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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
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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% p+ M9 |: T) k' h
1.5请参照本章例题,编写一个C程序,输出以下信息:
9 }5 j" ~0 B4 r4 X3 @# bmain()
" a' J% F; r- P: J. x$ _1 r{
+ s4 L, x, `. ]8 p# Jprintf(" ************ \n");
# U1 r# d W0 v6 Qprintf("\n");0 y2 v: W0 I. L
printf(" Very Good! \n");6 e: n& C* ?% P* ]* V
printf("\n");
k# O( K) @3 Y7 _& f4 L( Oprintf(" ************\n");% t7 ?: i2 A( w4 C
}
) u5 Z( j t& F7 \6 b3 {/ s1.6编写一个程序,输入a b c三个值,输出其中最大者。
; ?, V% I( i: H解:main()
- q/ k" k p8 L0 m{int a,b,c,max;
' f0 D4 W* |' D+ nprintf("请输入三个数a,b,c:\n");
- x" \2 `! G6 R# I3 zscanf("%d,%d,%d",&a,&b,&c);1 W5 A# L0 h8 Z. D3 o) R4 q- O
max=a; {' h7 n5 W \; O, y8 K
if(maxmax=b;
: K- U5 g8 {' |# H0 _5 |9 M; Y/ aif(maxmax=c;
9 e9 Z- V R6 ]: x5 ~; s7 kprintf("最大数为:%d",max);
5 U4 |% C e! n1 F' v" z, T}; S' y& v' |: Z; B' `
第三章
6 K5 D! D. N9 U3 j) S, G: \3.3 请将下面各数用八进制数和十六进制数表示:
8 [+ a- H" {( M9 Z8 `2 K0 L3 f(1)10 (2)32 (3)75 (4)-617
$ E0 Q% ]% x* g1 p/ E0 p: k(5)-111 (6)2483 (7)-28654 (8)21003
" H( v& B2 E! x, r3 U8 c解:十 八 十六/ q1 N8 p7 m0 s
(10)=(12)=(a)
* g; M& o9 E# T+ W1 |* X (32)=(40)=20 q$ a& q4 j/ O: Y% L/ S
(75)=(113)=4b/ i- w2 [' N6 w) A% {% O& j
(-617)=(176627)=fd97
( ~% [$ s% B2 H& P6 P( ? -111=177621=ff91; | e8 y1 I3 _, k+ P
2483=4663=963
6 o I! [6 D# b% _, b) Q% n -28654=110022=9012
& e) o! v* L- w0 x# i# l+ H$ Y8 V' | 21003=51013=520b
! s: W5 a$ [# p2 \$ z( B, \3.5字符常量与字符串常量有什么区别?
* b/ n0 k" P6 l6 ^, h( Q3 h* W& C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
. q9 @: X! K6 I% o+ ?. ~. S9 ?9 C5 H" l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'.
" I; t/ v' ?" }8 u: Y3.6写出以下程序的运行结果:/ Q4 @0 N- Z# O4 F3 |; x5 H
#include
! Y, K' a' T7 ]5 B+ Pvoid main()# N. _! e4 `% Q5 a. L3 @; S
{ s+ T+ F- w0 L
char c1='a',c2='b',c3='c',c4='\101',c5='\116';
& k9 s+ ^/ s6 Z" {# Wprintf("a%c b%c\tc%c\tabc\n",c1,c2,c3);
0 j. a) i3 G) B, E1 K' ]3 Jprintf("\t\b%c %c\n",c4,c5);9 h R! ]$ J2 g3 E8 v5 v
解:程序的运行结果为:
; P, p/ h3 K; x# V W3 O3 c3 X) kaabb cc abc* p/ K$ H" o- D1 i$ W0 j
A N
# Q/ V* H1 \& N# U8 v3.7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,+ n, X0 p/ _1 }2 D2 Y @
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".
; I* t/ K! v0 V2 P2 J& j# s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
Z6 h/ @+ ?6 h7 z: n' I- S输出.
4 g) v+ t0 I: G( n8 qmain()
) Y# |. r4 z( t{char c1="C",c2="h",c3="i",c4='n',c5='a';
& e3 R4 g0 H. V4 H: i1 N% vc1+=4;1 p8 B* [2 q6 m) Y
c2+=4;# P' y2 ]% N; }1 [: v U0 R
c3+=4;# k& \$ M3 k# H$ z6 `2 w
c4+=4;; Y$ x# `( m, g" g; P9 d
c5+=4;
6 f( e" Y3 ^* p& r9 Hprintf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);. M2 W5 V9 O5 T# Q7 S; R. m+ X
}
! ^ W! a: M' m4 [( i3 N" M3.8例3.6能否改成如下:
4 V- O! B, x3 `5 c3 C# H#include
% @$ g! p, w. U( Svoid main()6 x! z/ e! h; v* x
{8 g5 h7 M9 v4 i( H1 X
int c1,c2;(原为 char c1,c2)5 R4 m# }# ?, Q; |# T
c1=97;! U6 }/ _: r* Y, \
c2=98;
! V8 p9 {+ ]$ _% x% F `printf("%c%c\n",c1,c2);
( t0 \3 L4 v' V; P4 ~1 cprintf("%d%d\n",c1,c2);
* Y8 x' k/ r8 f2 L k% `}4 `7 ?/ B p9 e
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.- W. j1 ^# n( C; ?4 `- M9 \
3.9求下面算术表达式的值.( E7 W; i" g6 `9 X
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)/ u! o; V$ X: y/ q
(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
1 j7 Q; T$ C2 g# ^3.10写出下面程序的运行结果:) A' R( I' z1 ~4 t1 N3 F& U7 a1 t$ d
#include# `, G, ]" q3 }! k1 \* \2 q4 E
void main()" I$ Z% G @$ D$ N+ C, W
{* m. u* T- n, a
int i,j,m,n;
) N+ N& l" I9 d0 D2 T/ ei=8;
. E- g) Y+ L& z' L4 F+ o8 N5 h ~j=10;
! M( |# G/ K0 a+ }m=++i;
# O5 \3 u6 q/ F; C! ?n=j++;
) U2 g( [# r' `4 i( Zprintf("%d,%d,%d,%d\n",i,j,m,n);& e' u1 {2 H ~& Z0 x e! Q
}& ^4 v3 S3 c6 _2 r- M! N- G
解:结果: 9,11,9,101 l2 r6 c7 s7 f# J" p* F. ^% R
第4章3 m. m4 G. W& U. C8 g+ n/ f9 U
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9 s/ q( s/ w0 q4 p$ 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:
9 v8 j3 b, `- `) ^; ra= 3 b= 4 c= 5
( D& o. g6 e1 v, X$ F2 ix=1.200000,y=2.400000,z=-3.6000002 ?* {" {6 o) L
x+y= 3.60 y+z=-1.20 z+x=-2.40
& N! T. c* H+ S9 c/ f, C! ku= 51274 n= 128765
4 t9 ~+ ^% |) C2 Zc1='a' or 97(ASCII)
6 }' l3 c3 _5 g V! ], Xc2='B' or 98(ASCII)1 X2 Z1 m A! l8 c; q
解:
( D# H8 Q' ^+ L8 R6 ~; p% P+ Pmain()! ` d. ?2 F1 A$ x+ L5 }, g, Q
{
1 i4 H" l: {, x6 K$ K& R* ]7 W# eint a,b,c;
; k3 T7 \. p' w' w6 @+ F. Slong int u,n;
+ ?, z/ r' p c0 xfloat x,y,z;
2 p0 P0 c4 z2 \) Q! F1 C6 O' x& lchar c1,c2;
* C- m# `5 L* ma=3;b=4;c=5;1 c) D% X+ S2 H/ x
x=1.2;y=2.4;z=-3.6;
l5 {$ q* G0 X$ R& |u=51274;n=128765;
. r: T( L: U5 a8 F ^c1='a';c2='b';
% N$ \' n+ ^# H. dprintf("\n");' b' S* s6 }' r# A1 W
printf("a=%2d b=%2d c=%2d\n",a,b,c);
, ?; E# L- H3 p3 _0 Sprintf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);' p9 z& t0 _- F% h2 l) ~# B
printf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);# z- s3 J1 C" o6 N+ I2 n0 g* B
printf("u=%6ld n=%9ld\n",u,n);. i( ?4 b2 E: Q; l3 t7 H/ V+ e
printf("c1='%c' or %d(ASCII)\n",c1,c2);- i9 n( u/ r% Q V _; r
printf("c2='%c' or %d(ASCII)\n",c2,c2);7 c% O) s4 R, B; _' F n4 P
}
+ P% H5 x1 n, D% L( m# m$ z4.5请写出下面程序的输出结果.
% k' l6 g' e: ^ a' Y$ H结果:8 Y' T1 P/ R" ]4 l( c
57
7 e: r1 j" _' F& J$ s 5 7
( A( k! ], I! D# g x, f* ?67.856400,-789.1239622 Z8 F. M$ a2 ]$ b/ g/ }
67.856400 ,-789.123962
+ F6 O; ^/ T( h" \# V& b2 X 67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962 U7 E9 W! t# U7 k2 u2 B e& H/ W
6.785640e+001,-7.89e+002
5 |& U* q) d. J3 m+ b0 c& ^A,65,101,41
0 K$ g) l+ m: K/ M! }0 q% X1234567,4553207,d687
" P7 Q# i% v5 ]0 z" `" l$ k65535,17777,ffff,-19 G4 ^: a# D8 r& z+ F
COMPUTER, COM3 @" d5 n5 a7 w5 [
4.6用下面的scanf函数输入数据,使a=3,b=7,x=8.5,y=71.82,c1='A',c2='a',3 _- k" ^ y9 {; Y" a& V; k! s" \& R: Y
问在键盘上如何输入?
5 _$ N4 F9 { `! ^* Wmain()
' V* c. P1 ^. ~8 K9 w{2 ^, v# O; g, g! p' A6 k0 R+ n
int a,b;8 w+ O% W2 q/ Q9 O+ |; t U
float x,y;
: Q/ z' D+ n* J) q& q5 C" B& |char c1,c2;' J8 d% m* s1 P# h4 f# k+ F: }3 L4 W
scanf("a=%d b=%d,&a,&b);
; M- t& n# H g# j+ f G8 Hscanf(" x=%f y=%e",&x,&y);
. m7 k- `2 c" t T+ j) rscanf(" c1=%c c2=%c",&c1,&c2);. R8 `$ x9 o' O2 t
}0 v2 s. [ ~1 ^1 Q- ~2 C8 D) H c
解:可按如下方式在键盘上输入:
4 e U) S# M5 p" Wa=3 b=7
. B. p$ y7 _) ^- z2 b, F: x, lx=8.5 y=71.826 c u* L- _& M
c1=A c2=a
9 o) W3 D% S1 L: @) D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, F( _+ D& W; |7 C
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& [9 x7 i" F H' C
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+ ]* ], V6 j% k7 R5 v7 F* C
初步概念即可,以后再进一步深入理解.' `" ?5 }$ N8 Y1 ?+ c' [
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=-1 q* K8 | Y; y9 O$ k
3.75,z=57.8,请问; u q* R# x2 }- E' \$ j `
在键盘上如何输入数据?9 S3 ^7 X8 O3 W& }
scanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);1 Z5 O: W% g0 a" G6 q2 J
解:
+ m P* Y9 f( O# p2 m2 ^main()
; u4 X: _6 b3 [# X4 K/ X& B{
. s( i o# ~$ R3 u7 gint a,b;
: b* w* ^0 }9 l, Y2 O4 @) j" Pfloat x,y,z;
8 J9 l8 V9 j+ J" g4 ^char c1,c2;9 _3 Y8 g; Z- v8 ^# `5 x( w! v
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);3 H$ T! `) o7 O
}* ~/ r3 ^4 l1 }# h3 M
运行时输入:, P, x; T Q" Y/ e1 p
10 20Aa1.5 -3.75 +1.5,67.8* }* r; ?, e: x6 k
注解:按%5d格式的要求输入a与b时,要先键入三个空格,而后再打入10与20。%*f
2 B, W+ B% q: ?是用来禁止赋值的。在输入时,对应于%*f的地方,随意打入了一个数1.5,该值不
% w& T; P/ m! L( F会赋给任何变量。
- Y! Y! k6 Z! n4 B$ a& b% u4.8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,
% L1 R3 k# |- F9 R- o$ E% P3 g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
C% _ J! c: P' V4 M) E程.
, \9 N: O! j' z# h0 o% }# P- }解:main(); \2 X z$ k7 w* a
{
! z1 X+ ^+ a, K* y% Cfloat pi,h,r,l,s,sq,vq,vz;
+ P2 i; \7 v U& f* D+ M4 api=3.1415926;
3 j- s# g5 f3 k1 U* \7 i M5 Bprintf("请输入圆半径r圆柱高h:\n");
" E* S- Q, R9 Ascanf("%f,%f",&r,&h);$ E! L& [( X ^' R* L
l=2*pi*r;+ }/ G- e' b% S- A7 F
s=r*r*pi;
5 p2 e, u6 z0 n5 Nsq=4*pi*r*r;
9 X5 I( ]# E, |3 v" G; R% Q, Y3 c7 yvq=4.0/3.0*pi*r*r*r;
4 l) R! j- o. b0 ~; fvz=pi*r*r*h;
* T" ^/ X* r2 T7 p9 W( ~& Oprintf("圆周长为: =%6.2f\n",l);
- `$ p/ d4 J) I# xprintf("圆面积为: =%6.2f\n",s);. l Z% Q( k' s7 }# s
printf("圆球表面积为: =%6.2f\n",sq);7 e2 V4 d1 d1 w% U" u7 ~
printf("圆球体积为: =%6.2f\n",vz);* M# R! E d9 B4 g, X' Q+ T0 {
}" N- U [$ M8 a# A) s0 l
4.9输入一个华氏温度,要求输出摄氏温度,公式为C=5/9(F-32),输出要有文字说明,% |& t, I6 j2 B0 G( ~7 |& z6 W# R
取两位小数.
* [; o2 C/ Q: _" s. {- c$ M解: main()
3 D$ P* U+ a1 ~( c8 r7 K' y{
- E' h) }, [& M1 N8 c lfloat c,f;
9 g" `) l" |4 J# a! U: kprintf("请输入一个华氏温度:\n");# n% t8 m6 ^) [6 S
scanf("%f",&f);9 I% ~4 [+ ]# [4 }2 F4 H
c=(5.0/9.0)*(f-32);
: }+ r1 ^3 v* ^/ d. \printf("摄氏温度为:%5.2f\n",c);
+ I" R& C. `* U}
1 E L& _' r6 T% @第五章 逻辑运算和判断选取结构( [: z" Y) G4 ^7 I! b/ Y. m; ]
5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
4 c9 Z, X$ }, w% R1 fmain()
5 x" T$ @& H. \6 J/ z# C{$ O6 Z9 d) x9 N
int a,b,c;; |2 Y6 D. W& u* f) Y$ N
printf("请输入三个数:");: q8 a6 ], q8 x# r* q
scanf("%d,%d,%d",&a,&b,&c);
- v4 M6 \' t! _/ R, Q2 {if(a if(b printf("max=%d\n",c);
/ }& g( R- `& [0 k- l6 X0 m else
' @: A- [3 o- [5 |" b* w printf("max=%d\n",b);
" o' n+ J7 Q! h/ L0 S3 w+ d* X% N. belse if(a printf("max=%d\n",c);
+ N) ]3 V5 h: X/ M. \; E else. s; U. z( p7 e5 A% Z
printf("max-%d\n",a);$ B; v& ~# v4 z$ H6 f1 `
}( u6 o. Z6 @, r" G5 S; }
方法2:使用条件表达式.
! p3 ?7 P3 J/ D) q# M) A6 Vmain()1 g2 f+ C5 P4 Y9 u* c: Y; i- x
{int a,b,c,termp,max;
8 {9 M7 I2 \6 f: w printf(" 请输入 A,B,C: ");
. F T9 T0 v2 }1 W scanf("%d,%d,%d",&a,&b,&c);
6 P# d6 G0 r/ Y, Y; Z A8 F printf("A=%d,B=%d,C=%d\n",a,b,c);
0 i: }# g; u: \9 q! O temp=(a>b)?a:b;
$ B: B8 M3 H: b3 g6 e- `1 f" X) R% w max=(temp>c)? temp:c;, L4 N5 ^+ Z, p
printf(" A,B,C中最大数是%d,",max);
$ Y! E# l/ J9 B}
6 c m" r/ _+ O! f7 L+ `5.5 main()" l5 i1 C5 S9 l: ]$ D
{int x,y;& a0 ?$ p8 m. R& L* X
printf("输入x:");
# A5 i$ S7 a6 M" `scanf("%d",&x);
6 t7 Y1 K6 \1 Z/ A6 p! G) Fif(x<1)6 _6 Z D3 P) a; F+ H
{y=x;
% r& Q! C& U" H+ p0 ]) K. V printf("X-%d,Y=X=%d \n",x,y);
8 _) U. L: b, a A }
, v, ~' N( L0 v6 E5 { Z, n4 Pelse if(x<10) Z! k; P' w9 E5 `
{y=2*x-1;
) g6 e* i" _7 C F+ p+ J6 v printf(" X=%d, Y=2*X-1=%d\n",x,y);9 j) P6 f& d+ ?" X) c! {( p. x
}
$ J! Q1 y; C w5 Z2 {else# t, y- N7 E! z
{y=3*x-11;
: Q& `2 T/ R; d9 d9 \ printf("X=5d, Y=3*x-11=%d \n",x,y);
7 ]6 e h5 ^! c5 K; w# }( b }
1 h2 u- D5 T# r$ g1 I}
7 \ S+ `7 N3 k" W(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! 4 T3 }0 D2 e$ ]( \
void main()
0 [( s$ x7 f0 {) w. H{
. }2 U" r; t0 q" Jfloat s,i;
; ?% h% p9 G- y& o. T H! }& Pchar a;" `/ j' l/ [3 u5 Y
scanf("%f",&s);
! N w8 e7 j7 {; T& W0 dwhile(s>100||s<0)
+ Q l/ ?9 V% I* S8 E! G& U{
0 [% x4 w2 {. hprintf("输入错误!error!");& ?9 i1 X7 T% Q) [ D" G
scanf("%f",&s);. y! \0 c& [: c0 g4 b
}% K1 W+ N. n7 ~) {9 p/ L. l; _
i=s/10;
9 T* w: [9 \! L1 U, Z- W; n( \switch((int)i)( s4 y3 v! Z! ?/ f
{
7 X7 p' ]" x0 i2 H- l9 G7 V4 Dcase 10:' {' u+ o% l* y
case 9: a='A';break;
- x/ Q$ o8 k2 C2 Icase 8: a='B';break;
& g3 j3 M# z: N1 K) ecase 7: a='C';break;# [5 _% g0 {) J
case 6: a='D';break;0 o6 q: |3 L3 T# w" Y
case 5:7 r6 E. T+ S! `6 P! b7 X, K
case 4:
# N A! I% c2 S! z& B/ a3 Ecase 2:! u! S& ~. ^& p) d" y
case 1:
( o1 K" A- L2 S3 V) @; e2 bcase 0: a='E';
3 l; A- H, ` c}
( L, s0 _+ p: l# T! v# r- Kprintf("%c",a); J4 b2 e4 L- J, T
}7 L/ l( O" Q( e; y9 e3 e% k, e% d
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.
