 K- f" G6 F D; F
int number[10];
* Y; ^! O6 f y$ a6 V{int *p;2 P {: N) ]$ l& B- H
for(p=number;p printf("%d,",*p);
3 ]* ]9 M- l: r0 S6 J* t1 X printf("%d\n",*p);
3 V$ z: G* ?" f j; a$ M}! b4 D6 m, H: l: {0 d+ t
10.4
$ ]" T7 b: E/ m6 \/ Lmain()- x, `( y$ O$ s8 q3 u- B
{int number[20],n,m,i;
3 r4 {9 o3 @3 V scanf("%d",&n);
# B; T0 U/ a9 z# I( f scanf("%d",&m);, }+ X4 [# T- j+ O) ~0 A
for(i=0;i scanf("%d",&number[i]);+ Y8 @/ R+ k3 n+ W9 P
move(number,n,m);1 F7 R5 K0 `& v' @9 V' b N
for(i=0;i printf("%8d",number[i]);* a# k4 `, Q& c9 [& H, i
}; Y% @! G4 W( x. d! G
move(array,n,m); c& B% I6 q q
int array[20],n,m;) b3 U2 _' t' P9 K' h( V
{int *p,end;/ V1 q# \/ _ N/ A; }4 n4 |- X# N
end=*(array+n-1);
, p6 k5 R: E$ q: r# f3 a for(p=array+n-1;p>array;p--)* U- `. A: R B/ r0 S
*p=*(p-1);! U7 S" c( ^ T% {1 b
*array=end;
7 _; k/ ~- D+ V. e: X m--;
) z. m" L1 E: v3 Z' W6 D$ f if(m>0)move(array,n,m);
) h* h" F* r6 V. N8 u}
c) r& f. ?2 Z. T/ o, K" |8 L8 p10.5
5 v9 Y0 } h& e#define nmax 50- \) K/ d9 z9 `% ]) P
main()
; S) n4 L$ N& y: b) O ^! t% B{int i,k,m,n,num[nmax],*p;( H; R1 V; l9 C2 h. f6 S
scanf("%d",&n);
+ c! M" `3 P7 |5 r: S2 Y p=num;
: T1 ?+ g0 \% x; c @! ?/ {0 M% W for(i=0;i *(p+i)=i+1;' J: A7 i5 R. v' f0 o
i=k=m=0;
2 G: u6 C4 K$ s% r( H while(m {if(*(p+i)!=0)k++;
0 R0 u8 g) n( E/ D6 P) V8 D3 B8 Z if(k==3)
& V. E2 j+ Y1 r) H4 }$ y" X( E {*(p+i)=0;, ~% j1 B) h4 L: L% r: \2 }
k=0;& E1 A' C$ h9 u! R( W& ]
m++;
4 o, Y" Q) ~* _& A }
& X& Y m1 o# ^ i++;: }, N& G4 Z+ L
if(i==n)i=0;
- ? c0 X4 W. y1 w }* j$ k' q" \ c# w
while(*p==0)p++;
. R/ d/ y5 e- ~" S. M printf("%d",*p);. ?' v+ X3 R) r& o1 ?& y! a
}' G* f) c' V4 J& T3 t8 f
10.6
/ Y2 E$ u3 s6 S% k5 }7 E& Zmain()
. J( Q, q/ W: U{int len;
1 I: N+ j9 I) R" b: v0 D7 j char *str[20];
- B' u1 f; b$ Q. k0 B scanf("%s",str);( \8 I0 _, Q" Q0 z9 c
len=length(str);
/ s- m3 _5 S- b! ?+ [) T p5 q printf("\nlen=%d\n",len);; j6 Y- H, f" I# n6 i
}
( C- P4 i! Q8 Glength(p)2 [/ }5 h1 \% W5 O
char *p;
- z/ G% O; R+ Y3 W% p9 P) m: \{int n=0;. X/ X- l# c$ @! |4 x
while(*p!='\0')- Q: n$ I! N q1 T- K: N
{n++;p++;}. T7 S0 R" x! g6 `. x! [. W
return(n);
0 e5 {8 {1 _5 Y' F}
, ?/ N% c% S* Q10.7& F% Z) w$ S9 l$ e* ^) Z! Q
main()
9 n" J( k8 E+ M: y6 q* J" a{int m;
/ S% y/ J1 W% ~$ O" X" j char *str1[20],*str2[20];
9 y: @- N! t/ m: K0 f& t scanf("%s",str1);+ d3 c: M4 `: b6 L1 t% [9 ^3 _
scanf("%d",&m);
) W8 ]* {6 s7 K5 ^! E! B if(strlen(str1) printf("error");' K+ e) w" J* k- ~( e* N
else; a7 u9 K/ K$ ^$ V; x4 R$ j/ U
{copystr(str1,str2,m);
6 g# m6 I$ H: ~* {6 G printf("%s",str2);1 Q( Y* V# R% n% B i+ T; u
} S% P2 n. |! n1 K' w$ a! n
}
7 ~! W- Q! Z' a/ }copystr(p1,p2,m)
8 C/ U3 C2 [! z4 Y, w0 V% Mchar *p1,*p2;* I% c, }9 K0 O/ `7 S$ j
int m;0 f) R7 P+ ?9 y
{int n=0;
" `7 s- E8 t6 R while(n {n++;p1++;}8 c9 J; h- g ?, K1 Z
while(*p1!='\0')
0 m: ~" ~ u5 h& ]! i {*p2=*p1;
2 y+ J8 ]* P1 p- m' v p1++;
# ]: l9 q" `& ]$ T. u b p2++;' P4 K* z- g. L
}
8 X( Q) V. C) X& t9 Q8 ?3 u- D *p2='\0';" o) x# z* e1 Y7 T2 H; M
