- 在线时间
- 28 小时
- 最后登录
- 2016-10-15
- 注册时间
- 2010-2-26
- 听众数
- 5
- 收听数
- 0
- 能力
- 0 分
- 体力
- 2905 点
- 威望
- 0 点
- 阅读权限
- 50
- 积分
- 912
- 相册
- 3
- 日志
- 11
- 记录
- 13
- 帖子
- 56
- 主题
- 5
- 精华
- 0
- 分享
- 27
- 好友
- 16
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
3 J P% r4 C" e0 U1.5请参照本章例题,编写一个C程序,输出以下信息:
$ Q9 f. Q! I9 U* p: H6 K/ F, Xmain()5 v3 q7 Q u: W/ v1 P
{* Z4 f2 l# p# s
printf(" ************ \n");. m5 c) D+ o, `3 E2 P
printf("\n");! P2 a/ m4 K: P9 X9 u, t
printf(" Very Good! \n");6 z5 i/ O1 R( A4 d3 D
printf("\n");: `; ?; Q0 w" h8 f) S, \/ s& [
printf(" ************\n");
! N5 a% M0 ]( O P- O3 I}
# k$ N$ G2 g* l1.6编写一个程序,输入a b c三个值,输出其中最大者。2 C" g8 b" Y+ w H) U w
解:main()
2 \" {/ ]! z) L0 Y( X' R N* N{int a,b,c,max;& O2 g+ \# d5 m3 \* l) W
printf("请输入三个数a,b,c:\n");0 X1 N: k' _# E+ p# x3 n
scanf("%d,%d,%d",&a,&b,&c);
' H/ C( k, L& |3 I2 A1 Hmax=a;
4 T2 i& x. L, o6 `7 lif(maxmax=b;
* x+ m4 Z- U8 w. {6 N, yif(maxmax=c;
" |0 Y4 m1 [6 X& e% b9 cprintf("最大数为:%d",max);
% X( l+ u% ^0 p. y}
0 M* {, D! x% M$ n# q/ k9 @第三章2 m6 c0 |% w% v) V
3.3 请将下面各数用八进制数和十六进制数表示:
1 w. y" Y" R6 n1 E' B# [(1)10 (2)32 (3)75 (4)-617% ]7 t7 K" S# o. J
(5)-111 (6)2483 (7)-28654 (8)21003
+ b' T, N+ _- U/ N4 ^! ?解:十 八 十六
1 x1 W2 {$ t# F (10)=(12)=(a)
+ m* o: o9 A4 ?5 z0 O (32)=(40)=20% m% i7 b. L4 }* ]+ S. B' A w1 b
(75)=(113)=4b
) j; P. u' K; K (-617)=(176627)=fd97
) Z: q" z! z) o1 o5 } -111=177621=ff91! {: b, c& F; M& f# d
2483=4663=963$ ]0 K9 G# G$ t; a8 | R) F
-28654=110022=9012
/ `. j. f* V6 e" T 21003=51013=520b v% o: z# g6 _+ }$ P
3.5字符常量与字符串常量有什么区别?6 b2 N$ x; Q# }9 K6 l
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5 Z: v# Q. Y, A% }) W( u3 O- O- x# K
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'./ R- q( b; V" x* ^/ G. i/ t o0 g
3.6写出以下程序的运行结果:
7 c1 k# |3 u6 G; m7 {6 Q#include
. W0 D4 X1 r; m! l- g5 d% a% Pvoid main()+ F* c# ?4 H# C$ t- X1 _
{
) K$ x& x1 i& Ichar c1='a',c2='b',c3='c',c4='\101',c5='\116';
& V P, V8 O' ^7 O6 F9 pprintf("a%c b%c\tc%c\tabc\n",c1,c2,c3);9 m8 E* P* m3 `8 w0 ~
printf("\t\b%c %c\n",c4,c5);5 b) m' X2 Z1 `( F1 X7 s
解:程序的运行结果为:
" ]; C: n3 [) C6 v" m# p7 P# b3 ]aabb cc abc' q6 h9 l* C) O! E5 Q* ^
A N
5 t/ |. N {+ u' Y5 W3.7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,% K- `2 o9 j" ^. }/ @- e
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".$ @1 ^7 v' l& P3 ^
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
, U+ }. b. n* N; j( D- ~输出.
e. Y$ ^1 K% q. Smain(); N. t2 ~" p+ d: h9 a
{char c1="C",c2="h",c3="i",c4='n',c5='a';
$ x& ~9 N* f) w1 v. W% A; p7 \c1+=4;1 M( X8 j3 |( N& t' C
c2+=4;( H) `" C! D g0 b/ V1 G
c3+=4;
2 a$ Z& ?+ ~$ Gc4+=4;
g! E5 K1 I- f0 vc5+=4;8 q5 A; S2 l5 X% M. m
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);
6 T' Z Q7 o0 _ t* k}% y8 Y* N* N# r, Q& R9 p( z
3.8例3.6能否改成如下:
3 ]1 E3 j) l% M2 M9 l9 O#include
: `( R, n8 B: Y) ?) o! _. f1 Jvoid main()- y# w1 k0 w6 V& t; @/ \" E
{
9 U! |0 p, d* ?/ g$ Mint c1,c2;(原为 char c1,c2)6 H P4 s( i5 T1 N! t
c1=97;1 n0 H" c( H$ f+ V4 W
c2=98;
' d* g- g. c8 Lprintf("%c%c\n",c1,c2);( p' w b) o" V9 P
printf("%d%d\n",c1,c2);/ b8 t0 ~0 v/ v
}* D$ F0 Y, R9 R; w! c; u
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.
% m) J, G8 f( B9 ~; ~, e# {9 X; w3.9求下面算术表达式的值.$ M0 H% Y7 t" ~! G
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
! v1 b+ W' `9 D8 i5 x4 f, l# [5 s(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)% ~; G% f6 E1 d1 A, ]0 H
3.10写出下面程序的运行结果:- L R- h" B+ A
#include
z! l! o# O R: p2 ^. `7 `void main()/ Y* p& q2 P% J% t
{
5 ?8 u5 g$ A( E2 [0 p, Wint i,j,m,n;
& v) |+ M6 X, a0 Z* c5 ai=8;
2 j5 c/ ?0 ]' B* sj=10;0 S F4 D, \) ^" h3 z2 I7 [! [
m=++i;% Q3 I5 ^; z m# V9 }* j' F$ B5 Y1 v/ Q; s
n=j++;
& I+ j/ h& [ D- m7 f5 g/ Zprintf("%d,%d,%d,%d\n",i,j,m,n);$ _6 I2 J" Y& L8 C) O* s
}' i# Y' j# N* S: x5 X3 [* F
解:结果: 9,11,9,10
* `& T1 e) E: ?7 ?% y% o第4章. g' H1 h4 _. p; Y
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, A0 O. T6 Y$ `. I; u: R
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:2 S4 H1 N: e1 m. i3 y& _0 x
a= 3 b= 4 c= 5) D: g6 u! {/ f5 _5 Z
x=1.200000,y=2.400000,z=-3.600000
0 E: i: q' S4 R9 `x+y= 3.60 y+z=-1.20 z+x=-2.40: f! G" ^: H+ G' F' M" U. J
u= 51274 n= 128765
8 H2 v; ?1 M% i& S+ cc1='a' or 97(ASCII)
9 O) i: ~- D0 p0 f& tc2='B' or 98(ASCII)
" ^5 I6 M K+ [: I& l解:
2 n, I/ u' z+ V2 z# F' Bmain()5 [2 E* E+ i- }- z/ P8 ]
{+ x2 ?4 P t, I6 h: q
int a,b,c;6 ?& y1 U0 @6 s$ p
long int u,n;$ F, c, u+ U7 w/ [
float x,y,z;& y4 D4 E+ m9 p' ]5 N& X
char c1,c2;) v' s- b& ?9 g; O( K% j9 C
a=3;b=4;c=5;
) h% m6 M1 Q2 A$ \! C6 _. ~" Tx=1.2;y=2.4;z=-3.6;! k- ? r7 i, g/ N1 }
u=51274;n=128765;
, E/ ^2 l- y% p: Q# Ac1='a';c2='b'; N$ v# a/ L& R" ]
printf("\n");
5 D5 ^ |6 q% i2 ~printf("a=%2d b=%2d c=%2d\n",a,b,c);
3 x; p! _6 r' X% y- _printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
. E4 j8 {" T* P* [2 k: fprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);/ T8 @0 O3 K a
printf("u=%6ld n=%9ld\n",u,n);
: E8 [: k/ [+ [; B* J7 \! Cprintf("c1='%c' or %d(ASCII)\n",c1,c2);
& P6 q4 j( Y: q: d5 U7 k: Qprintf("c2='%c' or %d(ASCII)\n",c2,c2);2 `$ Q) ]+ @. _. Y- f* z5 [
}
, ^5 _! {" }6 O7 K# \* p q4.5请写出下面程序的输出结果.
. f$ ~% o7 |( ~ P; _: p结果:
" ?) y0 S A# ` K6 l2 r57 f9 S; S) X H' Q: E% J0 k: o; u
5 74 k, @- r' j3 P
67.856400,-789.123962. ?; U2 G& _& d p
67.856400 ,-789.1239627 e- r" Y- z% |+ a9 m4 a; X
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962
3 A( D7 n4 F: S3 j6.785640e+001,-7.89e+002: r. d! X+ K% t: y& Y D
A,65,101,41
: c. A% }. ^% n4 _1234567,4553207,d6876 p' h9 I& W6 h Z$ e9 y
65535,17777,ffff,-1
3 x! \: y0 D; ?( PCOMPUTER, COM
8 ?( @9 H# e; J8 N8 y& L4.6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',
6 O' H. z) r; q9 k( }" s问在键盘上如何输入?) W& T `) [* v. \0 Z8 |" f. S1 q
main()4 `- A& t4 R+ w" u0 Q3 q8 B
{
, A% k1 @) v6 F% Mint a,b;0 F" _* v& |* O
float x,y;) Y4 Z, E- Y! |% G- ~
char c1,c2;- T8 u7 |5 e) F
scanf("a=%d b=%d,&a,&b);, u O) O' [$ U1 |2 f7 J/ m+ W1 t
scanf(" x=%f y=%e",&x,&y);
- ]( R6 G. {' N% O. l w3 W: f$ _scanf(" c1=%c c2=%c",&c1,&c2);
3 C+ y) w& }, a2 V0 w/ X) o0 H$ n}
e% X. y3 M" o6 V解:可按如下方式在键盘上输入:% x& W! r5 M& x0 y. y, p! b; f
a=3 b=7
4 k% R: g! z- r. n7 k, }1 ^$ Sx=8.5 y=71.82
5 K1 {5 o6 J- w: M3 i1 H# `& }c1=A c2=a; D8 ]/ r* z- U
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
6 P2 d3 O) \ A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
5 p- o% H. P* k; o+ T# J. {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 Q( T1 v# U3 Z1 ]; u- a D. |# ^
初步概念即可,以后再进一步深入理解.. M' W" G0 d$ A# s, ]
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=- R, ]; L# h: f8 s; k% O5 f" x
3.75,z=57.8,请问. b8 g9 H I" N: X+ e1 r0 o
在键盘上如何输入数据?
- O' X9 A5 c% }) h- n3 Z8 qscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);' _, i; n3 a- _ m3 [1 j- N. o
解:
- h* t* P o+ fmain()9 |& o; X+ n% h2 R( m9 T
{# |2 r3 t b, q4 a+ S" {) P
int a,b;
: B& V9 v; @. B) s' B8 S0 _float x,y,z;* B9 P2 B3 O. d% a* @
char c1,c2;+ a1 v" B7 D; p8 }
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
7 J. O. V( ~/ u* `3 [( D0 t}
$ N8 C5 W9 c( j( e% }运行时输入:
4 }. s" j+ b8 V5 ]9 U1 `10 20Aa1.5 -3.75 +1.5,67.82 k) S9 R; e) R w
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/ h; a1 S- r7 T2 f9 U: p
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7 C( n3 o9 u5 Q" F2 a: |0 @$ z
会赋给任何变量。
1 i: e% ^4 V) Q7 q* n4.8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,; F8 m' O( Y/ J* z4 w+ N6 n4 q
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8 B8 g3 i% j2 }. i! I
程.# I! A) ^; {5 `
解:main()
9 W* f! h. }5 N$ e{0 e, [, l+ ]% i, R
float pi,h,r,l,s,sq,vq,vz;" V# A( d" `' e) b6 d
pi=3.1415926;- c; K1 `, Y1 I/ m; E# F( o K. n1 D
printf("请输入圆半径r圆柱高h:\n");
2 ^3 ?4 P# C& C, pscanf("%f,%f",&r,&h);
9 t% ~ s' y( E2 L$ _6 ~! Wl=2*pi*r;4 V4 w& x1 N9 S/ _( y
s=r*r*pi;
: V1 V& ~1 Y4 a" v0 ksq=4*pi*r*r;6 M! i3 T- C8 C4 d! ]
vq=4.0/3.0*pi*r*r*r;, b" I/ U1 ?! |" u1 {6 B
vz=pi*r*r*h;7 L, k7 J- e( g. Y( n
printf("圆周长为: =%6.2f\n",l);
- ~) ?" r( B( a: ~printf("圆面积为: =%6.2f\n",s);
l4 i/ o" B) ?printf("圆球表面积为: =%6.2f\n",sq);
9 y1 v5 R6 r2 `( M" Q0 \3 Uprintf("圆球体积为: =%6.2f\n",vz);
9 z1 T1 ]6 f, k8 C0 u+ \( Z}
/ z" H: N8 s! x' O3 ^. o4.9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,+ w$ v6 C3 F+ Y' Y- c" k, v$ J: a6 w
取两位小数., l- n4 [" [3 Q1 n# G
解: main()
4 J# j' Y a9 x. i( F{
! U7 {4 b8 ^* g* Tfloat c,f;
6 Q5 h2 r5 j1 F* j2 @% I" H Fprintf("请输入一个华氏温度:\n");7 h7 O x, T0 v& J
scanf("%f",&f);9 z5 X1 C( b( O3 l$ V. V
c=(5.0/9.0)*(f-32);3 @; A( q0 {0 c. B y) b; y
printf("摄氏温度为:%5.2f\n",c);
- b2 s& Y8 G# _! w! y}, G+ |3 z, k6 h
第五章 逻辑运算和判断选取结构
3 I* e! s/ z& `5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
7 x( C$ V- [2 G2 R5 T; Qmain()
9 n9 @ s8 R6 v8 O+ q0 q2 L* v{; G$ q! m O2 [
int a,b,c;" W$ ^$ o, c6 [- ~! D! p2 X3 n
printf("请输入三个数:");
* y; Z" T6 T, C4 B8 ?7 V W+ C$ q) k; Uscanf("%d,%d,%d",&a,&b,&c);
& t/ P5 n% @* F; R5 sif(a if(b printf("max=%d\n",c);8 v6 J6 z$ K6 P( A! a! s% m
else
1 a2 ?6 u: `4 o. B5 H+ B! h printf("max=%d\n",b);! r! W4 ^6 `6 k9 v( R5 U" l
else if(a printf("max=%d\n",c);
/ G% J8 [5 b) S1 z# Q c: ] else: N+ `! N6 u9 h# l ^( c6 {
printf("max-%d\n",a);
6 f- B- F/ n! X' Y}6 |) a4 ?( U9 M( `$ ]# m% H
方法2:使用条件表达式.
5 b4 Q# Z; S6 j5 F2 {main()
7 S; A8 u: W; c" L! F. Q t$ I5 q, n{int a,b,c,termp,max;
+ P2 @. j" E* q% U9 ~$ ^- d1 _4 o printf(" 请输入 A,B,C: ");
/ K! P2 f- B4 i4 | scanf("%d,%d,%d",&a,&b,&c);. B0 s0 f% D, [8 Z" e4 I" y) O
printf("A=%d,B=%d,C=%d\n",a,b,c);
5 t; A. n) U. O4 ^8 \ temp=(a>b)?a:b;& I# z( Z2 x- h7 k6 Z
max=(temp>c)? temp:c;
( Q. C# ]5 T2 X- F$ F2 }8 U printf(" A,B,C中最大数是%d,",max);4 i5 S9 w( ?5 _0 |2 B; n. K) W
}. o7 z; i" [. L- ~ P
5.5 main()0 _! H) O5 N, c4 _# q' J
{int x,y;
5 K4 P) Z- O6 t9 w7 u, Y, t( vprintf("输入x:");
2 _. K0 L, I( K3 _7 a( p- ]6 Tscanf("%d",&x);
9 F" o# a! J- Zif(x<1)
3 c _( w8 y' I1 S+ i) @ {y=x;
; g* s; g% _$ }: X2 B8 _* p4 Y printf("X-%d,Y=X=%d \n",x,y);
3 h# H- R! V% q0 h- g }7 u+ k A, h$ y6 Y( b3 v
else if(x<10)/ [9 |' h7 R9 g. c' Y9 ?
{y=2*x-1;
! ]. w1 n. l* }5 e5 q printf(" X=%d, Y=2*X-1=%d\n",x,y);
8 @& F* o. r& U }; }& }) B: E1 N) y
else
" p7 c r- D. M; `' R {y=3*x-11;
9 ~2 e( A( T: }9 M& {# \ printf("X=5d, Y=3*x-11=%d \n",x,y);. s' K0 u- Z* U& C, z" D- y i
}. _( z: u7 \% p, x3 J
}
% X4 P G; G5 m2 x( }(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!
: X( o/ u- M% O9 i3 A5 P, zvoid main()0 v. I' ]& I& ~! A1 w% c
{
- U- m5 K: z; Y# vfloat s,i;4 y3 s( Z2 K' n
char a;0 Y2 P9 p* p( A
scanf("%f",&s);
8 g0 u3 S* A4 {( C _% t9 K1 Lwhile(s>100||s<0)
8 z& ?+ T# |5 v' F3 f3 X7 t{
$ k, e' {7 C7 I: lprintf("输入错误!error!"); z7 x8 T( z8 a
scanf("%f",&s);
6 {+ |8 A0 s) \5 Q M# j( j}- \2 P7 W2 G" _
i=s/10;
' V3 Q' U4 P9 I& Rswitch((int)i)
! @! s, Y0 V/ C1 B% w4 @{- N1 A U9 b. C3 V
case 10:2 x( n$ E6 R" G8 m$ ^% o8 T p
case 9: a='A';break;, h5 b* p0 t8 R" M2 |
case 8: a='B';break;
+ I# \0 p% T" x# X: Qcase 7: a='C';break;
' D7 {5 V8 p! D9 w" g5 K. Ecase 6: a='D';break;" }$ A; Q& r2 b) ~) u" F4 I, d
case 5:5 V2 Z" i6 {: j ?. ~- K- z
case 4:
! B. J" G+ ?% I0 D$ v3 e" d6 m, t2 Kcase 2:
Z( T0 P5 {2 {! D N6 Ecase 1:
5 N7 L! F& \) o x9 a. Dcase 0: a='E';
/ Y! ^9 j2 d# a" Y: Q}
/ r9 r/ \+ _9 s2 Z' eprintf("%c",a);
1 l: w6 T3 x d- E6 J}* ]/ W- u" p8 C1 ^) }
5.7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3.4 W% A( I1 i( _, m
按逆序打印出各位数字.例如原数为321,应输出123.