8 D0 b5 h4 |# U" `6 G- T# m. G按逆序打印出各位数字.例如原数为321,应输出123.
: a* J/ ^, D, P3 h; Fmain(). @2 I8 {, a3 X' r
{0 `' G1 }! p3 V1 q- _0 @
long int num;$ ~# G- X2 a; N
int indiv,ten,hundred,housand,tenthousand,place;
4 h" W3 t! l1 n: @5 I& l& M printf("请输入一个整数(0-99999):");
' O5 Q2 s* X3 l+ \2 J6 h5 Q" Z scanf("%ld",&num);. C1 X Q3 u: R5 n# k# h
if(num>9999)$ Z5 W6 r B6 }! I" O
place=5;+ F0 R3 j" J9 M# U, P+ S5 G
else if(num>999)
% z6 ^+ c$ f* s- n place=4;) G K7 O. H( w8 f. I8 J6 O
else if(num>99)
0 l& r. o/ I+ X7 }- [* \ place=3;4 F1 y9 e1 |- `
else if(num>9)
' ?' [& l$ [( i place=2;- T! T ^ J3 G% t0 b# z+ l
else place=1;
5 `! b* y- D, p1 I- [1 zprintf("place=%d\n",place);
% z' {9 ^" b2 K2 W) X! ^0 oprintf("每位数字为:");/ ^' Z9 B9 b2 `
ten_thousand=num/10000;* Y( Z5 p6 k3 y. T% b2 J
thousand=(num-tenthousand*10000)/1000;
/ Y. C V5 f8 D) Q) ] Fhundred=(num-tenthousand*10000-thousand*1000)/100;
3 \% X x1 D4 d$ Tten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
* I5 S( b* H/ h+ {) t) y: |9 Gindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;, S& ?: h) I- C7 W% E5 H' D
switch(place)9 J: L( w* s7 h- s* I& ?/ _; q" I
{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
7 {. j6 Q+ X. F' | printf("\n反序数字为:");
; b* U3 {, L o) D- R0 R$ Z printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);
1 J8 u9 i2 C3 ]+ u4 M' @ l/ y break;
6 X7 [ v( y2 h, j) e g, Kcase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
" _& Y; X3 h7 e printf("\n反序数字为:");7 ~' J7 b- R: x3 S
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);
( B& B9 G8 O3 @1 k+ k break;! ~. K, \$ D1 f' c5 j( k
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);2 X0 }: _( M; `+ ?
printf("\n反序数字为:");
/ Z+ w A! t9 v7 M) @4 h printf("%d%d%d\n",indiv,ten,hundred);2 u7 Z2 h0 x$ r) p0 |0 a& P
case 2:printf("%d,%d\n",ten,indiv);
. \) D, y. n1 c" m0 e printf("\n反序数字为:");$ _3 h- {* r) z+ _& w
printf("%d%d\n",indiv,ten);. D8 }$ o( j7 O0 B$ N
case 1:printf("%d\n",indiv);
& _& q" ] |5 i1 |- Q. [* m) r printf("\n反序数字为:");
; ]2 B3 d. q1 V! n printf("%d\n",indiv);4 ?; H$ z2 Q$ D8 a
}3 C. |- r# S* @2 J/ v! C
}
' U. v9 U5 K/ i+ m* L# ~8 m5.8
& y4 t0 D5 D+ y1.if语句& {1 Q2 H+ v9 a
main()% O* N' z2 G& C+ v! e, I z x
{long i;
0 Y6 B5 C, }/ P; f$ U float bonus,bon1,bon2,bon4,bon6,bon10;
) O& z9 X7 R' v2 ~ bon1=100000*0.1;: d+ s8 C: j& {" h, n, B( Q
bon2=bon1+100000*0.075;& {* t+ P( ^' j. ^
bon4=bon2+200000*0.05;1 Q& f" j8 e7 f; [: j7 D/ T
bon6=bon4+200000*0.03;8 u9 ?" F8 j- m! x
bon10=bon6+400000*0.015;$ L9 j) I+ |0 M- ^
scanf("%ld",&i);& S/ G7 J+ C i2 \3 f4 g% `
if(i<=1e5)bonus=i*0.1;
1 k9 ^$ i' U; J! D else if(i<=2e5)bonus=bon1+(i-100000)*0.075;9 L5 Z# o. y9 f2 i: r
else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
: M1 v2 @8 N( C& F0 S; N9 B else if(i<=6e5)bonus=bon4+(i-400000)*0.03;( i+ }' U/ ?; M9 V6 T. c6 G0 p
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
& x& S* e) V0 o8 T' R7 e else bonus=bon10+(i-1000000)*0.01;' y# f$ @4 G p. p ]
printf("bonus=%10.2f",bonus);
' S& f- c3 [0 E0 ?. s/ s: t& X0 ?0 q}, k' C& E# U! T
用switch语句编程序) k5 _5 X; z' \, ] ]; L3 P
main()" }, Q5 Z6 R: K P# r3 t' @
{long i;4 \" U3 x m" ~+ O1 I
float bonus,bon1,bon2,bon4,bon6,bon10;
7 C+ N: N( e+ L2 j7 e% N int branch;; ^4 l% ~ K4 H
bon1=100000*0.1;
6 w- k# n1 L* A+ d, L' B4 L! l bon2=bon1+100000*0.075;
1 ~& x$ a% k% z- h bon4=bon2+200000*0.05;9 D' X/ z4 m+ t6 M" w7 J2 _, w$ V8 s
bon6=bon4+200000*0.03;8 M+ U2 z# [2 P w" `6 Q" A
bon10=bon6+400000*0.015;2 o$ }" d5 _& v4 _
scanf("%ld",&i);8 o) ^) O8 C: R5 E5 t
branch=i/100000;& x8 H+ w" d7 F7 ~) J
if(branch>10)branch=10;
$ M9 d& y5 t, y, ]' r5 t5 m) q switch(branch)9 k: b; j+ v# z7 _ P" b# {! t
{case 0:bonus=i*0.1;break;
& K- {& |5 [+ x/ _& k# v) _; | case 1:bonus=bon1+(i-100000)*0.075;break;
3 [7 S9 j. i" R+ d0 {1 B case 2:; i) n" ^4 J$ m( h7 @. r9 w
case 3:bonus=bon2+(i-200000)*0.05;break;0 {8 y% Q: H' ]( r& f
case 4:8 t5 b7 d# } [8 H9 h, \; G- @
case 5:bonus=bon4+(i-400000)*0.03;break;3 p8 o) X) H8 f8 ?& b6 |$ j
case 6:
" e/ L6 e+ w4 S7 ~0 A case 79 M4 O1 D' ]0 S. T s$ z
case 8:
( s' ~) s* M9 n6 o( ^ case 9:bonus=bon6+(i-600000)*0.015;break;% K( ^8 F2 f8 I$ I3 v, U0 M, N
case 10:bonus=bon10+(i-1000000)*0.01;
+ W& y/ e9 T4 B5 o5 @ }
6 C3 S" X7 T/ K3 a printf("bonus=%10.2f",bonus);
7 }8 h0 M; b w/ s% o} ! |0 p# t' [/ A8 l% W- |
5.9 输入四个整数,按大小顺序输出.0 s. H9 U3 i5 e. M4 o+ `( J
main()
( g" X: U8 U7 ]0 @% S5 d {int t,a,b,c,d;
5 O: G1 Z- I9 z$ T R$ h" T- P4 X printf("请输入四个数:");
9 s1 K5 @- p- ~$ ` scanf("%d,%d,%d,%d",&a,&b,&c,&d);
9 @4 \7 k: u+ C, J, A" H5 v printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);; z" {5 v& `3 f$ ]
if(a>b)
) r1 E0 c9 V* T! ^9 q6 s4 \ {t=a;a=b;b=t;}
e; f; j9 ~- h% v8 k" G if(a>c)" `/ z) v1 v5 k9 z% ]
{t=a;a=c;c=t;}) |/ c5 ~1 k `! `: T
if(a>d)" W. Z# V8 A% H& z* Q
{t=a;a=d;d=t;}* G y$ L; _5 V# J* A* q0 y3 M
if(b>c)4 w- i$ s5 j' ~' ?4 M2 d1 U
{t=b;b=c;c=t;}
' D5 o. W! }0 U if(b>d)$ Y& }) a$ y7 c) [ t
{t=b;b=d;d=t;}
3 h4 j6 g. U+ S+ h" m if(c>d)
+ O3 K+ C2 l8 X1 c! H {t=c;c=d;d=t;}" u3 ?5 J6 I" U8 f
printf("\n 排序结果如下: \n");" H' V, {/ D7 h
printf(" %d %d %d %d \n",a,b,c,d); Z$ p% `. Q+ z7 t- p
}8 S7 X' N- X3 r7 D4 M- `
5.10塔
# q4 P8 L4 W7 Q Xmain()
; u4 Q, \1 s; v- n5 s: z{/ B m$ p; ^, X5 D' N
int h=10;
: o$ k; ]! L4 K% ^' ffloat x,y,x0=2,y0=2,d1,d2,d3,d4;
6 a2 o/ _( n. J8 }printf("请输入一个点(x,y):");
* G. y2 h# `, n! n) mscanf("%f,%f",&x,&y);: y/ l( x/ y* |! R
d1=(x-x0)*(x-x0)+(y-y0)(y-y0);
c$ J. U4 a y3 H3 b) ud2=(x-x0)*(x-x0)+(y+y0)(y+y0);
8 v1 y* _# I; ~% Q5 ]: v. Pd3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
" `! a3 r% [7 yd4=(x+x0)*(x+x0)+(y+y0)*(y+y0);9 C- y, ?# B9 W! X. E, P
if(d1>1 && d2>1 && d3>1 && d4>1)- C) M3 a' i" `# e. j4 k" u% b
h=0;
; R1 k# a# Z# J! V8 hprintf("该点高度为%d",h);& Z3 A2 ^& X5 k( C
}
, B& I3 T0 K8 @' X6 N4 P% `0 |第六章 循环语句 L' A& p5 s% ?9 T
6.1输入两个正数,求最大公约数最小公倍数. m$ [! @( X' F% ], T
main()
( @7 s' P1 S' S: z H{
5 T g" y% o. Z) i) v. k1 ~int a,b,num1,num2,temp;
2 Y$ b+ g5 F! j, @5 T) K4 Mprintf("请输入两个正整数:\n");2 R+ v7 X4 I; [& I4 A u
scanf("%d,%d",&num1,&num2);1 l b- F1 ~" q2 A5 N/ Y- Y" m1 O
if(num1{4 A3 O: a5 f# a* h7 h% x
temp=num1;# ]2 t- b" c" d u$ B
num1=num2;
" c6 P" j2 o$ x4 Nnum2=temp;
# b; z9 ^9 [/ }- r0 c6 x}
, v" s V& ]5 ?a=num1,b=num2;
' Z0 y# _2 ?3 o% G& M% [4 \while(b!=0)
5 R3 }$ J" A9 F5 _9 w* E& V" @) O {6 _* g7 j& g( z3 {- E: z
temp=a%b;
# O; M- C! b) `$ Z6 E! j a=b;( _2 m. q, c5 E. `
b=temp;
. _ G. S& x' X/ Z, e0 U# W6 v }
4 q9 b% m% p$ w' [4 \printf("它们的最大公约数为:%d\n",a);
& j* x B4 F- U1 l& _, Z. m& \, ?printf("它们的最小公倍数为:%d\n",num1*num2/2);- X: D" N4 o* p( m9 D5 n# k+ H
}
9 r6 M1 q9 r& v1 ^3 |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.
8 s7 s ~6 o) p8 V解:1 o7 \& ?! J) |+ O) B1 X
#include < >
7 T# C9 S! x( h5 F2 ~, umain()
7 D9 c! g& U/ ~0 v4 x{
$ H: \4 F8 G# M) B- d, x1 i% h( gchar c;4 M6 c- B+ u: ^9 C& b
int letters=0,space=0,degit=0,other=0;3 D) e0 T: C3 P" V V ~+ s
printf("请输入一行字符:\n");
. r/ F& _ a1 L. p& X( cscanf("%c",&c);
' O9 G" b* L1 ^8 r4 |* \while((c=getchar())!='\n')/ h/ E c9 R& d1 P7 L$ N
{' p7 U. T, u& D0 N% p' M" L
if(c>='a'&&c<='z'||c>'A'&&c<='Z')5 |4 w6 p8 c. x
letters++;" F7 |' ]9 D) T6 U* {5 b
else if(c==' '), e! b& ]6 N( j: J8 H$ n( G
space++;: R" a- l8 u& s) E4 c: x
else if(c>='0'&&c<='9'); ~2 g9 M) J2 ~, U
digit++;- V" ]% M* m6 u8 h
else" v& V! R/ R7 C2 l' F: Q# l
other++;
9 w1 t+ I: P/ d; p}
- E; B2 Q4 |- [printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
$ D V _2 ~" i, _# Bd\n",letters,space,: D6 A" x9 d2 K
digit,other);% K0 z8 @3 }, {0 m. [
}3 B/ |; ?5 r: U: X
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.* E/ J9 I0 i6 |; N" E
解:, I6 @* u4 p8 F# ^1 b
main()
: T& Z8 n; I; A% ]$ V' P{1 y$ Z! x) I) n( @. i
int a,n,count=1,sn=0,tn=0;
. U& u+ c, ], T0 P7 k& k+ @printf("请输入a和n的值:\n");1 I8 ?) b0 \7 ?- {9 L
scanf("%d,%d",&a,&n);
$ A/ l/ ?4 @2 \, Xprintf("a=%d n=%d \n",a,n);
0 h9 t2 P7 M4 a3 z; @$ M0 h6 dwhile(count<=n)( ]* i# Y. T1 P, F
{
: e7 Z9 g/ }& [- f( q4 D+ j4 s4 ptn=tn+a;8 f/ U: A: |3 m S3 S- Q
sn=sn+tn;- y9 T. H$ t( C9 Y. |0 D8 I" Y
a=a*10;# M+ L: ?& x4 a) g8 d8 \9 ?/ [
++count;) a6 ^- s0 F0 J
}
; z2 [- |0 g" G& _printf("a+aa+aaa+…=%d\n",sn);
+ R" G4 o8 _ U8 }- s n}
$ ]3 F6 @) a9 I: V$ s: g3 z6.4 求1+2!+3!+4!+…+20!.6 z* J0 E1 B; }( S/ k3 k
main()
1 d0 w- G0 x: ?7 @{
( O- r+ a- x/ ]; M' @4 y2 [* d. T6 \float n,s=0,t=1;( Z! Q1 h5 z4 {8 h5 b t7 `" b- u
for(n=1;n<=20;n++)7 B: Z: l/ P3 {0 d+ E& X, t
{2 b9 m& P3 w5 i+ P' B3 g
t=t*n;) i9 P( c1 B. G# m
s=s+t;5 T3 Z! z7 ~+ ~- s7 L$ w
}
B' c2 Z, C, a; @printf("1!+2!+…+20!=%e\n",s);
$ g0 e }) G( k$ Q6 L' J5 X* o1 z}4 G* z" ^( K& i1 Q8 `9 p+ _8 O/ H
6.5 main()& A9 O" s, |; n: G9 D2 o" p
{, T. V. k4 X% ^" d
int N1=100,N2=50,N3=10;
2 i4 b# F# `! q+ W9 ^/ T' Xfloat k;7 s. e( F* v8 |3 U2 `" p
float s1=0,s2=0,s3=0;- K# U& y8 {0 b8 z2 Z
for(k=1;k<=N1;k++), W! Z+ z6 t, f$ v
{& u$ F: X m9 O; O/ o" K4 [- B
s1=s1+k;
; y& \" F4 V/ q3 U. H" ?}/ I( j8 h: r, M9 P3 }. I, N2 o2 r d
for(k=1;k<=N2;k++)3 c. K8 Y/ h+ T. a% h
{
5 R* b7 l6 a( p, R$ _1 Ps2=s2+k*k;
) m Z7 h, M# U" [( K}$ e" P$ t5 v; W
for(k=1;k<=N3;k++)
$ b/ H, i& m% _{
p/ _1 c- O! x; e' H/ A! `s3=s3+1/k;: F! N1 c) z* Y, F4 ?: N
}
5 f* v4 L5 o0 f, R4 K0 h* H/ Y9 _printf("总和=%8.2f\n",s1+s2+s3);2 `0 i; v0 o) U( i) n4 t
}' _/ ]) N( U7 I9 l) B
6.6水仙开花2 B7 ^0 e0 X. x6 L3 Q9 ?/ g
main()+ V2 T/ S. v+ y3 B
{
" D+ I; s: u; F# w- vint i,j,k,n;
' S# P6 G1 V1 L" `6 jprintf(" '水仙花'数是:");
1 f: C2 F* e& l/ w3 S% k; F% ofor(n=100;n<1000;n++)
+ t7 J* J! H; e% f- c( M{1 @! w! k6 z3 h5 s/ a
i=n/100;
' M) a. d% T/ ~j=n/10-i*10;
6 N, s$ n+ |8 F0 D$ r# ^8 \5 ek=n%10;: t5 n5 R/ D/ J d# |* h
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)6 \6 W7 I V2 A8 i4 D$ K
{" `- w3 u: S; A: ]1 U! I
printf("%d",n);# a3 _4 v+ z. t; K% Y
}
9 i/ ~; k6 ~( `/ X}* e% I1 a3 T9 d5 T- S
printf("\n");
+ l7 w+ [" h3 i4 u# ^" G}, ]. y4 z" A8 m; G2 j
6.7完数
0 f& X: e6 E8 a/ J0 w$ [main()
Q- F% b4 Z* b#include M 1000 u# X5 G |8 X6 G4 l6 y
main()9 P: \0 A$ W! N1 u" M3 z9 |5 n
{
$ c' j! T' \3 M3 g% oint k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;
F: n2 n2 C7 _int i,j,n,s;
$ g9 V* r; b( M; E. }0 f$ X% Zfor(j=2;j<=M;j++)
1 m- [4 i" c. [8 S. y{3 D4 S6 h. s4 f- G5 r- X$ |+ t
n=0; _- m0 L- x- G# |0 e; U a( A
s=j;
2 _6 ]1 F- Z, o, o! h, U* gfor(i=1;i {
( a/ z* H! a, _( p2 m* ~* c1 Kif((j%i)==0)3 {9 W2 f% f' T% X; f; ^) p* K
{
! \0 Y; U% Z+ S- p; P if((j%i)==0)" V2 [4 ~$ N; E, \5 ?; M
{
+ ^" M% m3 H! x, }4 J n++;
& p5 V! a; w0 ?& ?! f s=s-i;
) h6 W% \! \: ^ switch(n)6 C' i! I% L1 o+ b0 O
{
# a; t3 K1 g: j d3 Y" C O9 G! k case 1:
$ N. o8 ~% s# s+ [6 Z) x k0=i;
6 ~" i: N, Z' J5 _) _ break;
! |, g& W; @2 ? case 2:' U* Z. k3 T. q4 @( R) d8 S+ V
k1=i;% v7 ]8 M8 E: K) P
break;
) z( w6 K/ z- }9 }( j$ Q9 c case 3:
7 f6 S3 q {) ]0 E4 _: Q; C l k2=i;
0 v. {" l$ c) q; u break;( N; d7 t6 R9 c
case 4:) X7 @7 G' n! D9 l1 ~1 s7 y g$ `/ \' `
k3=i;
' b0 i6 D3 i4 W5 S7 b break;- d3 j3 V- Y/ D2 p
case 5:
$ R8 V. p( F' u8 W0 V- v* R/ ? k4=i;# c" s( A* u8 j5 x/ O
break;
) V/ M* P0 s# `, a; M' i. D5 d case 6:* `- ]; Q! B4 X1 o5 e E
k5=i;
0 Z6 v5 d+ }1 N! p/ \7 V$ | break;
/ V. {9 |& l# C6 ?- @3 m case 7:
; b1 R- J$ d5 Z2 ?; C3 f# y; g k6=i;0 p! |& ~ R8 j6 }
break;
4 Q& d2 g! v2 g$ } case 8:
& R- A0 ]: P) Q: m" b k7=i;: N) Y9 U0 G n q8 R$ b$ b/ x
break;
% j$ {0 Z! @6 F- f7 y1 ` case 9:
% E" }" a( z9 q* K: z6 r# e k8=i;2 \# Z" |$ a6 w4 k; g2 H5 c: x% v# D, F
break;
' N. h' y1 @) |2 U% C case 10:
- Q! |$ G/ E; h. T, r k9=i;
9 l7 @3 p4 T; B* ]% ^! ] break;% Q, y' m1 N) T4 Y: M% Y
}
" h6 E U$ a7 [" ~, K5 ` }
8 P3 G/ t: B# Q- b) R! f }
7 _) F4 |, r4 z$ ~$ Yif(s==0)- t7 f8 ]- m: W2 o
{6 H+ h7 H3 f* {. G) Z: o7 W) @
printf("%d是一个‘完数’,它的因子是",j);
' V8 a) c3 e" c4 q$ bif(n>1)" F6 Y' Y; Y! B) F5 R7 A# }
printf("%d,%d",k0,k1);5 R5 s0 `' B( h ?