}$ l1 P* E7 n% [6 \4 X, m
10.8
" S K& \1 O/ a U* e; h7 I2 H#include"stdio.h", W) h: J( t) p/ V6 V( t8 C5 M
main()- F4 i2 C6 P" g3 }% _9 c: F% g' W
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;+ R& b/ W9 O* ~" J
char *p,s[20];
. _# ~ e0 p, O2 T" K for(i=0;i<20;i++)s[i]=0;
5 D1 h2 P- C% _! Y i=0;9 L8 S' q8 v' n
while((s[i]=getchar())!='\n')i++;
% @: e1 k6 w- J6 ~7 J5 I3 ] p=s;
/ @+ ]- P8 W. n! w: G" \ while(*p!='\n')
+ M) w% u- b& | {if(*p>='a'&&*p<='z')
& N9 E! D9 K: d1 X( E+ e7 p- s ++sle;
7 R9 A/ z1 S2 f. {; H+ f j else if(*p>='A'&&*p<='Z')
$ W; Y; f/ G6 W" @ ++cle;
: w5 y9 K6 c8 `& ^5 u3 ]8 I; F else if(*p==' ')
$ }6 N. r, h$ R5 z ++wsp;9 O& I3 t" C& a8 b
else if(*p>='0'&&*p<='9')% T) n, P* M3 F9 C/ X
++di;
: ~7 |( \7 P1 x; C. [* e8 w, v- ? else" d1 \! p9 p% c
++ot;( _4 K! u! v$ I, L1 i7 @
p++;
- O4 k$ Y* z. }' c/ E/ T3 k }, \; T1 ]7 ]7 E) L! ]
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
2 \" D( a$ ?, } k6 [* B}/ L( c2 U9 U x6 X
10.94 q( W$ Z) r k* E0 @8 F
main()
2 U8 I# f& ?% x4 U% w y{int a[3][3],*p,i;. E w# m" [2 |! s8 `; Y8 q
for(i=0;i<3;i++)
, ^% D! S, W) H/ m& L$ S5 F scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);
3 i; u/ K7 O! z9 o0 R3 ~ p=a;# \1 g! P0 T- N
move(p);! b/ b3 V2 \# X' O- d; y8 o# ~
for(i=0;i<3;i++)5 K2 J* O, y. o; z% ?
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);
$ g4 a6 ]% ^ E} b8 d. M1 Z9 D$ ~6 J1 M
move(pointer)# J' ?6 Q: D9 O
int *pointer;/ v# {1 r0 y, o8 n
{int i,j,t;% l2 ~5 M4 q. f. b! \
for(i=0;i<2;i++)- g$ G* Q! \3 y0 {! z1 s
for(j=i+1;j<3;j++)
: X* w1 h# X) K/ ~# C/ Z& I {t=*(pointer+3*i+j);
6 L, w# F9 t7 `/ ]$ b. z *(pointer+3*i+j)=*(pointer+3*j+i);
7 D; Q) t" j9 e3 a# }3 x5 H *(pointer+3*j+i)=t;
8 T* {& @9 x4 b6 {+ P9 O$ x$ B }3 M% |5 d& [: |
}" ]. e! D5 a8 n- z, U0 K/ Z
10.10
: F9 g/ _. f( \7 ^* K Smain()
' m& s& A; p- [# t2 e: l{int a[5][5],*p,i,j;
6 C: }5 Q5 n1 N% H. n' W3 U for(i=0;i<5;i++)+ o b; T& ~5 K$ T" C: g
for(j=0;j<5;j++)
7 r2 o3 o9 B2 G8 g scanf("%d",&a[i][j]);" E3 y# j3 n7 M
p=a;
* U2 ]" c$ r& {! J$ {' Y0 g change(p);6 m, J; ]- Y; U& D' A6 [1 {
for(i=0;i<5;i++)" i# w* `- C' q( ]4 {/ i
{printf("\n");* C; f9 F8 |5 h. O% e
for(j=0;j<5;j++)/ x# y+ g4 Q* G9 _( G
printf("%8d",a[i][j]);& `9 o( @2 o3 c- ]; Q
}
$ s4 G/ e. I: r}2 E, {9 z. z3 Q7 H/ u: H. W
change(p)
% n* ]8 d! J: s5 g( Rint *p;
' w: M5 R/ L& h7 c{int i,j,change;
: c. _; S& F$ C" p. g int *pmax,*pmin;' @ `6 o, q. ~, ^# g7 V
pmax=p;% I! o! D# }5 y$ l3 c( T
pmin=p;
* ~# {; P( W! v: m& F& f; H" ?, f" T for(i=0;i<5;i++)) E# c0 {& }+ p. z
for(j=0;j<5;j++)
* N9 H9 J) b% B4 X. t) g, n) j8 f {if(*pmax<*(p+5*i+j))pmax=p+5*i+j;. n& I- W' n; |6 `
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;6 [; P2 }9 Q# c/ v% B
}
! w3 l, X0 b0 R6 y change=*(p+12);
9 ?' v6 Y0 X- E9 J( I# k *(p+12)=*pmax;
) r3 B. h/ u& {6 q8 Q *pmax=change;; [7 k- H4 p% f
change=*p;
u3 d+ I6 u- ?0 x *p=*pmin;