2 p& j' t2 O/ G, s9 H% ~% T6 ]main()
4 A7 d( [! `- Y% \! ?. | {
) `& U3 b2 m) q6 a% G1 k long int num;
5 s% d" g4 k( S3 h* v; }, Z int indiv,ten,hundred,housand,tenthousand,place;# t2 o2 @: h: N* P
printf("请输入一个整数(0-99999):");
# M6 D( S) y: j5 l9 y* v scanf("%ld",&num);5 a+ ?4 a. o- {, J+ {8 s7 O
if(num>9999)
0 u7 {7 c, P2 J! t2 N5 a+ s" n place=5;
/ D! [' I! F7 eelse if(num>999)% u: D6 q9 @" m8 u& n. o
place=4;
# ^2 q4 b3 w9 y, d' c- z+ ?else if(num>99)
4 [0 i' r, D+ V place=3;( f+ X0 A C- h6 w
else if(num>9)
" ^* m- m; Q+ K8 d6 T place=2;# q( M0 R7 V* l5 p7 z, a. Y
else place=1;3 }9 @" n, {# R9 O7 }
printf("place=%d\n",place);
# F0 {6 i" a& i8 J! wprintf("每位数字为:");
* X$ Z, T' R# ?3 y8 }# Kten_thousand=num/10000;* S9 d: l9 ]) z
thousand=(num-tenthousand*10000)/1000;
+ \! V2 g/ A) y& rhundred=(num-tenthousand*10000-thousand*1000)/100;3 B" N, ?8 V4 m/ U
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;7 u* L$ F+ j7 B" e3 p
indiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;/ }, p! U2 P$ \
switch(place)- H9 P) C5 U t. V
{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);, p0 R1 w* Y6 x% z3 b
printf("\n反序数字为:");% g' f; h8 T0 l- A$ p
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);5 @ S4 ~3 b. _) @9 h8 Q7 m
break;
3 b& Q* I8 S; y) ^case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);; Q( \9 T: y* ?
printf("\n反序数字为:");' Z ~2 w' A; ^' B( _/ h
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);8 T) E$ Z2 ]3 ^9 e( ^! d0 e: V
break;4 R5 f3 h) f9 V0 g! U& Z& h
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);
$ B! V/ Y2 f7 f3 j4 V3 I. }; Q2 B printf("\n反序数字为:");( o9 ?# t' z# |/ Z2 J8 I5 ?% v
printf("%d%d%d\n",indiv,ten,hundred);
+ x! S% {. E, D( J5 y/ Z# bcase 2:printf("%d,%d\n",ten,indiv); J w4 M8 V( {4 ?" u8 D
printf("\n反序数字为:");
/ m2 |8 m- [$ G- j3 x0 v6 U printf("%d%d\n",indiv,ten);
. b. o" c3 N( k& P7 ?5 Scase 1:printf("%d\n",indiv);
& x/ s, [) P% ?/ S printf("\n反序数字为:");
) H% \0 `% R8 D4 ?9 C9 l% g0 B+ c printf("%d\n",indiv);
7 O* Q+ G7 f {3 }* t1 i }; q4 K! k+ p- d! a5 T) _6 P1 ]
}
& o/ l c, d C; C0 O( [5.8
" t0 u' Z' A [* z1.if语句8 E ?) W! t* e4 m( _, C$ Y
main()4 G, T2 \( j! r% e. d9 z
{long i;: s2 a G! z% G5 z+ L
float bonus,bon1,bon2,bon4,bon6,bon10;
3 P0 o4 g# R" G bon1=100000*0.1;
/ A7 Q k# L0 f2 c& F6 K: | bon2=bon1+100000*0.075;
4 [! s; @: p* M bon4=bon2+200000*0.05;- M4 I8 [5 s+ {! L
bon6=bon4+200000*0.03;
& n6 ~3 C* [2 P6 `# w- y bon10=bon6+400000*0.015;
' J9 z* I/ e# o, Q" } scanf("%ld",&i);
, ?+ Y5 O% _' z6 x8 t if(i<=1e5)bonus=i*0.1;4 W% \6 G0 O) l/ X
else if(i<=2e5)bonus=bon1+(i-100000)*0.075;" I1 U# b' z, J' |" V! q: I
else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
: {4 x! m2 S& J; W3 c else if(i<=6e5)bonus=bon4+(i-400000)*0.03;
# M2 K+ W% E% i+ w else if(i<=1e6)bonus=bon6+(i-600000)*0.015;' [4 [: t# s. ^3 C8 K
else bonus=bon10+(i-1000000)*0.01;8 i$ Q) u0 {2 E$ v
printf("bonus=%10.2f",bonus);3 ]/ W; |$ a8 s5 `
}
* H4 `3 O$ E/ G( P; P1 C0 @3 }用switch语句编程序# H$ \, B# T: j) y" o' [1 v
main()- \* P; i5 q6 y
{long i;) S8 c) S* C6 j1 T6 r2 ~9 i
float bonus,bon1,bon2,bon4,bon6,bon10;
7 t% Z5 w5 t8 h int branch;$ l/ X4 r1 ^; A# D% J; {8 V
bon1=100000*0.1;* u* R* x$ X$ j+ {% @
bon2=bon1+100000*0.075;9 o% V+ w0 ~% @+ C7 E
bon4=bon2+200000*0.05;
3 G$ O5 B! [, q- @; | bon6=bon4+200000*0.03;
8 a M4 k$ i R O% h- |0 @& E* ~4 R bon10=bon6+400000*0.015;, v9 s" v, j+ u( c0 g1 O) u3 _
scanf("%ld",&i);& }" @$ A4 N( K
branch=i/100000;
5 i+ s; u% J' i) m9 x) |& [ if(branch>10)branch=10;: _2 J0 _' C8 P& T& v$ B9 g. k
switch(branch)
d8 W3 q; G8 k+ b, ` {case 0:bonus=i*0.1;break;
6 {. b' Z) X0 g case 1:bonus=bon1+(i-100000)*0.075;break;
" P( j0 u, M% q0 ?9 e( { case 2:( o, u. w2 H$ ?% b7 T2 `4 t2 S
case 3:bonus=bon2+(i-200000)*0.05;break;& Y/ n+ G8 @2 H# A
case 4:5 Y% N' F9 g" W
case 5:bonus=bon4+(i-400000)*0.03;break;
; Q; `5 q7 w6 I) ?7 h: B+ j case 6:
* u+ D% @1 l N# g case 7
6 ]8 }! D0 L- n( t) n case 8:
1 a/ {* C, c8 `& L+ V case 9:bonus=bon6+(i-600000)*0.015;break;
( |. c4 N" g# } case 10:bonus=bon10+(i-1000000)*0.01;# {- ~* e! t7 P' L; {2 l
}" ^3 ^) `# D, p+ t
printf("bonus=%10.2f",bonus);
8 c5 w# p+ X3 d} 8 W9 c7 g/ i/ M# h+ t
5.9 输入四个整数,按大小顺序输出.
/ t/ L# K" L2 b# a1 b$ X- ]- i! }0 xmain()/ t3 [/ I/ Z0 s" e7 q" ]
{int t,a,b,c,d;
# q5 T( T4 r n8 q printf("请输入四个数:");
' K/ @, ^: e6 r/ r scanf("%d,%d,%d,%d",&a,&b,&c,&d);
. m' y0 h: T: k' S printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
: g. p' [9 a5 Y! D0 u! k if(a>b)
' z& d- Z3 `+ \# m# R+ O, h9 Z {t=a;a=b;b=t;}
1 {. W9 M" r* B- ^ t3 A if(a>c)/ I. M6 M/ {$ ]8 t4 ]
{t=a;a=c;c=t;}
3 T7 _$ m r6 I" s: i3 F( ~9 d if(a>d)
C5 A! f# S2 C9 ]4 Q- G9 p% @ v) S {t=a;a=d;d=t;}
: w, V: {! s" D9 e1 R4 b if(b>c)
' G& n/ J2 q" h& t: b J {t=b;b=c;c=t;}
5 N) Q7 Q) V" y. j% X if(b>d)
% p4 f. f0 J% A& ^* z# L( I {t=b;b=d;d=t;}" T7 h4 j$ n; c/ J y' i4 o' U% ]
if(c>d)$ k) d, Z4 N6 m ]+ R' R2 n
{t=c;c=d;d=t;}
) U' Y! [5 X* J% s! aprintf("\n 排序结果如下: \n");% H$ H1 B0 n* m6 a& v
printf(" %d %d %d %d \n",a,b,c,d);2 K% s, Q- `. t5 B
}; v& K, ?4 i/ K$ q) e
5.10塔
& h3 ~2 b6 u4 Y' y0 Nmain()
: C( @2 H2 [1 ^5 }% `! |{
5 B. k8 g0 }3 p" u! r. L6 nint h=10;
% r- u1 w8 q7 a0 p9 [float x,y,x0=2,y0=2,d1,d2,d3,d4;
" E* U' f, o. }$ Y3 _1 rprintf("请输入一个点(x,y):");
& n/ d- b y3 i$ qscanf("%f,%f",&x,&y);* F: ]/ v+ q9 s. |& G" S
d1=(x-x0)*(x-x0)+(y-y0)(y-y0);
' r {+ _ A0 W. }5 v% ]# Z2 fd2=(x-x0)*(x-x0)+(y+y0)(y+y0);0 z/ n4 t$ s1 J4 A; i4 S1 e
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);: q7 ~+ t" g/ J
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);2 I/ g! \% b; ^8 T
if(d1>1 && d2>1 && d3>1 && d4>1)
0 o0 ?7 x$ c" W- \& Fh=0;7 T1 t" d1 S- h
printf("该点高度为%d",h);
% f5 T5 a/ f6 {& y}
6 ?8 A1 t* ^* z1 O5 i7 b- G! f, a( u第六章 循环语句
5 v1 K7 \' O* y! @6.1输入两个正数,求最大公约数最小公倍数.
$ z4 H3 ]; o* ~$ amain()" G" L G% c2 m, u9 ^. ~
{& ]/ r' x6 q2 m4 K6 Y
int a,b,num1,num2,temp;# _( s, u& D2 \
printf("请输入两个正整数:\n");# t5 g$ L) X U
scanf("%d,%d",&num1,&num2);
+ L9 `1 Q8 b$ }- wif(num1{
2 B2 H4 a, b, u* [: b) Otemp=num1;
, \7 t3 Q' z2 N2 L$ Z. |num1=num2;
, @% g* R6 n; }num2=temp;! R* d M# A7 L
}. w) b' v E6 O% H/ k
a=num1,b=num2;
3 F/ L$ W( X% bwhile(b!=0)
' `% |* M% p* o$ u {
9 L3 [! Y. h3 E4 u3 M k temp=a%b;
6 F7 m/ ~+ p0 j( c3 X% {6 K a=b;- [$ N1 K/ i: c% J- I2 y, q
b=temp;
$ }- m) J7 o3 B. L, T. o$ M }
- ^% l" a/ y$ @* Uprintf("它们的最大公约数为:%d\n",a);
2 C, }' v/ L* @% c( T3 n E, [+ Rprintf("它们的最小公倍数为:%d\n",num1*num2/2);
+ O2 M4 {+ i- L8 I* i- m# j5 B}, k' O$ J" N! a' h1 u6 \' G
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.$ U! w" k1 b* U" L
解:
, q* m6 X& c/ P2 T$ W% |; `#include < >
7 Z) [+ X0 ~$ Q8 z5 Qmain()
$ r9 i' k& R/ j' x1 a# W; R5 _{
+ \7 ~2 {5 f% B; e7 v8 wchar c;
& M) q0 W: l4 `# T1 L1 V Uint letters=0,space=0,degit=0,other=0;
+ s' L( H5 N0 n( qprintf("请输入一行字符:\n");8 x% u4 {# u# h; G1 g
scanf("%c",&c);
" I6 W! X, X3 c4 Z" Hwhile((c=getchar())!='\n')
0 I! }/ [ F# o{" B6 ?' b( p: K& j* ]& y" H3 e
if(c>='a'&&c<='z'||c>'A'&&c<='Z'): |4 d5 N1 R; q
letters++;) g0 m5 T3 k, Z7 x' M+ W, p6 R
else if(c==' ')( |, {5 E; i, h( O: h
space++;
4 H3 {, ~4 g, D; `else if(c>='0'&&c<='9')
) n. ~0 a1 c# |) ]# j+ J8 ] Ndigit++;
3 Z) a, R R- ^2 B' W- k+ d/ ]+ \else
1 U4 S! P+ r0 b h; O# D0 u: @other++;
* H/ A Y* g# `+ ?. g}% c# ?9 H3 [- [) k! ` S
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
9 J! w! Q- f+ {: k0 L5 o) ud\n",letters,space,
) t0 `/ O+ u' ?$ k+ z" Hdigit,other);
9 V4 E& |! k3 d8 z9 p" x+ v! T}2 \( S; g$ G: ?' v! | N1 c3 k
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.& |+ Y$ \# d/ Q
解:
2 H( U0 }: @6 o5 t. lmain()
" T7 ]" U0 P- _% z0 G, D- ]8 c4 N{
/ b/ L5 U0 I- M. o* U7 O, w# Y hint a,n,count=1,sn=0,tn=0;! n+ J0 d" e7 c7 q4 \* J/ z L9 W
printf("请输入a和n的值:\n");
, k* o. y0 t) e& g2 Z; @" @scanf("%d,%d",&a,&n);' ]0 g2 `# m! z
printf("a=%d n=%d \n",a,n);
4 a+ T- x; i8 L% ]8 `3 Hwhile(count<=n)
) z* _; K. |! b2 n+ t) P{
2 ? x, [# L! V& `5 T$ H; Htn=tn+a;# g H# Z! p' a: ^
sn=sn+tn;
1 [- T4 C$ G8 f+ W9 s; Q1 Ra=a*10;
3 u( D1 i& m, A4 c: m1 L' e4 G# W++count;
1 D3 Q% q Y" F7 A b* b}
2 T( G! v& W) e* b2 u" z- O, Eprintf("a+aa+aaa+…=%d\n",sn);
+ w6 ?! c, m" ]+ {; Y( D}7 _" F1 T v# {5 z8 u1 g
6.4 求1+2!+3!+4!+…+20!./ w& R# V; [! M1 \$ I9 j) X
main()0 c S5 [% {0 x! N7 x8 b7 B8 V
{
2 j% m5 v; i& z5 j+ c* q& S' b% P4 afloat n,s=0,t=1;8 L- i2 @% l0 U4 |( @, k1 a
for(n=1;n<=20;n++)* N3 n1 [- S" N* `. e
{
P7 {' c) a! W. v7 h% Ft=t*n;2 B5 |" c t- t( W- u3 V
s=s+t;! g1 T1 G" p/ A9 c1 c% W
}% o# t+ f% P( p5 [/ |8 a
printf("1!+2!+…+20!=%e\n",s);
. w( p2 K0 c, H3 \3 D}' Q. ]4 y) X) @. D/ G! S
6.5 main()
5 p8 E, t" V4 b{
7 ?, J+ P) i1 ^% q- t) O& C! k7 @5 vint N1=100,N2=50,N3=10;
+ A z8 N/ a c* q- k1 i7 hfloat k;
% W$ x2 G$ a( M$ Kfloat s1=0,s2=0,s3=0;( s2 D0 H7 x7 T/ V9 X2 t$ m
for(k=1;k<=N1;k++)
9 p. B8 \- ?, S' [9 L{+ Q/ N3 N3 p% Q$ e# B' l
s1=s1+k;+ M8 E" U0 F0 p) ^7 ]
}+ E* i0 f5 C @$ v
for(k=1;k<=N2;k++)/ n3 m4 ~6 T; ?
{# _( |7 l4 s S$ I# M
s2=s2+k*k;$ x2 v5 v& o$ q$ R( x- ?1 x
}# k$ o! D; D' w. T. b* U0 j5 M1 \$ z
for(k=1;k<=N3;k++) Q+ t4 W) H/ _& H& z$ s
{
5 x8 \0 g2 F) F9 fs3=s3+1/k;# W s; q1 R- M& @1 \5 d, ?
}
: W( _2 s) e) M1 xprintf("总和=%8.2f\n",s1+s2+s3);
% a e* Q0 d: O* o+ F( `/ H}* s2 K' C/ H/ X0 [0 D! E
6.6水仙开花# I! H% E1 r& o' R4 S3 P! x8 L, O! z
main()2 _" d. ]: G$ n' d$ j' v
{ q2 f K2 Y$ w: ?/ w% a
int i,j,k,n;
9 W4 U! s; j$ z1 Nprintf(" '水仙花'数是:");' I3 d( x& M9 ?/ w$ h- J6 _2 r+ z
for(n=100;n<1000;n++)
. J, @3 l% u7 e4 ?" k) K{' G+ g: t- b& z- j+ f
i=n/100;6 F' r: N0 L9 @- ^' `( Z
j=n/10-i*10;9 m5 P9 ?# A# |* g7 q
k=n%10;
6 J. [5 k. f [" q( j9 c4 ~# [if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
' \# o9 m" _/ q{
# B/ X9 G8 b4 c+ d; Oprintf("%d",n);' @0 j+ e6 K7 G' j9 t& ~" A
}* A. c2 _7 d6 X J1 I0 r& [
}+ @' D2 S N' `& {/ h% ^
printf("\n");# h& I/ Q& W' I
}: q. [1 y/ Q- k3 r. u
6.7完数
v% g8 }* B9 O8 m3 rmain(); _: h1 d( ? q/ T' }* |
#include M 1000, a0 D' n5 @* ?
main()
, L( n; }& c/ c: y1 C{
8 A. e% f0 W8 Q5 w& Wint k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;
. Y6 y8 @( t8 Q2 ^int i,j,n,s;8 Q( _& [+ y( r. a6 H
for(j=2;j<=M;j++)9 W% D: u* B0 G8 n! w, I
{
# Q7 ^- S/ J; ^; x$ L6 ~2 w7 rn=0;* @2 T% Q7 H% V( W# R3 y
s=j;
3 T9 \- ]) K1 r! I. Afor(i=1;i {+ }" {3 N- o' _ n2 ^' @. F
if((j%i)==0)
- O! t; `$ N% e2 b9 g# H% K* j {+ ~$ ^1 O( h# o ^3 F
if((j%i)==0)
5 k/ W0 V( T+ l: g9 {: S& p {+ T! ]& ]3 I* {3 w
n++;
: R# R) g! l: s2 B. ?* O0 x% s s=s-i;
6 [$ ^% ]: {" L; d' \ switch(n)
* j. N* ]4 [0 e; @' k {
) c8 Q- D8 x: x V$ x* m case 1:
, F5 ]1 k5 S" k+ r% Q1 n k0=i;
- v' q. o3 H! w0 g, F break;" G1 j0 r0 _" a5 \# L6 L2 r+ d
case 2:
- A+ C/ w4 m) I* c% z" \1 |; l k1=i;. h' V/ ?" ^' @2 y! A4 C1 U
break;
& p2 f3 J5 S2 Y% |1 k# ` case 3:+ Z! h- K! S' p2 W; k$ A
k2=i;
; T) I* M- A" l( m break;2 o8 z9 e/ Z' m7 R9 }0 G) r7 G
case 4:
* m4 H3 Y8 F. [! O& n6 D k3=i; e$ q2 Y$ A; p3 x" t
break;
$ [' R6 [5 [9 G0 q1 g; ] A case 5:/ X% p/ d8 n3 M% h9 V$ J/ `* d
k4=i;
5 t3 ?7 U- @, `6 Z break;) d! p0 m! h7 r3 w
case 6:" ?" h9 a/ O- g' x
k5=i;! m' [+ i+ v W/ \. k
break;
U7 e* F+ K: ]. P( d! Z% D. q8 i0 m* K case 7:
" R% u+ J/ X# G O1 w$ g k6=i;/ E: K" c p( C/ X7 h
break;, o# g6 Y; T7 Z3 X0 S3 x
case 8:
2 R1 w1 _4 U* V k7=i;
+ k2 [2 G& @% |+ L/ i break;
8 \0 B# M, l8 ]. B7 ~6 p: m case 9:
8 L+ V% H4 E1 C0 I( I% ? k8=i;
4 c' M: b( H# l- n break;
$ _/ v# ], R" t case 10: D ?9 ]8 E. \$ n0 E$ {
k9=i;
2 u" F$ j# _+ e, ` o break;% Q% j G# J7 b6 z& v0 Z3 j5 d
}
( }/ s: B/ ]% W$ N+ m- s/ F1 f; B5 U) G }
& v. i( V# z* A6 ]" Y# q V* Y, T }
( l( s" U5 Y) n# [8 yif(s==0)
0 p( m ?) A/ f+ p4 m {
6 ^( T; c% p h9 E: b5 Cprintf("%d是一个‘完数’,它的因子是",j);* O2 w, R7 Y) j. s/ ?3 u2 r
if(n>1)7 b+ O1 k5 s2 v
printf("%d,%d",k0,k1);/ V t9 o9 |0 n( v3 e$ w' S" T
if(n>2)
5 M' j% j$ T' N1 \! C& @, I printf(",%d",k2);4 G% n- O7 r% W* j
if(n>3). W( V( h/ c8 S# }: C( v* `
printf(",%d",k3);
. l2 t9 q, E) ^. wif(n>4)
) U& N8 g/ Z4 l! Z printf(",%d",k4);
. p9 d0 O0 p1 ]$ y6 e, p9 d3 xif(n>5)
2 G) G U5 M# E printf(",%d",k5);4 O% s6 s k7 Z0 s$ k
if(n>6)- c6 h& C z. ~5 Z5 s1 }
printf(",%d",k6);) r& \/ {+ Z; K3 s/ h
if(n>7)' q& V$ t1 i; E% ~! l1 h; P
printf(",%d",k7);* j: J3 D$ c7 W
if(n>8)5 O% W* b* H5 }' |7 q
printf(",%d",k8);
[ J7 m$ V/ d, Z- `) ]% Z: f4 |if(n>9); K: e( z8 |8 a$ Q" _6 P
printf(",%d",k9);
% ?/ ?+ r- Y+ ` c7 |9 aprintf("\n");
0 S( L, y6 S) R9 Z }
" D" F$ k( E. c}5 I U5 R* r6 t) ~" j- V' F2 J: \7 n
方法二:此题用数组方法更为简单.% I! v" S& K k
main()6 I) M7 ]) L$ o7 z
{
& ?% y" k* ?. G1 g* rstatic int k[10];; `1 e7 z+ X2 q
int i,j,n,s;. L% ]" b5 s. r3 ]/ d
for(j=2;j<=1000;j++)3 m c* w8 ` L! y
{0 b4 G0 ^! r& Y5 x
n=-1;: \# a5 u) |* E! t# v9 o
s=j;( d( O) q9 @ P+ J! H) S d5 J
for(i=1;i{
& Y/ i) D! ]/ b/ cif((j%i)==0)
, U/ G6 B8 s: @% h! X8 m6 y* H{
; ^" t3 p. }+ _2 y, H3 J. }; }n++;
$ \7 K! e- |# A& q( u9 v4 c# X5 ns=s-i;, B" B! d+ q2 I0 d0 _! q* c5 b
k[n]=i;
' G, \ p: D0 {& K }+ h" f) J6 s5 [7 a% D Q
}
: _2 ~" ^% Q* y# Z( L& Fif(s==0)
, H+ d v% t. t, J* _1 Q3 T{
' ~% Y7 N2 y! ^- O6 D; Bprintf("%d是一个完数,它的因子是:",j);
) i3 a8 B9 z- e% l' z1 ]for(i=0;iprintf("%d,",k[i]);6 r% K6 J$ g* p6 G' m% {' O( Y
printf("%d\n",k[n]);, [( G6 }$ p! ]5 H8 \: G/ N A$ q
}8 |- n7 ~8 C/ L' L9 v. S/ l9 }
}5 T, P" G/ h/ p2 L Z
6.8 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.