if(n>2)
/ J. p X3 Z7 R- @- S# a1 ~ printf(",%d",k2);/ q% f7 y+ D: D0 P* U4 R3 J: i0 \
if(n>3)/ f1 i+ }' @3 p( f# H
printf(",%d",k3);& D6 Y( f, F& o! h- Z
if(n>4)
; B& k/ A- @" ]$ {4 C S. q1 v printf(",%d",k4); `' V9 d& \4 b/ _/ K
if(n>5)
$ [% a/ {! u! R# N* S printf(",%d",k5);
$ r. d' t# X& U: p7 j( r7 y" Eif(n>6)7 c9 c6 ^, i' m& }1 k" y6 ^' e; C
printf(",%d",k6);
% h/ a! @ x. H' yif(n>7). F/ t9 P$ U6 f
printf(",%d",k7);
" }8 H' r0 { g. |if(n>8)" K& l' F& D' R. B
printf(",%d",k8);
: n- Z. n$ ~5 K% C4 S; G7 s( T& Kif(n>9)
) h. A/ F2 g( r% d, { printf(",%d",k9);% d; C) M* a, k5 |9 E
printf("\n");; R2 Y& F7 ?5 D8 C. J
}
9 L8 I/ m: E; Y: {( c$ x7 N}' c$ z/ ^* e" x( [/ V2 ^) W
方法二:此题用数组方法更为简单.* a, K2 O& x1 `6 v0 [2 Z
main()
+ F0 S. L! z' m/ W& p' K4 q( s{
7 d5 t' \+ a3 A" M$ p/ h* ^+ kstatic int k[10];5 R5 e: U; m! O( w- T8 g k
int i,j,n,s;
6 j- r* P0 X; F4 { ufor(j=2;j<=1000;j++)
9 R2 {. h' O( ]: q% q8 }{% s( s; J3 e( ?+ H5 f% i# k* G9 G
n=-1;$ N, \. M7 D) C8 g
s=j;
7 ?' z# ?4 N2 P! Sfor(i=1;i{7 p% r0 T! j! r
if((j%i)==0)0 h F6 n2 `( b
{8 R; ?4 V6 f, ~8 M, i O; v' E
n++;
: F( {8 i2 J3 {; ls=s-i;
@! a: A4 g0 M) K1 G' _k[n]=i;
) ~# r9 Q1 L1 N }
$ U% C s2 ?5 e }
" y* Z' L* [" |+ ^/ oif(s==0)# S* Q! D+ h% P
{
0 w6 T" W* R# ~' D8 ~printf("%d是一个完数,它的因子是:",j);2 _8 S7 R6 a- X# w, Y/ Z
for(i=0;iprintf("%d,",k[i]);- g- v! P% P) N1 M# p+ b( B
printf("%d\n",k[n]);6 u0 T* h+ s9 u6 K+ _3 }
}
8 X2 I( b; I7 y6 y2 E2 v}
. u) K2 ^) z0 i2 g+ l1 Z6 Q: K6.8 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.9 v6 Y+ z! @* S; z9 v( l9 O
解: main()
1 Q# j5 d+ T6 s) ~! @{9 u3 |" ~" \8 }4 F4 e4 M2 V+ v
int n,t,number=20;
! [& }( Y* g% b+ s" ^float a=2,b=1,s=0;
+ w5 h, E" H( i9 V! n# Cfor(n=1;n<=number;n++)- g/ U/ [2 M/ L; t
{4 m& t2 j8 F' g3 v, l- b- J
s=s+a/b;
. s4 w' R4 W7 W7 E Qt=a,a=a+b,b=t;
+ S/ ^# v4 @; h1 m3 u( c}" T' l3 k: Y" D! |4 j. E2 L
printf("总和=%9.6f\n",s);
9 n! T3 V& A$ T: Z}
2 K$ n m3 p5 D( {1 K# n6.9球反弹问题
" \1 M3 A! M- [8 P( ~, |main()4 e# H2 j# [& {1 y
{4 d' e" N6 S0 I7 `$ a
float sn=100.0,hn=sn/2;
: f$ ?5 w& ~, f( p0 X* Pint n;
) \4 h! \2 J8 o0 c7 c) u8 o# z$ yfor(n=2;n<=10;n++) p! L7 m# |! e4 L( M' M
{' s$ z0 `& o, j
sn=sn+2*hn;' N, k# V. w) D0 ?
hn=hn/2; t8 a' Z+ x" D* w6 S
}: ^$ q% l" j" w7 ^- _
printf("第10次落地时共经过%f米 \n",sn);/ ^+ _/ z3 ?( G2 `- k, y0 @4 D
printf("第10次反弹%f米.\n",hn);
" e" L5 j3 D$ p' }: p8 o0 U}
+ L# p% D6 t: T: b/ u6.10猴子吃桃
6 C9 w9 P1 K& Pmain()9 q. p- l/ V( @! E" e
{0 j3 Z2 {% e5 R; x2 D' C
int day,x1,x2;
3 x9 o! \+ b$ P4 L4 t! Y% T4 gday=9;
8 F$ V. ~5 i+ X+ Y+ ^& ex2=1;
6 m' T0 j! s& awhile(day>0)4 \$ V! j* A# q4 }: }& _* E7 |0 T0 }
{
- y8 T8 @3 }1 N5 D9 yx1=(x2+1)*2;
6 d7 H( d4 n2 f/ ]% @5 L& ?8 vx2=x1;
+ `3 L1 w6 [# k4 y# U& kday--;; @/ ?+ h Z V
}
1 }4 y8 ~1 B# {! h) j5 _" ~printf("桃子总数=%d\n",x1);
' U( H: u0 _, L6 ^- b}0 f6 d) }" K9 d8 M0 f! W8 A) v
1 m0 M! s0 p+ T3 @4 L& ]) |6 Q5 j6.12
r4 P) v6 L2 H4 a; N! L#include"math.h"$ B6 P# c, Z) z% ?9 ~
main()
0 }9 m8 Z3 f. n0 @{float x,x0,f,f1;' q. F2 g; u P1 C6 K0 I0 x
x=1.5;
" I- f! S. ?. A4 O$ O do
5 g$ Y1 G8 c. ] {x0=x;
3 T* D# A; f2 L f=((2*x0-4)*x0+3)*x0-6;( ~& {" ^* f2 d9 y( z2 ?
f1=(6*x0-8)*x0+3;
7 `) I9 U; i; F5 g1 i& @4 g: i x=x0-f/f1;8 n# G/ D5 K) B& r! b
}5 v1 q' t5 j3 e' S7 @5 \7 P k
while(fabs(x-x0)>=1e-5);$ Y9 d6 M. _0 d! f! C, f' D9 c, Q
printf("x=%6.2f\n",x);2 w- T8 E- ~' \1 o+ O$ H+ g/ H$ h
}5 L" i' B+ n0 O/ }1 O
0 f8 I, J# M, y' b7 M) U
6.13) }( p8 K O& M+ \( E: u
#include"math.h"8 I; u+ C$ z1 F L4 ^* j* }9 L9 f) `
main()6 C% C" z+ d- X) Z. p& F- {* ~& \
{float x0,x1,x2,fx0,fx1,fx2;
1 p; g# O; _. B4 z5 u do
9 r$ Z9 C( B" ^/ f5 ^- ^: |' B7 y {scanf("%f,%f",&x1,&x2);" D# F9 G8 d- O
fx1=x1*((2*x1-4)*x1+3)-6;1 C* b4 L8 U v
fx2=x2*((2*x2-4)*x2+3)-6;: l( `/ Q1 ^6 E. r ]. D0 O4 a$ Z
}
N" p7 E3 r3 @& j3 d* ^% g while(fx1*fx2>0);
8 Q1 K0 S& H6 O% O6 m" P do
( L5 |% ?- X8 t3 c1 m3 v {x0=(x1+x2)/2;
% n( b/ R( V9 l! _! o fx0=x0*((2*x0-4)*x0+3)-6;3 G" T; Q0 H) _. n+ y% A$ h- N$ ~
if((fx0*fx1)<0)% c) N* ~2 s( r! Q! f: I
{x2=x0;8 h7 p [/ x( @4 i- \
fx2=fx0;* }2 l, j4 @+ e3 p
}& _: o* g+ v; A4 L9 d0 [
else- S1 l8 b- O4 N3 j( O
{x1=x0;/ A8 `# g2 y( r8 Q
fx1=fx0;+ \0 R3 W; u8 N% |3 M& ^
}
/ m, h$ D2 J, h }& M! j+ Q Y# O. l3 ^* c
while(fabs(fx0)>=1e-5);
6 n& b' n& q/ P; R2 `5 Z2 F printf("x0=%6.2f\n",x0);
6 `3 C5 S7 b2 X2 H& [0 @; R0 D' |6 d" }! h}
! c+ N: z+ ]7 L T5 D0 [5 i& W6.14打印图案5 p- ?; C* o, X+ H$ ~
main()
+ k( g# L" G6 U" B& F& e3 l k{int i,j,k;
3 C& F7 f2 `( q for(i=0;i<=3;i++)
; Y: I* {- D2 x/ g7 g+ ^" W {for(j=0;j<=2-i;j++)
: _& S" ]5 Y4 p: C+ I/ e! u5 q printf(" ");
3 j3 k' {: E4 C) c/ ?8 E for(k=0;k<=2*i;k++)
% z% V" H; a5 S! V$ ~ printf("*");( U" M4 {# ` `
printf("\n");9 n: `3 Q2 O9 r) K% c, a
}- Y. _! i2 X+ {4 r* L9 P) }+ |
for(i=0;i<=2;i++)! d+ h! E, }( z1 f; W
{for(j=0;j<=i;j++)# m4 v* a, T9 _% b m2 b7 [' l
printf(" ");5 }) O; @9 D" j$ ~
for(k=0;k<=4-2*i;k++)
, p5 Q( z/ l9 ] printf("*");
! F0 H) R! t$ R printf("\n");! o: H3 v4 a, T1 F
}0 F" P- M- L t/ {" x
}
& q9 P; f0 }+ W+ U0 J* ?6.15乒乓比赛
5 O, x- L& I3 j0 a. ^, ]+ Wmain()
2 |% {' k1 u# p4 t, @$ p0 i{5 u( M/ s# ?2 i% b( t# p
char i,j,k;
) U6 [' Q: J2 n2 H7 Sfor(i='x';i<='z';i++)
* [0 c4 E' m- Q' t% }: d/ w3 efor(j='x';j<='z';j++)* L7 M0 j/ R2 y) l+ _
{: u- H2 L; g) ?" }- { [8 [
if(i!=j)# ^* ^/ G. h: i2 p/ S C+ l# A
for(k='x';k<='z';k++)0 D% [1 f; c8 W- H, R2 N2 T( o" y6 h/ Z
{. H8 ?' r1 y* U2 f3 C7 o
if(i!=k&&j!=k)" c7 ^3 i0 L# J
{if(i!='x' && k!='x' && k! ='z')
S$ v( ~$ U, K T0 ^+ pprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);; C- M4 k$ ^, |3 z, U* T
}, [% M6 L" P( Y9 y" `
}
1 x* H/ W1 e# y3 G3 p } Z, a3 |" M2 `6 E' t5 h/ C3 F
}
0 W+ m9 p. J; f, ZC语言设计谭浩强第三版的课后习题答案7 c Y4 |9 m( T; ~7 y+ v
7.1用筛选法求100之内的素数.1 G* N* A8 J/ g8 t& w
#include+ l* H5 t4 n5 d# ^3 C# W( `
#define N 101/ [* r9 W+ b) A% [& X: u: a% |
main()
6 \- I% \7 i9 d8 ^) A9 c- N{int i,j,line,a[N];- N2 U9 \9 y, n" u5 S# Y
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
0 G* O; V) q" u6 f; o if(a[j]%a[i]==0)
: z" V# ^' v& [4 y# a' \ a[j]=0;+ }& x' i% I' S% ?. d% H
printf("\n");& T! {! m9 K# \1 k, l( G
for(i=2,line=0;i{ if(a[i]!=0)
8 L+ c E- f( R, F1 Y0 l {printf("%5d",a[i]);2 {: s2 U/ a0 A7 t8 n& k/ ` f4 S
line++;( a5 T) V7 k1 j+ {. n
if(line==10)6 Z2 \' K# \3 l! {1 O1 w; I
{printf("\n");/ r1 t( Q$ Z6 F. M
line=0;}
+ a0 x. ^; o& G, k" _8 s$ | }0 c. }2 S& z1 w% T( T
}- ]! }! q4 x" W7 z- u i' `' t) b3 ?
7.2用选择法对10个数排序.
% w2 M7 Q% |1 @#define N 10( b: i$ |$ n @6 ~3 ]1 `, d
main()$ {+ F! ]- \, U
{ int i,j,min,temp,a[N];
4 x) k3 Y) q! h4 Q. P3 ^printf("请输入十个数:\n");8 J- }' e4 H) c# d2 e z
for (i=0;i{ printf("a[%d]=",i);
% V& H! K0 S, ]+ f! q) `% D! x" s scanf("%d",&a[i]);# ?# W* x, E/ I8 S7 G5 U% V
}
1 g4 w- _6 H. Y* Gprintf("\n");/ W# b: ~4 l8 |2 s- G4 [& s# s8 F, ]9 r
for(i=0;i printf("%5d",a[i]);
& e& L$ O: i0 _% W" Y. u% c2 Y Aprintf("\n");
4 p3 m _6 W/ k5 U4 P/ Jfor (i=0;i{ min=i;
+ w( G& _0 Y$ @0 u3 O for(j=i+1;j if(a[min]>a[j]) min=j;0 Y) X+ X) t5 X; X a2 ~
temp=a[i];
1 w* c: _! o& c3 O4 B i9 i7 o6 R a[i]=a[min];
, Y. N. z5 a" u1 L/ t a[min]=temp;( a/ z! z8 w: v
}
( I; T0 C- N- U6 wprintf("\n排序结果如下:\n");
2 @: Y- |$ Q4 J1 h9 e+ u. e0 Bfor(i=0;iprintf("%5d",a[i]);; W1 p0 @. J7 i0 n3 O' I
}
5 r' C1 Y9 u% E7.3对角线和:
" D4 ?) ?3 }1 o0 b2 x6 c+ smain()
p4 S2 x0 b, y5 o{
$ k" m/ F! M( v# ]3 Lfloat a[3][3],sum=0;+ \- y6 _4 S" V* C- G3 s
int i,j;# M5 d6 ~/ _- r3 P3 S# ]
printf("请输入矩阵元素:\n");! K% M3 Z& y' ^
for(i=0;i<3;i++)
) ?. X! r! w* [) ^1 M for(j=0;j<3;j++)
- p% X' s6 M9 ]/ K! P scanf("%f",&a[i][j]);
) d+ c5 x& r9 }! C for(i=0;i<3;i++)
0 u0 ~! q2 {4 q7 ^/ S sum=sum+a[i][i];
8 X( o! O5 D; `1 m printf("对角元素之和=6.2f",sum);+ D3 R7 s4 F+ w" y w& r
}0 q% K# x' J) ]8 X
7.4插入数据到数组8 U R& B: y: v7 D
main()" q2 Z( G8 p! m6 @7 F/ \+ N
{int a[11]={1,4,6,9,13,16,19,28,40,100};2 k8 Q7 O% ^' d, P' T+ L
int temp1,temp2,number,end,i,j;
6 B2 Q w+ r; A- |5 j# Sprintf("初始数组如下:");5 \( t" N! M5 f/ _2 m
for (i=0;i<10;i++)
- c1 m7 B% E$ T, Q8 V5 F0 [printf("%5d",a[i]);9 B) }; b% P. {: Z
printf("\n");
8 n0 }0 K0 }* \9 y5 hprintf("输入插入数据:");3 ^ a) n$ U4 f1 m0 x, A
scanf("%d",&number);) \4 E$ [" i- G- m9 m8 d
end=a[9];
3 Y& U- W6 I( |) r/ vif(number>end)2 ?2 n5 D' J% H6 i/ U$ d
a[10]=number;
- d7 t3 Y/ }, o# x4 X4 ielse' F* ~6 d& w/ _: \" F
{for(i=0;i<10;i++). A `' b; Y# Y5 c: o& N
{ if(a[i]>number)2 W! M1 {5 G, A
{temp1=a[i];- I- c4 I8 t2 a; ?6 p4 |/ s6 J: A% @
a[i]=number;* B- J/ o# X; V: j( E
for(j=i+1;j<11;j++)5 Y; u# ?4 E% f
{temp2=a[j];# P9 I6 c1 A1 V; m8 j5 ^
a[j]=temp1;- s2 k [) ]5 \# q% c0 I
temp1=temp2;
6 z0 E9 Y1 _; h3 T }
2 V8 X6 n6 O6 ^* ?8 i" q break;8 V/ k, \' `3 l
}* f* r6 O# Q# o2 B2 p
}; ]& B# ?2 D" c" a
}: B# G& {& W+ Q: G
for(i=0;j<11;i++), [, U% [( D6 r, } h
printf("a%6d",a[i]);3 E$ l5 N0 {7 b# J2 o
}9 P: p% ]' h1 M2 Z1 n" D' p% J
7.5将一个数组逆序存放。
3 I5 C9 H4 {0 ]! \$ Y. s#define N 5/ e& }, W4 j/ N Y
main()
- o+ f3 }0 y% Q9 i+ p) Q{ int a[N]={8,6,5,4,1},i,temp;( |1 s% z) x0 R* }# V* J# r
printf("\n 初始数组:\n");4 c. R! q5 t# Q' I$ ~; c' A8 P& s
for(i=0;iprintf("%4d",a[i]);; C: ]/ X& A- b9 a1 d4 n4 d. H( |+ U
for(i=0;i{ temp=a[i];. t+ v. I3 C/ r7 `% h/ L
a[i]=a[N-i-1];
5 }: k5 U$ E( v" Q1 p9 E. f a[N-i-1]=temp;
% q2 \: e' {' v: r}
7 W8 i/ E) W+ L2 ^" lprintf("\n 交换后的数组:\n");/ d( o& T$ u7 ]6 ~2 P6 H' J, C
for(i=0;i printf("%4d",a[i]);
2 B1 {: c& A8 U1 k" r& |" p}$ q. T o+ ]. n: C. _
7.6杨辉三角8 z4 X7 J: E5 e1 X5 m6 O; r. E
#define N 11
8 O$ t6 v, y! O' c) _& f) lmain()
; E7 h0 h4 b+ W+ Z3 |# W7 ^' E{ int i,j,a[N][N];% G% k- ~8 r6 |1 @$ m
for(i=1;i {a[i][i]=1;
% g: A/ X' Q- e, W1 R. Z6 J a[i][1]=1;
- [$ n2 q' W# _& z" Z" I6 Y2 ] }
! P+ L/ C6 a1 X/ t for(i=3;i for(j=2;j<=i-1;j++)8 C0 ~/ w+ v5 A9 a, N0 l
a[i][j]=a[i01][j-1]+a[i-1][j];1 S6 t' U, F# _- p
for(i=1;i { for(j=1;j<=i;j++). e. j# b5 @. e5 B! U9 p
printf("%6d",a[i][j];
# w5 G: `* Q8 ~* v! N printf("\n");
" a. r8 E$ |+ F: r2 @! u! K }
; Q* ?9 ]! y& R" C- S4 i printf("\n");" U+ {- g- i4 y* k
}
4 s4 O3 h" S- O8 n3 \4 N7.8鞍点
~" a: X' ^$ x#define N 10
) R; w. Q1 I% f5 \5 `% m6 D#define M 10
2 G( C1 w% @' r6 ~, E1 `# Q/ Jmain()2 s Y/ `5 E4 O5 _# F
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;. i& W- }8 }" e
printf("\n输入行数n:");. r4 s8 t6 p4 z+ U
scanf("%d",&n);
" H# \0 I4 L3 k% e' @5 B- ^; M printf("\n输入列数m:");
( V: P) L0 t1 y" T3 \ scanf("%d",&m);) P( Q7 m: m4 r8 I* N
! [2 J8 r" M8 s7 P for(i=0;i { printf("第%d行?\n",i);
. r( \) u5 E9 R- L, m3 i* A# t for(j=0;j scanf("%d",&a[i][j];2 d5 E' j8 t! t1 p* d& @
}* K% l% L: i& a2 M
for(i=0;i { for(j=0;j printf("%5d",a[i][j]);# g, F- B& v! N5 {1 F
pritf("\n");
! o, o( U9 M: U$ @ }9 O: ]& z6 [+ ~6 s9 S7 F
flag2=0;
8 D E0 N) Z) C+ v$ I. R2 m0 p2 A0 d for(i=0;i { max=a[i][0];( g% U b% C- f( \- a
for(j=0;j if(a[i][j]>max)
" P4 w9 |# L" M* _7 m- m6 j { max=a[i][j]; B r9 W7 B0 v m V# f$ T
maxj=j;1 p( y9 B4 ^, U; d! ?