( r' r. i2 \6 `: n4 o8 ^9 H/ r! } *pmin=change;6 E) G/ n4 ~- R# r7 ^4 f, p
pmin=p+1;
6 x( t. s; G- G/ p6 g- P- ^ for(i=0;i<5;i++); \3 W% J3 ^1 y% ]& ^4 k
for(j=0;j<5;j++)9 h/ Q9 j) }7 G$ I+ Y3 A. y0 o
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;! x# t3 `) U& F/ r$ f
change=*(p+4);
, `5 g0 h8 S5 Y3 _ X3 j *(p+4)=*pmin;3 G: ~, v+ U) T
*pmin=change;
: g9 D& ]7 H- i. o3 L4 g& \. [ pmin=p+1;- e( Q \5 h. l, `4 y2 d
for(i=0;i<5;i++)8 T# G$ y6 Q0 r
for(j=0;j<5;j++): \3 k6 d% ^5 D* S6 P" ]8 I% G( ~/ {
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
: P9 d& O/ V3 o2 s7 h* b4 y% g pmin=p+5*i+j;
- U- k9 o# O0 v2 {3 }0 y2 v2 C% {0 K change=*(p+20);: J* D7 _/ h, n: Q4 r) Q
*(p+20)=*pmin;. m. z4 b- \% E1 M' y' \
*pmin=change;4 c$ \4 L; v/ m! R. E) v7 o
pmin=p+1;; G- v9 R5 ?1 }& u3 V, d1 s
for(i=0;i<5;i++)- l: T8 m/ o; b- v2 W6 U
for(j=0;j<5;j++)" d! K6 f, w! Q( T1 Z
if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
! o) h) M- a4 J e1 Z &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;( j. U- y2 ]: x2 T6 y7 n' M
change=*(p+24);
' u: x* ~2 H4 H *(p+24)=*pmin;
' Z+ q- \( x' M) h# [ *pmin=change;
* d+ Z) R5 D* Q}% |$ D; y( U* F! n+ Q$ W
10.111 h' q) N# H' \6 m2 C
main()- a( ?" g0 P4 u+ |0 s/ B+ I
{int i;
+ w' T& P0 A$ g0 m char *p,str[10][10];
1 {: _: C1 r4 r) M for(i=0;i<10;i++)3 D0 ^+ t$ i8 Z
scanf("%s",str[i]);
; W4 p9 Q3 g' @! z$ a) i p=str;7 z+ h: e) ~ {: j! S
sort(p);
! m5 w5 U; n" v6 _3 K for(i=0;i<10;i++)
. |& `9 e* m1 z9 b printf("%s\n",str[i]);) l+ h! q5 u3 L
}( c' e$ b3 k( `: o* n' n: T
sort(p)
( k* E; E, f Z3 [2 |: Bchar *p;
* k! b( i6 b% a$ @: e! w{int i,j;
7 v$ X. H7 f' n' V6 f, d char s[10],*smax,*smin;
1 k4 L a- r# K; ` for(i=0;i<10;i++)1 j: i! e2 P" h& t
{smax=p+10*i;9 }- G3 c/ ~- A4 {0 {6 N
for(j=i+1;j<10;j++)" V) P7 u* b8 R" k- y
{smin=p+10*j;" v* S/ S# ]" X' O
if(strcmp(smax,smin)>0)
0 p3 Y8 R7 K$ X% I) w {strcpy(s,smin);% D5 r1 G% v$ W) ]9 k
strcpy(smin,smax);1 p% S9 h! A: c9 M3 n! d7 A& n2 x( }: S
strcpy(smax,s);
/ z; }& ~5 \6 q7 w- ]4 {- d9 X# q }: U2 t* u! B) c) D2 X" t) m# o1 Y
}
7 v6 ~- ]7 V6 z" a8 X$ o) y } Y: @+ P$ P7 a3 j* I+ v; K8 l
}2 l5 U; l3 V& J8 J o8 R; c
10.12
8 I! N3 L# I( ]3 m% \#define MAX 20
3 Y8 A8 x" M/ k" l, ^5 n$ k( gmain()7 a a' v2 \. o( B, \8 v: t) y
{int i;
' {, c$ c3 h0 p! C3 l1 Q* ~ char *pstr[10],str[10][MAX];
; E" K0 q1 o" f3 ~1 n! E& ] for(i=0;i<10;i++)
- B" ?2 ^: q$ \! N pstr[i]=str[i];
+ D6 B! A) d- x% I. ` for(i=0;i<10;i++)
4 u9 H. u, Q0 B' Z scanf("%s",pstr[i]);
9 A5 G7 k# a8 P F( }" I* w sort(pstr);- d! f; w5 f8 t
for(i=0;i<10;i++)$ K7 y$ Y: v& z# q$ }9 v
printf("%s\n",pstr[i]);* l, k2 _4 _3 k* w( K
}
7 T( M6 N7 b) v6 L" n) rsort(pstr)
: ^# |8 Z( d; X$ M8 \char *pstr[10];
5 e% J5 \! i; h( Y{int i,j;
* ?' Z7 M( j0 u B8 Y+ m( _/ u* i char *p;# C6 v% s8 v: ?4 i" p
for(i=0;i<10;i++)
: o* Q# F# |5 c# D/ C0 d3 ~+ F {for(j=i+1;j<10;j++)
J6 C) r$ ?" L( P9 S0 f' J( k3 s {if(strcmp(*(pstr+i),*(pstr+j))>0)% t p$ l3 Q4 N, f4 ^
{p=*(pstr+i);6 m! p4 p* A! v/ }: K+ ?