1 D$ ]( J( U# N4 |% i" v解: main()' D* i+ m5 ?& f" m% O- g7 x( r
{" O5 Y. P* |6 [ a
int n,t,number=20;- _7 Z2 b$ A I/ b$ X+ S" ]6 L
float a=2,b=1,s=0;/ \' k: f" z; E h F
for(n=1;n<=number;n++): |* V; d. U% o J! u3 j, |
{5 j& Y: r: S' U9 F6 w. D
s=s+a/b;
2 o0 ^9 R1 s; X/ Ct=a,a=a+b,b=t;
. Z1 r- }' J' J. i& S4 `}
: v8 D8 p4 s/ i! W7 ^3 Xprintf("总和=%9.6f\n",s);
# Z# s* K& H1 Y4 [2 m}
2 |3 s2 W* j+ B6.9球反弹问题8 ~8 A! ]- z7 A/ j, `8 W& I1 i
main()
( O1 V7 Q0 p' s: h{3 \* I* O' }9 Q/ H
float sn=100.0,hn=sn/2;
1 h! C6 s- a i8 A3 x0 {int n;
+ d/ I q$ \1 d6 S& i/ E: s( ]7 y D) q% Ofor(n=2;n<=10;n++)
! m* B- k9 O& D{1 M2 E$ x. t* N/ z
sn=sn+2*hn;
% K# |8 y9 u! |! J0 g3 y+ G9 Dhn=hn/2;" i7 `0 I1 {* f: W" h
}# v) X& R# W G8 a y
printf("第10次落地时共经过%f米 \n",sn);
3 D6 v0 \+ v4 E- b. K, M% Kprintf("第10次反弹%f米.\n",hn);% ]5 x, l/ A. \- U! O
}1 e0 A0 i# Q+ v& |( h
6.10猴子吃桃
& [$ p* L3 G2 \8 P/ tmain()9 l; g8 Z8 f0 f1 H$ A+ `8 z( M
{7 |% i* G$ d+ _3 ^1 W2 G U
int day,x1,x2;
1 F) M9 k- L" y! Fday=9;3 _' W" }: |0 |, K6 |
x2=1;
- r% t* M/ F2 l2 X, J# _8 _while(day>0)! z3 Z1 X9 z! a- B# ]- w* y8 O
{
. M' X% q4 j/ a* }x1=(x2+1)*2;9 W1 Z ?9 f* n
x2=x1;
# d# j; J- @5 L/ y" V9 rday--;" t. Y8 G- ? q2 b* T4 U- {- ^
}0 t5 c% Z* B6 F* X& }. J, D% ]4 j% M
printf("桃子总数=%d\n",x1);
2 i' I" u5 W6 T* _}
5 D7 y5 B6 E6 Q; M' a' E- `" v* o% E1 Z8 Q# r) L8 e
6.124 E5 ^8 t8 C/ P( T( V
#include"math.h". S8 o1 |* d( a! _5 ]
main()
' G/ _8 E3 j9 r4 T6 B8 A{float x,x0,f,f1;
3 e& a T5 S% J0 n x=1.5;/ b7 S# O% a/ b) B: y4 O M
do
' P& W# `, i( ?# J i( s/ N {x0=x;
/ E+ f% h5 ?! `- H f=((2*x0-4)*x0+3)*x0-6;. v7 U! [ F/ Q1 }
f1=(6*x0-8)*x0+3;
6 p* o- j* d- i& V, i1 W3 O4 N; N x=x0-f/f1;2 x8 T, H- P' h) g- b
}
3 Q) c3 {4 `# ]% O) U while(fabs(x-x0)>=1e-5);
0 @; A9 S7 K+ B7 E/ Z9 U printf("x=%6.2f\n",x);1 H5 Y: |- y4 E( g
}9 x# R5 t0 A5 E& n0 d3 m5 d' Z
3 P2 Q! W- O9 Q9 v- x% _$ \: _6.139 ^; @* G, {, L
#include"math.h"" @, l( r/ z) X* D+ @* ]0 ]1 j
main()
* z+ F3 \. w2 W9 q3 k: k{float x0,x1,x2,fx0,fx1,fx2;; u3 K- v( R* c m
do
4 e, C2 d( W2 b; i/ Q# i9 p2 r {scanf("%f,%f",&x1,&x2);
$ G* U7 q# B( L# G" K+ I fx1=x1*((2*x1-4)*x1+3)-6;
. `5 Z/ r, ^- T: k# |# R fx2=x2*((2*x2-4)*x2+3)-6;8 N# f6 M8 d6 t! p9 q/ a
}& k0 h% l* `" M+ y2 L1 a
while(fx1*fx2>0);
/ l# P( I: T& N( \5 q do
! K I, M9 l' r; C( i# m: e {x0=(x1+x2)/2;
" p9 f; Z, u6 W( Q" i3 U1 o fx0=x0*((2*x0-4)*x0+3)-6;
3 c, h# g& k$ _7 y x if((fx0*fx1)<0)
, @/ a4 _1 R) c- X6 { {x2=x0;+ Y4 e* I+ C9 J3 K- y
fx2=fx0;
: i G0 P6 h# j1 p5 ?, u }
2 h& k- h3 H z# Y2 ~ else
7 o' b& p) N& i# V {x1=x0;
. z5 ^! S0 t7 U2 j/ x% y' f/ _5 X a fx1=fx0;$ E/ b& m* P* O+ f7 b
}
. P. _4 s& Z* A: P$ |& g k }1 l/ D; r8 o# f8 ?, y1 l f
while(fabs(fx0)>=1e-5);
& m: d' D# y5 [' B printf("x0=%6.2f\n",x0);
7 r v l5 c# ^. p3 a& `+ Z}
" Y7 C0 r9 m6 Z: d+ c5 A6.14打印图案
) e* C W/ [; Y3 `, D) ]% }5 nmain(). N* J5 ?% n& V }8 L5 I
{int i,j,k;
' g! y" ^) q; F5 b6 }# T for(i=0;i<=3;i++)& s2 g. F& a% C3 i& O% S
{for(j=0;j<=2-i;j++)
& M6 i( h$ o4 i/ F( } printf(" ");6 K7 S! r/ W* k% w- P. J2 V- k S
for(k=0;k<=2*i;k++)
; R; b4 C6 {) o, B/ E printf("*");- T( f1 C& [8 I$ w+ q! s
printf("\n");
- H; w" r+ b0 { }
, K" l7 y& i2 M for(i=0;i<=2;i++)
% Q( D; ]% y& i% \; _% s {for(j=0;j<=i;j++)
) B- R3 j; {7 X# I$ b/ P& L printf(" ");' B' a* r. L, z. I6 o8 F2 A! s: b
for(k=0;k<=4-2*i;k++)' O6 a2 i, u4 k2 k3 T
printf("*");
W, B8 S/ U$ F0 c, `4 x; _ } printf("\n");6 ^3 q$ X' H& c' `5 }! ^% z( O- b
}
/ s$ \$ C) ]# h: Y/ x/ C3 ^}
+ z g9 T* N" W# y6.15乒乓比赛5 I# M6 \' c9 C$ {
main()
& l8 C) t m8 J4 |: P1 c{
: v" ?: V$ `& ~, q# S4 U7 dchar i,j,k;
6 \0 r& s8 E# e; L' ^* c, Rfor(i='x';i<='z';i++)
' ?( `8 Z( F8 Yfor(j='x';j<='z';j++)1 ]( w0 y. t3 h+ Z4 I
{
# F- S5 F: X& b9 N, sif(i!=j)! f/ G& E) z. b' E+ @
for(k='x';k<='z';k++)2 Q: F5 q) V/ g6 y% m+ `
{0 _' I1 L% e9 T
if(i!=k&&j!=k)
" x2 j; P A' d( T {if(i!='x' && k!='x' && k! ='z') Z6 ]5 Q" {+ _; a- N7 p( \
printf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);
% c7 y* G! V1 {% p( \ ^, S }2 [/ z( v& Q: S
}
9 q6 c+ {% A1 I& b `/ r }
1 o4 S7 g9 {0 A% G' T3 v; Z}+ w) d5 f+ h. I7 G) @9 R# _1 y
C语言设计谭浩强第三版的课后习题答案7 i, d' q( s2 w! I
7.1用筛选法求100之内的素数.
8 D9 G4 j% R6 I/ |4 w4 i5 K% `* `#include
4 @- ~5 x4 A' Y" h, a3 ?#define N 1010 I5 c3 i) g4 v1 ?6 k
main()7 C n+ o2 j* e0 P( L4 V/ X
{int i,j,line,a[N];
* v1 e2 A: J: j) f* yfor(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)( N# J }# C2 Q9 H. s+ [
if(a[j]%a[i]==0)
& |2 g* A, B- y! U E a[j]=0;
7 p8 y$ o& D2 j5 aprintf("\n");
. d0 I* m4 w( V- Rfor(i=2,line=0;i{ if(a[i]!=0): v$ `6 q+ r7 I; N7 E4 x2 f2 B
{printf("%5d",a[i]); }& x9 n: q, C# l
line++;7 b3 I' a& {3 ~0 c$ c ?
if(line==10)* m; R1 G( c" q) Q8 k2 i' K
{printf("\n");' I: u+ s; x; o( P K5 c! k8 y
line=0;}
6 x; B# ?, p# V1 q& I+ ~. c( T }: X0 p q4 g' \) p3 |4 `# \, T2 }2 @
}
o& e4 o S& [1 a7.2用选择法对10个数排序.7 o8 \! g# v ?0 |3 p
#define N 10 l! C ?7 {1 A8 B2 f
main()
* ~1 r4 G t5 x8 j+ C{ int i,j,min,temp,a[N];* |$ P$ P1 ^, z$ M3 A% v# O- p. K
printf("请输入十个数:\n");* u6 q3 D" b' m/ Z! _' r+ @, X
for (i=0;i{ printf("a[%d]=",i);: Z# O( K' u, R5 o8 ^
scanf("%d",&a[i]);( d4 q/ h% P: f, w5 @
}" p; S0 k; b* j- p1 z2 X; _
printf("\n");4 V& @) H8 T _" L% |) P- X# {
for(i=0;i printf("%5d",a[i]);& i# t3 x5 b a: J# s. j2 F8 s
printf("\n");8 G+ _% F; J/ H: m
for (i=0;i{ min=i;! w( l5 w3 K, c8 u' c6 q5 C( Y7 K
for(j=i+1;j if(a[min]>a[j]) min=j;
+ n7 P+ z6 t |% P temp=a[i];0 B$ o. U1 x x! A# X% Z" W- Y
a[i]=a[min];5 H& F5 a( y. L7 {- o5 x
a[min]=temp;
V7 _" j! c7 v6 u1 g Z- y, d# @}2 {# m: w4 ]7 G0 ]0 N
printf("\n排序结果如下:\n");
& i% r z8 [7 d4 }. R- `for(i=0;iprintf("%5d",a[i]);) I0 U/ N6 P$ {
}
' w7 d- K, ^5 S y; P! K( z' B7.3对角线和:
. O. a2 [0 M8 ]" [0 Pmain()
|" h( }+ R/ y0 }# j4 p7 W$ S8 `) L{
! @! Z: F* j2 _% n$ d) J( gfloat a[3][3],sum=0;
! |5 v5 ]* l# @1 |int i,j;
: E) Z) E/ [& B% m# Eprintf("请输入矩阵元素:\n");9 j" F' ^, j% @4 C$ t3 E1 A: D
for(i=0;i<3;i++)
8 l% |, B9 q+ O% t for(j=0;j<3;j++)
6 w0 J5 \4 c) Z8 |6 P8 G scanf("%f",&a[i][j]);" Z- h) y; [/ B- T+ A
for(i=0;i<3;i++)% V% C7 T! `/ V$ j7 ]: y7 }0 S
sum=sum+a[i][i];
& R4 u. U- \2 `; i! u6 v l printf("对角元素之和=6.2f",sum);
, {! k1 i+ I6 i. ]: y- }}6 H; ~- S3 r! w5 o
7.4插入数据到数组5 j* w7 h) }5 Q) \- e
main()0 A7 i1 b2 `& _3 f" _
{int a[11]={1,4,6,9,13,16,19,28,40,100};( Y$ D3 O9 J! D
int temp1,temp2,number,end,i,j;( ^5 O4 e4 p- X2 p& C
printf("初始数组如下:");
2 t: ~' K, q% [- Z1 O$ ^for (i=0;i<10;i++)
0 X+ l. {' B/ b% `! F1 x) vprintf("%5d",a[i]);5 S; X) H5 E/ j) M
printf("\n");
7 q: ]0 Q. F+ X# s% \, T3 {printf("输入插入数据:");
: [- _, e3 m: vscanf("%d",&number);
5 F8 g4 N {& [& eend=a[9];
9 x% M( Y9 m m/ [4 k7 eif(number>end)& G' c# F8 t3 i: O6 I4 k
a[10]=number;, g) Y7 u9 o# o- k7 n8 R* i
else( U& J1 N' a# A5 X( p& J( ~! `
{for(i=0;i<10;i++)
7 y+ C9 J1 A, W { if(a[i]>number)$ [; i- D& G* c0 A- T' b' h
{temp1=a[i];
! m3 \1 f4 t1 r; n% s: i% U& q a[i]=number;
4 s& k) H1 T! ?$ ? for(j=i+1;j<11;j++)/ [1 y3 U0 {. O( `' r8 g4 V
{temp2=a[j];! I3 n6 N1 b; {% F. \
a[j]=temp1;
/ |" a: f' L; L6 l! C6 u9 J% n temp1=temp2;. n& m; ^; X7 u$ E1 B# m6 S
}
( j3 \8 q" A0 y1 R break;0 l& `2 M) E0 ]- g( G
}
" q& j1 f, {& d9 z% l, S' e }, i1 W7 I" n; _' q: v
}
: j8 C9 X9 X4 h1 B& H for(i=0;j<11;i++)
0 ]/ E) O0 Q2 W/ [% b, u printf("a%6d",a[i]);
0 K# d' Y" O; G/ T7 S# f; \}3 `4 B: W& E' g& R& v* ]7 q1 T, {
7.5将一个数组逆序存放。5 K, R! o) E, V, k
#define N 5: S( }- G4 K: D* g! P7 ~& {# U2 g
main()2 F/ d* e& w' |
{ int a[N]={8,6,5,4,1},i,temp;! q8 @0 J, I& d x- O
printf("\n 初始数组:\n");
: N' X" U% u r5 Vfor(i=0;iprintf("%4d",a[i]);
! v4 f6 y) j( ?for(i=0;i{ temp=a[i];
. g- Y% A F- L: p' s a[i]=a[N-i-1];2 S4 l) e' w5 n, V
a[N-i-1]=temp;* M8 g' k% ?. N& F! l
}
( w5 w+ \% C: C( a( K6 Oprintf("\n 交换后的数组:\n");
$ I* O( X" x# ~2 I9 Q4 T0 vfor(i=0;i printf("%4d",a[i]);4 d2 J6 h- X. D' R$ V" Q) z+ V2 d
}5 O! y% B% k) Q$ d
7.6杨辉三角 _- [- R3 E. q% |5 E
#define N 112 q2 d) m3 w( K9 S, T/ p
main()
* ]) z# y8 y6 N5 R* _{ int i,j,a[N][N];
3 @4 M1 j* \0 @4 H4 p, o" M" A for(i=1;i {a[i][i]=1;
8 h: n, \7 }3 G; h( V$ @' \ a[i][1]=1;
% C# p) u6 G( n1 Y }
6 q% l' n8 {) \1 F for(i=3;i for(j=2;j<=i-1;j++); f( z( J! x2 o. R, K1 r3 E
a[i][j]=a[i01][j-1]+a[i-1][j];: @6 @9 V# c7 x$ n- R4 G. w: W
for(i=1;i { for(j=1;j<=i;j++)
3 `( q5 X+ n! K' J printf("%6d",a[i][j];* I1 Z4 D3 z; |8 h# p6 l
printf("\n");
2 }( X. v* ?, @, j/ ?9 G }8 |1 d0 f! r5 Q( l
printf("\n");
& {' S6 ]/ Q G2 v}, q/ G# [& g6 _) p- L
7.8鞍点
: n4 z8 F3 l1 n$ K% j#define N 10+ ^ v) f4 }+ d
#define M 10/ {: F/ i4 O4 \ c
main() k2 j& [7 l+ T" U2 [4 h" z
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;
( S$ k2 P% }. d/ h printf("\n输入行数n:");, R2 w5 z5 N4 D' T4 ~; y
scanf("%d",&n);
1 I+ i: o2 o$ n printf("\n输入列数m:");
, U7 _& h# E+ K# E% i+ a scanf("%d",&m);
- S2 {" S% u5 N% j* j. ^ , {3 P8 q0 D, l0 r6 a
for(i=0;i { printf("第%d行?\n",i);) S+ B5 M- V9 c0 V) f7 B
for(j=0;j scanf("%d",&a[i][j];
/ y# t- p9 W8 j! V5 e* c p }3 i }4 }& [$ ~' W Y6 V1 E8 z/ ]
for(i=0;i { for(j=0;j printf("%5d",a[i][j]);6 ?( v7 s* A% w. d) I5 I8 {4 l
pritf("\n");/ c' s p" O3 ]' ^5 {" y
}
; a/ I" o3 ~" n; Y flag2=0;$ r* M; Z: u1 c6 f% J; X2 S
for(i=0;i { max=a[i][0];9 y9 q; h( [+ y5 S1 s
for(j=0;j if(a[i][j]>max)
' l/ n$ j/ f$ F6 }. V' R* M { max=a[i][j];
7 f0 j* r; m1 H( F maxj=j;+ ?& |8 c& l2 ^/ z& |
}
1 k+ a1 v7 u* L- H1 q# N% { for (k=0,flag1=1;k if(max>a[k][max])2 ~) E/ |3 ?. M6 E0 R
flag1=0;( V1 {9 c$ L2 r0 d* ?