}
6 H* t% r% S, Y; k" P/ | for (k=0,flag1=1;k if(max>a[k][max])
2 g( M# t. a# R! d flag1=0;
% r7 t0 n& k+ B0 M if(flag1)- b6 J* H0 j# V* c K% z
{ printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);2 G# J. {( P5 t: `( u$ S
flag2=1;. W+ t9 m6 L0 f* ]& q$ f, |" Z
}( Y% y+ r% ~! `! x7 M9 L) m
}
, X) g4 s. N1 Q1 G, T7 mif(!flag2)1 ^: S, A' H; M) q
printf("\n 矩阵中无鞍点! \n");5 _8 x9 H$ ^! |( u6 g$ e$ u; a
}
! Z9 s9 ^# b) X9 M) u0 N7 y ]( A' V
* O1 Y; J; s2 F; `7.9变量说明:top,bott:查找区间两端点的下标;loca:查找成功与否的开关变量.' O9 g3 n- e7 S& G2 c; k0 I
#include/ L/ m1 b( q! }+ {5 U
#define N 15& W( V; {+ g1 B$ ]5 M! i( c+ N
main()$ Z; h$ ]4 V7 v0 z
{ int i,j,number,top,bott,min,loca,a[N],flag;
& P6 K2 ^1 {% u* p8 ?* ` char c;9 m' [1 U. U. c9 v3 c7 w& G
printf("输入15个数(a[i]>[i-1])\n);& D+ n0 B/ g) Y Z ?" p( f# ~
scanf("%d",&a[0]);
9 J0 r! S W$ N6 D& G. B/ h i=1;" A' V w/ O1 n1 Z' R# B8 ?
while(i { scanf("%d",&a[i]);, V" l" Q1 U6 k& M
if(a[i]>=a[i-1])4 T" t1 l7 t3 S. T* ^/ P; A, |$ G
i++;( `( L2 _8 Y+ M1 i0 |9 E& L& m
esle4 I* Z2 J! j' L
{printf("请重输入a[i]");
1 A- D* z' {/ Q3 {5 [; N printf("必须大于%d\n",a[i-1]);+ D4 k @" h2 s" `2 w
}9 C) p/ y3 Z7 W" s
}
+ G6 J0 g0 G0 A) r8 [, l, g/ g+ r printf("\n");
! \* @8 p* O) ^+ Q. | for(i=0;i printf("%4d",a[i]);
4 P$ X$ d7 d9 Z" l5 D/ _ printf("\n");
6 w' W! J" M+ f% B 4 c Q; z9 u" W2 M
flag=1;5 c! p I7 ~, |3 d
while(flag)
5 @$ O8 X( T$ P {
( e1 Z1 o3 M, y6 ~' \- Q printf("请输入查找数据:");
- q" q+ C4 ~! t- R# Y% u scanf("%d",&number);/ R* p/ C; e4 M) U( C9 d% s
loca=0;; b. ^+ T5 J t* @1 J
top=0;" [! I, k, U9 D4 E- N4 y
bott=N-1;8 c* E4 p/ H) z# p4 R- d
if((numbera[N-1]))
. ?4 s& D- \' ` loca=-1;+ [$ q. R2 C- j& _
while((loca==0)&&(top<=bott))& F' o' v% Y$ f
{ min=(bott+top)/2;
9 T+ N1 G3 T6 R# d if(number==a[min])
$ b3 U. { B2 y { loca=min;/ `# J7 s$ Y4 c0 M9 C3 O9 V
printf("%d位于表中第%d个数\n",number,loca+1);
2 k) F8 }2 Z- C }/ i) j1 m" }) z3 K* T) v" V
else if(number bott=min-1;
5 G' e' X W* H* c& J else, n# p2 P p) r! X! y4 F$ ]
top=min+1;5 ]! H2 E7 W7 Y) w# z
}1 z2 a$ L9 v( }7 T* _2 F1 l' n# f
if(loca==0||loca==-1)
{1 o$ Q! j& T4 H. X8 O8 B# Y printf("%d不在表中\n",number);
6 m# m" `" h1 H3 k1 m; z printf("是否继续查找?Y/N!\n");. z2 q' X; u2 M
c=getchar();
. z! d( k L2 ^( Z2 A1 F$ w if(c=='N'||c=='n')
! H5 [/ H! n. N n. l flag=0;2 g X5 ]* Q* p1 {# G: O
}
) [9 i' t* S2 P" d* h; J+ h6 e}$ @- @/ s4 Z; t6 F5 X y) r" p" B
9 C- s$ ]: B* Q- \7.108 D1 t2 H' y% |8 `& C% }
main()3 m3 V2 e) v+ e* r6 A; V A
{ int i,j,uppn,lown,dign,span,othn;
/ u: l$ b9 i4 m( V5 d0 | ?% i7 f char text[3][80];$ j( ~5 X& @! }: W. q
uppn=lown=dign=span=othn=0;
1 S2 p. I0 A$ ^ O for(i=0;i<3;i++)1 O2 V* R; O# @! M) }1 b" a' j
{ printf("\n请输入第%d行:\n",i);" w F/ W' n9 o( w' v4 u% m( l" A
gets(text[i]);0 |( r7 ~" e" }: u3 K
for(j=0;j<80 && text[i][j]!='\0';j++)
5 o' ]( o( B' h {if(text[i][j]>='A' && text[i][j]<='Z')
. Y# y# t7 u+ B. v- Z7 Y uppn+=1;- B/ n/ m" a( i6 _% a5 g
else if(text[i][j]>='a' && text[i][j]<='z')
. S3 h: H1 j" b: {1 A1 ~5 O lown+=1;1 j3 O; e$ Y }& s
else if(text[i][j]>='1' && text[i][j]<='9')
$ a- ^5 Q/ A/ f1 S dign+=1;4 ^! e9 o1 k |, U5 R
else if(text[i][j]=' ')3 p1 o* r2 U8 d+ ]) @& d' ?! S; c
span+=1;
' [" ~/ ^, @! z$ S6 p& T else
( @4 s8 X! F4 g. X othn+=1;
9 R' u" A0 R: y: ~& \9 B' M }$ r- r6 S& p2 h g# s; T
}/ c* c+ j" m+ s
for(i=0;i<3;i++)
$ B% h% t# V$ R2 I V' T printf("%s=n",text[i]);% R) [3 O: ? t' l
printf("大写字母数:%d\n",uppn);5 i' j. n) e+ Q: e r
printf("小写字母数:%d\n",lown);
" k% j9 N# d& W$ I4 X4 e% C( K printf("数字个数:%d\n",dign);: a- |$ I# u, o/ r, E
printf("空格个数:%d\n",span);, v4 H5 O; H0 }+ G0 t# [: z
printf("其它字符:%d\n",othn);
: ~7 l9 T( U' b- p9 E4 a0 m}
* K. d2 ` h; n G
. C. r& J, t3 C; S
. k5 B5 A5 p1 \7.11% i" c3 I4 @% N3 V1 Y( y7 \# v
main()
* @6 \3 I2 o/ P+ h* M- |. x" a {static char a[5]={'*','*','*','*','*'};# F0 y+ b3 t. t" n
int i,j,k;
7 Z' w' ?: v- G- c1 U char space=' ';
2 ~7 P' W2 n* s( {% i3 v, r for(i=0;i<=5;i++)5 h0 x, g( Q4 P& I: k# @/ M
{printf("\n");& b3 t. U4 `+ v" w9 [4 p- L
for(j=1;j<=3*i;j++)
5 U% v9 D0 g8 `0 B4 l! H9 F printf("%lc",space);
$ ?' J; M% }4 ?7 B) P7 J for(k=0;k<=5;k++)- {+ I4 ]4 M) i6 F& D; T& W
printf("%3c",a[k];8 p6 A9 S9 [' r4 c
}6 C( O& @2 {% J; T5 d+ t, V( `/ C
}
5 Z. j- g( t6 i1 |9 B% u7.127 m: a: b1 {0 t- r
#include# @% V' [7 w9 r- [7 e" J
main()
( p) l. z( v/ w# [{int i,n;
4 ~& a9 D) A i: n( k% Q B6 n4 r: [ char ch[80],tran[80];* Q% I2 c6 S: h! d* L# ?
printf("请输入字符:");' f( `9 i4 W' v5 h
gets(ch);
( D/ N2 l: I* F/ xprintf("\n密码是%c",ch);
4 v+ `8 o- n$ b5 `+ V* Hi=0;1 p! \* l7 z3 F
while(ch[i]!='\0')
, a0 j. }/ I1 t. w6 Z6 S: M{if((ch[i]>='A')&&(ch[i]<='Z'))+ m. L8 [9 u) o
tran[i]=26+64-ch[i]+1+64;
+ ~ Y I: @6 J) D" ?else if((ch[i]>='a')&&(ch[i]<='z'))& x/ n; ?5 t2 K) H7 f4 c7 C
tran[i]=26+96-ch[i]+1+96;" k+ d1 N6 T" W& J r4 P5 M
else( M9 H' `3 o7 C7 m6 ~8 Z* T
tran[i]=ch[i];/ G7 z) w2 f8 O% m' X }* v
i++; Y3 x& P( ?9 h3 L1 A% D
}
6 j" u* R% G# |8 {n=i;
( ~7 ^, z% ~- x: N" v6 z" dprintf("\n原文是:");; R. @' y( {: T9 ?4 C" S5 Q
for(i=0;iputchar(tran[i]);
$ h5 P" A* G% u, S}) v) S+ Z$ D* F8 Q9 y! i, h# E1 I
7.13
- N& k8 N" M5 A2 m! {main()
* g& r' U7 Z7 ?9 ^. g1 V {
0 K* Z S! r% ` w$ [ char s1[80],s2[40];
# E! |$ ?# V; l8 H8 N! T int i=0,j=0;
& x% X) e2 d2 R2 W printf("\n请输入字符串1:");7 _: f4 f. w }" E3 U' }2 p1 Y1 x
scanf("%s",s1);) @1 z) ~& ^: o. U: g/ }
printf("\n请输入字符串2:");+ p5 f) S, W2 a1 k
scanf("%s",s2);
0 U. H3 V0 L/ {0 l. R8 s( v while(s1[i]!='\0')& E& f! c8 Y% G' m5 X! m
i++;
/ j7 N L( k* Y" Q- M( fwhile(s2[j]!='\0')
% B. V+ n& C/ w0 b s1[i++]=s2[j++];3 u9 M3 b5 e% S: x: K8 E+ |
s1[i]='\0';
3 |$ m$ h3 o# `+ N9 N5 ?1 ?# kprintf("\n连接后字符串为:%s",s1);
* T% e/ Q+ t0 ]& P) v: x U }6 M. ~6 Z' I* h: ^- f$ r5 @
! N. O/ w5 ^' n" u2 |+ R
4 ]& B5 a5 \$ w0 Y7.148 v3 g1 Y7 d1 [
#include
8 T; o7 s+ C4 X0 ` Wmain()* l' a# v0 z$ c+ b0 l% c" E
{int i,resu;( f# f1 B6 u# c$ D
char s1[100],s2[100];
W; R9 N, B0 \/ K8 k; f printf("请输入字符串1:\n");
! ]0 M0 k8 y( V! e9 ] gets(s1);* r* {2 _; f3 e* M" k+ {) G$ M
printf("\n 请输入字符串2:\n"); C- v7 i; [& G( _" `! H6 x
gets(s2);1 V! \9 b e; P. p, X) ~2 u9 L
i=0;
& E$ s! j& w& o' r' H while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;
7 I" n$ R! s4 z# {/ T if(s1[i]=='\0' && s2[i]=='\0')resu=0;
' P, x* c; |+ J* y( O0 m8 z( A' c# G! h else
7 ]" @" M* G0 y- B+ P* _ resu=s1[i]-s2[i];
+ h1 [+ ^: d7 b% P" [7 o printf(" %s与%s比较结果是%d",s1,s2,resu);# `5 z; Y3 ?. F1 I. s8 m0 m! M
}
& { c4 ^8 @ H" M7.15
8 i( `6 Q& d( S0 f: t1 O#include
0 s2 |* X0 j$ A3 c1 o$ D; }1 bmain()
4 Z/ V5 [" {; Q {' D' e, {% q) k, I+ E/ m
char from[80],to[80];- l* u' R3 e2 N- N+ ]. o
int i;
2 p& J Q4 r5 P }' ]& D* D printf("请输入字符串");
& D* E! y/ v" S5 }5 a: } scanf("%s",from);- P5 o& h" D6 z2 g" G& z
for(i=0;i<=strlen(from);i++)
0 y- M# ?* d. p6 `' ]+ @ to[i]=from[i];
2 x6 D: g' [" M5 b9 q printf("复制字符串为:%s\n",to);& t( F- U- G% [3 Y6 I, d1 x
}' a. L& _' Z8 u% H9 C
# B$ r0 y, A2 M, G3 `* b% K& {- M! }$ G$ _8 G- P) m
第八章 函数5 p/ w) w o9 V( o6 R: U- @
8.1(最小公倍数=u*v/最大公约数.)