*(pstr+i)=*(pstr+j);
' {7 }8 ?$ R J: R: l- ~; Z! _ *(pstr+j)=p;7 T/ ?& v0 r9 M
}
5 t2 ^, I) r3 [8 \ }
9 }# i3 ^$ t; P0 L }: W7 y. P; c6 n0 L0 X5 C* e, r4 a5 r
}
* F, |. f' r2 ?9 i; ~10.13
* r {' r% {" ~0 C#include"math.h"
0 {2 B. ?% |3 @' s( o+ Mmain()
0 U' U; Y2 W/ I3 K8 X4 {{int n=20;
2 C6 |) T( ?4 V, O5 Q6 e5 Y4 ?2 d float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
1 G, S" }, \( y" J5 |* ^" n scanf("%f,%f",&a,&b);
5 I6 f( [7 O) K( ? scanf("%f,%f",&a1,&b1);' k( U3 C4 N4 z4 }4 x# W# f7 ]
scanf("%f,%f",&a2,&b2);
1 s* a# K- m3 J- ^7 T6 Z p=sin;
) Q+ U0 b3 `% H8 q8 V c=jiff(a,b,n,p);
& q/ ~5 _1 v* G) p' c printf("sin=%f\n",c);
# j% n ~5 t$ k- R! q0 j p=cos;! J! J5 [( m6 g0 H8 |8 ^. b
c=jiff(a1,b1,n,p);
$ V! W! ^! w9 k* w: ? I printf("cos=%f\n",c);! a. {& a5 r# s
p=exp; w9 R" W. a- H
c=jiff(a2,b2,n,p);
2 f7 X* j* @* l- O2 ?' d printf("exp=%f\n",c);$ a- q9 z, b7 N( p( e7 |8 p
}" z. i! s0 a# M1 y- N- A& `
float jiff(a,b,n,p)
e+ W4 b! G6 N( a4 c" d1 N$ Xfloat a,b,(*p)();# e( I; D2 B$ T5 ]
int n;& D% O! D" i2 z) }
{int i;7 o3 S+ }5 y% R0 c; X: M- {
float x,f,h,area;' S- V4 `8 G7 @
h=(b-a)/n;
7 X3 p' X8 g: r( K5 F- U x=a;+ N: J$ B9 Z! b3 K8 k. j' m
area=0;
5 X. ^+ L I5 n. Z for(i=1;i<=n;i++)
' v7 ^: v* K: | {x=x+h;/ `4 n) f* u% M& r0 s
area=area+(*p)(x)*h;/ Z- M6 ?+ q7 [* {6 E3 u: Y( g$ W
}
. p5 G* W, ^% d/ }' |' x Z9 | return(area);/ Y, E B6 Z7 B% X
}9 Y+ m# j* o9 v4 Z
10.14
, u8 ^0 y" T" ^7 c3 {! _0 }main()
2 y6 E0 }2 y; d* N{int i,n,num[20];
% S+ V- R8 d/ \$ { char *p;6 w* _2 K( A" t f# ]- ~
scanf("%d",&n);
) r& u3 _1 W+ k/ z+ v, i( u for(i=0;i scanf("%d",&num[i]);: l5 y; }" `: H# p: X0 P
p=num; ?0 j, ^% f$ s
sort(p,n);
. S5 o- F; i3 F% j: ~ for(i=0;i printf("%8d",num[i]);
+ X4 F, z3 h. x/ I! s) v}
& h, g9 l; T# T0 C% w/ bsort(p,m). B7 D/ J/ m, c4 o' Y. x3 L* i$ R! y0 P
char *p;
& t8 D% `, r4 Z- T xint m;# ]4 A% J9 I( g' G L3 Q
{int i;$ u- Z3 P' }; B6 ]" j4 W' }. U( ?
char change,*p1,*p2; Y- q" _" a! u- c" Q! C2 f( g
for(i=0;i {p1=p+i;
; |* G9 Z7 s, x' X6 _ M6 x1 j p2=p+(m-1-i);& f$ f% g2 W" Z8 H
change=*p1;
0 M. Z+ ^& f0 Y4 ]7 x* ^ *p1=*p2;+ ~( x- h0 |% A
*p2=change;4 v' U& M) M0 g; A/ B7 C: l# H6 m
}, J. E# v& W1 C' w4 {
}. F2 e# n, X) m0 J8 j& l2 H- I0 t
10.15
" _" y/ m& J9 K2 W- j0 t2 P$ h/ @0 Qmain()
, c+ T$ ^) ^) g. I{int i,j,*pnum,num[4];' f- {2 u9 o5 ~, U: S
float score[4][5],aver[4],*psco,*pave;! N; B+ o6 B; t' y5 H& U
char course[5][10],*pcou;2 A9 N- i$ Q0 h: Y) M& W
pcou=course[0];
% L$ c W! G3 f* F/ t9 I1 b, h6 p9 y for(i=0;i<5;i++)
$ o4 [5 \5 F, Q) s* l# G4 e scanf("%s",pcou+10*i);
$ |' R& b) n) f- f4 m printf("number");% \. V2 j Z2 |+ @3 @
for(i=0;i<5;i++)
0 s- r2 I1 X! S% E5 `: L7 Q printf(",%s",pcou+10*i);' {: n: I- I$ X) X( I: R) R
printf("\n");3 c. c3 v: r7 `( @: _( I/ _
psco=score;9 u& U; m2 O$ ?2 A" T" C
pnum=num;
5 N) ~1 U( n+ P- J! }+ G4 |5 G/ S for(i=0;i<4;i++)
( O& T4 ^, ^% t# v% S {scanf("%d",pnum+i);
" E! z0 c6 p% A& P u for(j=0;j<5;j++)
. Z9 L/ C# y. }; n9 Q/ I scanf(",%f",psco+5*i+j);
4 h3 p0 {. ~5 b4 A0 M } C6 x; I {: e0 ^
pave=aver;; j1 E; l, m5 h! U