if(flag1)
7 Z5 [0 J G3 z! {) c/ | { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);
+ |" g# h7 y X5 a. x+ F flag2=1;% Z; s& U7 H7 u2 v: v" w
}
! B! p4 N- I, H}
( G6 [3 J1 c; T* `* t" mif(!flag2)2 P' r$ p, X+ y" o7 y+ {) M1 u
printf("\n 矩阵中无鞍点! \n");
: S! G1 V5 h# i n; |" C3 b}
! X: W$ s8 b U
% L& z, e4 e) ?1 a2 m7.9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.+ x6 `8 j* k- W/ H: Z$ w/ p& i
#include
* |* k4 n# [9 j. C8 Z) g& M) x1 G% E! u! E#define N 152 y, e3 ^ m8 G
main()5 j: r9 Z! ?& G- t( w: s" o
{ int i,j,number,top,bott,min,loca,a[N],flag;, W# ^8 F! I' M$ d3 |
char c;6 z; E5 m& E w
printf("输入15个数(a[i]>[i-1])\n);
7 T+ t5 W6 B2 Z# X/ o$ k" ] scanf("%d",&a[0]);; k+ C; G: S' D& p. M3 a
i=1;
2 f% g- C3 R) S- m3 N/ [ ^ while(i { scanf("%d",&a[i]);
, f! p, H7 T4 P. l* m8 d1 D if(a[i]>=a[i-1])1 j: V. k2 B1 h6 [# Z/ I8 k
i++;
6 \1 C. E5 Q' G) W C! I% g9 k S/ ~! d0 R esle9 |8 m" P" S3 ^$ ~, C% F! ]/ l
{printf("请重输入a[i]");0 ]3 }! H; m+ ]- N+ U" m1 @
printf("必须大于%d\n",a[i-1]);, _$ t. ^0 w E+ V7 a( \: a
}+ d" h2 m) L( i' T
}7 @3 u+ F5 ]' O- v9 K2 u7 \ G
printf("\n");- `# M+ G3 Y2 a0 S& k
for(i=0;i printf("%4d",a[i]);
% h4 {8 u/ h& Z) \& K printf("\n");
, P4 c" ^) K8 n: \. G" B) y: h: b) r
% h: L" R+ y7 r- r' Q5 @* K: } flag=1;! `% L T) N. R3 g! x7 a5 V
while(flag)3 [3 B- P- w7 e1 N! Y+ _0 Z
{ s! x6 \, l# h6 i* S
printf("请输入查找数据:");$ X2 v2 e2 t4 O6 J1 z4 K6 ]- n
scanf("%d",&number);
4 Q) T' R: v; M+ r" K2 `! p+ Z loca=0;
8 x2 z- f* T9 A f- r top=0;" W [4 [3 g3 T# I
bott=N-1;
+ E/ z2 C% @+ |! U. X" v. {* j if((numbera[N-1]))
6 I+ A. a* v- _: x loca=-1;' B1 n# |1 k1 z1 e7 g
while((loca==0)&&(top<=bott))4 X+ s0 h9 p5 G9 M, \
{ min=(bott+top)/2;
& Z8 q5 S" Y5 s" {) S if(number==a[min])
2 F1 }3 x: W6 z0 M { loca=min;! }& y# m. |3 [- ~! U
printf("%d位于表中第%d个数\n",number,loca+1);. t$ B" _" S! R1 |1 I
}
' h6 Z: T) e0 l2 o1 m: I else if(number bott=min-1;
9 k7 n* M$ s) J& R3 m" _2 r else
8 \0 L% `+ ?9 a, @0 M top=min+1;1 `1 z& ]4 _5 B% B3 C4 O: V0 e9 X) `
}7 s5 P; ^+ O7 {# v/ D1 |/ _
if(loca==0||loca==-1)
0 H9 f9 }5 B1 b% C- X2 o, M printf("%d不在表中\n",number);
8 z1 \* x0 \9 M' }, Z: V% J printf("是否继续查找?Y/N!\n");
j- t( }8 u* a c=getchar();
% E" i: U8 y3 B: `/ X if(c=='N'||c=='n')
* A2 f, n7 B5 k$ f6 l# J flag=0;
( p8 d o5 ? @. b }5 {0 S" o4 t/ f8 n0 }% N; V
}
' J7 l; X' Q6 T; K
2 n8 ?. v# _7 J5 m v) T2 m" h7.10
( t H3 ]* k- Z/ `2 K" `5 @$ A4 zmain()1 X6 n- g( O/ _* n. O7 }
{ int i,j,uppn,lown,dign,span,othn;' x# p; L3 }2 t7 {1 Z# @% Y( O
char text[3][80];
1 T9 k- U9 s( U: u# |& u8 { uppn=lown=dign=span=othn=0;' K( p" u" P6 w K0 C& `
for(i=0;i<3;i++)
; z. I8 _ y+ H5 m" K { printf("\n请输入第%d行:\n",i);
9 }" ^. L+ }1 ]0 h* G) s4 L gets(text[i]);
( H0 U! ~. n) S: D H( E for(j=0;j<80 && text[i][j]!='\0';j++)* I: a* e) |& ~ D6 E
{if(text[i][j]>='A' && text[i][j]<='Z')
3 H5 ` d/ [4 c. k% t* L uppn+=1;6 G* y6 ^ X }5 `, r8 Z9 ^
else if(text[i][j]>='a' && text[i][j]<='z')" w! j6 ]- M! C9 O, o. u
lown+=1;
. e! R/ Y2 a5 S7 N! W6 C else if(text[i][j]>='1' && text[i][j]<='9')9 e1 _! S5 ]0 \! e; a# z
dign+=1;# B" M; N& U; @& B
else if(text[i][j]=' ')0 l! J% ?# g+ B9 g% n, @
span+=1;3 [. e( R1 g# p) R
else
' @' j1 O5 ]$ ?( F7 ` othn+=1;
$ S' z$ \# H. r. l& g0 o }( \3 W7 x1 j6 r6 Y
}
8 s2 X: Y4 a- t; f: T4 q* O for(i=0;i<3;i++)! _$ D1 D0 [0 r! @# C4 A T
printf("%s=n",text[i]); f5 m2 `, H% I+ q( K$ D$ H
printf("大写字母数:%d\n",uppn);
, {# M8 B) [0 b- Q% l printf("小写字母数:%d\n",lown);4 t) v9 }9 K& ~, r
printf("数字个数:%d\n",dign);; j6 d9 U- A8 G
printf("空格个数:%d\n",span);
" |6 L3 e( d3 z$ L5 P5 ~2 ` printf("其它字符:%d\n",othn);* M7 n+ m/ {) K# T+ n. ~
}
( s9 `7 W' l/ z2 H
" t6 @5 ?5 p5 L: Z/ q9 S
$ h8 {$ T, @& I4 y0 V# Y& i* M7 D7.11+ K4 r7 N) F, [% y
main()$ e3 Q& H: d* W2 w. ^; a
{static char a[5]={'*','*','*','*','*'};$ a+ c/ U2 r8 k* ]# `
int i,j,k;
; {4 C+ T3 Y7 T& D# C char space=' ';
4 p) F0 z9 e9 E for(i=0;i<=5;i++)
% _3 P8 f% n1 v( ]# b( B K; u {printf("\n");4 m9 k# v! J7 |& i0 h- O- @! x# h# G
for(j=1;j<=3*i;j++)
0 R% T& q& `' C& c; a" M) ` printf("%lc",space); H$ g7 j3 V6 ?6 m; P5 L
for(k=0;k<=5;k++)
4 S. y' ^5 [2 e. s* F/ k! | printf("%3c",a[k];7 ?" C. N5 p4 W
}
' o# k1 [% ~9 e}
, f+ X) r* q0 G7.12
! W& [# X. u* A3 @; |* T#include
' ^% E' O6 X2 y& @main()
$ E0 u% [- q5 ?- C. d' b1 l- r, t+ c{int i,n;
, a% | Z2 ~* T. x9 f+ R char ch[80],tran[80];
0 t* u" R" a/ }* u6 h printf("请输入字符:");* N! Q4 f: Y# m( }2 }
gets(ch);
; F# D3 w }9 b, A, z& `printf("\n密码是%c",ch);
, G$ P8 } b* M( O3 l% p" q# wi=0;/ R/ R! d, S) b1 L' a3 ?5 k7 G
while(ch[i]!='\0')' V2 w, b6 l! @ H/ l: y: X
{if((ch[i]>='A')&&(ch[i]<='Z'))% S, d, K6 i. J, x$ C: i. _' v! Q/ x ?
tran[i]=26+64-ch[i]+1+64; }7 ?' ^" |0 n- g
else if((ch[i]>='a')&&(ch[i]<='z'))
3 @- s! R4 O5 p tran[i]=26+96-ch[i]+1+96;
. V* f5 d& K$ g/ d0 \1 Celse @. J- e8 i( E
tran[i]=ch[i];
k+ V; \# s1 ~/ [+ D i++;
! ^% e& h1 B# b4 k- @& z- g! t}( u8 v& O; s w3 D; q! l: _3 v- t& K% @- E
n=i;# }, K7 e( k% `: M
printf("\n原文是:");
* b% `7 P& K. r# V4 Q# nfor(i=0;iputchar(tran[i]);6 d4 E8 a9 `2 Q z" ~
}
) S, X( g/ z$ j$ I7 b/ w; D2 t7.13- i, E8 N; m* ~
main()3 v+ S) K9 p0 G4 H H. g6 K0 H
{
3 C: c- H; E! l; w. d6 i' v" _ char s1[80],s2[40];
3 l% }# ?! W0 i7 K' _; p int i=0,j=0;7 O D6 H; a$ D* j/ |" q7 H
printf("\n请输入字符串1:");
' i+ h- _ O& ?0 r" G9 `+ c scanf("%s",s1);
: m' Y1 K6 `6 c% \# e$ q# {& q! T printf("\n请输入字符串2:");
& `( t" {6 ^9 O% q q1 ?1 q scanf("%s",s2);% U5 a1 t" Z" p5 K( ?6 v
while(s1[i]!='\0')
% {' b$ ^8 D2 |3 [5 c i++;4 }* I7 w8 E" ?! U# f
while(s2[j]!='\0')
& M+ p" s- q: n! V s1[i++]=s2[j++];( j# i/ J! p- t
s1[i]='\0';
8 G) G# p( T8 _0 C1 o* R& Qprintf("\n连接后字符串为:%s",s1);/ r- J/ L% W. h' v
}
/ a$ T" E F5 R# Z1 k
, P+ T% o% z' S. q9 r( S3 X2 Z$ n. p5 ~5 K% F; [9 n" c( u
7.14
% a3 ^6 n \" f#include, v) Q8 q" u- ~! P+ Q( Y0 @; l. }4 I3 N
main()+ q/ O3 ?0 z' z
{int i,resu;
1 I# q( w" d( A char s1[100],s2[100];
. Z" m" F. \% j/ A( N% i8 y printf("请输入字符串1:\n");+ h! O, S1 R$ y8 @& H
gets(s1);
2 b1 \+ a b" o7 { printf("\n 请输入字符串2:\n");2 Z. u' o) f/ X# {: F
gets(s2);( Q- E Y8 {' }: L' a
i=0;; E2 k! c% {9 j9 ^2 y- S( l
while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;4 a2 j5 _& n$ {9 f5 C5 d
if(s1[i]=='\0' && s2[i]=='\0')resu=0;0 x9 w* l& K6 p2 ?. `2 G
else; m( m. O& M: s
resu=s1[i]-s2[i];
6 I% j' D* e- X' g/ H8 b0 k printf(" %s与%s比较结果是%d",s1,s2,resu);
0 [# O# j) | Y* x* z! X! ^6 V}2 ]. p' o& `9 q" n8 j$ n5 S" N
7.15
% T4 [4 ^9 z7 G/ o#include
6 t& ^( ^. g C( Q7 kmain()
) p- C& ^' C* S' w2 ` {
& v2 u% j7 D! _0 ~" u char from[80],to[80];
& m0 F! C1 E2 D+ x* i int i;$ F) E4 c9 z& d9 X
printf("请输入字符串");
' u; ?. d. |! Z* B# f0 S scanf("%s",from);8 p. n) N! X- E$ `* b
for(i=0;i<=strlen(from);i++)
; b& B8 H# k5 Q7 W% w to[i]=from[i];
: }9 J& s. m* I, R8 B9 H printf("复制字符串为:%s\n",to);* D/ w2 W t% [" L
}2 f+ `2 H% L" o& c( X
, w. Q2 N; k. B
& D- e p9 N, E$ r" q- u0 l% X第八章 函数# z$ P& O. X7 g
8.1(最小公倍数=u*v/最大公约数.)3 v/ V+ r- Z- g% v3 ?( U7 r2 Z
hcf(u,v)
2 n+ p* ]3 _. L6 H) Yint u,v;9 ~4 x* c' T4 |, L7 p7 a
(int a,b,t,r;
8 T# ?$ g& [; r' @! M# y; l if(u>v)4 v0 F7 M. b; u8 z" S3 {0 W
{t=u;u=v;v=t;}
0 T0 S; w, M' {% \( j a=u;b=v;
. \0 e/ K2 x' [3 [& e while((r=b%a)!=0): @4 U6 S( i6 R# k+ {! @( w) H! h$ K
{b=a;a=r;}+ ]3 m% [# o, l0 ?% M
return(a);$ P: J8 @- ~' G! ~9 K
}$ O/ I3 a: W2 z
lcd(u,v,h)
4 R! O4 O' e, ?( V int u,v,h;8 [& @" x5 L4 P8 W+ ?- i" S d) r! _
{int u,v,h,l;
h( I3 r& b( | x scanf("%d,%d",&u,&v);
6 T6 O2 d: Q; }/ ~& ^, s7 c h=hcf(u,v);2 [% R' M5 U& Z2 A
printf("H.C.F=%d\n",h);! j/ H7 l, r, c+ q
l=lcd(u,v,h); q6 P y3 K% [6 Y' ~# m
printf("L.C.d=%d\n",l);
& W; S# i. t; K9 f6 V }% ?, d1 A% N; G |/ p
{return(u*v/h);}$ h5 h( X; M- K$ H
main()
- v- ^* L" b$ V. N1 {7 E& V {int u,v,h,l;
: E! L" V4 U5 j4 n7 \5 v/ { scanf("%d,%d",&u,&v);
- F$ F; j8 |: Z$ x0 @ h=hcf(u,v);
; T' M; U9 A1 T" ~$ v printf("H.C.F=%d\n",h);) B' X1 ]- d* f ]2 ?$ P: `
l=lcd(u,v,h);
% i- v! h- P3 o6 m( p1 g printf("L.C.D=%d\n",l);4 `" H' I5 s( ]1 z! }$ q# c' Z
}
; M8 _$ J5 R) m/ F$ Q2 O; ^
- m- Y' C0 J, }6 m
! f* R- i4 h0 T) A1 b4 X: a& z, N, K9 A* \7 { k6 C' Y
8.2求方程根
* ^5 P& ~& N$ ?#include9 Q- E7 Y1 t/ x& d6 Q1 d- U
float x1,x2,disc,p,q;% U& F" d, B1 ?' ~$ ~1 S' A) {$ X
greater_than_zero(a,b)
3 z: \, c7 z F8 ]0 M$ X+ ofloat a,b;0 F9 F& `( a- k' r! F. N* L
{; ~' m9 ^8 @, z% n2 P
x1=(-b+sqrt(disc))/(2*a);
& s; h8 ?- c! T6 B: i. Yx2=(-b-sqrt(disc))/(2*a);
! m2 e! x8 E8 r0 f3 O1 h}
" g0 g2 y4 V7 g+ ~# b! @% Z( {2 lequal_to_zero(a,b)
! v/ t0 H# l$ m" [, P; kfloat a,b;. |3 [5 [& o+ R4 {2 {1 ]0 \
{x1=x2=(-b)/(2*a);} R U) X, P2 K+ R) ?
smaller_than_zero(a,b)/ N% j/ ^7 b' L# ]; B; D
float a,b;
6 N; T" r, p1 [9 L/ ?/ W' A/ [' q{p=-b/(2*a);4 S, A" C$ \8 [0 [5 G; g# P
q=sqrt(disc)/(2*a);
- d& u' j$ J- {& \9 } D; _}
' G9 M# L% r7 k" x; ]main()- o( l7 u: d Y6 J
{
7 Z9 S) U9 D; H" Q7 b3 y# V+ ^float a,b,c;
% m. Q0 C6 J$ E: t% Eprintf("\n输入方程的系数a,b,c:\n");, w! q" P3 t# M1 J/ L7 h
scanf("%f,%f,%f",&a,&b,&c);
5 _; z6 x" R; N3 pprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);* F/ F K+ y6 x* i
disc=b*b-4*a*c;
) ?6 o: X- Y0 Qprintf("方程的解是:\n");
; G! J1 Q1 X5 k+ ^* Fif(disc>0): @% }4 q' W, d( `; Q) P- W* c% ~( x7 b) C
{great_than_zero(a,b);$ _2 @5 P* X* B/ g: n, o
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);- j/ c1 }* b+ ?2 D
}- i* j: ]" E% F
else if(disc==0)$ f5 M9 w) M1 a" y, d, @
{
. J/ z+ I7 C: W8 qzero(a,b);. M0 m0 t" n @) ]) e8 V) d
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
4 p/ P* i9 T P, X; O9 H }* G$ f8 \2 |& H
else
) M" t6 v2 s, F' @4 a {$ W4 t& @% g+ o/ V9 f
small_than_zero(a,b,c);. u& V# L; o0 {/ ?+ x, ?
printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);
# W* h( l' O4 ~# R }1 k3 c2 |8 i& R# `1 K
}9 s7 m1 n4 [8 V7 B
8.3素数
( p3 W) h3 t! Q#include"math.h"
5 ~: H) E+ k9 b3 A8 [6 t- Qmain(); E% g. h) [& E% a& Z3 l: ~0 J; `# ~
{int number;+ ?/ l. d$ D3 q$ L
scanf("%d",&number);; Y0 X8 v; b9 f. a
if(prime(number))
7 C4 U5 b8 @1 _5 x printf("yes");- s" d, o& {+ w; N
else( b* ~) h+ c+ d4 M" G4 u! Y
printf("no");% K5 a6 C: m5 y) X
}4 U5 m) D+ W/ N) T/ d; i
int prime(number)
w2 k$ P }" Q5 {8 Xint number;) T) f9 U* w3 r, U
{int flag=1,n;8 d4 v+ ]4 `; D; I) Q2 W/ o
for(n=2;n if(number%n==0)
" T2 p% @0 r+ W4 v6 r( L flag=0;* Q5 F: D u5 B' d9 z
return(flag);
5 J* T1 B! [9 b$ T3 q}- S6 ^" P% d1 X4 J4 E2 N) j4 A5 p
+ L% V% b! T0 z- t* Q" v1 H
5 S3 ^2 ~- C! c7 ?/ j0 U' B4 q# g* T$ W. s, r' X5 z
8.4
/ t. ~1 M0 ^9 P3 h#define N 3
9 R% C& |' a- B- Kint array[N][N];
4 z5 R+ M G) Hconvert(array)* I& ^, e" W9 `7 L+ \3 }/ R9 d2 `
int array[3][3];4 `7 R8 q5 }3 h- P0 B
{ int i,j,t;/ o+ `( L6 ~2 S
for(i=0;i for(j=i+1;j { t=array[i][j];8 @1 _+ |1 p! ~- r% Z
array[i][j]=array[j][i];
5 J `2 m9 `2 g; ] array[j][i]=t;
& ~/ L1 h$ n- o4 M3 O+ h6 [ }' a" p, i d8 D7 S8 R
}
7 m5 m# t- G. d4 Qmain()% h2 a% H/ |# T; q3 m
{
6 c' N- ?2 \) s- ?: {$ M/ |. o1 l int i,j;
7 H& X% F! j W9 ^* m printf("输入数组元素:\n");1 J4 n; ~0 s; q' A* w! X
for(i=0;i for(j=0;j scanf("%d",&array[i][j];
3 l3 y9 k( @& v" _1 x7 j3 ` printf("\n数组是:\n");
) n! {( d& L+ v R4 b% Q7 W$ Ifor(i=0;i { for(j=0;j printf("%5d",array[i][j]);. B1 t2 \) i+ I* D/ p) ]
printf("\n");
6 k$ `- K& r. ?6 m }# d7 }0 p+ J( ~/ w! k5 }& s
convert(array);
) A: U; V0 r* x& N printf("转置数组是:\n");
: W& R8 E) Q# ?, E- }& [0 j; H6 s9 S for(i=0;i { for(j=0;j printf("%5d",array[i][j]);; T, B: {& h4 j4 V* ?5 a
printf("\n");$ y7 {9 u, N- ?& K# y/ b5 W, _
}
' \( F3 z$ L$ \! @} p7 e4 s. x" o" W: m' ^! F
. d U. K0 e; ^$ P3 x# ?5 C9 s
" [6 Z: y+ q5 e$ W2 f$ e% e: N! M8 P( X. t
8.5* c/ t/ ^8 @. G( `; B
main()" h2 W' X7 j% W
{
1 U& @+ _6 `2 jchar str[100];
: _( u2 O% O; i2 V; N printf("输入字符串:\n");# m! R$ X& R) m, ^
scanf("%s",str);
x2 x% f D8 _% k1 J4 F3 |: V inverse(str);
' T6 m. _& H+ k& x* f1 J printf("转换后的字符串是: %s\n",str);
, z }0 z9 j5 o& y; j/ ?}
4 n* c0 h- U- W6 vinverse(str)1 h- ?0 O3 W0 |$ e) v6 }$ R
char str[];6 } k' b9 |% ~+ P5 q$ n
{
, u0 N; s# K$ o- g8 a char t;. S8 t' Q1 Q+ [ c& N7 e) O+ ^
int i,j;
$ w$ G2 E5 }% w" V6 D1 ~$ M for(i=0,j=strlen(str);i {# {+ |, v8 i' ~4 e, m7 h
t=str[i];0 l5 V9 F3 \/ }0 V9 Z
str[i]=str[i-1];
: X: W/ Z8 g; c% @( q0 H5 V str[i-1]=t;
; Z- F, q, o! {+ v4 I) X: f% v }
5 ^( y5 O `) B2 p1 G3 s& e}: X7 [) y; V7 d0 n4 [( d
) L D3 `1 m- T( F3 P( N" e5 c# M, o: ]$ Z
5 j2 y' @2 D7 q$ }& A* l& A
8.6, A8 @4 a: \/ x$ C: K; L( Q
char concatenate(string1,string2,string);5 v+ S$ \6 \- h1 ]8 E
char string1[],string2[],string[];9 ~6 W' L2 \6 ?! A
{
' X1 O! M4 A1 F- f5 _# g9 n1 Oint i,j;& U, D3 E: z" Y- v; s) r+ L, `
for(i=0;string1[i]!='\0';i++)) l$ I) y, u" u0 z. o* l1 |$ {
string[i]=string1[i];- t9 _( p0 f" f Q( z
for(j=0;string2[j]!='\0';j++)
6 k s/ R- ~7 P% u2 F/ z) q3 V3 H$ E string[i+j]=string2[j];3 [; U6 u- g* E5 r7 k0 x6 w+ A
string[i+j]='\0';: Z; S0 V% M. |( h
}
; w/ S; M& l; J6 p' }. m4 w! ?' Nmain()8 c* C, M2 B: m+ F* ~6 z5 g/ d
{* x/ n0 K* u4 K1 T- |6 ]' g1 n5 p' [
char s1[100],s2[100],s[100];
/ S( y6 a1 c: f printf("\n输入字符串1:\n");9 z2 Y$ {- o) ~( G) e
scanf("%s",s1);( q1 o+ ^/ P" L8 A0 C# L
printf("输入字符串2:\n");
l+ ~# u+ c1 q) w) J5 B3 {4 V scanf("%s",s2);
/ Q8 f- j2 ~5 I% q concatenate(s1,s2,s);
9 k; q1 _: g V }' k7 i printf("连接后的字符串:%s\n",s);- q" u M4 M/ Z$ T$ x1 q& e0 ]
}, e# E5 l6 ?3 b& j
/ }# z* f9 J0 B3 z
6 s; d/ ?3 ? k+ y7 Z
8.8! t2 F+ i( p! E- v- h9 N) @ e# Z' |/ @
main()3 n. E% p% A( p, T/ K, O
{ C1 A/ i( L) o
char str[80];& c2 y- ~: x" ~* M0 K1 B
printf("请输入含有四个数字的字符串:\n");
6 b/ K4 J8 z) s, K, C$ A8 @4 W scanf("%s",str);
% ` A" ]7 W- c( i. c- `8 ~ insert(str);
* ?: ~% r" \8 b& J7 g}6 y/ T# A$ T8 A. g- F
insert(str)* p# N/ T# p- X- U4 d$ H
char str[];
) j5 T5 @2 p9 _% h2 K{1 l' s' p. @3 {8 {# O- s. ~1 P
int i;
) v! h- l* ?6 g+ h for(i=strlen(str);i>0;i--)8 q8 n% p& W* w4 S, ^
{ str[2*i]=str[i];+ o9 [8 e: t, ~) ]% `/ ~' \
str[2*i-1]=' ';
* Y$ v% I8 f0 s) N& T0 X: O% k }" [0 ^% u6 `8 ~: A5 D6 v
printf("\n 结果是:\n %s",str);
% g/ t0 o8 n9 c" K }% A" I- h6 O' k
( M; [' l7 o4 t8 A1 }- `8 C k5 p' V
1 C* d9 V+ y4 L* D& w# W [# F7 B3 ?
8.9
( u, z" H" `& ]8 j$ R: `: ^6 n#include"math.h"/ _/ c s% {- y* [* T4 q
int alph,digit,space,others;
; S+ ^) I( U' @! |main()6 \) j" s" L3 A4 ?