7 I; l$ h& @- _6 s N4 nhcf(u,v), a, R, l9 d+ _9 ~% P6 d! F
int u,v;
3 d, N4 j) h' ~) S+ j+ u(int a,b,t,r;6 [; K) S8 M0 v, k" _6 p% H
if(u>v)
( X; ]' h# ?/ B; b) n {t=u;u=v;v=t;}
6 n& W: t& y! P4 K# ^* G% Q a=u;b=v;) h# @9 T& F& V
while((r=b%a)!=0)
7 K2 x) k0 Q0 w! Q( {; k {b=a;a=r;}
* H# ~( ~8 I! I7 o+ i# [2 Q; L return(a);- I9 \8 s' I- I4 V, S
}( `9 M& {+ f* S; x" ^8 c9 ^
lcd(u,v,h); {6 y, `' l N; C5 r9 Q
int u,v,h;- x- b) q2 F, o/ z! B& Z0 r
{int u,v,h,l;: K. \% D S$ M" p9 X1 {; h4 Z
scanf("%d,%d",&u,&v);3 q( t; T/ R! d7 |( i- P
h=hcf(u,v);9 `7 ?9 W5 b$ h# f2 d
printf("H.C.F=%d\n",h);, U& n0 n" m9 o
l=lcd(u,v,h);
$ P, E: b% Z6 J6 c3 c; c" T printf("L.C.d=%d\n",l);
8 X I2 J( |4 k0 i, |+ L& g/ m }
% x+ w1 Y; Q; \/ @ {return(u*v/h);}1 f% S0 F6 [* d9 m y/ }* A. m3 Z
main()
+ i d+ |" N, }) l: T& { {int u,v,h,l;
" T/ q$ |# L3 f scanf("%d,%d",&u,&v);3 ^5 e9 L% p6 N" k
h=hcf(u,v);+ _' T, ?6 I0 `
printf("H.C.F=%d\n",h);4 f, d3 k* [8 N2 A) I
l=lcd(u,v,h);, X2 |5 P# y* L! j; G7 w4 H
printf("L.C.D=%d\n",l);# J6 A2 I( D# L0 {) D8 H
}& T/ u% j( o! {# w! N) U
+ M2 ]7 [1 `8 _/ D) O
- W+ A5 M% w5 w6 L2 U: v* s+ U; c9 I: p
8.2求方程根
$ A, e3 i9 d4 F* I' U o#include
, |- R& r( }& E" Tfloat x1,x2,disc,p,q;# G7 Y8 O: }7 H- p
greater_than_zero(a,b)* V" _/ o8 H8 e( {8 Z- M- U
float a,b;
" G4 R% r1 P# S- D7 V{1 _) H2 e' D- D# x ?' K
x1=(-b+sqrt(disc))/(2*a);0 t! W/ d3 d# N" v
x2=(-b-sqrt(disc))/(2*a);
; I* {. O2 v9 J' s}. e! @5 J4 K* _2 e( C
equal_to_zero(a,b)% V; [) z r- V8 Y8 E( c c: N, {
float a,b;
2 g5 c1 C4 p/ l5 P) |& @{x1=x2=(-b)/(2*a);}
$ X( G* M& `9 ]/ c# |3 ] Jsmaller_than_zero(a,b)
6 I5 l" {( D- A* s, i! X9 Cfloat a,b;
# ~% t( B" h" K* Y" z{p=-b/(2*a);* g g# A7 g& W/ U( t! m' j5 Y5 J
q=sqrt(disc)/(2*a);% p+ e1 u( j# O1 j
}8 m! ^3 t0 \6 S1 W* u Z/ [
main()
5 f- u% x' L$ H9 y{
: _$ T% Z" Q( C2 Rfloat a,b,c;
* ]6 H7 n& o; e) r5 u0 @5 ^- J8 c! k9 wprintf("\n输入方程的系数a,b,c:\n");3 `/ Y4 P: P+ |8 i% B4 R: Z
scanf("%f,%f,%f",&a,&b,&c);
9 w3 @0 [1 i& J$ ]0 z0 w, uprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);
0 C. |% `) U" N; o$ Ydisc=b*b-4*a*c;! O0 \! m6 R* p9 s4 v
printf("方程的解是:\n");
4 ^, n; d4 l% T3 E( cif(disc>0)
5 C$ G$ A- Z- L6 L N* B{great_than_zero(a,b);
, N5 W9 p5 u9 T) i; [printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);" i4 ]& Z8 [" l! L! R0 o1 O& c- G
}
# k& m2 }+ _" I) e. ~else if(disc==0)
; a) r) {* |$ s2 y {
- N0 ]2 } f) [zero(a,b);3 X, l3 h0 P0 W, `) z4 m& d. |
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
3 `! Y8 a$ `' K" h8 ?( E$ q }
$ U& ^# \, X2 q k3 Z9 Qelse
( x( M- L" n R3 B c {
, H0 R. Q M+ D9 g0 ]0 Q3 P small_than_zero(a,b,c);
5 F$ V# c; Z/ k- [# U% ^' ~% S printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);- U8 z9 A4 Y1 `+ C
}
2 B+ w! j' c0 `5 _. q9 x}
1 F/ g6 E O0 O( `( `/ \8.3素数# \9 w6 W" I( p( O8 S5 N0 x
#include"math.h"6 j. A" H6 o9 q1 l6 L
main()
1 H: [: q8 A0 N{int number;6 L! j2 F, o" \, v T/ e
scanf("%d",&number);
; ~* Q5 k8 J' b. Q& D% ~ S if(prime(number))
& j3 ^. A% R; n% l* E( S2 f printf("yes");( O0 @- r+ ~: x: e
else4 M/ f) B0 d7 |* v N
printf("no");
, \* p$ g& z+ [3 E1 ?}& a' v$ Q" Q" [* Q1 Q
int prime(number)) i* T$ }! ^+ y$ p, s n7 C- P
int number;
! r7 B2 Y; A4 D* G7 T{int flag=1,n;
- {: N) U; h1 o( [ for(n=2;n if(number%n==0)6 W: G9 g* |: t" L9 [4 w
flag=0;9 k; I) b; ?" K% [4 ]- r
return(flag);
' W* B+ O. |+ A; c}
% x; K' z4 y: w, C: M: w" F, @. V
& a* _. j% [( z: C5 H) ^3 l
! A# Q% O( z8 @. a, v4 D, z" L8.4
: |% f, z! D) j9 O* }#define N 3# y- ?+ C1 \# k+ }, i' Y
int array[N][N];1 T/ h7 J2 K* B0 B" i/ y3 U
convert(array)
6 B# a- y0 x9 S6 z! l. K0 Jint array[3][3];: k* Q! Y" n. [
{ int i,j,t;
+ K6 |3 l. f; Q+ A& A" v9 L! T for(i=0;i for(j=i+1;j { t=array[i][j];
0 M7 p: E9 W g array[i][j]=array[j][i];
" r" I: N5 O( U4 I array[j][i]=t;5 ]5 p3 R9 N+ }$ M$ ?0 b
}* B( J- G& o- t( G: M
}
8 F) N0 b4 j, h- xmain()! e3 x2 h$ G$ u0 U R6 m
{9 w$ S( e& M g5 Y3 Y P7 J
int i,j;
- E$ Y: @8 U3 J% y6 P printf("输入数组元素:\n");
. d" B A& L2 }& r( |* I for(i=0;i for(j=0;j scanf("%d",&array[i][j];
" b2 ~( \" z) |& ~( b l printf("\n数组是:\n");7 O3 P1 f; t$ @# t7 a1 N$ H
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);2 e# {& U$ B; V k. A6 s
printf("\n");% v& U& l: l* O( Y) [
}3 N( l# W1 O( S
convert(array);
. x2 X. D$ R: M7 F. T; ^" Q printf("转置数组是:\n");9 j# E2 A$ Y* H% A
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
, [2 ?; R( L# B; `+ J8 O printf("\n");
: Z2 a/ | s2 A }4 C3 \" K. E& o' ?9 }2 K1 w' x& i) ~
}5 u/ N Y' J. @7 j0 F
2 ^1 X! [8 ]' P# n9 K, n+ }3 v+ M2 [6 p* D; h
: \, L0 i* o0 B; e% E% p7 O
8.5
2 g4 W) O E5 M7 }6 H2 w0 Lmain()8 S9 G# v5 v% M
{
, p# _. S$ e. b2 kchar str[100];. D6 c/ |% N0 j: ?7 q
printf("输入字符串:\n");- X3 E" |- C" W L2 t& d' W" c! ?
scanf("%s",str);. Y9 J. ^) O! E* ]0 T
inverse(str);
' V% R+ U+ G# u/ S8 U8 A printf("转换后的字符串是: %s\n",str);. v) o! n3 u) S( J& `" ~. v( S
}2 l" P+ Q/ J0 P, T
inverse(str)
8 P( Z% ]% A9 d3 ^/ L$ ochar str[];
3 z, g+ _ N9 O. ?* L) E; J% _{9 }9 B- K1 ~3 b& H5 z
char t;& }% A6 K- K4 d& Q
int i,j;; y3 l* ]* k& O) c8 {
for(i=0,j=strlen(str);i {
4 o4 l! W2 G7 u6 t3 {& B) }6 t0 J t=str[i];
7 t" p3 Z. U% X- G* n str[i]=str[i-1];4 R) z" D) j: Q8 n
str[i-1]=t;
5 ~. t$ [' z4 ^& y- u }: E0 y. ]- b& _! ?
}+ o) x8 ?7 Z' r7 J* W1 n
: J& E4 \/ M n* [. N7 ]8 S+ u
. L* d! j# X" F
) _7 q5 i& d+ ?2 o2 ~. L8.64 f' k& B& d* D4 H
char concatenate(string1,string2,string);! H$ t% K9 {" `% D5 g1 ?; p3 B( C
char string1[],string2[],string[];
. ^5 u. Y7 P+ ^4 Q m0 M{6 }& T: U" I7 R& T. g9 I
int i,j;
% @) u n0 c' J% S+ Ffor(i=0;string1[i]!='\0';i++)1 [! s8 G% Y) U* m
string[i]=string1[i];: z6 l# O3 ^6 h, C
for(j=0;string2[j]!='\0';j++)
+ V3 k& }& I- ]3 {+ y string[i+j]=string2[j];! v/ C" ?$ E! t0 |; W, r6 D' J
string[i+j]='\0';
# ?6 D8 U$ q4 \# C4 B2 ^}
4 A) [8 T1 o2 y1 g' b! [- omain()* M2 s; F" ~7 V- L T1 }' y5 P: W
{/ O0 |1 W7 u( q5 Q- G( `7 B
char s1[100],s2[100],s[100];
! u9 Q+ D# }. C: U. x* y9 s2 D printf("\n输入字符串1:\n");, r" s- ^$ I, h( o! k; Z5 @# x6 d9 Z
scanf("%s",s1);
" m) e) v- z5 [' V4 q( ` printf("输入字符串2:\n"); h% x$ b2 ^. U( [1 [
scanf("%s",s2);
4 ^" ^- ^ ?- j# l8 ? concatenate(s1,s2,s);
% H" ^+ k, r7 Q printf("连接后的字符串:%s\n",s);1 _; c- u' }+ [+ Z
}8 ?; V1 f9 [% w, d3 n" V
+ B' r N. U- M' e& u2 Z3 s7 |0 S# x5 o
8.8. r" ]8 Y. C/ l! P- r1 b: b
main()
! a z( E' K0 Z{
/ Q! j7 b4 T( c char str[80];& y) G, r# R; F. Z1 J/ ~( `
printf("请输入含有四个数字的字符串:\n");
7 g' b, U# Z5 R9 r/ V. E4 e scanf("%s",str);
. S! J: L* L0 Q) u% F2 K( K insert(str);6 ^4 w( j8 Q5 X8 C' Z- Y
}/ Y7 x- x; B6 i# v# g+ W
insert(str): y. l t& K, a3 q) S6 R/ o# Z: @
char str[];0 S o1 z6 H$ d+ P0 [
{7 e* I v! h! d) m2 \2 _
int i; {& _3 E' T3 l& K9 `
for(i=strlen(str);i>0;i--)1 N# [/ U+ \ K# f$ ?" x
{ str[2*i]=str[i];
2 v$ ]0 a9 ^! q7 G) S str[2*i-1]=' ';% y( P7 V: i6 \& N2 z0 e
}
- q: V8 }7 J" Z printf("\n 结果是:\n %s",str);7 V; ^, \( h% K. B) U
}- r( Z0 R0 p) l4 p4 g& d4 y
! U8 ]% f; ~* i# h2 f
5 i+ r) N/ K! z: [0 T. ^" ^+ s" F5 [- F0 T: {% l" x
8.92 R/ n+ I4 l* G- l3 B
#include"math.h"
( F# |4 |( z/ Gint alph,digit,space,others;
3 D( Y8 Z& q* u+ fmain()
) f( Z& F1 t6 y& Z, G9 c% [{char text[80];
|& W1 M) U" j6 c! c* V gets(text);
1 G' o b8 y* H& e alph=0,digit=0,space=0,others=0;- }# ] Z6 z5 ^+ ]
count(text);
# x, J- S6 Z$ h' e( I printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);
- B7 q/ O% r3 Q}
& N/ @& l/ s5 |1 gcount(str)
* @. d5 |: S# j3 p0 }char str[];
( X3 F7 a5 p7 g0 j{int i;. c( z2 C+ K( C# R% L7 j- \: W
for(i=0;str[i]!='\0';i++)
1 f. Y% a) m+ \, Q$ R if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z')). f6 ~5 ]4 K: S1 O
alph++;, l! P- h# v# f" @* t! w2 Q
else if(str[i]>='0'&&str[i]<='9')2 `' N5 N/ ^' V9 v2 C
digit++;* P' I# i, n: y3 V
else if(strcmp(str[i],' ')==0)
3 A9 Q+ A7 V* a space++;
( c, ~# r; z& h1 Z4 p else
. k1 s# Z F4 V6 F others++;
# a6 t- k! u5 D/ }" N8 ~6 ?}+ k P- Y8 j9 k/ w
2 C2 Y B$ ?( n, ^8 D8 t
! x: S9 A W! T2 N4 x" e6 E8.10
; W7 F8 T2 O3 O' i2 r" w7 ~! ]int alphabetic(c);: G/ a1 M+ @) B/ r7 U
char c;; W$ a$ A$ T# w% [
{. M2 S& b3 _( }6 {
if((c>='a' && c<='z'||(c>='A' && c<='Z')). O4 ]* t' Y* a( n" n' k
return(1);# {/ R% L% V' n" c/ t& e5 }
else
' ? E1 Z7 z+ l: I& n M3 p return(0);4 b# S$ Z- S9 |
}- t' I- ^3 A3 `1 A* m5 v1 I8 h
# m$ ]9 H7 F \8 ~int longest (string)8 E! P& b% T3 i6 ]
char string[];6 O, Z- h. v% ^% C0 y
{
1 _" ]. k* t5 ?& `' y int len=0,i,length=0,flag=1,place,point;2 V9 B$ S: a, t2 T, ?
for(i=0;i<=strlen(string);i++)
4 i Y- A5 J3 Z- y* @4 K if(alphabctic(string[i]))3 e* q' T+ \& s* b- P8 ]; k( L' w8 n
if(flag)
: ~+ Z$ p6 B: G4 P; J Y! t {* R3 }! W! R3 @7 N1 I3 W
point=i;+ J5 d. C: h% i! A. U6 I" D6 Z
flag=0;
: @! n7 R% k" n x: S+ u& g( t7 Y }
$ O9 u; V7 g- ^. a$ h2 ` else
4 S0 X- h( e! v7 \ len++;# s6 P2 G9 X$ S- P3 u
else: g0 }1 i f5 b% y0 [0 i
{ flag=1;% l' _+ b8 {7 n0 t* n
if len>length)
8 B" @0 G5 T6 d7 B6 r {length=len;4 n$ f2 [) S* W4 y2 N
place=point;( x0 U/ \ c) w" Q+ k |* X
len=0;! V N' {. g. p: c6 \* v1 d0 c
}
0 K8 L4 M7 \" j. i }
& s" B& u2 ^" Y F S E9 s return(place);
3 l3 F) R+ u6 m. d }
. k3 G# X$ X0 c2 V) Mmain()
! {7 }* t* z' u3 f{- _( o$ R: \9 n/ v h/ L# a7 L
int i;& a- U( t) K" K$ J, _4 ^9 I' O
char line[100];$ n% I7 X% W s7 E0 T
printf("输入一行文本\n");( v! A! m7 L( h. r
gets(line);$ d2 C, S# m6 c
printf("\n最长的单词是:");
3 b" K9 H! e u6 W4 z# gfor(i=longest(line);alphabctic(line[i]);i++)% d/ M( y) i9 \; z( B
printf("%c",line[i];
8 `6 r2 m5 _2 z. l. r* cprintf("\n");
& B" ]8 F! U( W. i- N7 I8 R. ^}) @1 @2 S2 n! c8 W2 M
0 G( `* S/ M8 J8 S3 f
7 S, E# V0 M$ j
; K* w2 } \4 n$ v9 a9 w" [ D8.110 k* r& H/ ^) @+ D% c5 j: }
#include5 {- p1 q) G. T+ X; g N
2 Q# U& X( H. f6 w% [6 V#define N 10' S7 i3 t3 A% Z' w
char str[N];9 K. ^; w8 z) v/ ?' p3 k
main()
! o" D8 d1 E% n{
; f5 V; z0 e; _2 J8 ^4 Pint i,flag;
; J9 n4 ]9 {) O4 x8 X4 Nfor(flag=1;flag==1;)
4 S( i7 l# C( l{$ Q6 q+ H) v& U } q- C
printf("\n输入字符串,长度为10:\n");
0 g# e6 C; t4 v6 G. B5 M7 B6 h scanf("%s",&str);
% o8 S9 r7 l! I) g if(strlen(str)>N)
4 n% s7 b! P4 E* C/ x printf("超过长度,请重输!");8 e3 _6 a! t9 ?0 J# q
else
, q- L/ j3 o! O* p flag=0;
6 {) Z! |: W z" x" k4 {}
B) i; A7 ?+ s4 Nsort(str);
7 @/ G( \7 U, I/ j+ r3 V- Uprintf("\n 排序结果:");
2 i2 M$ a5 ?5 Q) Afor(i=0;i printf("%c",str[i]);5 [! c. n1 u& n1 {
}
) x& Z4 \- h9 z; G! W6 p/ jsort(str)' m* ~4 {, S- R2 K: W
char str[N];; _. s: l T0 f, l8 f, }/ ]
{7 B' F0 e* g. g( j
int i,j;, J( p; \/ \; i$ w- V1 s; m" r
char t;
! M0 }7 l1 }; U/ i) sfor(j=1;j for(i=0;(i if(str[i]>str[i+1])
4 ]6 u% a- p) E1 M- ^8 d { t=str[i];! \0 H$ ~& C3 b# p( b
str[i]=str[i+1]; L' c5 U: U6 e) z
str[i+1]=t;
/ U2 f8 ]+ i, Q1 |5 a }5 m, x0 u0 W! C3 D& B7 @
}1 z6 w }" Y2 J4 S" w0 J/ @3 g O. f
8.12
, V- _6 y8 O9 O) Q#include& r) ^. Q" _8 }; z( {
#include* Q' ^ @- ^* h* m( \5 g
float solut(a,b,c,d)$ m9 A6 }. n+ K3 b& _
float a,b,c,d;
* ?) F% \/ q- q" `) R3 f( S% N* U& i{float x=1,x0,f,f1; w: K' }+ ?. K5 t5 l
do
1 \ h$ J2 |" `: [$ \0 U {x0=x;
" X) n" C; A* I/ ?# |: S f=((a*x0+b)*x0+c)*x0+d;
( F9 p0 B# Q: ?* T$ e$ H7 Y5 x. _ f1=(3*a*x0+2*b)*x0+c;
0 {' G5 i& S+ v! M$ X# G x=x0-f/f1;' y% w4 Q9 N1 q! `+ t7 X
}5 M% ~: R/ @8 B5 u) Z, c
while(fabs(x-x0)>=1e-5);& Q9 V' { U7 Z
return(x);- _6 S2 O& i- y2 O/ x0 B
}
e! w7 F" X J$ gmain()
6 F% T A; B5 |- q( \{float a,b,c,d;$ a! z: R9 x/ h0 _3 S+ ]4 E* r
scanf("%f,%f,%f,%f",&a,&b,&c,&d);
8 }9 g- |4 ]/ h6 g2 z7 d printf("x=%10.7f\n",solut(a,b,c,d)); H# \. `- X0 E* v2 i& n
}
. q) w" M! u' e7 w% `* L7 l- l8.13
& B, I1 f3 ^' S0 @9 D8 p#include r# `; x: z; l. ]0 k E
main()
$ o6 ?3 D0 T' g! s$ ?+ k{int x,n;
+ C2 j, g0 o7 c4 K float p();6 Q' B; F1 R. `* }+ Y0 M
scanf("%d,%d",&n,&x);- u$ a" u( s2 F* @; `! S
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
1 G% K. c2 d) _" W: l}
" D3 a/ B0 C: d8 efloat p(tn,tx)6 }$ x5 ~1 u/ p+ U7 J$ ?' g" V2 T
int tn,tx;
5 K7 ?, s6 T) D; n, \+ }{if(tn==0)
8 }, E$ {& `6 C( x8 i R return(1);
5 ^! E5 E4 |* u* |1 s5 g else if(tn==1)
- O9 C9 Z, _) y( c- @ return(tx);0 a/ s6 q" b6 V% c5 I( p
else
/ h$ k' E* H! O; l5 \5 ~' j6 H return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
) v8 N! d* X& e; s7 w4 F. g" q}
8 [/ P' e) K3 \! C# m5 w5 F8.14' M6 {) O5 n% N' ~3 `% B
#include "stdio.h"
9 i5 ^+ {$ `5 j% `#define N 10+ A9 q9 }4 k- [9 V* [9 i
#define M 5- l* y& z" V& X% V8 O
float score[N][M];8 y0 L9 v4 L# P* s6 O
float a_stu[N],a_cor[M];, t; [- Y8 i- g+ W7 X6 v
main()
% y7 i6 K0 N* y( f8 [ f% y0 V{int i,j,r,c;
1 a1 G. e0 {& s4 u4 Q* Q3 U float h;- K5 ?( r6 n& }# E
float s_diff();0 }* Y. @8 V! P. b
float highest();
3 x" d$ o; A. U1 S6 t; b r=0;1 ^7 k: O7 S$ l0 ~2 m: J: j
c=1;
* S+ g. w/ [0 t" J7 l2 n' x input_stu();; x6 ~) e3 Z4 h4 }
avr_stu();4 x$ Q/ I1 L9 \
avr_cor();
: B: W5 y3 {1 n& f printf("\n number class 1 2 3 4 5 avr");
. W& \ J$ w8 u for(i=0;i {printf("\nNO%2d",i+1);
8 b2 A! W$ P2 j& f* ?7 e; ]6 \9 t for(j=0;j printf("%8.2f",score[i][j]);
6 \! c# m. @# o3 b printf("%8.2f",a_stu[i]);
& r. s* B4 o5 Z/ y; n, h8 ? }. y; [ s* N7 c- P4 l; r
printf("\nclassavr");
7 Z9 G c! g' q1 x. z- i for(j=0;j printf("%8.2f",a_cor[j]);
7 m4 t* \1 K B0 A! Q& l8 x: H% }1 [ h=highest(&r,&c);
/ y" |; @. T' c printf("\n\n%8.2f %d %d\n",h,r,c);$ H" {& O2 E. D" j6 E
printf("\n %8.2f\n",s_diff());) c8 s1 l5 t! @' }$ t
}/ J$ c# J) r8 m. C1 d* r
input_stu()
) c: x$ N2 p' K/ Z1 n8 u# a- k! ?{int i,j;
' p" X; C9 N) t" F* v float x;
/ ~" Z6 \1 H# d: x' j for(i=0;i {for(j=0;j {scanf("%f",&x);
& U; s: X4 ]& o/ \" u' T score[i][j]=x;7 g, v7 o8 D' K6 u. ~" F* V0 g6 _9 r
}
5 Y5 p& u6 R! y9 L/ S" K* X }
4 l4 G$ R% n0 X9 Y7 u q}# F: d1 L w% o
avr_stu()7 V9 J( _9 G7 Z8 C, U
{int i,j;
' X4 g- G* i, L float s;
- q( r& o: W: Y" n Y2 h% A for(i=0;i {for(j=0,s=0;j s+=score[i][j];
% X' X% `: |* b* M; `8 I7 d a_stu[i]=s/5.0;
$ P' u) V' A3 X+ X" c* N }' l9 [# Z' n) h
}- w+ f6 M/ L4 z" }. Z7 C
avr_cor()/ a8 E% F6 |8 b
{int i,j;4 g8 _$ G9 }- d/ R! S5 j, X
float s;
$ t$ D4 E* ^8 G/ ? for(j=0;j {for(i=0,s=0;i s+=score[i][j];5 J* y, Z2 ~* z! |5 l
a_cor[j]=s/(float)N;
" u: }6 m6 `1 ]- W }: |( I' y% D5 Z
}
, _) C: u7 c9 U5 r7 ? W% Q. Gfloat highest(r,c)# j* X+ V0 ~, d9 y8 g) Q7 S, V
int *r,*c;9 O( m8 T D1 X
{float high;# t& k3 W/ R' o# e
int i,j;
4 i6 N# v! [2 W% Q3 g; x- Z* Q/ M high=score[0][0];
0 u( l5 ]) B9 t2 o for(i=0;i for(j=0;j if(score[i][j]>high)
9 r. j+ i- S! z, T# o; E {high=score[i][j];" H3 F3 q' x [5 {4 V8 i( O
*r=i+1;
: K' U0 H& W% f# E8 e3 k *c=j+1;
* n# Y% l* C$ P }1 t. H. s& S4 }( k
return(high);
# n _/ h* N! s" s; P, W8 b}
# U" n) H( s# l8 d- P, o/ S5 L4 _float s_diff()% f' I( H8 U/ b
{int i,j;
5 X6 n5 D9 x9 h% ] float sumx=0.0,sumxn=0.0;
7 o5 Y7 K7 W. P0 s for(i=0;i {sumx+=a_stu[i]*a_stu[i];
" T' N+ [7 d) F c sumxn+=a_stu[i];+ T# l/ k; K. D% g2 R' L& I. }
}
7 Z3 G8 f# \8 S# a0 z return(sumx/N-(sumxn/N)*(sumxn/N));
- R8 l3 h+ y& m( ?. @( V}1 Y. ?' D8 j( |! B+ m4 w" U
8.153 }2 o6 n) u8 _8 [. [/ h, g+ o8 e
#include4 H# K N/ }( j$ p. r1 ]
#define N 10& _/ H% E7 h, F
void input_e(num,name)
2 |- [6 `- B) _5 [. c* iint num[];
: j/ m) n" }( O- ~% \. Ychar name[N][8];
1 _$ k0 Q( {( u* a1 y3 E7 J{int i;
, P% z0 z4 N% G k4 e3 v; q for(i=0;i {scanf("%d",&num[i]);
0 g4 F# [ Q! s0 ?9 M gets(name[i]);
$ P. ^; |9 @- f( O1 S }
6 |4 `. _9 B q$ t4 k}
+ a0 p( l9 G$ g3 F$ O u6 n, tvoid sort(num,name)$ d& D9 g3 Z) C2 U
int num[];% s) [4 @) R' i( ^; B" Z/ L, |/ \
char name[N][8];% }0 E. h9 d! w+ O: d6 L4 @
{int i,j,min,temp1;. D3 D8 _0 R( R. P- ?# K1 i
char temp2[8];9 s$ b* [/ G9 g1 D
for(i=0;i {min=i;
_& x5 w6 ?8 K* C for(j=i;j if(num[min]>num[j])min=j;
+ J, [" {5 T$ f } temp1=num[i];
& D7 [. J$ l: C* a | R num[i]=num[min];
5 B3 B/ @9 D9 u: }8 U( [# g num[min]=temp1;2 }; p0 \! g4 Q% r% Z
strcpy(temp2,name[i]);$ d! L+ _0 F7 [: o6 e5 c7 ?' ?