printf("\n");
! W% y6 W/ q) O8 V avsco(psco,pave);" y+ _: L2 y4 u0 G' R
avcour1(pcou,psco);( Z* B; P- A* T' y6 ~8 X1 J, G- C
printf("\n");" i$ \( l; N9 w: }& u
fali2(pcou,pnum,psco,pave);9 Z3 m3 R& D! }& T* I$ c! R
printf("\n");
7 m+ m0 r4 ^& s6 ?/ t2 x2 x good(pcou,pnum,psco,pave);* w- s4 O! ^% k9 q
} R2 V: e0 u# `9 k. U1 n
avsco(psco,pave)
3 l; G X; q: N- Ufloat *psco,*pave;
6 y. @! Z8 J+ q( Y4 t# w- H8 I$ k{int i,j;
! d* I+ p1 s' f; l2 z float sum,average;0 L% G6 a2 n8 C7 Q
for(i=0;i<4;i++)
: I& Z" `/ P$ O& O3 S {sum=0;
' r% X9 R2 }! W* ^ for(j=0;j<5;j+)" G) u0 I8 }) F
sum+=(*(psco+5*i+j));
- F# W( Q# R: c% W# @8 n! q; T average=sum/5;7 b1 ^, z* T- A1 Z, f) k
*(pave+i)=average;
, l7 k. H9 W% l5 Q/ W& f0 @5 @ }
. H9 n6 t; o4 W5 |! }}/ B% h2 D7 c$ u, N! U
avcour1(pcou,psco). l6 ]# C0 r3 T$ ^! g
char *pcou;
. d, u1 n4 w" k/ u( ?float *psco;
# N- d" g: C5 C# t" y9 o& ~% ]- `{int i;+ t! k; e0 P$ O; l8 W! l0 `
float sum,average1;. u) M' ^; L! o7 u6 q- Y- Y
sum=0;7 {- K Z7 w, ~
for(i=0;i<4;i++)0 |" S: l% ?- D/ ]
sum+=(*(psco+5*i))7 Q! S) P; _! i$ [. n9 e. }1 ]/ A
average1=sum/4;- F, B# [& Z9 R5 t# n7 _
printf("%s %5.2f\n",pcou,average1);+ h' n, w$ @1 T
}
) G' m" h/ b/ g+ D' {0 j( G5 v& ufali2(pcou,pnum,psco,pave)* }; |4 C7 b. A! U9 g; N' S# ~
char *pcou;
" q2 G1 g( U0 `1 l5 n) [4 pint *pnum;7 _) L( \8 m2 {$ f e2 [7 k5 _
float *psco,*pave;
: j- R& }( a& D{int i,j,k,label;
' c9 q2 p" K" f* K8 U) X" V* p printf("\nnumber\n");
8 H; N" g( G4 k for(i=0;i<5;i++). _$ o: M+ A$ ]
printf("%-8s",pcou+10*i);! M) |% U+ V/ Z* a# r( V! C
printf("\naverage\n");, Q0 F1 p% s; D. d5 A9 L$ h
for(i=0;i<4;i++)* {& J' a a6 O$ h# A
{label=0;
! ?3 R5 q8 B$ \" K. N. X" p4 i for(j=0;j<5;j++); P/ d' }6 X* h2 x
if(*(psco+5*i+j)<60.0)label++;
8 j z" g9 s" k if(label>=2)# {7 j+ X5 ~# o
{printf("%-8d",*(pnum+i));1 \6 n1 U) j, @3 @7 `) T& E+ P
for(k=0;k<5;k++)/ Y, y& l, r- M# }
printf("%-8.2f",*(psco+5*i+k));2 H- K% ~8 e* j$ k
printf("%-8.2f",*(pave+i)); m# u" M8 `, m" D6 L$ ^6 K
}
1 j3 T2 N8 \( o9 q4 w, x) H# C }( ]' P; q/ \9 |: L
}
* u: q% N' x# A# F" ?8 U8 C: `good(pcou,pnum,psco,pave)
! U7 o* A" ^9 i8 J3 k# m# Fchar *pcou;
. g3 x& A3 @, [3 i( A. yint *pnum;) v: @: C8 x/ L2 D
float *psco,*pave;7 U! q5 Z: Y! Z n8 v4 z
{int i,j,k,label;! D' O* j# P; P
printf("number");
+ e, \+ D9 o4 q, W( T' u2 { for(i=0;i<5;i++)0 Q: H0 W f0 d1 X3 V( g- n8 j z& b
printf("%-8s",pcou+10*i);
2 f ?, o7 l$ `& ]+ f: F# z0 S printf("average");
2 h' I+ V5 s P* {' x" U, W for(i=0;i<4;i++)
9 G- {4 f7 f0 M" I {label=0;
& W! p- N6 Z! E0 \# `2 v! f4 M for(j=0;j<5;j++)
. K' @. d$ y8 P+ I1 Y! t: N7 D if(*(psco+5*i+j)>=85.0)label++;7 a) ]1 W" _, L( @
if((label>=5)||(*(pave+i)>=90))
9 J8 K! \8 [( p- E9 U' ^' \ {printf("%-8d",*(pnum+i));: |3 A- q& ]5 ]+ M0 k: Q: m
for(k=0;k<5;k++)
3 Y3 W) v. \4 S. {) Q& x printf("%-8.2f",*(psco+5*i+k));8 p$ c, z9 `/ _2 n
printf("%-8.2f",*(pave+i));) X, Q5 ~5 i9 \- l$ g" R# _
}
3 C( G7 k8 x$ z* W0 y* z, @6 |0 s& Q }4 t- I# |7 j- T( ~) t. F
}5 Y1 J ^5 ~ ?2 z% U2 v+ {6 m( q
10.163 n. ^9 e" y+ q3 A7 h0 K4 T: F; ?