{char text[80];, }' p0 z9 c9 R+ f+ e* \
gets(text);
; O, B! ^4 i' J/ Q$ P+ x) w alph=0,digit=0,space=0,others=0;
; c" }( y |. j& g, U4 f count(text);
7 P" ]3 r: S) S" g* ~ printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);9 S/ l: q9 J Z9 z' _
}' z1 f* u/ [' U& ?' J' c# `
count(str)& l l8 D5 t/ U* j) X/ ^% Z2 N' `
char str[];
9 _' z5 J' o3 i+ j& r{int i;
. s7 C; V8 H j0 | for(i=0;str[i]!='\0';i++)7 A8 t( Q/ R" p. R: l: L& [1 I8 E
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))) o5 @: G& _( ~9 O2 q
alph++;
* w+ Y& K( e& ?9 [7 T R4 O% f else if(str[i]>='0'&&str[i]<='9')) Z7 g7 v7 S! R. h) D: N
digit++;2 r8 T" C) ^9 l+ F
else if(strcmp(str[i],' ')==0). F2 r; W# }3 U% j7 K) Z$ b* ]9 }
space++;
& Q d( S0 T# _& y4 i3 O else# q: B1 X7 m$ L( H
others++;/ K8 _( e& H9 y/ U
}6 B* W. \4 F% ^* ]5 p0 W
, s2 Z. N* p! G
/ o3 x* ?9 w' O$ c
8.108 U2 J$ Y( U$ |1 R
int alphabetic(c);
x: G. b, B% L. g9 e( ?char c;
9 ]% i+ g( O: ^% {9 _; C O{
v) s- N$ o1 r5 M* k: L9 V. i0 W6 A if((c>='a' && c<='z'||(c>='A' && c<='Z'))
; ~+ Q4 }+ @; T5 T! J6 ~% O return(1);
$ i& ]2 T1 ~: `& V else
& [, K. H; j8 O3 A' j$ ]0 Y return(0);
+ k& \( g4 O0 k; u/ n}
# {& b( v: C- P* {) v
( L, P8 n( `- Q8 D3 gint longest (string)
- m! v4 A, Q5 u/ J; d d# g) {char string[];$ s5 n$ ]2 N% Z) Q4 }
{
/ ?, P4 ] I8 a" b# P& t/ b int len=0,i,length=0,flag=1,place,point;( ?6 n* T( m& _/ T+ T" P
for(i=0;i<=strlen(string);i++)
) R$ ]1 a: y$ t3 D2 o' r2 D if(alphabctic(string[i])); Z$ Z, K5 y& F# s, H
if(flag)0 S# I M8 Z% c3 B! N* E9 K5 t
{
- \. ^8 q, C6 O) V9 ` point=i; h# w' F# c9 u& S) P
flag=0;0 }! w) Q, b8 |( e/ V4 t9 [: p
}
0 M. A# L5 a. n. M% I7 v else
) |* g. z% x0 m" S/ c# ^ len++;
3 ?2 j! D7 D. U; c9 ~- C/ ? else
: R, q" `1 L6 X Q( a- z+ b4 _ { flag=1;: j. L: x7 O* s! J3 @
if len>length)' l3 g# g' t. x+ p4 d3 V8 L
{length=len;4 b7 ?$ l# X, A6 U ?5 `; a# S
place=point;5 a0 _, t% z: u2 [ G: H7 E
len=0;/ r4 L1 a( R* K( P- q E& }7 i! j
}" |3 H% r6 ?: O6 n+ D. i
}4 L/ @ i9 P1 `4 s8 H1 J: W( J* ]
return(place);5 C1 i, c/ B( T& g5 y
}
6 h/ y# K* j& R+ m: m- C: U- |main()+ a! E, ^( P$ f6 R# y$ t0 T3 L8 [
{. J- b) M* H' d& D9 P
int i;4 S5 o6 p7 z9 H
char line[100];; s; g/ H* J1 S: k& q
printf("输入一行文本\n");6 w0 r' L4 r* Q
gets(line);
0 C/ S% @6 O+ l3 S; {printf("\n最长的单词是:");
9 S$ V) g6 [/ mfor(i=longest(line);alphabctic(line[i]);i++)2 X8 w1 p1 [/ I3 c7 W
printf("%c",line[i];% R5 F8 v+ l$ o ]$ R6 P/ s- B
printf("\n");
5 R {( |6 {; O& S( z% U}, Z& Y3 U# s" c |( a. t
2 |2 p' }9 b- o9 m U! k+ \4 z
; p S! B/ C- E, T' u h' l; Q
4 ` P" L P5 ^8.11
* n9 Z/ m' s6 E$ [4 Q% c8 m T#include
+ `8 s% `( o! h1 S$ X$ h& J6 t+ i# s9 }
#define N 10) T2 \& e( j [7 }" P. y
char str[N];! C, J2 y4 g- a( g: a6 v1 w/ m/ D
main()
9 w' v. s3 k B& T# `{8 U6 _6 D4 O8 v3 }+ A
int i,flag;
; y, T& K, _. N8 \5 C0 A- yfor(flag=1;flag==1;)
- F( ]7 Z6 c+ x& T0 s$ A) O{, R7 G. T' s. s. A
printf("\n输入字符串,长度为10:\n");
9 B5 L* x, ?+ m) g% W scanf("%s",&str);
+ @0 i3 B+ M# j' ^5 _3 A if(strlen(str)>N)
* ]( a8 h( v2 D$ h V* A/ s n printf("超过长度,请重输!");" K0 ~3 l: a; F' |. b
else
5 `7 X! C, v2 G$ ~/ ` flag=0;
5 ^1 F0 \& E- p% g Y5 Q}8 K u+ e( C/ r5 A
sort(str);
0 q' j0 Q4 ^3 A1 Aprintf("\n 排序结果:");2 w: Y5 e! Q2 t% |( t& s
for(i=0;i printf("%c",str[i]);: p) v' X" @3 C
}4 r/ ~% t5 }3 p' f B
sort(str)
3 c2 ^* g! d9 M1 W% tchar str[N];
8 B; N+ v( G7 c{
( G" _3 G4 K8 uint i,j;+ L( e$ j% n9 b2 n
char t;* t% \2 e2 g: u- y
for(j=1;j for(i=0;(i if(str[i]>str[i+1])
4 k ]( q+ M1 R, p; p { t=str[i];6 D# a, |9 [7 ?0 l1 R9 B! W
str[i]=str[i+1];
6 t: B! L& y- V' b# w" l- @8 j str[i+1]=t;
; Z' b+ o5 A' \: } }" n( l4 o& A& C& E6 {, A
}7 |7 M7 A7 z" X. P
8.12
7 n# J8 N' P1 ^9 A#include; E; d6 {2 |3 ]3 H9 N, v
#include
5 [8 P* D' o) s$ I9 R) Wfloat solut(a,b,c,d)- y7 W, _( P" @3 j8 N
float a,b,c,d;$ O! c+ C- B; \' B/ c0 o% A! E
{float x=1,x0,f,f1;1 o* H; C/ r+ j2 o
do
! Q$ Z7 @2 Y/ Y8 L; | R! h {x0=x;
; |7 Z0 v* V7 j* ]3 \2 _- A f=((a*x0+b)*x0+c)*x0+d;/ H& x8 ?" B( I9 \" ~; l) K; S
f1=(3*a*x0+2*b)*x0+c;
n \- R2 y% @* n8 N ]/ a8 k x=x0-f/f1;) r5 j2 K7 j5 n% D* y' d w
}7 D& W7 ]% H6 }1 _
while(fabs(x-x0)>=1e-5);
- `$ J* W0 G+ l return(x);& U$ @5 h* Q, K, f# D* g1 z8 b
}
1 p, ~3 }1 [( F1 P+ p, Z0 I4 lmain()( C9 e7 t0 m9 m! X
{float a,b,c,d;, F7 k6 Y8 X0 i6 J `+ L4 {
scanf("%f,%f,%f,%f",&a,&b,&c,&d);* H2 P3 D _# F- T- l
printf("x=%10.7f\n",solut(a,b,c,d));; X5 T6 H( t+ K! N/ G5 @: `* i- ~
}
' B9 A& n5 R! u: q& {8.13
$ I) M9 N/ V E$ n4 ?' _#include
5 N( E6 n6 ?" O: r- B6 ]main()
6 K) O" r# J! \( J{int x,n;: B ~4 \& @0 g; t! C! P* r
float p();
* e- D \- m% I* N" ^ scanf("%d,%d",&n,&x);
, ~! [& I) f5 [" k# R6 E printf("P%d(%d)=%10.2f\n",n,x,p(n,x));6 f2 |5 g6 _" r1 e
}5 ~4 ]6 ?2 [* y0 W6 W, P4 T9 ^4 ?2 }
float p(tn,tx)" p7 M6 U( ]; M- a( h7 F
int tn,tx;
- t6 K( \7 ^) Q% ]# S9 F# _{if(tn==0)
3 e$ P4 B9 c. K return(1);
* @7 ~( K6 w' A5 A3 b& p; I% w else if(tn==1)1 y0 x1 {. z: e/ Y: \
return(tx);
$ U+ T. g' r" |8 }9 H else' s c. W- d5 Y; u
return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
0 C/ l9 m. X- p}
, P* K( B, _/ m3 d; O4 x* S. v; E" ~8.14
, O! ]1 s6 P& K#include "stdio.h"# G* W2 Q( }* s/ }+ D1 @
#define N 10+ {+ I+ L6 ^! Y. ]4 R
#define M 5
# O( g. [0 m+ m- k- e3 afloat score[N][M];7 Y. x; Q; O; F+ Z6 n
float a_stu[N],a_cor[M];! F, {& P0 N* w0 m
main()" p# X+ i& U, f, `4 u/ N: z7 }
{int i,j,r,c;
, E& X1 Q: Y e: }& E( n; Q2 a, j float h;
X2 S" c) Z) q j8 b! a float s_diff();% r4 |, [: }8 Z% }6 Q$ L
float highest();
& Z! V2 s C2 r$ _- F r=0;
+ l( y5 o) G2 _6 W5 Z" A/ ?- k6 z4 W c=1;
; t. H6 Y7 o2 A* p4 {+ M: @ input_stu();8 C0 ~% f6 G& y# N5 f5 @5 [. x
avr_stu();
& F; p% n) |4 r avr_cor();
0 P3 ~# o6 O2 V1 L printf("\n number class 1 2 3 4 5 avr");
' ^! ~$ Y: b9 S( b2 S' x5 T for(i=0;i {printf("\nNO%2d",i+1);9 X& l8 }" m8 h6 @8 z0 W
for(j=0;j printf("%8.2f",score[i][j]); B7 N1 e4 _5 Y2 n! |" |7 ]- H/ L9 ~
printf("%8.2f",a_stu[i]);. U0 ^# d/ |. E# j& B
}# E- S+ ]2 R- H* n
printf("\nclassavr");
! I- }+ B: }- D# s! S4 I for(j=0;j printf("%8.2f",a_cor[j]);0 J; {/ [: Y& D2 \- _
h=highest(&r,&c);
3 Y! a/ n" A4 S/ L7 s printf("\n\n%8.2f %d %d\n",h,r,c);' z+ H* H* {9 P& c0 \# k/ V
printf("\n %8.2f\n",s_diff());/ ?; U" S; z/ p/ x2 @, m
}
+ E# m/ Q5 ]% K+ Uinput_stu()- E/ n8 r2 j+ `6 W) K
{int i,j;; J b N2 [0 \1 I+ Y4 b
float x;5 P* K8 l2 ^$ ^# i, X
for(i=0;i {for(j=0;j {scanf("%f",&x);
! c7 ^0 U! H1 i, e4 I1 M4 S score[i][j]=x;! R+ [* _3 T, |6 W* Q% b2 W7 ^3 Z
}6 N: y2 U5 ]0 R+ v0 o
}
4 _! {/ @# c& r$ j}2 K2 \" O$ s% w/ o; C& t6 ~
avr_stu()
' H# R$ }% g$ y' M- E* x{int i,j;5 F0 Y, g2 ?4 c8 Q
float s;0 l9 v' u0 T9 d) a
for(i=0;i {for(j=0,s=0;j s+=score[i][j];
1 w8 X! e- A8 ^% c# p* r a_stu[i]=s/5.0;$ V( p5 |4 {, v% S( A; {5 Y2 }
}* Z6 C- X, }+ ~8 P) K6 v
}
& q/ s ^. _' O u0 I6 \9 tavr_cor()5 Q1 f% v$ C% n8 i
{int i,j;! P+ {; ]$ W! k) T6 C/ C
float s;
+ d8 _; h1 s, e! c1 ~ for(j=0;j {for(i=0,s=0;i s+=score[i][j];8 ], {* M* R& r- F8 w& L1 @/ P: \: g
a_cor[j]=s/(float)N;3 c- X& t: U1 Q/ Z8 }# K
}
; I) v$ r2 ~) a$ V7 ]}1 n0 P) P. J& k% Q5 w4 R
float highest(r,c): r6 Z, w- e# {+ u% ~
int *r,*c;* O: x1 W8 K" O4 C2 l! A+ i
{float high;8 X3 U5 m7 O) T. T+ A
int i,j;
$ D! ?* O7 S( D1 e! h/ p- H7 o3 P; C high=score[0][0];
- D3 m" E, g& X) s* R, d u E2 s for(i=0;i for(j=0;j if(score[i][j]>high)
8 P0 L5 J5 Z: D" w8 u {high=score[i][j];
+ X: g. H& b( D4 n$ o *r=i+1;5 j$ L5 \1 d( `$ i& h0 Z( y
*c=j+1;
! N2 K) A# r( [2 D }
$ v. z. ]9 l* Y& V return(high);+ J$ Z+ z/ T, W
}2 B! P! m& A1 y! }( z2 r) d" t" v1 s+ U
float s_diff()
3 S; n. [! Y. ]4 N& R{int i,j;
9 I- V" p' [2 m) b, A9 Z float sumx=0.0,sumxn=0.0;8 j: y' O. T( e" `- X
for(i=0;i {sumx+=a_stu[i]*a_stu[i];; u- B, ]: p; Q1 s6 w
sumxn+=a_stu[i];
4 q5 C6 Q7 Q+ ~( r }' s% K$ i0 C( j6 S2 `+ e3 T/ V6 H
return(sumx/N-(sumxn/N)*(sumxn/N));
3 r: y0 ]- @- w* P}
4 z" I. |( l# {" r- E! c1 y% A5 _8.15
, f7 J0 v% I p9 W#include
& G. T9 a' }. W, [#define N 10
: C5 I/ T. I1 s+ i" g% Nvoid input_e(num,name)4 W: C$ y" v& \4 ~. c! t7 g
int num[];
+ h9 u0 L8 B5 c1 e, }) wchar name[N][8];
( r; Z2 g6 b( t( z. G! j: p{int i;
( \+ f7 o6 n) R for(i=0;i {scanf("%d",&num[i]);
8 h& l" Z9 V( _5 |/ M) ^ gets(name[i]);
9 x# Q8 x) R+ N1 j1 z; m% e9 {/ F3 o" @ }1 [" u2 t. i; n, W6 ~
}
) W$ j6 g1 ~' @/ P! c) j7 L8 Qvoid sort(num,name)
: Z; }2 C# V7 o9 O R% `/ Bint num[];
0 l; x: E; }' Xchar name[N][8];9 k3 [2 r% @- o1 o1 o
{int i,j,min,temp1;
3 u1 l; \4 [& {0 c8 ?; b! G* l char temp2[8];
) D0 m& m k4 o- N8 { for(i=0;i {min=i;" Z9 L: M) z8 k+ k6 x6 N, U
for(j=i;j if(num[min]>num[j])min=j;
, P+ O$ a0 S8 V! G" Q temp1=num[i];
; i5 E* O, t" T, r# u num[i]=num[min];
! j0 C* d& ?1 Z! [' Q num[min]=temp1;. W- D9 |( w. E5 t: E/ y
strcpy(temp2,name[i]);
# c( x. {0 I7 J) y5 p strcpy(name[i],name[min]);$ r# [( M/ h3 f& R, ?7 a
strcpy(name[min],temp2);
# z/ w g0 f/ P7 X }
2 A' y6 B6 }, p- T for(i=0;i printf("\n%5d%10s",num[i],name[i]);
/ g, o( R4 w: q' V; l}# ]1 _( O5 |1 D7 p9 @8 @" A
void search(n,num,name)
# h% L+ A: J# \6 S; G) [int n,num[];0 p+ r7 R5 O: `% I: n! h/ J
char name[N][8];( q) Y. X# E7 z- I6 v1 Z
{int top,bott,min,loca;, n0 o* `4 a8 n' Y% t5 F+ K8 S
loca=0;
* G& b$ q; \% Y1 ~+ M top=0;
% A* L9 W, v) ?& p! v1 `% F! q bott=N-1;
$ G4 e4 \: n, {1 l if((nnum[N-1]))
* A2 H3 A' Z5 ~; |# @" I loca=-1;% q& a* h& c( a: d9 P# ]% F
while((loca==0)&&(top<=bott))
3 f0 |) A# T0 R1 |1 }/ X* p {min=(bott+top)/2;6 a0 K' V6 D! I1 F7 f4 R
if(n==num[min])
- U( K# l, P) `' p$ q {loca=min;
& |: F6 C1 d0 A t printf("number=%d,name=%s\n",n,name[loca]);
l" b4 A9 [1 W }% N/ W [) a! Z I# j4 S: T
else if(n bott=min-1;
; }( W) P; Z) G else
$ }/ W. u( |$ f j0 `0 x; o top=min+1;, f) b9 W7 i2 e5 ]4 u: l$ N
}7 l* y5 X2 F, {: u4 {
if(loca==0||loca==-1)
5 b" m2 K# @" \) y8 ? printf("number=%d is not in table\n",n);
' e+ K6 v% x: X& N K4 m9 i} F( j* @) t6 ~+ [
main()
( n I) s/ X. v5 ?; ?4 o{int num[N],number,flag,c,n;
3 D2 B( V- E. r char name[N][8];- t$ m# V) l; |" g* }$ J
input_e(num,name);
* ~* S: u# ]" {! Y/ V sort(num,name);
+ C, m4 D: t# r Q for(flag=1;flag;)
j# q8 M* B3 t+ a {scanf("%d",&number);9 D2 F4 T3 P$ A @( @
search(number,num,name);) E) G+ ^# h' j" v# c
printf("continue?Y/N!");
' C/ K0 m9 l, }. e1 |, \$ } c=getchar();
# B- r1 ]3 ~( C2 D$ N if(c=='N'||c=='n') m! a$ h% p! }) p8 r2 V" a% V
flag=0;
! ~% N1 A( h( x }
) }( D9 k, f; W5 S7 C. j}
: o+ e6 _' k3 t0 l/ ]6 x& R: k7 T% {% n. N3 E4 K
8.165 d! r9 m; N: ~3 ^; o/ h; x, v
#include
6 x* x. p$ G$ P0 C" w#define MAX 1000
: |1 Y( Z5 a7 O( ?main()
* x) T0 a8 y. Z/ h{ int c,i,flag,flag1;, c6 P, y6 _: `1 U* m3 f& u
char t[MAX];8 u3 i* ]* @& t! h
i=0;& ]! H4 Q- J8 E$ U; H
flag=0;
$ R7 C! D- U5 X1 E. z( a flag1=1;
! X7 v6 e5 }3 q9 B/ U3 E5 X/ i/ P printf("\n输入十六进制数:");. Q5 E1 h* `8 L' m( t4 L Y; k
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F') H: ~1 j* C* B1 h2 `4 |
{flag=1;
2 H. d9 C q1 q X t[i++]=c;
, d" T; H: Z* s; Z }
) Z$ f$ Y" H* ]( l' d else if(flag)" B, x& n# l5 w( f" f0 B
{
* C6 F7 O) d3 o" s t[i]='\0';, g: l% R% M2 H
printf("\n 十进制数%d\n",htoi(t));1 g% \$ |( I( y2 V- a1 S P5 F
printf("继续吗?");
' C \% q; K6 L' U) w9 F5 v c=getchar();' s. j2 T. H9 j# |& o1 x
if(c=='N'||c=='n')/ D, Q% o; ?8 L6 J0 ~
flag1=0;
# Z; v5 x+ I. v! y- V0 Q( J else6 W: z* B$ f1 \0 T) A7 D& H) @
{flag=0;/ l1 T) |4 S& W1 T" A- I5 o
i=0;
: j5 [8 C+ Q3 J( Y# J& h printf("\n 输入十六进制数:");" {, k; T! @' }+ ?" e B3 P% a# T
}
( d4 B+ M X$ e7 g' @}
: o }1 M. F9 E, W" B( _7 m}
, m M7 q. k* q}: @9 a5 C4 \: j0 ?1 F
htoi(s)4 X) @; |5 V3 ?9 b+ b# {$ b- u) Y
char s[];
0 e$ t/ w" a4 W3 P{ int i,n;0 g! p" n1 K c- d3 K5 i- x
n=0;
7 ]$ n0 }: L! ]8 | for(i=0;s[i]!='\0';i++)0 @2 Q7 G( y! Q" N" N: g
{if(s[i]>='0'&&s[i]<='9')
. n$ T5 R3 s. P& A* V% u n=n*16+s[i]-'0'; j% h; K! Y5 \; f/ U* m. Z0 W0 H
if(s[i]>='a'&&s[i]<='f')% c6 _. A# n& u/ H
n=n*16+s[i]-'a'+10;4 f+ B/ R2 V4 M# S
if(s[i]>='A'&&s[i]<='F')
; c5 g! }6 l9 a7 q3 i n=n*16+s[i]-'A'+10;
1 b8 Y' U2 N R3 q, a' I! n$ O" V9 Y }1 e `% k; Y7 p8 x* q' \& E8 O
return(n);
1 W5 Y# G; _. ^- h' ~; E}
6 {2 v: v/ }1 W1 Q @7 P- h5 g, W# A: N8 [+ F
5 b+ ], U% e3 }: `3 c
; e8 A( n1 [/ t( _3 q+ U8 f% P8.17
" Q; W. w8 J. u+ v( M#include
7 B: F0 S2 v: U1 ^' N7 yvoid counvert(n)
/ w: y. _% m2 t+ G" `" q6 W5 Aint n;
' M4 h) ^+ x3 i+ W. N% w& G! Y{ int i;
0 H, J m2 a, F, t if((i=n/10)!=0)
4 c6 F7 e* Z: Z1 P$ N3 o( v- C1 H' C convert(i);* M" Z# f6 b; D% j+ y: j. _/ b
putchar(n%10+'0');
: U* }9 F7 [( H6 o# \}7 E& \8 ^/ J4 r% X t$ p
main()
8 z- m" r9 d2 y+ @{ int number;0 S K& Q" @" x4 @* C
printf("\n 输入整数:");+ \9 _( Z& c9 F- I, P; ?