strcpy(name[i],name[min]);) t, o$ q1 L8 P. l2 B- |
strcpy(name[min],temp2);3 s Q6 V0 W+ @( x! a' S! v
}
k! n! r: m) I, z; v for(i=0;i printf("\n%5d%10s",num[i],name[i]);
q3 \5 }2 Q" C+ g) q) `}
7 T/ f5 u& W3 r4 J- tvoid search(n,num,name)
# G6 {6 P3 Y0 }9 y8 `2 P2 Aint n,num[];3 x6 G: R7 i% l6 N) [7 c
char name[N][8];
9 l3 _2 I* x* E' e4 {$ ^{int top,bott,min,loca;
2 a7 F( c9 s e. d. @1 D) M& e loca=0;8 r' d8 V" G$ ~! G5 d$ K
top=0;% I t% s" b5 x m/ M
bott=N-1;
$ r) L6 M ^! C4 S3 K2 ] if((nnum[N-1]))$ A1 Z3 t: E5 `! \, U9 R. w
loca=-1;
; w! F a) [5 [, F" m5 I& ], q while((loca==0)&&(top<=bott))( V8 C% F8 d: E
{min=(bott+top)/2; Z) T" q: J4 }, X. ]5 E
if(n==num[min]); i6 D3 v' Z8 F$ S
{loca=min;
8 `( U/ f4 t! j9 C+ m( w/ M printf("number=%d,name=%s\n",n,name[loca]);
. b a9 a6 Y4 x2 z" N' c2 J& L }
3 G6 D- s( j4 P# }5 g/ `8 B+ a else if(n bott=min-1;
* R% G9 W2 w, g& A1 d else
3 V9 f2 O! T$ l0 z4 A u) \ top=min+1;
( L# j7 R2 W" r) @1 Z$ s! ]5 F }1 ], h6 G% M4 w) d8 M$ C+ \2 a( c( x
if(loca==0||loca==-1)
: C! O7 G" x1 h# D printf("number=%d is not in table\n",n);% P# {! h( u1 {7 i! s, x3 N
}
; x Q- y0 a p0 G9 c" nmain()
! ]" b& X3 X2 i8 ?" [{int num[N],number,flag,c,n;
4 t+ f1 u6 S/ f char name[N][8];% i7 H, m' P- H( J4 ]
input_e(num,name);* v" h5 W3 p( ~) F5 H8 z
sort(num,name);- P: ]2 C& @' z& [
for(flag=1;flag;)4 w+ ?% P6 q" `/ g
{scanf("%d",&number);# i3 @( K0 ? i( m( l
search(number,num,name);3 u, a7 w) n1 {
printf("continue?Y/N!");
# e7 T, v- V1 N c=getchar();9 _5 P9 T) g! z4 E, R
if(c=='N'||c=='n')& U7 o- n8 A: h, u2 z+ A1 H+ E
flag=0;
( e8 e8 L& g& M5 `5 _ }
* T( `8 [6 I* `, x, i8 W( v# W9 E$ _}* H& l* E$ @; H5 V; `; U
* U1 y& Z( ~+ {& o O9 T8.16
3 K5 |( J2 A3 r% Q8 Z" ]# m2 T#include
: L0 m( a( H/ x8 I#define MAX 1000
- q! r) d8 ?* G* }' mmain(): x: @) L; f$ r
{ int c,i,flag,flag1;: [, J5 i4 R% v" o) b
char t[MAX];
, S6 g3 n" u4 S" S# x" Z i=0;. `6 L2 h# I- p0 l0 d* w/ x
flag=0;
y* Y; U/ Q$ A' L flag1=1;, a7 Q3 x& j6 |3 x& }
printf("\n输入十六进制数:");
( T2 J( s0 g& B, h( X while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')9 u' k( d4 P- l' k8 I# P2 [1 S. q
{flag=1;2 y8 B2 F( F. f4 v( E3 K2 G
t[i++]=c;
1 d( n8 V: J4 h' [" @$ N P } s- L8 k9 F* ~
else if(flag)
& o2 d- k. Q& E: F4 v# b" F8 P {
: _+ ]3 `; s/ K& z: p7 I- h! _. G% i t[i]='\0';7 F* g( H: i8 y3 C
printf("\n 十进制数%d\n",htoi(t));
$ @# [5 y8 E9 N printf("继续吗?");% i) `+ V/ g" u4 ?0 i, t) X( q
c=getchar();$ q3 o) g6 I. v& u2 c, a3 j( ]' R
if(c=='N'||c=='n')9 W% |/ M: o8 u% I: b
flag1=0;% ~% T- P, w" j
else( a& |' O. q" Z
{flag=0;
: M* I# L; j% U4 z) m5 } i=0;0 h4 `& V9 @5 a! n- i2 m
printf("\n 输入十六进制数:");
# A2 Z* |& ~5 x6 d: F& Q5 {# r4 ? }1 b7 R6 M( @3 k$ i
}2 p: A0 z; J. A5 O3 U
}3 v' ]- ], b) e: e/ C+ b4 p
}
' S# ]. W' `' `: F/ v9 m) rhtoi(s)
4 Y& B, \" \8 e3 a, {" m3 xchar s[];
+ ]8 |' G6 O3 n3 m* A: {{ int i,n;
6 p' g& M7 m3 V, [ n=0;
( J. }4 s% I l for(i=0;s[i]!='\0';i++)/ V7 m! d. ? w8 c8 T
{if(s[i]>='0'&&s[i]<='9')
9 q$ u4 D B" L n=n*16+s[i]-'0';' L2 A6 O) P8 f2 k! K O/ I O6 ]+ U
if(s[i]>='a'&&s[i]<='f')$ g2 O) C( B) Y" T1 H2 r! K- l
n=n*16+s[i]-'a'+10;
2 X+ r: K, ~& T/ D# W if(s[i]>='A'&&s[i]<='F')
5 O( N* Y2 T- U; }$ r n=n*16+s[i]-'A'+10;
) `8 r/ e0 ~; [4 }* S }& i9 x& }5 ]" {) p- N, L; |8 @! Q
return(n); n" _+ s! E1 b2 p* E8 P+ T4 ~$ E( ~
}
% q& [, c% i5 P( F+ `
% C! z3 K3 K. K( C
: ]0 V2 m9 p. m9 }$ T
. }8 _ E/ d' b4 w' C: H) B8.17/ ?5 a4 E2 G" h" d
#include
8 Q& I6 k7 }7 g+ q3 ivoid counvert(n)
$ U5 E$ Q% j6 A' Pint n;& W2 }/ z6 I8 x u4 M" z* ?: a
{ int i;
0 c5 K/ W! l! Y2 ~; i if((i=n/10)!=0)
7 ~1 i4 U: o6 n. p% a convert(i);
0 ^+ `' V. K5 P' r4 b, y putchar(n%10+'0');1 X0 b3 {2 z0 i' I' ?( P3 C$ Y ?* U
}
$ r G8 i! y4 J0 ^2 f6 U6 ^' n5 k8 xmain()
: t/ O M2 g4 J{ int number;
! i; [! U/ ]% l9 s# p, v# z% ]. ? printf("\n 输入整数:");3 a' V" I2 m* N2 u" n% \$ k9 p
scanf("%d",&number);: R( t# {# `! n. f Q6 u( g: P
printf("\n 输出是: ");
& t/ n9 k ~1 b. M. z# ~6 b. h if(number<0)
7 y. t8 Q# ^4 [7 p { putchar('-');
; V0 X" |% ~2 a number=-number;9 G x7 r. c7 V7 z* }1 Q0 U
}
9 v% S) u0 ^+ E/ m; `% ?$ sconvert(number);
3 N- r+ ^) V% o, k. q}- U: O; j$ `" z' D% I
# p7 \) w4 b3 w; ]$ L
7 s) F, k, N2 n! E& K( r! z# Y
. X$ j# {, r% J
8.186 i5 B4 F! G2 a' d# t8 ^# \
main(). X4 w& Y- O0 r( n U% e. _3 k
{9 x* c. X& b$ c3 j% v
int year,month,day;" J3 a z; W4 F; D
int days;2 g8 G) B2 }+ N' |6 H K: `; @
printf("\n 请输入日期(年,月,日)\n");
0 q9 D( H9 T$ Y9 f( c2 s* q$ ^% I4 i scanf("%d,%d,%d",&year,&month,&day);
3 V' s! b/ S5 a2 @# [ printf("\n %d年%d月%d日",year,month,day);- F) I0 c& n/ [% \
days=sum_day(month,day);
: B' s& M/ Q; o/ x2 G if(leap(year)&&month>=3)
# \) o4 P$ a" o" {4 J days=days+1;( L0 f* r5 g- \; k0 y
printf("是该年的%d天.\n",days);; q; T( X/ |7 y" Z
}
0 n. P! z) i0 L/ b/ M3 I static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
8 A/ P: }0 R; X int(sum_day(month,day)
5 N* m1 a5 R3 R6 j! D int month,day;' }5 ?; r4 k& S; } B. g+ Y
{
s3 R! e* ?3 o: h; S& |0 X3 J int i;: z4 W" Z' s1 y2 C$ K
for(i=1;i day+=day_tab[i];' c& g3 _0 G3 P* E; A. U
return(day);
% O4 H0 }. x: s; Z @% A1 X0 l$ L }+ G1 C% b+ X* x' p/ v6 ~
int leap(year). t/ W, a6 E) U4 N ~9 d
int year;( R8 \6 y& r; J. H4 ?8 Q) f1 l3 m
{
7 l ?$ l" c( u/ F( R int leap;, k, A' z# s' Y4 U: C8 @
leap=year%4==0&&year%100!=0||year%400==0;9 _6 o0 h- N- b. F
return(leap);, A) S. B2 a9 o, n6 F0 ]4 S
}
8 H! s' V. E- U' [1 F第九章 编译预处理
5 r7 w5 S6 b" E+ C$ j% Q2 K9.1$ l4 t2 i7 r; W9 w% t
#define SWAP(a,b) t=b;b=a;a=t+ I ]1 b' {9 n/ A8 y& A* a
main()
1 W3 O) _& ~0 |1 n9 G{5 o% e: ]* |8 T8 U+ G( @: Y) ^: A
int a,b,t;! _& h$ W1 k( y6 _; y' A
printf("请输入两个整数 a,b:");5 ~/ M+ X! Q6 w$ K8 Q/ t! d) y; P
scanf("%d,%d",&a,&b);
% M' ~0 k# G* }" w% I7 f# p) Q' ]SWAP(a,b);- Q# D' M- H* M$ X
printf("交换结果为:a=%d,b=%d\n",a,b);) E1 H* Q) e& B( a0 V# F7 P, c
}
7 \8 O3 `; V' ]& y. g
0 x2 d+ `3 Z7 }/ Y. |
x; W- y) g3 {6 P/ B6 B# j! }9.2- V c) S- g3 ]. u' g
#define SURPLUS(a,b) ((a)%(b))9 G# Z" \, V5 Q+ Y$ q
main()
0 j# i0 e4 X m! u$ k: x6 b {
% L" ~- j& n) ~$ q" h) `" E, H int a,b;" E( P( s- r( K5 z* k: { {- }, Q
printf(" 请输入两个整数 a,b:");
2 b0 {6 z6 N, G; [( x: C; X5 T scanf("%d,%d",&a,&b);6 F* X+ ^& d+ n$ U) P
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));
% V, J" N( L; Y& k# z0 A7 ^- M3 u }8 k. @ G& m) q) y/ i( X1 r
?3 h6 D' d. q
0 E; k8 S& L7 s
9.3
, J& g/ V# h9 L5 ]- T0 s#include6 g1 u5 t5 U# o- K( g! s
#defin S(a,b,c) ((a+b+c)/2)8 ^! @' r% M% K3 k" }
#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-
, G5 \ H% h3 r$ m7 o, _c)))
) R3 [! v' \/ v$ v6 r( X# `main()
( i3 f5 P7 p- Q# {9 m {
2 C8 H5 [+ \( H' y) B! B: D float a,b,c;; T0 ^: T1 l# ~
printf("请输入三角形的三条边:");" s" \9 @& }) q# k9 P( t0 } J
scanf("%f,%f,%f",&a,&b,&c);
9 u6 d# }4 O+ z: u2 f if(a+b>c && a+c>b && b+c>a)& Z% _( C1 O- K+ c: e
printf("其面积为:%8.2f.\n",AREA(a,b,c));; F. |% i; U' \( n( M/ K+ r {. r
else6 m: z8 C8 E3 t) j' `5 V. C
printf("不能构成三角形!");4 |* |2 c8 b, K3 }9 y
}- l. Z' Y: p5 A( t
# M1 f K+ d. m6 Y1 n; r$ O }" {) O4 M3 x+ T; _
0 d( o" O7 @- d. ] f" G, Z0 \6 @9.4. ~, f4 A0 x6 Q+ R* A2 {3 H6 O
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)
1 E/ [6 k/ P) ~main(). Z8 q8 ]% N; U
{$ \. S# K% @7 s
int year;( K: r+ I" K$ o2 M- X- ?
printf("\n请输入某一年:");; w9 @7 `! o: A6 y8 v
scanf("%d",&year);
_' G' o) K+ N% c if(LEAP_YEAR(year))4 i7 c! P( l6 u4 n1 s
printf("%d 是闰年.\n",year);
% a9 U4 j! Q) Q* {: ]6 V/ z5 u* c2 N. | else
: t9 V( s5 \. H) I5 ]7 W printf("%d 不是闰年.\n",year);
4 ]" c+ h5 e! w }
) j) z, w2 l# @; {; H; D4 O. h) W, K# w" y* _- D
7 ]- \- r6 W" @" K7 P1 K
5 o, |" j7 y7 M9 E) }) y9.5解:展开后:
6 `" T" t& _0 s# p3 fprintf("&#118alue=%format\t",x);
- h0 N) C, Q3 ~& N4 mprintf("&#118alue=%format\t",x);putchar('\n');0 Q6 w$ b3 \, |/ y
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');% k/ K2 l6 {2 Z8 J$ R3 c0 ~$ \& y
输出结果:
- e# q- ^/ c5 b& r. v4 ~, A&#118alue=5.000000ormat &#118alue=5.000000ormat) d& w& O" r2 f- H( G/ K6 u. {
&#118alue=3.000000ormat &#118alue=8.000000ormat
9 p3 T4 H! S' j) ^& H0 U( z
0 p& i2 v6 I! s- a! A x# g. C: p2 ]" P# X0 S0 H/ e( f
9.8
# n4 ^7 ?) G6 {/ O' w' W4 Amain()- A3 |3 B: }5 Z/ C
{8 ` V! i# K8 {4 B
int a,b,c;# p, W# Y6 I4 b6 V2 B8 R
printf("请输入三个整数:");- C, m0 L+ b4 ~5 O* S+ O8 N
scanf("%d,%d,%d",&a,&b,&c);
" n7 A4 R9 q2 w7 U P8 l printf("三个之中最大值为:%d\n",max(a,b,c));2 p: t. T: F: n4 I/ T
}. U3 T/ E4 L, [! b& w- s6 O
max(x,y,z)
1 E" i, A# X4 h' t( R" c* n& A int x,y,z;$ a& A& L2 y( R; d6 R
{
3 J1 U2 m- x$ p* \ int t;9 v8 J; C/ l" b. k3 f! Z }8 `
t=(x>y? x:y);
! `1 k _4 X5 Q, T) i/ P v) h return(t>z? t:z);
( ?6 _9 U) b, l; g, C" R }* b% s7 ~( \* ?; e r `" m
* j" s I/ \/ a
# M/ q/ b! \) S
& f8 P9 Z' J" Z( z9.10' j5 X$ W- m- U) b& ~
#include/ l0 g1 U! f3 X* {: L9 j* l
#define MAX 80
3 d5 s. w6 q: c; Y( r#define CHANGE 14 \' R+ r; {$ A2 |7 S& r4 b& d; F; `, ?