#include"stdio.h"
A: ~6 \( H, k/ f* l( Emain()
8 h% Q l* ]% ?{char str[50],*pstr;
# K" Y' k5 K6 x. h int i,j,k,m,e10,digit,ndigit,a[10],*pa;
5 H+ `- W! ?/ ]( ~- x* ^( j9 K/ K gets(str);( q7 b/ h+ I/ W# z9 e3 w
pstr=str;
7 F1 X2 }) R6 v4 m pa=a;
- r5 l, @' m# R2 S ndigit=0;$ ?" g0 i! U% K& f3 ]- U
i=j=0;
* m; e2 j9 J2 {, A5 E+ `& @9 T, G- r* Z while(*(pstr+i)!='\0')4 ], [! D* w/ T* L$ e# t
{if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))
( f$ p+ L7 P/ h- n' e$ F j++;0 |1 o1 }- C0 w; z# w" A
else
$ ~- c+ e, R' z9 J! o4 S {if(j>0)4 b9 T" E) ^* ]3 b$ V& v
{digit=*(pstr+i-1)-48;6 V F. o. J8 s% @
k=1;
- ~7 c. s/ [' ~ while(k {e10=1;
. C; q) @3 o+ M. f3 i for(m=1;m<=k;m++)
# Z) P( P) Z6 {" n) ` e10=e10*10;
) {& n D5 d$ t digit+=(*(pstr+i-1-k)-48)*e10;
. p: v" T5 c c4 F# I, ` ] k++;
6 [- ~ |" o" } }
2 O: [3 H6 C. c2 v *pa=digit;* h$ p& D/ u0 d5 W, C
ndigit++;) z6 e9 k# t j5 m$ y
pa++;
/ P6 V% E1 A1 w8 V$ [, l l j=0;
$ q8 w1 p p c2 ^ w }
1 D1 e6 Z) M0 `9 M* q" |9 i }2 s( V3 t& v2 g6 L
i++;
) v. ]! A: s7 O4 m( z# T6 e- R4 O }
: F3 U0 P& B0 V6 F% ] if(j>0)
1 }$ [; s( ]% u5 M {digit=*(pstr+i-1)-48;# o+ S0 ~9 |0 R5 q0 ?5 P; s2 E
k=1;5 \# j6 i, a3 H2 a3 l( Y% t
while(k {e10=1;4 k$ O, [+ z, O' N3 L
for(m=1;m<=k;m++)
1 U7 ]) q/ f1 }) C e10=e10*10;7 E, e W9 F5 }* e6 f2 p
digit+=(*(pstr+i-1-k)-48)*e10;
) U% E" j7 C, o4 ~2 Y/ E" u k++;
: X8 ^. q% x8 g }
; w+ r4 ]0 n# y4 G1 t- A9 g i *pa=digit;
+ P' C% [' A+ S1 u9 R ndigit++;
+ b# k; d2 [8 z5 g/ x: |$ N* ? j=0;
6 V5 k' J) l/ Y' V6 V- |. v; h0 P } ' ]$ s, F( I4 F5 d
printf("ndigit=%d\n",ndigit);4 Z! n& b ^8 w r u" R! b( K- W
j=0;# Q1 t' N. k& {8 E- Y x
pa=a;
* ^2 H r1 f' E* A# G/ J for(j=0;j printf("%d",*(pa+j));
0 }$ }8 Z( v6 T! b% r U4 |8 j}9 d7 T! y/ [7 x
10.17
# U! g1 }: G( I( e; Mmain(): ?) B: n/ t2 y9 {
{int m;: L K& ?8 L! B1 D0 a
char str1[20],str2[20],*p1,*p2;
) w# i# w+ d/ ~8 v scanf("%s",str1);
: D p9 E( z1 ^# S% L scanf("%s",str2);
, {) P# Z' U5 \# o0 m' A9 x p1=str1;$ k7 S+ |) H3 |
p2=str2;
* t0 Z6 S, b: L" I m=strcmp(p1,p2);
; `9 U g: Y: h# H' E printf("%d\n",m);+ h+ m' I7 |, ?( w* d7 M( W
}
5 }- f4 v, e) n( R, N$ f9 `strcmp(p1,p2)
: T: e/ ?3 N( m5 f% z! Pchar *p1,*p2;
L [, @) T$ c8 Q{int i=0;
& }4 |- b7 t1 g; P while(*(p1+i)==*(p2+i))
. s2 e" p& k- |' d2 p) d* v if(*(p+i++)=='\0')return(0);& F2 @5 u; H* y
return(*(p1+i)-*(p2+i));" k9 z! L1 T# ]2 b* T+ ]$ f# c
}
6 n3 V2 s7 e R6 P10.18
* o# M6 C2 I9 j7 |) u# v2 B; \main()' K+ N" O& L+ l1 t
{static char *mname[13]={"illeagl","January","February","March",
, ` @3 ]" f( V. i "April","May","June","July","August","September","October",4 g. z7 N0 g3 A& W- u
"November","December"};
. b/ q' r& p- y' { int n;% j3 z* X$ M- r4 n; _6 J, u
scanf("%d",&n);
8 i f/ D1 r0 w8 B$ P7 i5 m4 r" m. U7 s7 [ if((n>=1)&&(n<=12))# c+ A3 y, }9 F: [6 d/ \
printf("%s\n",*(mname+n));
5 B0 ?9 a" F) j8 c- I6 w else
/ x% U' V: M! m3 ~+ ` printf("error");
8 J1 X7 M9 f( l7 Z0 [2 b}
: t2 g& I" q7 L; Z10.20' x$ f! j. h- ^, \6 q
main()5 D5 y$ [' d8 k; {7 H/ @3 F1 |
{int i;
4 V$ {* q! b2 Q5 c char **p,*pstr[5],str[5][10];
1 |9 X$ R: F5 y" T for(i=0;i<5;i++)
) J7 E& L5 K5 A pstr[i]=str[i];! ~% A( w6 d" b
for(i=0;i<5;i++)
. m9 {# V0 n3 b$ F" B3 \ scanf("%s",pstr[i]);
% Z4 Y- G) a5 J& s% m. P: G+ j7 N; x p=pstr;' _! D- g& v8 l
sort(p);8 w1 X* d0 }+ h$ N$ \
for(i=0;i<5;i++)5 C4 L# I3 W- `4 k8 c* z
printf("%s\n",pstr[i]);0 J5 W- n. z+ z* P) \: R3 F
}" Y# ]& t* m0 K) m3 m3 i
sort(p)
: G) p( n9 E1 g' Ichar **P;
% u5 }6 x- O2 m3 \! {( \4 v/ {{int i,j;' @. c/ b6 G. X! U' z0 z9 ~# P