scanf("%d",&number);
& F/ ?0 t% a7 F" ? printf("\n 输出是: ");# Z$ H1 N, h7 @! \& D$ X/ z
if(number<0)
g n7 f; o% p1 r s z ^ { putchar('-');8 |3 [# i5 W6 D0 H
number=-number;( _2 g; ?7 w- G# |! H. W
}
2 s' q* M! s" g: C' R! qconvert(number);; y0 d0 Z4 ]& p* i3 c/ f$ t
}
0 }1 D3 V$ s" ^5 V% V: |# O5 ^( H
% p; a H: l# U
3 q( _7 @: Q0 Z4 u! W1 u3 b+ M$ e6 u) o: Y) y( ~& f" o
8.18$ v1 X& H) C" ]! Q6 b, |% `8 V
main()! r' z) v) m6 T& N
{$ |5 j, C$ u S" e
int year,month,day;* q7 X8 M9 n$ d9 }& E9 ]& W
int days;
% b- q# d% r4 y$ p% G$ I printf("\n 请输入日期(年,月,日)\n");
. z9 C$ K7 ]" ` scanf("%d,%d,%d",&year,&month,&day);2 z: ^/ L3 U. j. q0 n
printf("\n %d年%d月%d日",year,month,day);
3 |3 }! ?( D% ^; I+ t2 k, Z2 f+ h days=sum_day(month,day);% F6 M% g( s& Q, z! L6 @/ k% s6 j
if(leap(year)&&month>=3)
T) `; ]# D- d9 A) y$ n7 b0 T days=days+1;$ [; z. C0 P& W _9 `
printf("是该年的%d天.\n",days);8 z3 ?( t1 o/ M5 }$ o5 F8 X
}
w9 z2 Q+ K! H! X/ S- j' ~ static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}6 u0 E' J2 @) z E* C
int(sum_day(month,day)
" J5 l2 W% W0 p int month,day;/ {: I* r4 Y/ k* P# ?
{
* C( B$ s+ M C' I! c" @ int i;; U7 u6 J& R+ T5 z4 X
for(i=1;i day+=day_tab[i];* [% ^6 n# W% M$ ~( z) T
return(day);
/ p3 [1 R( g+ v9 q4 z* w, K; x }
( E9 x9 a3 R r) G0 h3 D int leap(year)
7 S* ]8 r& I9 w2 s+ U6 s$ @* B/ D int year;* A) k$ u$ D2 ^2 e6 H
{+ ~' H( \& U2 v& U% N% k; i# S
int leap;
( P. P/ S0 H4 _ W leap=year%4==0&&year%100!=0||year%400==0;
# L ? R. w9 ~6 l7 ` return(leap);; k1 u0 J- {4 Z, n2 L" A7 o, M
}4 Y, l) q1 A# o, _' e
第九章 编译预处理
/ Y. s. N& m- {9.1; l7 Z; V' B1 f+ p
#define SWAP(a,b) t=b;b=a;a=t: A+ E; h. z9 L: T3 e
main()% H2 T2 j( i* r
{
7 v& O E# Z/ s" Mint a,b,t;
3 W4 |, ]9 ]4 ]$ o5 _2 y: M+ n; Gprintf("请输入两个整数 a,b:");
, R* P' Z2 A+ s ~scanf("%d,%d",&a,&b);( b( |" Q0 s$ p* k; y
SWAP(a,b);
- I2 H5 `: J3 F$ aprintf("交换结果为:a=%d,b=%d\n",a,b);1 A% o) N4 Z5 t( p
}
2 a' X/ ~6 r# b1 Y; K$ P 6 P) b/ g- ~6 q9 B' O# s2 K- |% u& {
! Q, Q3 b4 f n. h% s9.28 j8 _7 O; J) S! v
#define SURPLUS(a,b) ((a)%(b))+ D2 l5 G' ^6 _3 B6 t
main()& F( `2 A4 |3 k! V% E
{
& L Q+ E7 l' a$ \. P8 h6 _ int a,b;
- o: P6 @+ b9 f( Y4 g) A# G) ^ printf(" 请输入两个整数 a,b:");
4 d! l* g, b3 @2 [0 z scanf("%d,%d",&a,&b);
0 {0 O- D3 n; H8 kprintf("a,b相除的余数为:%d\n",SURPLUS(a,b));
' p0 K8 x" I; j$ ]2 [7 x, ^ }
+ I C1 n+ F, s+ e. y$ w2 M1 F$ |4 E8 c" r$ C- }9 \
/ P6 i. r3 S7 Z& F2 n9.3$ m6 g5 M! z& k8 m- l9 g
#include. r4 K" j: C, q
#defin S(a,b,c) ((a+b+c)/2)
' [ \" s ?8 O' F#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-% S+ B5 j! _& M) S
c)))
, y# a9 K i0 c- m& J9 x! d5 Wmain()
$ x, O! r, o( i) o$ t2 I/ ? {& h# e7 T! ~% q5 H; F
float a,b,c;
1 H" m9 E/ Y7 G! t7 y/ m3 _ printf("请输入三角形的三条边:");4 H h! \ g) [4 }
scanf("%f,%f,%f",&a,&b,&c);
1 i, L$ b+ C \- ?1 [ if(a+b>c && a+c>b && b+c>a)
* h. A/ @9 f, o% l# F0 x5 q b printf("其面积为:%8.2f.\n",AREA(a,b,c));
3 W$ ?' b& j' ` p else
9 K8 G4 a5 G$ o( y Y printf("不能构成三角形!");
; ^* z f1 n# ~% d" Y7 v }
2 x) _$ ]8 q2 L5 z6 F
. f4 ?" o7 X. D/ K: m4 r1 c- H4 P/ ~/ B
( y- v$ \+ j+ R4 y8 U+ p; u5 U7 l; e2 J
9.48 c+ `; h! z& _- ~
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)
r' @( y: j5 y1 K9 Omain()
3 r" F% T( Y- G" u8 p {
& ^+ f& Y4 B/ s int year;
( B1 o: v4 v2 S+ J; Y( Z+ |# t9 W+ V printf("\n请输入某一年:");
4 C+ q! B, F% u. T6 Q3 d- } scanf("%d",&year);
# @9 n) `( D/ d2 |/ x" b. S+ x' b; k if(LEAP_YEAR(year))
. G* f7 { b2 O2 x printf("%d 是闰年.\n",year);& M0 C- V- O3 ^0 I) H$ ^: I
else
& \$ ]* a& Z- B printf("%d 不是闰年.\n",year);
. v. r0 V5 ^9 v! s }
( N- B6 h7 C, e* a3 f/ C s0 L; W, A) I
: Z5 r+ _( S8 t) v6 s
, i- ~' @% {( k; ?. G9.5解:展开后:( f: l1 D1 n+ ^4 L( J5 K" ?2 \
printf("&#118alue=%format\t",x);5 | m9 O; A" W; N% m/ T* U/ s
printf("&#118alue=%format\t",x);putchar('\n');
: b1 r6 ^6 H) \: {printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');
( g! \+ E/ |0 }* B/ d/ n$ B输出结果:
) i' a$ c* h% C& g$ J&#118alue=5.000000ormat &#118alue=5.000000ormat- p: o3 [7 x: x$ V& f# X
&#118alue=3.000000ormat &#118alue=8.000000ormat' [% Z( e' d* _3 p& Z3 I3 v% ?3 O) v
; F8 U" P* n& T! y; y! z9 v) E5 m
6 s& V! M. ~* _5 Q1 h9.8
/ ^& q3 I9 b! Bmain()5 Q8 f: w7 e" d9 y3 K: _) d
{
4 [1 ?8 p: T( u' L# f& \. L int a,b,c;* s [! ] p9 V) j9 S
printf("请输入三个整数:");9 y; f. u9 p! N3 E5 K' N
scanf("%d,%d,%d",&a,&b,&c);
3 W$ I" A" O( [ printf("三个之中最大值为:%d\n",max(a,b,c));
, s% l9 q9 x" q& _5 S' p }' {. o: W7 @" n/ x: e: [
max(x,y,z)5 M; P# q' D V9 s- c
int x,y,z;0 X6 X" z: [8 T% ~0 w) K7 {4 q
{
2 x7 J( k4 k, w% _( A int t;( h3 X8 G z1 b9 J
t=(x>y? x:y);
3 q4 {8 M5 @: ^* V return(t>z? t:z);4 \* g' U0 ^9 J+ g S
}
% W' c+ T0 E! r& C/ f: E+ t5 }& T" R6 u. w! q/ U
' v$ }# r, i% x! ~
1 S% {$ `' `; Z: f8 t. ?9.106 c* p, }; {% y5 z; o+ |2 L
#include
* F' o& w# t3 F% `8 D l( d7 i4 }#define MAX 80
# Z! |6 E% w& ~ Y |#define CHANGE 16 N$ P6 { c& _/ A8 O9 k* ^% Q
main()
/ V& d z. w( ^4 ^{
+ p% \1 k+ c' g7 ^7 R1 f% B: P3 W char str[MAX];
& P6 r# C3 O) K9 K/ t& k1 H int i;
6 M+ w& C3 I0 k1 \3 u- M printf("请输入文本行:\n");
0 ^( O( l% M3 Q* ~; Z$ L$ |7 { scanf("%s",str);
9 |7 K+ T1 @% \+ g2 @* j; }6 A #if(CHANGE)
2 K% K4 Z% Z( T$ R- |) U9 g {
2 J6 N4 J0 F o9 x0 L: ? for (i=0;i {
9 p/ c9 v' i+ d2 g) B if(str[i]!='\0'
+ l f5 |6 D! l- K2 h& |7 n if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
7 e" u) S% ~9 G$ W+ `7 l' B: @ str[i]+=1;' ^2 ~+ N0 j% x6 F6 h
else if(str[i]=='z' || str[i]=='Z')
2 R% ?2 q; V1 E/ }. q8 {$ c* P) ^ str[i]-=25;" I( R7 l. A8 x# J5 T3 ?
}3 ], X' |" a' j& y8 O
}4 {: [* L/ z$ @# `: B, T; e7 u
#endif8 \) h2 P# X) q! K
printf("输出电码为:\n%s",str);
: \% e9 _: z9 f z% H; J}
0 {7 r) P) }2 y/ u, d; K& b第十章 指针
+ d4 d9 i% g6 L+ q& M10.1. G _/ Z& N$ E
main()7 Y% c. e1 U2 m4 n& z
{int n1,n2,n3;
1 B7 H9 x0 g% h" }. U/ I int *p1,*p2,*p3;
4 V2 ~- i( J ~: V; n- z+ V8 e N scanf("%d,%d,%d",&n1,&n2,&n3);8 t: N7 f1 @2 r6 d
p1=&n1;4 J8 F8 g8 M( o4 {$ E1 C$ ~, m% f
p2=&n2;
& I& P0 g6 T3 j( y& F8 c p3=&n3;
0 q% @7 J: H. _3 v% n if(n1>n2)swap(p1,p2);# _0 B7 | H$ \6 O: k0 S6 n: x
if(n1>n3)swap(p1,p3);) X4 i! c6 E& x0 E) X
if(n2>n3)swap(p2,p3);
! C2 M G/ f7 D3 a; J+ @ printf("%d,%d,%d\n",n1,n2,n3);; S9 d, c; N( e
}) Q2 A J; |6 A1 b% F* f
swap(p1,p2)9 f: }) w t+ K
int *p1,*p2;/ W4 _0 K" f7 ] s: _8 d$ P
{int p;- o- U. T7 d7 g& Q
p=*p1;*p1=*p2;*p2=p;7 ?3 w! a, v U2 Y9 l6 \# m
}
1 N& \/ B) w$ {; D. C10.2! V+ @) Y9 p! C' Q" }3 s
main()/ F9 T7 M5 ~6 L3 B: ^+ b o) ^
{char *str1[20],*str2[20],*str3[20];# s( w% I- M1 g
char swap();' ?* J& Z; ~4 {8 o1 Z Z+ i
scanf("%s",str1);+ X4 E ?4 R6 Y7 G' ~7 e# a
scanf("%s",str2);$ J2 D* o! x8 s! l& A2 ?
scanf("%s",str3);
, T D3 A2 W. U( t) V+ o Z$ Z if(strcmp(str1,str2)>0)swap(str1,str2);
5 C- u; m a* Y: Y6 I* x# | if(strcmp(str1,str3)>0)swap(str1,str3);
) \/ o- a+ G5 C* ` if(strcmp(str2,str3)>0)swap(str2,str3);8 ?. f9 W& S4 ?5 o9 e
printf("%s\n%s\n%s\n",str1,str2,str3);
6 |6 {9 M! s' d" H/ V/ R8 c}
- Z6 o0 z) `+ E3 |( G9 c$ Mchar swap(p1,p2)
$ D3 d' P& G- |4 G8 T! lchar *p1,*p2;
9 a- K: I7 z8 F# r3 ~6 L{char *p[20];( x1 M7 ?: I* q/ @8 f
strcpy(p,p1);9 e+ o, n# i7 O7 s
strcpy(p1,p2);
0 t! A! X1 i6 R3 Y4 q- q strcpy(p2,p);
6 x0 M5 W# C, |! H}
1 k! ^0 {/ J7 r3 H+ [5 w10.3$ f) }3 _* {0 c6 m0 T' t8 }
main()/ _, G$ F* z8 X c- z
{int number[10];- o0 h7 Q8 Q3 N/ x2 f
input(number);/ d2 b6 g, Z/ T8 L+ w/ X
max_min_&#118alue(number);' s2 x2 k4 i. S! P/ f
output(number);& T2 ^; n- a, p# R* [7 I% T8 [1 t
}$ \% f& c, S7 U- a
input(number)
: w; y* F/ e5 A' Y4 jint number[10];8 J% v( L& D) Z4 o2 E
{int i;
$ ]* ~; E G+ L# |% t: K2 A6 b3 d3 D for(i=0;i<10;i++)
% }0 [- c t4 J' L* M) V scanf("%d",&number[i]);! v$ w" T, e: c i: v, q, o7 l* c
}
+ ~) _, u9 ^2 k3 [; gmax_min_&#118alue(number)+ D8 H. ^+ ]) `1 G1 d6 t
int number[10];
9 n. G( o4 w7 H3 }2 Q{int *max,*min;
; ^, p, i9 H) C. T! f: i int *p,*end;6 G: g6 U7 Y5 P9 E6 e# L w) [/ _
end=number+10;1 i& D# G. M9 Q5 L K2 {1 n
max=min=number;1 p/ e4 U4 G8 C! P3 X# v N: c
for(p=number+1;p if(*p>*max)max=p;' Q2 g2 x0 W( i( f9 X1 O
else if(*p<*min)min=p;
$ D5 y# q' V) A1 c4 G' [ *p=number[0];
4 |* _+ U) q$ B/ x7 A9 c! n, V number[0]=*min;
" O( B X# `) Z% L. e8 s *min=*p;, {& p. F( O: I3 Q
*p=number[9];' R8 s; S7 c0 u1 n1 o
number[9]=*max;
0 e/ h: a( F; F% [/ y! \% Q *max=*p;
: j0 G/ N6 P# a: i return;
1 b8 H1 N0 z( R: ~7 X+ f, ]}; _) {. r& z; Q; \- ^
output(number)8 w7 c1 R2 ~% M4 J }
int number[10];
- G) P0 A( J. v: ~ I0 x{int *p;4 \6 N% C! k- f
for(p=number;p printf("%d,",*p);/ L3 Z# z7 d! P; k4 a( o3 R" L
printf("%d\n",*p);/ x" c- }3 }9 e- d/ w& |
}
! M! S1 B4 Q c" v( R. S. B10.4; D6 @: A; g% T0 H6 w6 _
main()9 r5 j* A( `9 N2 o \( V
{int number[20],n,m,i;9 W K( J+ y( i0 U
scanf("%d",&n);
; Z; J: n D4 U. G2 |( G1 |8 r scanf("%d",&m); O! r" W+ g& } V; h8 k
for(i=0;i scanf("%d",&number[i]);4 U7 s3 A7 S( z4 d
move(number,n,m);
8 F, f5 A( d8 O' _$ B for(i=0;i printf("%8d",number[i]);
- v: H5 f2 @1 P4 Z}/ ^+ z$ `7 e) `" K
move(array,n,m)8 l: y1 L9 {! U) L- Z* m
int array[20],n,m;
; E0 ?; E$ o7 R* X{int *p,end;9 {$ _' d) E, L- e, M" U
end=*(array+n-1);
! E" B2 B4 V+ ^0 D. N; m for(p=array+n-1;p>array;p--)
% n- F2 Z: S, p' f5 ? E *p=*(p-1);: d. Y3 p' F% V/ W' p! c. O8 t! [
*array=end;0 o" c6 c. e! |- a) _9 }* K# ~
m--;+ Z) V1 y! W) o6 W
if(m>0)move(array,n,m);
# c% c' ]5 d n3 `- u! [}- s) | ?$ L% S7 e3 t0 z+ Y3 L \* L
10.59 [5 e3 A5 b* y6 v* G
#define nmax 50
4 w6 C. n" S$ i) w, G( X) a3 Pmain()
0 X) P8 g' { ?! w# z{int i,k,m,n,num[nmax],*p;
: W# b0 u/ { U2 r( j4 h scanf("%d",&n);8 v/ _# Q Z9 f& Q3 E6 P
p=num;
) Q9 ?$ q7 ^, R& R for(i=0;i *(p+i)=i+1;
$ _7 V% k: I4 L4 T# ?; _9 U8 M& s i=k=m=0;' M+ c. B5 d7 V- d+ f% V2 s3 e
while(m {if(*(p+i)!=0)k++;! D! S. [0 F- T6 {. f3 G& h
if(k==3)# R B( x* K2 [& w5 r
{*(p+i)=0;. @# k0 Z! F$ r; I
k=0;
" C* Y$ D; A3 d& m8 ]8 N9 E m++;0 z# i; K/ O5 u" y+ g- E6 p9 i
}* d' |# ~( d% T% w& W* |; Q
i++;/ I1 L$ W+ g2 P; `$ J O* }) L; r
if(i==n)i=0;4 V0 p0 q7 f% l, c& s
}8 }9 V3 R6 p/ C7 G+ A1 k
while(*p==0)p++;
! K; i; c& I! |% e4 W. M: d7 E printf("%d",*p);
) q) e3 l7 U" V$ D6 x }" n}* v1 u, ]" l; G2 r, |) P- `
10.6
( @5 J6 x3 g; [3 G. @7 N5 I! imain()8 i2 S0 y& y! y4 ^- n% V
{int len;
1 Y- u& `& n8 ] char *str[20];; z/ E5 w% r. O' c3 o V
scanf("%s",str);% Q8 u7 y4 L( \- k. u
len=length(str);
( @0 i% x+ Y5 {& {4 t printf("\nlen=%d\n",len);
$ c$ A1 Z, M4 x+ T0 B}
3 t( n' ]4 V" {' ]$ s6 dlength(p)
2 O+ f% A; q, h5 c& g( \. vchar *p;
% a+ m1 h: L- X8 A; T9 C' B, m+ O{int n=0;
7 x1 _- |% v; R, t9 [ while(*p!='\0')8 d5 v6 T8 e/ s
{n++;p++;}8 V' P+ k/ B; I
return(n);
8 L7 ]! d: X& y$ n$ q$ Y, [}
+ F& G) ^: ?# k+ j* S10.7
5 z$ O$ R9 d, k0 amain()7 g3 o @9 W) J" Z
{int m;
; C. R8 I! m; e7 F7 S; H# i char *str1[20],*str2[20];' K! e; f: d4 p0 t q. C$ |, l- a
scanf("%s",str1);
f5 [8 s Y% _ scanf("%d",&m);& L' J- i4 e' {2 ]1 b
if(strlen(str1) printf("error");! @# I- F* l3 Q g, e$ ?( z
else6 ~& L7 W( h6 F- m5 ]! h
{copystr(str1,str2,m);7 U6 |) \1 `% ?