main()
( c4 m i) C5 [4 ?- a) C* w* V. ^{
* ]; _8 b. x7 W' n: k char str[MAX];7 u5 q' [+ ?8 Z! [' p0 k# c3 y
int i;/ ^- B9 c% F% o/ n
printf("请输入文本行:\n");; {7 l# w& }$ r8 {
scanf("%s",str);& w0 q: i& c; c& s, L
#if(CHANGE)) W, [5 R/ w$ ~, v# S
{" [% z8 I, s. t& ]( i" s- O
for (i=0;i {9 Z9 [/ N7 B C. k5 J! Y
if(str[i]!='\0'3 E$ p4 b. m7 N1 B" J2 k
if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')+ c; x$ }1 g" u% v. c$ V; t. T( O
str[i]+=1;- u* O' w, i8 F! F3 ]; V
else if(str[i]=='z' || str[i]=='Z')
5 z& l+ P6 @/ s* G$ H. Y9 C! D str[i]-=25;
* w" j% P6 q& j2 ?- h! B }$ k' W+ x' w& v8 k7 b
}0 y2 z, ^+ z0 i/ C9 a5 r; W
#endif
' ^ z/ T$ i0 q) L5 Uprintf("输出电码为:\n%s",str);7 b b8 U4 G& q v3 [
}
( _+ X4 _5 |! i第十章 指针
" ]: O- ?" R# Z) m10.1
6 |' i3 a) V$ ~! S% W4 lmain()7 j9 r) U5 ]6 M
{int n1,n2,n3;
: ?5 x* A" v& S; Y- v: M3 y int *p1,*p2,*p3;
/ h/ D: M0 E7 @% t/ D8 r scanf("%d,%d,%d",&n1,&n2,&n3);/ o! L" h* x6 B( |' Y1 ]
p1=&n1;
$ ]5 B1 l( R8 a0 N0 T' U p2=&n2;' U+ d: ~, D; g [7 C, P5 k
p3=&n3;* ?, f* ~6 {2 i& U
if(n1>n2)swap(p1,p2);
3 J6 L" v% A2 c3 S if(n1>n3)swap(p1,p3);3 ~, j/ [. N# x" L1 c8 ?3 M, e
if(n2>n3)swap(p2,p3);; A) E3 h4 i/ {
printf("%d,%d,%d\n",n1,n2,n3);6 X: F( z; s. N: F3 M1 U' _
}# d6 n# e( `( Z6 z4 v% p
swap(p1,p2)
* T/ d- t/ q& Tint *p1,*p2;3 }- H+ N2 o, i3 K7 _5 D; n3 W
{int p;
/ s7 G& ^; L3 k* q2 s: N2 p p=*p1;*p1=*p2;*p2=p;0 S) P! H, i! ]0 ~6 i* y4 O" N4 Z
}
! t; X+ [3 ?; m- R- g10.23 @( N p9 b0 b7 Z, T
main()' E! x: J. P" E; e( N X) g
{char *str1[20],*str2[20],*str3[20];% i* Q- f$ g0 N- F' Q/ c2 |
char swap();8 ]$ i) s( q, x: U& L5 X' F5 C
scanf("%s",str1);8 O* x) E3 U) U! ]8 m3 z# V& S3 X
scanf("%s",str2);' c+ u$ v8 D! r4 D6 g
scanf("%s",str3);
! J Z5 @/ o3 _ if(strcmp(str1,str2)>0)swap(str1,str2);1 ]' S$ V# w$ G, C
if(strcmp(str1,str3)>0)swap(str1,str3);
. d& u" @. Q8 t6 V if(strcmp(str2,str3)>0)swap(str2,str3);
6 g2 t! \1 w# W8 i# e1 c printf("%s\n%s\n%s\n",str1,str2,str3);
1 W! f- K( u: Z) v# L! \}
2 Q6 o- g# ?6 r6 z( ochar swap(p1,p2)
: D' g: P+ V3 l, J% B0 k6 M; }char *p1,*p2;
: M: i1 Q& C# ^ e# z, U{char *p[20];2 X9 i0 B1 R8 V9 I& ^! k
strcpy(p,p1);# N$ N* L8 I1 J. k- t0 y
strcpy(p1,p2);( w; |2 S, h$ L; ^, x8 Q
strcpy(p2,p);
3 d4 V* \1 {6 c9 [9 M}/ u. ^0 Z9 I( A/ U: p) ?7 p2 j ]+ w- P
10.3
1 @( _6 I. g. j) j/ ~0 V. x$ Jmain()
. C% _0 }/ Q/ k8 L, F3 V{int number[10];2 Z- ]3 t. t% J5 J
input(number);
' O5 c) k# Q, I max_min_&#118alue(number);/ K7 o0 F2 \$ v9 G! p9 q
output(number);9 T, M2 O7 C. Z3 a4 b0 ]- {, H
}1 E9 s, X4 m3 W) x9 j
input(number)1 R4 e+ {7 U8 N) Z" f
int number[10];2 s: Z4 K( j, X+ R" A
{int i;
$ {4 G8 \! H: }- f6 F for(i=0;i<10;i++)
) z. u l# ~1 _) {8 q O, c scanf("%d",&number[i]);
9 I$ [4 X1 U' a! L}
, m" D4 j+ r1 g$ F+ k/ lmax_min_&#118alue(number)
1 b! @1 E$ A2 b/ ]6 m9 ~2 j' N7 a& iint number[10];
! x% A2 H; u- c* D* A{int *max,*min;
: x9 O Q# u/ B int *p,*end;
1 [' L5 v, `& ?! Y end=number+10;
2 R. U' Z. ~4 l# E+ ]& f# V max=min=number;
$ T F8 k5 v% S7 U for(p=number+1;p if(*p>*max)max=p;
0 p2 s- p- G. A. c9 x" X: P" i else if(*p<*min)min=p;
+ p5 K4 H- |- I, v) j *p=number[0];( s J b7 Y5 y* Q9 F8 S
number[0]=*min;- P3 d& p5 E! B' ~
*min=*p;! E7 t% X7 y% W4 P: W% k
*p=number[9];
/ p0 _# s$ A' f* I$ Y$ G% W" L number[9]=*max;
. _+ Q. V4 t' w8 j *max=*p;
8 }' m) j, O3 G2 A return;
2 q. z' Q! v& Y7 e1 l6 X9 A}
5 r) p8 K" F( Doutput(number)
4 A6 e5 Y/ x. _$ P" h C( h' [int number[10];/ O! d9 D a5 D, Z w. [
{int *p;
0 I4 p( I2 l1 X0 Q* D for(p=number;p printf("%d,",*p);
$ G) a: M1 Z5 J }0 M1 O printf("%d\n",*p);
& e0 q2 _9 O: ~( L}
9 J s6 _1 T' T- p10.4
/ C$ K4 e& [5 Z0 ~main()
. l! U+ q5 S+ p, `5 h/ V& G \+ s: U{int number[20],n,m,i;
! F m; I. _& a/ T+ N. V scanf("%d",&n);
1 }. m$ ^" Y3 A8 t7 H scanf("%d",&m);
1 ~$ \2 N1 V0 S, O9 j! ]8 N, A/ i; y for(i=0;i scanf("%d",&number[i]);$ f2 C% u1 j1 O0 G+ `/ ^3 L
move(number,n,m);
& Y$ S& K. e1 K4 A for(i=0;i printf("%8d",number[i]);
5 W3 N8 X. k8 k& F6 f}
/ g; L) |: Y1 | e& xmove(array,n,m)
; [! |' Y2 q* zint array[20],n,m;, m9 B; e! Q$ r7 X" S+ j
{int *p,end;
/ [/ M) R q8 { end=*(array+n-1);, M3 l: j: _+ a o
for(p=array+n-1;p>array;p--)
2 b1 V! x; s F$ `& n$ i* | *p=*(p-1);
' V8 s* @$ C) p0 ?7 ^ *array=end;: i% {# r4 F2 {* o( t1 d1 K9 Q" S
m--;& ^# r7 A; [) k
if(m>0)move(array,n,m);
. y& M L, D y}
% M8 {7 m8 ?# v10.5
2 ]" j% j' H2 e# v9 F: z, t#define nmax 50
! v5 d0 A7 f5 u6 F* F# Ymain()
) y4 @" d* _8 T6 \$ p{int i,k,m,n,num[nmax],*p;2 t! s# }8 m/ d n
scanf("%d",&n);" ]3 o; E7 m; Z
p=num;
- w" F9 X8 n' s, B6 p! d( `% a for(i=0;i *(p+i)=i+1;. ]1 J( m. }" w5 C
i=k=m=0;
2 _ ]5 A' y5 P while(m {if(*(p+i)!=0)k++;0 a5 b+ k% b: M% I# f8 {& B
if(k==3)+ W' v# J4 x+ g9 F. S
{*(p+i)=0;
+ i* @6 x/ q3 `0 l8 ` k=0;' {: w2 N# x9 j2 m9 C5 X3 {( a& _
m++;- R3 [$ h/ N' o( m/ }! ~5 a
}' y e3 U; H# {* A0 B% _8 W* Z
i++;
% i; u( c- S! {1 X- W if(i==n)i=0;
. y& L7 ]8 X) R- Z. F8 w }+ b: |, E% t) `% a' P) p3 u! e5 w
while(*p==0)p++;# M! g( O* F9 E. P$ D
printf("%d",*p);
/ o( Y" w7 W8 q: s}9 L; o8 b4 ^+ C: G
10.6/ C- J$ K3 d( U. q; v9 I
main()
5 V% M. L: n' v( ~* C& b{int len;
, E0 ]* E5 I5 V- L; p; n8 g char *str[20];
6 c8 j- `- x' T2 B( y6 B; x+ }3 A scanf("%s",str); G$ F$ ~0 h3 B: A z
len=length(str);
6 Y1 m6 m, o& e- M! J3 v+ s printf("\nlen=%d\n",len);( |+ `& [" b9 {8 N8 h* d4 y
}$ R) t+ i' \" J' J* D
length(p), C8 [9 @/ {; }& L
char *p;
# T/ P6 }" Q; k( f; Q$ D0 m{int n=0;1 W. z+ q3 d+ a5 j9 X6 {
while(*p!='\0')* `& Z; v ]$ z. n& ?6 w. [) N# M
{n++;p++;}
/ ?( ? k o" X4 o: p6 O O return(n);7 P+ c4 P" y2 l& ]1 i' K
}
7 j* }" w" f1 c" T: c$ S10.7
& N8 T5 J+ F2 l5 X% ^main()
2 M9 ?' n) O" {) k& y5 }{int m; b6 L6 X s5 f6 k& @- N y
char *str1[20],*str2[20];& W0 ~$ ]; }, P' u0 e
scanf("%s",str1);1 l: m8 q8 H! A- z8 N
scanf("%d",&m);4 B( p5 e6 S- `' x& h8 X* t% h
if(strlen(str1) printf("error");
& Y! N% f9 z% i else; \7 S( g: }8 O% K; Z6 ^' T! q
{copystr(str1,str2,m);
1 c4 w0 O3 O5 ^# b, M printf("%s",str2);, S! D$ u4 X1 i* f
}
; @: W" {2 R8 o% p) _+ g}
3 e- C: Z! q& N$ n) m$ Vcopystr(p1,p2,m)
6 Y4 c! B8 N7 s) _2 R! F7 ochar *p1,*p2;
, G9 q) G `3 T* Y3 Zint m;
! T2 W$ K& w" [1 e{int n=0;" z9 J: K4 C9 _. N, D( a0 v# g
while(n {n++;p1++;}$ U4 o/ y6 s* b$ i# M% L. l
while(*p1!='\0')) p/ K" k! X, W# P, O# \
{*p2=*p1; u0 y J/ F( u
p1++;
1 X* R/ v* G. ]( `+ `( t/ K3 p: n p2++;
+ v4 t+ E" [: u- N. d" x8 X; J6 _& Z0 p }
. B& }7 |8 Y. a) g0 O. @ *p2='\0';" a% h" U0 Y' c4 A4 B
}
* b. W3 l0 f3 ?8 H$ ]& t D10.88 k5 f% S( L4 Z5 K/ i7 u
#include"stdio.h"( F; X, |* e8 `6 _
main()' M6 _7 Y: g4 R! T: e( m
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
+ s& Y7 S# ` B d4 E- ~ char *p,s[20];2 E9 r/ X+ w( Y
for(i=0;i<20;i++)s[i]=0;% |: C+ t: c/ m8 w, P
i=0;( r0 q; k4 Y! l* J# }+ Z# h
while((s[i]=getchar())!='\n')i++;- V" k) F8 s& m
p=s;+ X3 O4 @: _- C( I
while(*p!='\n')
7 X( N4 [3 c5 X {if(*p>='a'&&*p<='z')1 T/ `6 [: e8 J
++sle;
: S4 |1 U# X1 Z6 r3 g4 w/ c6 t0 S5 c else if(*p>='A'&&*p<='Z')
6 J0 D1 R/ G! Z$ D+ J ++cle;
% Q4 `; B- T Z6 M* E6 x else if(*p==' '): W/ [8 W/ ~: s( O; h4 r% o% P. f- A
++wsp;7 Z/ _/ S) O& ~0 Y
else if(*p>='0'&&*p<='9')
9 ]- m' B/ t6 E" I. v( _% ?% f ++di;
- V( Z: Y+ g$ ^ else; H4 ?2 U% F4 D
++ot;4 x# o" P, A! u- [. f
p++;
. c2 l4 v. w+ `- \3 i }
0 R! N$ h5 l; ~4 Q6 U printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
! |1 F' o T$ S. \) |5 ]9 r+ R8 i}4 ~/ W- p5 a, m, c
10.9( ?* K$ }) T# E) V% E
main()/ \4 Z9 {. D" w. B7 X% {: j
{int a[3][3],*p,i;4 Z- c+ k( T) b. b1 }+ U
for(i=0;i<3;i++)
3 G5 C' R- G2 @6 `. h# Y scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]); K7 K+ q1 p3 q- S& X$ n% ~3 U
p=a;! D8 F7 X, ?! I) k5 _6 H$ b
move(p);
4 B% \- e! r) R2 Q for(i=0;i<3;i++)9 D- l \ K& F% g, `7 j! n
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);) {; }+ E& g# V0 a% k' t+ G2 I: K
}
( c! |( e) D F# L6 T% r% ~. Ymove(pointer)' Z) P/ k2 D; C- q* v6 j' ]0 D
int *pointer;! { W+ e, H/ K- T1 S3 E9 \
{int i,j,t;0 Q% @5 `* M0 L7 _8 p
for(i=0;i<2;i++)& ` }2 c: F, D
for(j=i+1;j<3;j++); p' k5 N0 I: ^: L, P
{t=*(pointer+3*i+j);
' A) Y# |6 X5 }# k6 O *(pointer+3*i+j)=*(pointer+3*j+i);- |8 A' m! e( x- r/ _
*(pointer+3*j+i)=t;
: m. ~1 H4 V" K. Y9 O }) `1 ?2 F( [$ f1 ]( i
}$ Q; @- \) d: S/ c6 ?