char *pchange;* r6 Q- [1 I5 D3 Y: V/ N
for(i=0;i<5;i++)0 h' Z3 W8 r- b) a; L
{for(j=i+1;j<5;j++)
2 a* v ?& z5 V& d& v: i {if(strcmp(*(p+i),*(p+j))>0)( s4 i- n7 `6 Z k% y, ^: ]2 F
{pchange=*(p+i);: a4 B- H1 Z0 {
*(p+i)=*(p+j);
3 u9 x" w) b0 p" v9 d3 J% ]) T1 N *(p+j)=pchange;- @" f% [% H- v+ B4 r
}4 u& }+ ]; @# B9 {9 j/ e
}
/ t! }' K7 I9 R, {$ l5 J3 r }
1 y7 J; q' `/ J9 I1 R}0 S; I% e* q9 m
10.21 v5 P6 p! Y) z3 g; W; `
main()# M3 V5 f- B! @1 z, R
{int i,n,digit[20],**p,*pstr[20];
) p7 F) {, F2 s5 k$ {* L7 O scanf("%d",&n);% k$ x: K# i6 I+ J4 z5 |) d p% E
for(i=0;i pstr[i]=&digit[i];
9 H( |# I% X6 Y9 X- Q for(i=0;i scanf("%d",pstr[i]);
' k6 Y- ]) J# o2 t+ A7 Z! T9 ^) a5 B p=pstr;
2 D- s5 l: j! r, C% V8 O sort(p,n);
' o# s! O% l9 C3 r; e for(i=0;i printf("%d ",*pstr[i]);
2 m4 D1 B7 t, } m2 B( x}! `8 e( |5 K" U8 c. h' Z
sort(p,n)
8 t3 t* y* n2 Xint **p,n;- H+ y U- L! M
{int i,j,*pchange;) w: V3 }! Y/ n1 F# j. |
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))6 G$ o! P. E7 a
{pchange=*(p+i);" a& q0 N; a3 G! d$ l. [
*(p+i)=*(p+j);+ t: R: N2 V q4 |& C
*(p+j)=pchange;
+ m3 d5 W' |5 {9 J8 d1 w+ r3 M }
9 E7 e# t2 ^1 L4 Z3 O }
: j8 M) n0 S4 h6 X$ b8 [: [% Q& J }5 F" Y" P( l& a: t% l" N& S6 W1 |
}( ^1 [0 |3 H+ c+ t" ]. t, z5 [; N
第十一章 结构体与共用体& _9 j/ v4 p, g3 q5 Z/ c- O
11.1
1 {7 ^" A0 M q# m& u/ Ostruct
4 b& X# a' R$ k2 P5 D {int year;
1 d8 n3 B- f, }! H, g8 _" X' U3 o int month;! y( g) W, a7 }4 p3 ~9 J# f
int day;
+ N9 Q9 Z W+ I* M4 A4 A }date;' W' n& m R1 }8 M) w! g
main()
# q8 I0 r7 `1 d/ N; z' P3 ^) H{int days;! B& [: Q0 ]: b3 S- m/ s; Z
scanf("%d,%d,%d",&date.year,&date.month,&date.day);
3 H4 M# @: W+ `- p+ \ switch(date.month). W& l* f' X( h% _& @$ v5 V
{case 1:days=date.day;break;
6 p& ]. c0 [6 B- ?) u% G/ U case 2:days=date.day+31;break;
1 p. V3 e: K( h/ f$ @! {5 A" Q% f( h6 ? case 3:days=date.day+59;break;
: H- L0 Z* E# @3 N! B case 4:days=date.day+90;break;7 C) {9 f( R, Q. q& q! _1 i
case 5:days=date.day+120;break;% T! c2 I6 G. B! t) W
case 6:days=date.day+151;break;
3 ]) ]+ @- c p( t+ P case 7:days=date.day+181;break;: n% F7 @! r: o' m" `
case 8:days=date.day+212;break;
3 t o* F# G1 p( ]( O8 l" b case 9:days=date.day+243;break;
2 T V9 R. X* U9 r$ e! F case 10:days=date.day+273;break;
& J5 ^- T# N7 f6 G& [) B/ b# f case 11:days=date.day+304;break;. z# z' o: ~- \1 n
case 12:days=date.day+334;break;
* Q# c3 M7 B( F }8 m4 K" b* b$ u7 U" [9 h
if((date.year%4==0&&date.year%100!=0||date.year%400==0)
8 o& H2 Y/ f2 y/ @8 @ &&date.month>=3)
- B3 _$ {5 [# h8 q2 S6 M days+=1;( c3 P" ~# B3 q R7 K/ o9 q
printf("days=%d\n",days);% p0 q: ?. m }& M% i8 ^2 n
}
, V" E* Q& B* j, u% k, W) J11.2
: [' }3 b D z }3 y4 g$ x$ Rstruct dt7 U6 u( D0 G' m# e4 D N
{int year;# l% j% l$ V+ w6 c) p' k) c8 p
int month;
; M# N, T& n! j" k0 x5 f0 P int day;
8 J* U4 w" g4 L( [ T: [ k6 } }date;
: @8 R* }4 N* f9 M) ~+ M' E, D) Jmain()
/ Q& c& W L. H1 R1 ]+ I: f{3 G9 X$ s7 c6 m8 s4 {9 d, ]
scanf("%d,%d,%d",&date.year,&date.month,&date.day);
+ b% _9 B) |* S7 c7 y5 Y printf("\n%d\n",days(date.year,date.month,date.day));
9 b% F% w$ n! e, q: b% r$ {) u! G}. ^* [& Y" Z: H+ s) o3 l
days(year,month,day)
6 C9 T/ g; _! Vint year,month,day;5 w* A" M/ D6 [5 Y& A. d& R
{int daysum=0,i;
$ _& a2 C, I* s static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
2 b1 J' I6 b( h8 U. _. l for(i=1;i daysum+=daytab[i];
6 I% c4 E$ \) G b5 v6 r' y( I daysum+=day;
7 X" O: v- w2 O W- A: w0 H' U M if((year%4==0&&year%100!=0||year%400==0)&&month>=3)9 d p1 G' p5 i6 S
daysum+=1;1 O- Q* `" k* V5 h
return(daysum);