printf("%s",str2);
0 n o" S4 N) b+ }# J }
& S: x9 @* |0 e' v6 i" U9 r}% i1 U$ r* p( ~, ]4 r+ V8 e
copystr(p1,p2,m)
8 Z7 l' ~! K, \9 e, ^5 achar *p1,*p2;& q$ {8 Z! `- l; l. e
int m;/ R0 e# j+ F4 L# {+ O
{int n=0;# L/ O& |% c8 i3 ]+ Z7 T" r
while(n {n++;p1++;}
( X) P5 V1 p- H9 Y, R while(*p1!='\0')
: p r* H" O# f0 X {*p2=*p1;
! K9 u# a- ~7 l9 E" F# V p1++;
4 Q# f/ g. p+ d p2++;5 o4 j) G3 `" w5 N+ }( F
}
* F2 ]& e! e4 U& u* B4 A5 b *p2='\0';
2 H! k" H8 ?! v}/ K9 F% c# t. n/ A
10.8# e3 M) ^( s" j% W
#include"stdio.h"& q+ M& p M8 ~1 k
main()
7 K. k, L' M6 s% B1 P{int cle=0,sle=0,di=0,wsp=0,ot=0,i;- C& P. O8 K5 f+ P* g1 C2 _
char *p,s[20];2 p7 ^+ p) \ J' X
for(i=0;i<20;i++)s[i]=0;' d' {" V7 p+ M' N
i=0;' K# J9 ]' ~2 V$ I
while((s[i]=getchar())!='\n')i++;
5 F* m+ V0 S: Y4 E- H9 M' k p=s;+ F' t P/ X' ^- t2 q9 |
while(*p!='\n')9 @ H5 `' ^5 n
{if(*p>='a'&&*p<='z')
! J- M% u, `% y; O ++sle;6 O2 t1 e5 o# l: _# O
else if(*p>='A'&&*p<='Z')
$ K6 I4 r3 G$ F, v% v ++cle;
/ G7 v/ }$ v) k0 }; T' q/ s else if(*p==' ')4 T) I* \$ z- U$ \$ _8 K Z
++wsp;( M+ x+ o5 n, _9 n5 ~2 Y8 Q
else if(*p>='0'&&*p<='9')8 e3 w4 L# ~; f* T2 n% M
++di;. e. ~% Y0 \' e, _
else$ o, w r& p) _; q X$ Q
++ot;
- Z* ]' V7 R0 V2 c! H p++;4 p$ K1 A1 E Q+ }, a
}) ^1 a+ h/ z. N w: r
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);! n# ^' m# V3 o1 q S5 u6 v
}
/ k N/ I% w1 L, J- w2 c10.9: P y* ^7 Q5 h/ ?! A. U7 G
main()9 {" ~+ i, W; X- s6 u8 ?$ @2 n
{int a[3][3],*p,i;2 y& g; r* \: p
for(i=0;i<3;i++)
7 O# c; |; A n; J! n% D; S2 b scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);, R$ u& `9 }3 N0 P
p=a;, G' g p- j. `$ _- c- O
move(p);
5 m: F) ^, u1 t( s& D for(i=0;i<3;i++)
" b6 B& ?, Y: b* S printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);1 O& X- V! {. p) C$ j. C
}0 n; y8 r* j1 B' S0 C
move(pointer): q7 C& P! Q/ S1 k
int *pointer;. R9 _/ Y9 L8 Y4 X$ n
{int i,j,t;
) x- Z1 A, T. H# e' K0 p for(i=0;i<2;i++)9 S% C; z' x5 m
for(j=i+1;j<3;j++)8 W. S9 u' u9 I$ f3 S9 M& O+ a! q
{t=*(pointer+3*i+j);! O0 _* w8 l) c" a
*(pointer+3*i+j)=*(pointer+3*j+i);# B& b! z3 d+ { C! U- z/ \
*(pointer+3*j+i)=t;
! }+ y" z2 a9 f* [1 A }) V, _! Z6 O m- B. ]( m* ?$ v
}/ A2 D! R; h4 C' \* y$ K
10.10
; B- i% {9 ]4 ?8 A7 }0 X5 Emain()0 Z, v" q' a: u! \1 s& C
{int a[5][5],*p,i,j;" E* c; a/ G9 ^1 W m$ }
for(i=0;i<5;i++)
& S5 _/ Q& Z7 N& F/ O% B- I for(j=0;j<5;j++)
# z0 P3 E( H+ T4 ]0 e scanf("%d",&a[i][j]);
) m0 @ T7 B: t J p=a;3 G0 e$ _$ J1 R0 \4 v. M
change(p);6 A4 m }/ ?; ^+ i; X
for(i=0;i<5;i++)/ [9 y( V5 I" x6 T# K# N7 B' ]
{printf("\n");1 ]$ C0 t6 A7 N* x9 n
for(j=0;j<5;j++)
3 g: A" [1 F0 T+ h8 J; z printf("%8d",a[i][j]);
7 n% |2 P+ R" K; p8 z6 Y, Y }
! [3 V; P9 D) j' L' s& r}2 }' _$ n! a& R$ y
change(p)) w* |% }( Z* r# `+ L
int *p;+ h3 h' E1 [! t4 [
{int i,j,change;
1 p& `" f9 E& J: e/ H4 H5 C: W int *pmax,*pmin;2 s8 s8 E# B: W1 ?) Q( @" f/ Y
pmax=p;
" M5 A2 \; i# f( l" l pmin=p;( c; R2 E5 a6 S6 N& g$ h0 p8 L
for(i=0;i<5;i++)8 e( R/ W y& ~- @! z* H
for(j=0;j<5;j++)
% `; s/ E# i% b- v$ U {if(*pmax<*(p+5*i+j))pmax=p+5*i+j;
) M/ G. W6 e! ~1 o+ U7 H if(*pmin>*(p+5*i+j))pmin=p+5*i+j;
5 x! k; m' {& F/ T! ]) ?2 [1 Q }* T# @: w: h" i& }
change=*(p+12);
- B' D3 M7 V% p" y2 N& k2 [ *(p+12)=*pmax;
. L e" j Y7 O *pmax=change;
% r5 {0 W4 o/ } Z! E change=*p;. M% h: k+ m% ~
*p=*pmin;
, r) C$ ]0 t( |/ K *pmin=change;9 @- N) l/ C* Y; {) F
pmin=p+1;
7 K5 X9 l3 l9 M: L4 B; B* t for(i=0;i<5;i++): G+ o5 A6 l4 w. ?* [2 l( |+ Q6 i
for(j=0;j<5;j++)' _* @, H9 z) S3 K* I# t
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
: ?! t! C h; H+ d change=*(p+4);7 c; ~+ {2 l/ b3 _% {* M: K
*(p+4)=*pmin;! \7 x7 ~5 u1 j
*pmin=change;
! l. C" n+ x0 d+ u. _: } pmin=p+1;
& a$ q, l B: }" G( t for(i=0;i<5;i++)) z( h$ j w& ^" K4 {1 H5 l' }9 v
for(j=0;j<5;j++)) d- f$ m; c) U' Y
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j))) w3 O# [; s: }8 c
pmin=p+5*i+j;( c* Z& B, G. ?6 u* q
change=*(p+20);7 L3 ~% o5 e. v: T7 \
*(p+20)=*pmin;
) A' k: f- A$ z+ y *pmin=change;
{8 W! a& M. h. M; r pmin=p+1;
7 b6 D5 L* [0 G+ U for(i=0;i<5;i++)
! Q$ B( e) P( n( R2 t for(j=0;j<5;j++)
8 [* F- i2 E; \# A- E if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
6 k) a/ O0 O) {* Q4 ^ &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;& p% k7 E" Q6 \# c' [& a
change=*(p+24);
) ~9 {) x2 a, W: H5 [% } *(p+24)=*pmin;/ v' l& m7 S$ l" z1 K& G9 S
*pmin=change;
& K2 a! }% [; y1 A6 v2 {; [( ~' N}
7 h1 Z7 f8 J: D) E; c$ A10.11 b7 j: t4 r9 q. J5 d
main()
/ \$ r, o) t! u! d0 G; o- C3 h% Q{int i;
0 ^# E/ U* W8 I8 n char *p,str[10][10];
- g* M' E X" A1 P/ R/ p for(i=0;i<10;i++)
( n9 n/ ]6 o* x6 F7 ^9 B scanf("%s",str[i]);
+ L! M3 Q+ Y: \3 w5 w& c5 g6 [0 A p=str;& s+ {! O6 k( T# J* G5 Z
sort(p);% f, \* c( Y6 k* {8 e. d+ G
for(i=0;i<10;i++)) n v8 ^2 j: Y: w* i8 D
printf("%s\n",str[i]);, Q+ C e A, z2 @: J
}
8 y: a/ V. A2 d" Tsort(p)
; R3 R. ^* a+ Echar *p;; x/ `$ a( p/ v3 I$ _: R8 ]
{int i,j;6 X9 @8 w H# {$ D$ L
char s[10],*smax,*smin;
! W I0 h C6 E: D u5 V( Y4 x) ] for(i=0;i<10;i++)
, C0 p2 E+ N- V {smax=p+10*i;) T; t% v6 X' I* V; U+ T& \
for(j=i+1;j<10;j++)
: c" D4 _ ?2 z8 {. R! w% p {smin=p+10*j;0 }+ V X2 c$ p& c" h: a! _! \
if(strcmp(smax,smin)>0)
) E" f0 [* B( m) c {strcpy(s,smin);$ m7 d5 l9 F% {4 n
strcpy(smin,smax);
2 _! ?# e7 |1 Z6 V! k: e1 Q% {1 K strcpy(smax,s);/ ~9 `3 C' B6 |2 ^- N* a
}
7 \( a2 H/ q. i0 E- K" W+ T/ W }* S# O L6 L# H" a- d5 A7 W5 |8 ]' j. n
}
( Q! L4 r( o3 Q9 e0 u# j# n2 ?}: I, Z0 U: j3 P. F ^5 g n" v, h' s
10.121 T2 ?3 p4 ]# t2 b& Z x1 n
#define MAX 20
- [6 m3 {6 o3 f6 O5 L, xmain(): O/ k5 C4 J: F9 Z; D+ J. B' R+ S
{int i;/ Z* ?. G* d1 F" Z
char *pstr[10],str[10][MAX];
1 f w; q; I8 }( }' h, E, F$ }$ ? for(i=0;i<10;i++)
4 j, m* v& F/ `, ?3 K3 S3 n pstr[i]=str[i];5 x3 P9 t; m$ a; [
for(i=0;i<10;i++)
- @9 u0 _, U( D2 i9 y scanf("%s",pstr[i]);
0 U; {8 W) F$ G8 J/ Z- F+ z. h1 H sort(pstr);7 F" A% y$ p x4 v: Q) H- |! d
for(i=0;i<10;i++)
7 U% j5 }) X5 Q3 J/ n' R8 p% P3 i printf("%s\n",pstr[i]);% e) c& u8 o3 i$ u- {% p0 Q( p/ \
}
. V9 y2 |; D. @4 ~: Csort(pstr)
' v. Q3 H( {/ L4 J! bchar *pstr[10];4 E* q$ i) L, W! N# d* Z
{int i,j;
& K/ V4 f7 V* t) c7 R% i+ h char *p;
* ?$ L; r U6 N8 ?9 C, e. p: \ for(i=0;i<10;i++)- b" d( h+ y2 P0 n
{for(j=i+1;j<10;j++)
$ C4 R( A, J* R- J3 C; N/ b: h4 N {if(strcmp(*(pstr+i),*(pstr+j))>0)" q6 y( c. ]) B8 m* e% l
{p=*(pstr+i); v+ Y- P& _# f) B
*(pstr+i)=*(pstr+j);* O) }% o: n& J7 r* b* x5 [. [. n
*(pstr+j)=p;1 B: O, a; }6 I; O; N8 p% V* Q7 R
}
. d4 d2 p8 N% H6 Q. [ }: k6 v2 G& T }
% U# \& _, K6 S( ]# O0 k- I. u }
" u* }1 L6 D1 b}
1 }. Y- R7 |( m z) ^5 K10.13
9 C( W6 J! n2 n- s( H#include"math.h"
' j: \1 v/ u* y* O2 Jmain()2 F* {' @" ~( m- d# ~ e, H+ H7 \
{int n=20;4 X- `9 u9 G) A6 O
float a,b,a1,b1,a2,b2,c,(*p)(),jiff();2 k) Y( Q9 L( b; T7 K( u4 ?) j
scanf("%f,%f",&a,&b);: Y! G+ J# u2 o% ^; U S
scanf("%f,%f",&a1,&b1);
. q6 T+ i" M s# v& a3 r( B3 u6 G% v scanf("%f,%f",&a2,&b2);
" c5 j7 p' Z" j, H1 U p=sin;6 t- D, C7 _9 Q% M* _9 x- B8 B
c=jiff(a,b,n,p);3 g' c4 c+ I; [, |1 r
printf("sin=%f\n",c);
% o1 g# @% x( _6 r/ R p=cos;; R; G: Z8 L- e8 J
c=jiff(a1,b1,n,p);+ ~% ~8 @1 U( t( N
printf("cos=%f\n",c);
, `. U0 `: e, F/ ] p=exp;
5 M. o$ }( O% V. [9 @ c=jiff(a2,b2,n,p);
8 J% G9 a0 w' Y$ j- l- ? printf("exp=%f\n",c);
/ W! ^9 @. h- j. A/ N! S% J}. t: ~- p0 T8 a+ O* e) p8 d. p
float jiff(a,b,n,p)0 ^% A. O# k$ {' P1 h5 @( M3 u
float a,b,(*p)();
, @) A: p: {. B* Dint n;
3 f3 _4 z3 C- [{int i;7 g6 B) K; r( w: D0 h" `8 ^1 L( O
float x,f,h,area;% w. ]# a7 A0 V
h=(b-a)/n;
# |; @9 S8 o6 I x=a;8 s; u: C+ B Z' Q
area=0;
( \: ^. ^, P, f for(i=1;i<=n;i++)4 n+ r/ m- @( E H! `- p9 E
{x=x+h;% O% B7 r Q' |; U
area=area+(*p)(x)*h;) d5 ~% `8 g: u4 h7 x: t% i9 {
}
5 b+ c7 ~, }8 i! U. h7 `& P5 t return(area);0 P; f+ A; R2 a8 H2 s& {& U
}$ z$ N. Z7 j3 A+ o
10.14* H5 c5 E. J9 `+ m
main()8 g6 v. W+ u& }
{int i,n,num[20];
4 d( ^ m# g* i7 a# U5 ?- \ char *p;
/ Y" q }; ]9 t D$ f# D3 k5 h scanf("%d",&n);/ U/ \ |) h% j4 f) Z$ ~( R# @
for(i=0;i scanf("%d",&num[i]);$ J( ~3 W3 h& | L g, T5 ~
p=num;
% g! {+ u0 @% g sort(p,n);
# a. O5 q4 Q+ D1 P; }7 V for(i=0;i printf("%8d",num[i]);
9 C- G, t6 s$ p7 j7 G}
3 K0 w! P* K9 _, x$ w. Q* qsort(p,m)
# B' U2 j. s2 }3 L' A- d1 |char *p;% N- X" ]% L' z9 s6 c6 o; |5 Z$ \
int m;
a/ v2 x& S3 n& d; M. ~{int i;
' g, P5 K3 z% x9 ^9 n6 } char change,*p1,*p2;
! l( Q( t- D4 }0 R- U7 W, O" x- ] for(i=0;i {p1=p+i;& x3 O$ Z+ u; Z+ q- h- q* y
p2=p+(m-1-i);, v7 \) x) R4 C; O
change=*p1;7 u7 E) h; e2 {
*p1=*p2;
& s4 V) t7 J* j7 y' M *p2=change;
% r6 k4 B ~, B+ r: e1 I/ P& Q) {- ?% |, K }4 m" E" W: `" |3 k4 |( k
}* B' t ?. u, T; i5 E: f
10.158 J6 G) m; ^! C# K5 ]
main()
8 T1 t3 T! B" l8 W1 L T" d* ?{int i,j,*pnum,num[4];0 k( ^$ E( u8 ?3 y9 \" {# g* ]
float score[4][5],aver[4],*psco,*pave;9 ~* M, o" d5 _
char course[5][10],*pcou;
- x4 K! X6 G4 }0 a% h D" _+ l/ Z pcou=course[0];; Y# b. U+ \1 n1 R: e6 `
for(i=0;i<5;i++)
" d) @5 s0 d2 w7 P! f scanf("%s",pcou+10*i);6 Y$ f+ E8 W4 `
printf("number");, X8 P5 Y- J& ^$ i+ w8 F! ^6 \7 J
for(i=0;i<5;i++)
6 G6 y4 w% i' h$ d2 M+ |2 E, H printf(",%s",pcou+10*i);; a/ [2 u+ p! s0 K2 t
printf("\n");: v6 f( } T, {$ i! e. ]
psco=score;- }! N& c5 p. n0 ]7 `7 I: S
pnum=num;- q7 O4 I% H9 `5 \" U
for(i=0;i<4;i++)
. C/ {5 e& T4 f# E( v {scanf("%d",pnum+i);. x/ Z1 P$ S, ?3 |/ [( o$ B
for(j=0;j<5;j++) y# n3 B9 N8 l- O
scanf(",%f",psco+5*i+j);" `0 v5 W T7 d; e
}
0 [+ @/ F) q- X0 B& n' ] pave=aver;( I% t( @. b9 w& p1 k
printf("\n");- X3 d( i% |7 @" Z @
avsco(psco,pave);
" o9 L# y8 L' r O1 s2 m7 f avcour1(pcou,psco);
( X1 i" T g) w- @ printf("\n");
C3 b$ S% g; r- J; ?. E+ f fali2(pcou,pnum,psco,pave);( P5 M7 A, I% N# O4 s( T
printf("\n");; h0 ^( p' A& I; r2 B
good(pcou,pnum,psco,pave);# D0 X: P: B6 v
}
' Y% n5 M1 P0 H* P$ R8 E1 k, Pavsco(psco,pave)
2 }! {: ?4 s2 x" Rfloat *psco,*pave;
0 a, x- q- E( Y$ j# _{int i,j;1 I: j) W7 ?0 n3 [0 f
float sum,average;
* w; W+ n* @. P) I- n' N for(i=0;i<4;i++); Q B# q+ U! H
{sum=0;/ {2 g, o2 }. F) L1 y
for(j=0;j<5;j+)% H- t4 D& Q# K9 f: W) b
sum+=(*(psco+5*i+j));
, ^' ?1 d% q1 u4 B% w average=sum/5;
7 v! G% _% ?7 p& j *(pave+i)=average;; ]" X+ [( @6 I! x" F
}
" z: S8 V0 e* _9 S+ D+ V4 s% |}
# s& m$ A- o) i2 I0 X' davcour1(pcou,psco)
/ e& _* l+ j. O) kchar *pcou;7 f$ k4 G: P# G- H; a) s. i
float *psco;
' X. I) v& I6 }" D" h( Y{int i;
% l3 x; U4 M( H# b- G- Z float sum,average1;
1 q6 v J" z3 Q sum=0;
* e4 [( b* i' a8 K, h for(i=0;i<4;i++) Z6 S" t5 \$ M4 B% F) \
sum+=(*(psco+5*i))
0 t9 u$ u, ?- u average1=sum/4;
M7 G ^' `& }4 w7 `- Y" A printf("%s %5.2f\n",pcou,average1);) d3 h% B% z* y& E7 r. G* D& a
}! ]- R0 e6 s6 }/ Z1 ^
fali2(pcou,pnum,psco,pave)
8 ~& Y) q& y' Y. H9 y$ dchar *pcou;
: J# A- D7 u! a+ a, ?& L- fint *pnum;' G6 H# ~) c, z' X- b
float *psco,*pave;4 C+ ~0 m$ H% y2 b
{int i,j,k,label;5 J8 D; V7 u( f4 x
printf("\nnumber\n");
$ V4 ?9 v& X! n( J# Y$ `3 l p for(i=0;i<5;i++)
/ z$ ~. Y9 j2 j3 Q4 _( Z8 k printf("%-8s",pcou+10*i);& m$ X& o; v) G( B" D( s
printf("\naverage\n");( \+ B- @3 @9 U3 J
for(i=0;i<4;i++)( H% S" R' q( S; Y- K
{label=0;$ p2 x0 ?# B& R" r$ @
for(j=0;j<5;j++)* T7 @& a# l' Q
if(*(psco+5*i+j)<60.0)label++;, {' c0 P2 D2 G' `. B$ m4 l
if(label>=2) Y' B. ]; g$ t. k. E
{printf("%-8d",*(pnum+i));; |: K6 V1 S/ V% b7 m4 `' B& t
for(k=0;k<5;k++)# ?0 l* O: o6 v
printf("%-8.2f",*(psco+5*i+k));
{ j; ]. b# {( K B S0 s: S printf("%-8.2f",*(pave+i));
% L/ ~3 F' u# z }: o3 b3 N- Y5 D7 a
}) u# ]$ o" c5 c" Z9 w+ }- O% V
}" K/ c/ d" T; G: ~( B