: l; N; ^* l$ |3 L( ~+ o) v}
8 @- f* r) l4 ]6 Q+ P11.3
4 T- n" y( u5 m2 Y% W2 N4 F11.4
+ z6 O; x- g! [% g' D#define N 5
$ {1 a( k; q9 c: Y+ l2 j0 w4 J, ostruct student$ C7 c: h4 z1 B6 P2 d' V; Q0 r r* v
{char num[6]; T3 k% A/ j+ f8 _, A E# d
char name[8];1 `' W `: `( }' g
int score[4];
7 j# L5 t5 j* p! F. n' X }stu[N];; ?- W" L# w; H; z& f9 `) c* e
main()
& q. u- ]: C" e% Q7 y: W" I$ b3 i{0 v/ w3 d, c. y8 _+ ^ c# {
input(stu);
( A1 {7 U/ p/ Y6 i% B print(stu);
, j/ m: B: w- \1 O6 i; N$ _% j! M}
" e/ e2 Y6 _( _ q i* Sinput(stu)
; q- X" _! _7 \$ B# cstruct student stu[];0 g/ B0 k2 c5 S: k) d7 K
{int i,j;
- ^5 j' U* L$ x' J5 B8 E: |* Z b$ K for(i=0;i {printf("number");* `' w4 w" T8 @ L/ _$ p
scanf("%s",stu[i].num);
& H/ T2 Z4 z) Z" D( k& M" k) Q printf("name");
3 x+ Y) ?# _ i' U9 | scanf("%s",stu[i].name);( n* Y! ~ ]8 A: ?$ S2 E
for(j=0;j<3;j++)7 t" ]8 G6 ]" D% |
{printf("\nscore\n");7 j! w. z* |! r/ `( J7 Z
scanf("%d",&stu[i].score[j]);
8 {# m. R" ]7 p' e; M w& g }
5 _ `# h S4 x( H: N' c! L printf("\n");
1 a7 D1 u* ]* Z3 w# P }5 w+ Z8 Q( T; F! @
}
' J1 J- a0 K5 F( Gprint(stu)$ Q! }2 ~3 k) K" F1 K4 x0 `
struct student stu[];
w( J7 ]( O! ^* j. l+ z{int i,j;
# b! H6 }' } Q7 E7 u printf("\nnumber name score1 score2 score3 \n");6 v3 t6 q$ L3 {9 [( S8 G
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);2 M7 _% e# J/ ^. X
for(j=0;j<3;j++)" |8 f* R/ ?9 O* g7 S2 R4 q2 J
printf("%7d",stu[i].score[j]);5 v. t% o3 \4 U; M# l9 k
printf("\n");
' ~# O- k6 c/ [ }
4 v0 d& k8 s& ^' _$ G' j& e% H}4 u" A! l' A7 ~- Y, w
11.5 A' D- k1 E: N6 Z( A" o' `9 e
struct student
+ K6 Z9 {% ~) d4 w8 q6 Z" w' d {char num[6];7 ]5 c* ^( u$ g4 U! d
char name[8];
- N' T0 h+ x8 x int score[4];
+ Q5 A& o; y7 j5 E float avr;3 R* j+ y0 v! ]: Q
}stu[5];
* _1 u* F- e- O- W6 ]8 dmain()1 V( N- f+ q: y# K& b
{int i,j,max,maxi,sum;
/ M" n, Z) t' E4 @ float average;% S0 ^; z4 J% h8 C' }
for(i=0;i<5;i++)
4 X. a1 f8 j M# ?6 ~: m# r {printf("number");4 k* [! m* @4 Q3 }, ]' N0 E
scanf("%s",stu[i].num);
% ?: G1 c4 U8 m5 E printf("name");
2 q3 e- u+ }; h/ T scanf("%s",stu[i].name);% B- }+ j- @/ O8 U9 v+ e- u
for(j=0;j<3;j++)) K" \' s! n4 X9 u
{printf("\nscore\n");
- {9 e+ B9 p f+ K0 A* g0 [( _ K scanf("%d",&stu[i].score[j]); u) y, @$ Q1 b
}$ l7 z# ~- R6 o% K$ j& g' ~4 x" v: X
}4 D0 q) H1 l. Z" P, e
average=0;
% l: |! Z; C6 l" G9 V% P; [$ Z- L max=0;
. z% ?+ @. W5 I: M maxi=0;# L, ~' l# N* m) J
for(i=0;i<5;i++)' a. u* O! P# p- K# I6 w" k
{sum=0;& D3 g2 i5 z1 L: ^3 i
for(j=0;j<3;j++)& }8 A. k q. S- @+ k
sum+=stu[i].score[j];
6 _$ K, I. n& [) k% Y. f. J$ k- J E stu[i].avr=sum/3.0;
% Q+ I0 @/ |% H7 b+ O' I average+=stu[i].avr;) Z* [7 m0 z. h m) e
if(sum>max); M- w3 ~! y$ h: V
{max=sum;9 k( ~$ A3 K; w( {4 Y5 k
maxi=i;8 z! [- B7 t# B5 g9 H X: a2 w
}
, [' }% N: t5 P8 G! j" ^, ? }
4 W. G( j, x- v6 g6 v& T6 U average/=5;
' g3 l8 U7 O; \6 r printf("number name score1 score2 score3 average\n");3 Z3 Z7 G( N$ @) |% @2 C
for(i=0;i<5;i++)
, P; H1 r. L- b" b% x' h6 w/ h {printf("%8s%10s",stu[i].num,stu[i].name);2 i" [* U* {/ T' i' a. X5 F
for(j=0;j<3;j++): u' n8 @7 Y3 M8 A& |; C! j6 t+ I' Z
printf("%7d",stu[i].score[j]);* A5 H: i4 H+ D& }/ e
printf("%6.2f\n",stu[i].avr);
3 L+ I$ M& ^5 O8 y2 H {! o }2 ~% D. Y: @3 [/ n# z
printf("average=%5.2f\n",average);
$ e4 Q' }' ~' g9 C7 [' E printf("The best student is %s,sum=%d\n",stu[maxi].name,max);5 S1 G) w8 j) y% K6 L
}7 h6 a0 E! x { m8 P( j
" Q a: L) _% Z1 y; S; G
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zan
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