good(pcou,pnum,psco,pave)
) C, U4 Q1 D; n4 Q& U4 A# tchar *pcou;) {: M# K. Q; U& T" W
int *pnum;
: X2 |0 Z6 o( N$ _/ ?, P/ [float *psco,*pave;* \5 ~0 u7 F6 x. t" m8 m5 x- G- ~
{int i,j,k,label;
) J9 u: A# M* k printf("number");# r# G) _3 e8 }" ?7 y8 D4 y
for(i=0;i<5;i++)
) q1 v E& N3 F1 @$ \$ M printf("%-8s",pcou+10*i);$ B( X! h2 ^4 W' |6 \
printf("average");
/ x4 Q' y$ ^( q$ D5 V9 G for(i=0;i<4;i++)
7 F4 ?- U3 _6 [6 l# n3 I$ O% D {label=0;
% X& Z- l" J4 l$ [# @5 m) `( U0 I- \* d for(j=0;j<5;j++)' H2 U3 F/ i' C. u
if(*(psco+5*i+j)>=85.0)label++;/ D. W" [7 h3 m8 f
if((label>=5)||(*(pave+i)>=90))
6 S2 L- z3 l. n" S6 f( L) S6 Z {printf("%-8d",*(pnum+i));
5 `- i W0 {; E, J+ h1 O L for(k=0;k<5;k++)
' x7 Q( }' y& t5 @" y printf("%-8.2f",*(psco+5*i+k));
$ O0 l; A; q, _9 y1 s& z0 c0 @ printf("%-8.2f",*(pave+i));! r0 m1 |& ^, q; Q) r$ d( p
}
+ U0 v; A# ]+ p0 \( s3 f }
# \2 G3 |: j, A% c- N}
, [3 D6 A' ]% q( b10.165 f5 |+ C+ t2 U! S9 E9 N2 J
#include"stdio.h"
, r8 B6 I7 M+ Z Imain()/ j" {) s1 t: q2 x$ n
{char str[50],*pstr;
9 X6 l9 E4 D9 E, D, u p int i,j,k,m,e10,digit,ndigit,a[10],*pa;3 H. G- N' ~+ y$ Z+ p$ n
gets(str);
- W: d% B. U! \$ U! ~5 e* b1 |5 | pstr=str;. T6 l: b, x4 ^$ x3 Z/ k
pa=a;
. |5 R" k' B) R$ i8 ]9 ^1 Y ndigit=0;
+ b j ]$ W$ |8 J i=j=0;: ?+ u3 o1 L/ l; j5 I7 W& M; Q
while(*(pstr+i)!='\0')* H8 C5 ^( J3 W; R$ V6 \
{if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))
" p6 W: Q% s9 B# s# i j++;
$ L8 r5 S! A8 }4 o, ` else
" h9 w/ O- \+ a8 d {if(j>0)% F" |2 e2 k+ F: d3 f
{digit=*(pstr+i-1)-48;, A1 {" H4 R" z0 x% Y+ H9 z8 W
k=1;
0 C6 j p; `* `% ]) N; ]7 q$ V; i while(k {e10=1;
- W. \, j( l$ r- C" o- s for(m=1;m<=k;m++)
! T* b! X; V+ D e10=e10*10;
) w4 [6 l7 S' j+ D% F! ?' | digit+=(*(pstr+i-1-k)-48)*e10;4 v- L( p- Y4 }0 i ?' Q6 t
k++;% [) E# J) F( \3 c
}8 [' Z/ E/ D# p1 D
*pa=digit;2 j5 m3 c6 B9 z5 q' K
ndigit++;6 X9 g# b& ^# y3 R% n
pa++;
1 [ f3 W- s4 v, {$ P! U j=0;( o. K$ ?8 J/ h% ]/ K9 v' \* E' v
}
. ~. k9 j; J8 M, j }4 D' m6 h4 a- N8 s3 v( @4 R
i++;/ c* J1 {* @5 `3 O' w6 N
}
5 i& c0 V. q2 t2 s. e if(j>0)
" B, \: _8 G- Y- b {digit=*(pstr+i-1)-48;
3 q6 _$ l3 ~, }% I! C' m k=1;
' y9 n. Q5 F; X! h& e% O while(k {e10=1;4 k$ f# T: a/ |' [4 i0 l
for(m=1;m<=k;m++). s2 F3 k$ G; P9 r: f) Z8 j: x; _
e10=e10*10;
) F4 E* Q* u9 P6 q8 t. I" v" u* K digit+=(*(pstr+i-1-k)-48)*e10;
8 m9 M: ^/ p0 e2 E8 ~' Q* r k++;
/ H+ P* i- q( _% b9 E4 v }8 t& y7 l, _5 ?0 C; a5 S
*pa=digit;
8 m3 i! y" G0 G7 n1 I5 f ndigit++;2 w$ |7 z: v j( A9 D: d0 j
j=0;2 C1 x5 Y8 l9 i
}
! I5 r5 c# t; d2 } printf("ndigit=%d\n",ndigit);
6 n8 V$ A& Q9 }" _- T c0 f j=0;
! C9 s" J' H+ F pa=a;
4 C- c$ z- k* p4 a i. F for(j=0;j printf("%d",*(pa+j));
" o8 g# E9 g. ?+ p}
) e9 C4 S% d5 F10.17
P. H+ x( p8 D2 b" emain()
7 s/ o+ L8 ?+ x3 N& x/ ~{int m;0 _5 f( Q, J. |! c0 n/ }+ j
char str1[20],str2[20],*p1,*p2;
! x+ W" `( k k7 ]. T scanf("%s",str1);
; }* `+ n$ W6 O/ K1 r* J; N) I3 @ scanf("%s",str2);4 w R$ w; \% x+ p
p1=str1;
/ O Z6 P- i2 Y; f0 U p2=str2;
. {2 [" h9 n$ C& h m=strcmp(p1,p2);
$ B! w$ I0 x5 a printf("%d\n",m);
% A& Z4 u1 h9 q! L1 `}
+ q6 M3 e% F4 y! wstrcmp(p1,p2)
" ^' X2 _+ v0 |7 J) l- }char *p1,*p2;6 @ g5 m8 {/ f
{int i=0;! B& O7 M9 Y$ C9 G; v; |: j' u
while(*(p1+i)==*(p2+i)). L, @- p' T* M; H4 p Q8 o6 g! @
if(*(p+i++)=='\0')return(0);$ b, v% u$ @# y# C
return(*(p1+i)-*(p2+i));/ T; h- }! l0 C- x7 b) z
}
! w! U/ l- g$ N2 U: S/ E, Q% w10.18
4 E2 P1 y4 f) R0 vmain()# `- A! \0 W P/ w5 \
{static char *mname[13]={"illeagl","January","February","March",
8 [7 ?2 e# B/ W "April","May","June","July","August","September","October",
8 p4 j( y1 F- b# i$ r5 \ "November","December"};0 w$ |, y. [+ ?
int n;
, Z7 S+ L9 a' ~8 |+ ]( w scanf("%d",&n);! u b2 X* H" [
if((n>=1)&&(n<=12))
1 B4 ^! O3 G2 K5 ^- u! j printf("%s\n",*(mname+n));
' H" C: O7 J% Y) k( D else% g d$ }6 ^2 ~: A* g4 k* Y
printf("error");
& T7 v( d* d! P6 K- G, j}; z# l" I7 k* I: i
10.20 Z6 P! U6 Q, J) {; P
main()
$ w1 A, ^/ W) R7 K e{int i;
; `3 \% Y- U0 z% |# ^0 a- E2 p char **p,*pstr[5],str[5][10];. d3 z& A, z3 b, }% C5 R4 H- ?1 [
for(i=0;i<5;i++) s8 d6 V; i: e6 f7 \
pstr[i]=str[i];6 [% y7 S* ?! c
for(i=0;i<5;i++)5 t$ H- y3 C5 j# j0 \- _8 r( X
scanf("%s",pstr[i]);# I8 C1 e9 ^. e0 O% A
p=pstr;
: r0 ^5 G" P G) f2 ?* ~% ^1 E# I sort(p);4 X L' ?4 q- m0 s! _6 d
for(i=0;i<5;i++)' q9 f3 }. n0 o2 [
printf("%s\n",pstr[i]);
! l* n3 ~; ]! r0 q}6 p0 @: i/ S! n" G7 `9 ~
sort(p). T9 n( E" D' w6 F! ~4 _$ r: O
char **P;
4 _# s2 C/ a3 H! b& g* s9 y{int i,j;8 w* {$ W; [: V/ o
char *pchange;
; L8 S+ e) i" j: | for(i=0;i<5;i++); V2 p- l! ^: D9 E1 ^* P
{for(j=i+1;j<5;j++)
% K# n0 H% ~: A2 ^1 T& {3 j( i {if(strcmp(*(p+i),*(p+j))>0)
/ B4 K$ u/ v" ^( f8 G% L3 c {pchange=*(p+i);, M8 @# l; |3 O! @5 u% t6 R
*(p+i)=*(p+j);, F* f/ k- \/ e) k! g3 j
*(p+j)=pchange;+ H) Z7 @4 ~' U6 @$ A( g
}
/ u n2 ]5 t7 ]* C W6 |) b }" ^. P9 n7 Q8 \' j) L$ G
}
7 R n. z* b$ K, ?4 k% ?7 A}
" h) d+ a4 W$ q6 z+ [2 @/ D10.21
# [9 \2 x- A0 f: y9 a6 j# f$ hmain()
+ o; y7 \4 K9 f( C{int i,n,digit[20],**p,*pstr[20];
) W' H9 C$ f2 S; V6 \- s+ {) q scanf("%d",&n);
+ k0 N, d0 `0 [& V5 F: k0 Q for(i=0;i pstr[i]=&digit[i];
" |* I' ~: o Y' r# | for(i=0;i scanf("%d",pstr[i]);
+ t4 {; G6 e' F6 H p=pstr;* ], I; l! u) F, u
sort(p,n);# _* J6 D: J- I& c' W
for(i=0;i printf("%d ",*pstr[i]);8 n7 d$ Q, p6 F' I) F- M
}
* X5 g5 L+ U$ {* T% psort(p,n)
# m+ z4 R r4 Gint **p,n;, m" e$ o8 \+ t0 C u
{int i,j,*pchange;: z. D! p' r" Q1 B: I# Z
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))% Y! z8 a) u: u
{pchange=*(p+i);
: F- a' d8 o; o8 T' w7 S: ? *(p+i)=*(p+j);* S( Y* b7 h& c) B! A
*(p+j)=pchange;
& Y% O7 \$ z0 f& [- J. Z$ {! Y }/ n! l; P# Y6 }" j2 ~! w
}7 L& {, N+ o( f3 W8 o% s' |
}
6 B9 G' \# v- V [2 y5 N}
/ F D. y6 ~! | c8 s& c3 a, h第十一章 结构体与共用体
/ u1 `- q- [( u, v+ e" F11.13 }2 B: u, U4 m1 S6 N
struct& u+ U7 U; r( k$ N9 H" h! N! W
{int year;& y0 L5 f5 r" M3 u i/ p
int month;% g+ `$ @- _( D! R+ c8 j& X# I
int day;
- B6 c! _" ]" d5 N/ d8 S }date;
7 i, M }9 Y) C! g6 O0 R+ Jmain()
+ K& m4 ?+ n, _; Q/ [* Y{int days;, A3 W, Z' D3 u
scanf("%d,%d,%d",&date.year,&date.month,&date.day);. x+ _/ l) B: P" L. \. m
switch(date.month)
% a: j3 J% k2 v4 D- ]4 I0 I {case 1:days=date.day;break;
8 e- S8 E- W9 a3 Y& R+ F( x7 U, w case 2:days=date.day+31;break;
6 O; c1 p1 [: O, P* K' b- J case 3:days=date.day+59;break;8 \! y/ h' I+ i$ N0 h% }/ U
case 4:days=date.day+90;break;" G4 H% i! H, M" [
case 5:days=date.day+120;break;8 x h# ?8 _( H7 A: x( ?: a) F6 b c
case 6:days=date.day+151;break;
! z x9 D- ~/ J: |7 c case 7:days=date.day+181;break;
, k9 V0 Q8 l( | {" Y case 8:days=date.day+212;break;
1 r3 F( I$ l3 Y7 V$ g case 9:days=date.day+243;break;0 B' D& z* f1 ]6 ?- c: v k$ j
case 10:days=date.day+273;break;
2 @' F* t$ ?# Q+ V" s5 s0 {) L case 11:days=date.day+304;break;' A$ {% f7 c$ U7 Q
case 12:days=date.day+334;break;( d& W: R* D# q6 o1 h. H N: e
}
9 ~+ |$ V! U* k4 `: ] if((date.year%4==0&&date.year%100!=0||date.year%400==0)) M4 B$ l; `+ B, D& L3 c# R
&&date.month>=3)4 P$ }$ [2 \1 t
days+=1;& x, ]% F. R4 Q9 r
printf("days=%d\n",days);* W) O0 V9 D7 |: N3 I+ p
}" L6 G8 D1 D- H" z$ S
11.2
& v5 z! o% R1 Vstruct dt
; D5 N! p7 r; n1 O, g" V {int year;
: M2 |8 l/ \6 H% L& S( n5 f int month;
. Q. J9 w! u- ?4 [0 _, w/ q int day;! U$ [! k' }% {1 `' o
}date;& a- a9 m+ w; D. k9 w0 D/ a! z$ d
main()
2 |& k3 j m; h2 i9 q{8 R! Y3 r/ d7 @' p" E$ Y* N
scanf("%d,%d,%d",&date.year,&date.month,&date.day);" x% Y3 C o! P
printf("\n%d\n",days(date.year,date.month,date.day));4 O$ j8 ~* Y" _% A2 f4 {+ l* z
}' s o" c& K- f5 ?. I/ {/ H
days(year,month,day)) D9 d$ I6 H1 W1 g+ k
int year,month,day;- C" Q; B2 X: j( r/ R3 i2 T* p
{int daysum=0,i;
) @4 |8 {7 Y0 v; h1 T static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}, M% F% z, m5 ^4 e5 l
for(i=1;i daysum+=daytab[i];# V$ C+ U! m1 B% w/ e
daysum+=day;
& _* g' V* e7 g' G1 O8 R if((year%4==0&&year%100!=0||year%400==0)&&month>=3)" f8 `* i( I5 ^# R! n p
daysum+=1;
. R/ u7 w: h- _; L return(daysum);1 K3 f) L. A J" A5 _6 F% l
}/ j/ A2 W' p6 V; q# Z; D
11.3
0 P* B1 S2 O0 B. J% Q: A/ O3 ~11.4" [. U! ]# R0 S9 {; u) O9 Y
#define N 5
" D5 B4 g' F% E% ~( estruct student2 e* F% H1 X( I& r0 @9 E) M5 u
{char num[6];
$ h- `$ N( F& X# `* I char name[8];# N H% f/ J) N: \
int score[4];
$ ^& X8 [ l2 o: \7 H" a3 V2 j }stu[N];9 ^. @; q5 ^" h' d) G8 \# b7 h
main()
) i" S, ]- b8 B c: P7 m/ }{
2 [9 c1 t3 j- _/ z9 i input(stu);
. B3 K4 Y& H2 n, l, q. m# m print(stu);2 t! L: W) Z( T2 V7 U' i: T% j
}
' n* P3 r, p6 M$ L, U, a7 minput(stu)
( ]2 R! B! A- }0 _struct student stu[];
8 Z9 M0 r: X5 B{int i,j;7 H" }1 A; z+ G1 S" a7 c! @' @' J
for(i=0;i {printf("number");
( e7 s. V+ l8 k# i( J9 \- o scanf("%s",stu[i].num);
3 N+ ^/ n0 @1 K printf("name");
* y- E. V( L) S( E# L: d scanf("%s",stu[i].name);' h1 S) u2 Q! k% E5 x1 p& R; j5 ?
for(j=0;j<3;j++)
8 B1 R9 w* S* f1 n5 o* k* ? {printf("\nscore\n");
4 q1 M9 m! L4 Q# ` l/ ] scanf("%d",&stu[i].score[j]);
0 z! | X8 O+ Y/ O: N9 q" J }9 o' U( ^, {- d! F2 b/ ~
printf("\n");2 H/ M' a7 L# d' E
}
* u9 y, ]4 ?! u2 B}3 F- Y* b2 |: O2 F( n) O u& [
print(stu)
3 c+ ^) g) p' W& \9 u5 dstruct student stu[];4 R3 T5 u5 V1 G
{int i,j;0 o% o# M/ L7 o0 p- P; m) C
printf("\nnumber name score1 score2 score3 \n"); Y" b( R m! x6 y
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);+ m1 M _& {5 I9 r
for(j=0;j<3;j++)
1 B0 a* r/ i8 m. w: G1 C: p2 t printf("%7d",stu[i].score[j]);
+ `. `7 l( q* n printf("\n");
2 d& q- A7 [ ?( Z' N& M3 G }
- c0 h% H! {! @4 }8 ?( d}
" P. W) a# t2 L2 d; v' g4 F: w11.5% l2 l; a$ k3 i' L
struct student
, L5 w# s; H6 E$ {. ~" K {char num[6];
+ V2 i7 S/ E- r$ W. m char name[8];
7 U9 o7 e1 @! K+ c0 P int score[4];
6 }: C7 F( r: K" N( c2 g3 Y float avr;
& _, i7 }1 v5 t+ u1 o7 T( r; U }stu[5];6 f: x( Y" ~: l1 n5 |* R* B$ l
main()* y1 h' E+ H) H3 n1 G! Z3 ]) R' A
{int i,j,max,maxi,sum;
" R# m5 [" G5 ?6 K/ I7 j1 A0 u float average;7 x* U6 K$ n" R9 m6 b
for(i=0;i<5;i++)
4 @2 t, S0 U$ H+ l {printf("number");
4 a6 [2 A/ a8 R, [4 ] scanf("%s",stu[i].num);
8 E, E! C8 w# U) v6 f' r2 w printf("name");1 u* S5 _) H* \3 G5 j0 _1 J8 Q
scanf("%s",stu[i].name);1 ~! `5 ]& W* Y3 u
for(j=0;j<3;j++)* h$ k( T# V+ D
{printf("\nscore\n");
9 E3 p$ L' N9 E; D$ M/ _& r4 {& w scanf("%d",&stu[i].score[j]);
7 G# {$ s* {9 o8 a7 i }! h. [0 r: r6 D# K- Z
}' u" E- k' T3 n( r. P3 b1 t
average=0;4 @2 M8 {9 u1 K, B2 a
max=0;
5 O( Q5 t# i5 ] l/ ? maxi=0;' H3 m8 }7 N: P- G- i5 y8 r3 l
for(i=0;i<5;i++)" i% [7 ^: e: ?, a0 X, c* Y
{sum=0;
6 \0 u0 R v6 ?8 m; R% u2 C for(j=0;j<3;j++)* ] O# q% t1 J: l* y5 ^
sum+=stu[i].score[j];; W6 _2 e7 P& n$ \. K2 Z* `- O; P
stu[i].avr=sum/3.0;
! U* Y1 i( g" v7 ^7 `+ _7 U& _ average+=stu[i].avr;
4 F) o. L; r; C: W) ` if(sum>max)
9 W) P; p1 J% N* k7 R" \2 d( R/ H) z {max=sum;
0 m7 t- S% G4 e. n( L# H: p maxi=i;7 l" V! a1 H W3 T, C! q
}( O4 w! M6 J3 w r* ^8 a, j, U
}. t0 z" s5 z* T' `" @
average/=5;
; z- j- \4 g- i7 ^ printf("number name score1 score2 score3 average\n");
9 F) \2 v; A/ r' Q/ }6 y% [+ ?2 C. g. u' D for(i=0;i<5;i++). ^6 p- }5 f7 f8 Z4 q
{printf("%8s%10s",stu[i].num,stu[i].name);
' J6 y4 o' v( a9 Y for(j=0;j<3;j++); C, o8 i3 {, A, G0 V
printf("%7d",stu[i].score[j]);
# H. D& ~/ j3 ~7 ^" h H. d printf("%6.2f\n",stu[i].avr);, n; T4 b2 v1 a! e x
}
2 o* S& P' L+ n5 L- P printf("average=%5.2f\n",average);5 a+ J/ Q, e' m6 `2 A: L+ G
printf("The best student is %s,sum=%d\n",stu[maxi].name,max);
. U9 n0 K3 f1 U# c# L}; {- {7 j5 ^ ^' X
q, |7 T8 M4 m2 W# S# w8 S |
zan
|