10.10/ P! t4 Z' v! h
main()
3 P1 q! s) s& ^" k{int a[5][5],*p,i,j;
8 J; p9 b& ?; d' U! ], E- G$ L for(i=0;i<5;i++)
/ _* `$ s6 W1 D5 p/ \2 }- g for(j=0;j<5;j++)
8 s! V# W8 H2 Z, z scanf("%d",&a[i][j]);
) W( v. n0 \; M j' Z. x p=a;
* M- j/ g+ W2 E6 S change(p);
4 ]6 c7 o( T; m/ ` for(i=0;i<5;i++)6 L' N) m$ H3 `0 t5 m6 X
{printf("\n");7 E' B4 p# n( B3 `* r! z
for(j=0;j<5;j++)3 N$ G/ ~& G- U3 E, g( S4 @5 x
printf("%8d",a[i][j]);
# t! ~# s3 o( ?1 I6 b }8 H6 l0 }8 V. x8 {7 x" Y4 D
}- Q- v* a5 T- u+ d' e
change(p)
8 z$ t, a+ ]$ W. s; zint *p;6 V3 b, P$ p! c* H+ C
{int i,j,change;0 D* J# Q8 \9 U! h
int *pmax,*pmin;/ z% m+ N! Y: f& ]) n
pmax=p;
5 E+ W# m) r: ^4 K8 e) H pmin=p;
1 v6 b, ]- b! O. ]3 N% z, J for(i=0;i<5;i++)
6 e1 K- k5 [ O) h for(j=0;j<5;j++)* V5 i* c+ O+ O3 W! H( u% z# h
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;
/ d0 d+ N7 J5 D7 N; W if(*pmin>*(p+5*i+j))pmin=p+5*i+j;
! }% L9 f; w. L1 H0 U) w( S }
3 q6 H$ c+ L" e L change=*(p+12);
& @( s2 Q; Z% W% u. R" I0 d *(p+12)=*pmax;
) @0 `1 @) E, x$ `" V+ u+ G8 _ *pmax=change;
4 |3 {, \$ K+ L, v change=*p;
, ?- a+ n; n7 E# C* s, D *p=*pmin;
2 i' z5 r+ E. a6 ~7 J1 j' v *pmin=change;) P+ ^( U8 n `
pmin=p+1;
7 q( B9 p$ H9 F A5 a g for(i=0;i<5;i++)! j# D4 M L* q* c1 g
for(j=0;j<5;j++); P$ y2 X* ~% u, R4 v+ D) Z
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;& l2 P _# N7 t) A
change=*(p+4);. J2 G$ H' ]4 p. D- b- p
*(p+4)=*pmin;* R( c! k5 p* |" }# a8 m1 R- S
*pmin=change;7 @6 w: ^8 k0 f) ]! u
pmin=p+1;
" b/ s8 Q, M" N% h3 O' E2 W& g' t/ q for(i=0;i<5;i++)' H2 K8 X4 C$ X' k
for(j=0;j<5;j++)6 o3 P1 l) k* w0 @1 X
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
/ k; a- @0 [3 e% s7 ~! G pmin=p+5*i+j;
# {5 A) s! g5 p+ u) v4 I change=*(p+20);2 h9 P/ E% }: T* Y2 W+ O6 \# L
*(p+20)=*pmin;& q; ]# Y" W( X; }) F: {- w& J
*pmin=change;8 a( ^! \0 W* @6 x5 q0 `( u* |
pmin=p+1;$ E! o* E n8 w& W, V
for(i=0;i<5;i++)$ y B/ M4 f( L' q+ @; R2 [# s
for(j=0;j<5;j++)
5 Q6 \, N: ~% M; Y if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))4 p4 K5 q: o9 G' P+ ~. b6 Q
&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
# n7 Y" h- g8 s) v. s change=*(p+24);* y; y/ [9 Y" {) [( x
*(p+24)=*pmin;
1 \2 k. }! K- D( M6 q2 ^( O0 V; e *pmin=change;( C& r4 G5 x& m$ D; k" B$ e
}
' X6 _) L& h: r6 v; z10.11( t5 o; F( N0 c# o6 i
main()
* N: U5 k/ t r0 X% z/ h{int i;; D8 X7 o3 V# Y0 L3 q+ _' e2 [
char *p,str[10][10];
$ T3 k0 A; B7 A7 i for(i=0;i<10;i++)6 y3 i, f) V; K; j* T" B
scanf("%s",str[i]);& P' C, h$ K8 A% i( \/ {3 a% z
p=str;! H1 f+ S6 W7 n4 }
sort(p);
! S3 `; j+ ^: L$ Z for(i=0;i<10;i++)( D( \, v3 f ?1 v9 b* b' N$ F1 o
printf("%s\n",str[i]);2 ~, |% X! q5 k$ @4 m
}; f6 ?/ w" ?3 E7 }% j3 h2 ~
sort(p)
& x8 r; N3 ~* U: m6 @! S6 j% F/ Ochar *p;! U: s7 ?+ x$ a1 J( m! i7 q
{int i,j;$ S" E3 T9 b4 D
char s[10],*smax,*smin;4 H( K2 g2 ], e7 a: i4 H
for(i=0;i<10;i++)# i" _+ f# ?# s0 D1 t
{smax=p+10*i;0 ]# w8 F. ?( w8 P& i0 {$ J0 u5 y
for(j=i+1;j<10;j++)
9 W: ]; m' e% _, ]) F {smin=p+10*j;0 x- A( u- y- c/ D3 k P$ b5 |
if(strcmp(smax,smin)>0)
. P6 `2 J3 C% W- Q, v {strcpy(s,smin);
. p* O" b( ?! H8 z$ h5 M( v& j strcpy(smin,smax);
+ x- p, v+ r' e5 w4 P" w* p strcpy(smax,s);: R% n1 P; s. z7 \# J9 R# z$ {% M
}
! x; O1 e% a m9 n0 ^" ? }: ~6 |$ h) `$ N; [( N
}
6 B" c8 m* m/ m" a) V}
, F v7 L) E1 A, S( U' v @10.12* ~+ \! D8 I5 p5 S+ O
#define MAX 20
2 i# A, D: Z6 c+ ~5 i! D8 Omain() b: b2 a4 h" b
{int i;
9 s* Z4 q# l0 M5 k/ i char *pstr[10],str[10][MAX];
% E* L3 v+ G" l; j: k% r for(i=0;i<10;i++)$ b/ J% \; w! w! o
pstr[i]=str[i];1 }2 E; p5 F: b) [: h. S2 L
for(i=0;i<10;i++)
: ?+ ~( e3 M2 K w. a- U8 Z scanf("%s",pstr[i]);: |6 w$ ^' ~7 C% U2 k
sort(pstr);$ i) N/ w5 @) e; V$ |
for(i=0;i<10;i++)
+ S% x8 A" x+ B) p printf("%s\n",pstr[i]);
$ }! m9 V! f `5 o+ A( `( _}2 R8 h" U$ m# Z- J2 {
sort(pstr)$ B3 X8 Y8 N0 R! ~
char *pstr[10];
9 D. w' t/ P0 p) e6 L8 K% M, [{int i,j;
# U# P9 }& h, Y9 K! P# e char *p;
; k( \/ {5 h2 V for(i=0;i<10;i++)
# |9 m7 j% G( F, o- z4 t {for(j=i+1;j<10;j++)
3 k6 A$ ]( u$ \" I( \ {if(strcmp(*(pstr+i),*(pstr+j))>0)
/ j* U- i5 S8 f {p=*(pstr+i);
$ T% M/ @; q9 f) e- u *(pstr+i)=*(pstr+j);
# N d2 A- Q% B4 k *(pstr+j)=p;1 X$ O- K* \5 f0 H1 Z
}8 J+ R' { U8 Y) J
}
- n* t9 D5 X2 C) d4 K: x* {- I" G }, \5 \" @/ s# W
}- `' b2 A- @% U. l
10.13
, o8 @0 H: Z8 n/ x#include"math.h"5 ~5 |! T$ e" ]0 f0 f
main()2 F/ P! @ ~, O( X! e$ M
{int n=20;
6 c* Z1 d. o$ n# A9 B7 \ float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
; n2 c) T" Z: T- N c6 b scanf("%f,%f",&a,&b);/ i9 h. D4 P. I. G3 [1 q0 ?
scanf("%f,%f",&a1,&b1);8 ?' F2 v/ s( E% _) ^8 f
scanf("%f,%f",&a2,&b2);
9 V6 _- E+ v% x p=sin;' a C* {, p, k7 V7 C
c=jiff(a,b,n,p);9 N& Q# X3 X" d( g" g# H
printf("sin=%f\n",c);
. c. D2 z$ b8 s1 `" h p=cos;
: c" P; p1 w' p! Q5 L- G- g c=jiff(a1,b1,n,p);8 G# _5 ]" q! W% e; u$ d
printf("cos=%f\n",c);1 j2 H+ J' `" e+ m5 i$ `% M1 W
p=exp;
& O5 t, V, \! T( @$ e& Z3 ]( q c=jiff(a2,b2,n,p);( E; f6 K* R( B5 y2 _! }
printf("exp=%f\n",c);' B; h& C5 U3 d9 `! a0 w7 Q
}
# j% y5 T# ?* vfloat jiff(a,b,n,p)4 [6 x; h+ @ |: M
float a,b,(*p)();
3 E) E8 l. _2 x5 kint n;9 @# i n- j. f! k8 N! Z
{int i;. F$ b5 x, R8 P6 q; n4 P
float x,f,h,area;
0 z, I' _+ r, A& Q h=(b-a)/n;+ a; h( D/ W4 g
x=a;+ W( k; z2 `9 i* C# D& c
area=0;: ^0 W: t* o5 C. V
for(i=1;i<=n;i++)+ Z; w, _+ O) C! f( J: A9 T5 B! Y
{x=x+h; U* I7 R4 Y1 G/ P
area=area+(*p)(x)*h;
1 L2 ]5 u. H+ Q) T, [! n/ j }
2 V+ o, F% i; C& |# S, y return(area);
6 E' v% \4 I8 w# g7 n( k}" ~) y3 \& N) U
10.14
" u9 S( M, d% Y/ N" w% D. w. bmain()% H6 u+ I8 ]+ x5 {& _
{int i,n,num[20];
$ b. F k, R& |, u char *p;
+ w* U1 j7 R5 Y+ Z; H scanf("%d",&n);0 D5 L6 b+ |" |1 j- J2 S- ?
for(i=0;i scanf("%d",&num[i]);9 m' Z2 Y# i$ h% j/ S) @& T
p=num;# A1 H$ d: w2 b
sort(p,n);# @! ^, ]8 {7 @" S' |* E2 ` r
for(i=0;i printf("%8d",num[i]);1 k) X6 s7 s) n R2 p, W
}
8 }) ^ `6 O( x% lsort(p,m)
9 I8 z7 F; s1 ^; ~( V( Q) f1 ?% _char *p;6 v0 O) g2 Y ^) u4 F8 [# J2 ?4 ]
int m;
7 R! R5 r! p0 w- ?! ^) L; p5 W{int i;
9 A f1 A: ?3 @. _0 ]% F char change,*p1,*p2;
: F9 ~3 n7 X% V, d7 H r! Y* f6 C( x for(i=0;i {p1=p+i;& `/ O/ Q) L" B( M) X
p2=p+(m-1-i);; C2 \- j U# u& g+ a @7 X, r
change=*p1;
/ p+ v0 u, [. w" a7 Y8 N; m" u *p1=*p2;6 b# {. u- A3 X5 y7 |& x
*p2=change;
. r4 K j3 m8 Y/ f* Y0 D1 g) k! F; t }3 }5 \. e. j4 z( B3 C
}2 A) u) }, ^/ P8 m3 W
10.15
! Y6 l9 m( l) a% r9 n* Pmain()
]* e# u, }1 B4 o. T9 F{int i,j,*pnum,num[4];
/ Y2 Q% x. N1 I$ V2 g [ float score[4][5],aver[4],*psco,*pave;" g8 y8 d r, A# g# ^
char course[5][10],*pcou;
- x& H1 C+ A& S! b pcou=course[0];( E, p+ U& W3 \2 x, \# b7 Z
for(i=0;i<5;i++)
& |) M9 x2 M8 J1 u' C5 d/ z5 n scanf("%s",pcou+10*i);& U* i( c) p0 ~3 m# ~' J
printf("number"); d/ e* E7 ~, t1 o( t
for(i=0;i<5;i++)
( M8 o4 K2 ~# ~! f" ]6 p printf(",%s",pcou+10*i);% t/ n& Z/ U8 C. V
printf("\n");
* ^% c% V5 Z4 {, x psco=score;. h0 H( p9 F8 w& q! m4 ~
pnum=num;# y: I* ]! i" V! n( q
for(i=0;i<4;i++)
, |! y$ \1 |. w+ q" W3 r {scanf("%d",pnum+i);
) D* y/ s2 k( V4 K for(j=0;j<5;j++)
w. S. U# N/ s e+ W' h scanf(",%f",psco+5*i+j);
3 [' \: ]" T. `6 m( g9 f9 N }
8 W7 P# L! h( `; T pave=aver;
. X' l) w# f+ C1 U- p printf("\n");# R$ e$ {: U+ N' {
avsco(psco,pave);
G+ \4 |4 }" ] avcour1(pcou,psco);& ]: a$ ?# r9 z6 K2 q
printf("\n");
; o' M& y! p4 ]5 ~- V& w; w fali2(pcou,pnum,psco,pave);! Y3 t+ ?0 F X( C% r
printf("\n");
( E& q2 Q+ ^- S: h' C7 ]% Q; e! z good(pcou,pnum,psco,pave);
3 h7 l2 o, ^6 ?1 B6 A# n& Q}
# \" i! p* L; j3 V8 W/ Q6 {avsco(psco,pave)
+ T- d/ u6 W8 t# x/ Rfloat *psco,*pave;5 v* J4 B! _, |
{int i,j;+ V/ U$ P- A1 ?, F* f7 E- \9 x- ^8 o
float sum,average;$ W+ i1 e5 G' y4 s; g
for(i=0;i<4;i++)
; T2 k, d1 z! Y3 o& c9 x {sum=0;
1 g3 W G' c, B for(j=0;j<5;j+). a% n; }- r* n: g9 y; l7 k. C/ Z
sum+=(*(psco+5*i+j));
6 ?- W6 {. g. E4 q- u1 h6 Q, _ average=sum/5;# r9 j+ A8 A" X- P
*(pave+i)=average;+ D* Z! \/ m- `; A
}
& T8 ^% u3 {6 a5 a5 l}- q' G- m+ f4 m; X+ h, I
avcour1(pcou,psco)
; ~% X* z: R- G5 K Z1 f4 nchar *pcou;% s2 H9 H( }) k: N% x: B
float *psco;
' S+ r P1 p8 u9 k6 s8 D* g" I! A{int i;
) t4 k6 B# m9 f2 F float sum,average1;. E K) I* u! E9 K
sum=0;
. V" M( \& a q# u, {* K3 D: ? for(i=0;i<4;i++)/ l3 ^- b k- c
sum+=(*(psco+5*i))& d5 K& I9 e2 l! [" U$ u8 E2 Y) R# N5 `
average1=sum/4;
9 q% [$ q; l4 g+ h* J6 ~. p printf("%s %5.2f\n",pcou,average1);7 \5 q3 z5 M' |; k# S' y
}6 l; {7 t$ @/ J v& B1 o! `
fali2(pcou,pnum,psco,pave)
6 h: r9 y- T% v8 ~. J: fchar *pcou;
" _+ s1 ~# \; N( c0 `int *pnum;
7 e4 T! a: S; I$ ~4 Ofloat *psco,*pave;
; N, ?1 \- l4 q7 g" F9 n{int i,j,k,label;( D+ ?% F' R$ g0 y5 |: m) u
printf("\nnumber\n");
: S+ G# K0 w2 {/ c for(i=0;i<5;i++)- e$ d- {" P2 G8 ~8 f
printf("%-8s",pcou+10*i);
2 a, O3 I" R, c" {( i printf("\naverage\n");
8 |! ?% q1 v& o W for(i=0;i<4;i++)9 ]- [; N0 b1 U, Y
{label=0;
+ `9 Y+ L3 y+ K* A for(j=0;j<5;j++)7 W, O2 X9 @6 ?, P+ e! ~
if(*(psco+5*i+j)<60.0)label++;
$ q$ x7 N3 A6 [( C! G5 ^9 y if(label>=2)2 |* y4 `. j* {' K
{printf("%-8d",*(pnum+i));
4 `+ ~( e' o6 F2 G; E for(k=0;k<5;k++)$ X# n0 n& C% ]; l# z+ q# P
printf("%-8.2f",*(psco+5*i+k));0 o& g) ^$ P/ Q' |
printf("%-8.2f",*(pave+i));
. N9 B% |6 P# I" F }
, {6 }% `+ v- x8 E2 f }! @9 E) W( V- g
}( c0 s& @# S h. Y" [
good(pcou,pnum,psco,pave)
7 }0 a! a7 z# Lchar *pcou;" n' |: l' E/ m6 Q- H5 X0 l* T
int *pnum;+ t6 ]' r( t% U% o
float *psco,*pave;
0 _2 U, g. r; W# O( M2 ~( {{int i,j,k,label;
9 w, ~2 I0 |" b: s6 Z2 U printf("number");( Z8 g2 S! o( Y! I% @* h4 ]2 r. K
for(i=0;i<5;i++)
% e+ i/ E# i: B, {/ f" X i printf("%-8s",pcou+10*i);+ \4 k7 D0 K m& [4 h0 i
printf("average");! R/ L& f `- D7 ~
for(i=0;i<4;i++)
3 ?; N; t0 h1 y) y& n- C {label=0;
, o+ F9 y! F6 q$ f4 c( p for(j=0;j<5;j++)4 a2 k' L8 E% u) w* B5 g u: L- M
if(*(psco+5*i+j)>=85.0)label++;
( E& K- x( l% ?7 l L if((label>=5)||(*(pave+i)>=90))
$ w" [% f2 G) M, v {printf("%-8d",*(pnum+i));
) H2 \. B$ `( U6 I5 _% b! o for(k=0;k<5;k++); |$ m2 X$ E$ R& l8 }0 h
printf("%-8.2f",*(psco+5*i+k));! o; f8 x. A. T0 r. `& ?
printf("%-8.2f",*(pave+i));; D6 u# z8 `7 G
}
: r3 P. X) D1 Z! j2 Y& }; S9 s+ t5 D }
$ j0 y2 V& U j# Q! T}
7 {$ J. }4 \ ]! n" ^* s4 Z10.161 M- E' ?" j. i( ~, O
#include"stdio.h"
* w6 d* q/ e0 f1 G& pmain()/ d, G1 u" x5 o- F7 I8 A, o
{char str[50],*pstr;1 L0 i$ b: P3 g5 V4 a' U9 c
int i,j,k,m,e10,digit,ndigit,a[10],*pa;
2 q( p& t/ p5 a+ Q$ ` gets(str);+ d- z- s" W P5 i, |8 P* f
pstr=str;( y9 J6 Z4 ^. J5 }3 m
pa=a;
( B% f" P9 c: _' Q1 d4 Q$ ? ndigit=0;
1 j& l! }) H" A0 ~' p i=j=0;
0 i( L* J% [5 P8 p- k6 P: b while(*(pstr+i)!='\0')% m+ c% n/ n5 S
{if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))- C' b, Y6 _- X" r0 q& f
j++;* v5 u- [1 L% Z& f$ G1 |/ _: K
else1 x: E% d4 v* u9 V* U
{if(j>0): Z$ I: Z, X2 e8 ~; m0 M
{digit=*(pstr+i-1)-48;
7 R8 ?- D' v. T! ]$ X k=1;
4 E7 [6 U' M: v while(k {e10=1;& f1 }! v; ]) L6 Y2 {: \
for(m=1;m<=k;m++)% [ H4 R- g4 C) t' n3 [! r, _: h$ ]
e10=e10*10;6 V7 L' _4 ~8 g# z4 f
digit+=(*(pstr+i-1-k)-48)*e10;
) k; d. L& O' V# Q% o) X k++;
+ V% d5 f5 L \, ]# ] }
4 O4 }; N% i1 ?7 ^% V0 Q# _ *pa=digit;9 G1 f9 ? O' }/ t G
ndigit++;) ?8 K8 A" `/ v
pa++;
4 z# t, u8 i Q6 [0 b- q j=0;6 @0 T) v s2 _
}
! `, x( x+ N z }: K: u' A1 H- b7 z$ ?" h
i++;
* M X- t. o6 _+ Z. v }
) E) E: v, a; a1 l/ {3 f( H& t$ ^ if(j>0)/ H6 |' e# l u6 J' \
{digit=*(pstr+i-1)-48;
4 w- r* ]' w- l4 M k=1;
/ ^/ ?2 \9 g4 R5 ^" ]/ e/ K while(k {e10=1;, y( _ P( i* k( s. j0 u
for(m=1;m<=k;m++)
' S+ Y# |) r5 v- V% e& | e10=e10*10;
5 J0 \# I$ Z# M# i3 m" J% L1 h# ? digit+=(*(pstr+i-1-k)-48)*e10;
- O: k! }$ O: K1 l% [+ w k++;, N! y; k! k. i1 W
}
; B6 a! x' N; g0 d7 s *pa=digit;
! t# ^! ^6 a, |% R/ { ndigit++;
; T+ }, D: @* @ `2 D j=0;
: p& x4 y% f/ l3 X% f, j1 Q } - s3 q% g! g# X: t4 G" _
printf("ndigit=%d\n",ndigit);
0 g1 r* U K0 S/ z1 S- S j=0;% h7 |9 }% r" {; ]" t: `
pa=a;1 `) K1 V+ f% v; ?' m
for(j=0;j printf("%d",*(pa+j));- v3 \9 \' A/ u& z8 y j
}( X: q! f! f, S5 ^% S# J& _0 G1 f1 w
10.17* J. W. F' S6 c, D/ k
main()
) n+ M: i' C$ y# P$ n3 z' n4 L{int m;
3 v+ R7 _* s+ t char str1[20],str2[20],*p1,*p2;. Q; d( V# r9 e+ @
scanf("%s",str1);9 q0 o2 S3 C+ ~! Q( J/ _
scanf("%s",str2);
- Q; \/ F9 t" d# o p1=str1;
9 l3 L1 O0 z- f1 |) N p2=str2;* M$ v u; P9 t" E" I6 S8 w4 ~" h
m=strcmp(p1,p2);
4 _) M4 k, J7 }- x3 _* R# W4 }9 a printf("%d\n",m);4 U9 }: z, T f: q2 W
}
, {8 z0 P7 ^# H8 I8 S$ lstrcmp(p1,p2)' ]5 S, e& ?! y5 {) u( u
char *p1,*p2;- [, m) p# {9 n9 L- h) F% `* V8 w
{int i=0;
4 X5 c' P) a5 x! I3 I while(*(p1+i)==*(p2+i))
0 x! V$ j, S6 t2 p" t9 \0 O if(*(p+i++)=='\0')return(0);: `( A) X! y# l
return(*(p1+i)-*(p2+i));
- T! a+ R& V2 A- p! b5 c3 K}- |/ H4 S d2 m" N. @" f! i3 j
10.18: i. v) p3 H& v/ K6 u1 B
main() J2 @1 p) A3 J* L6 d7 c
{static char *mname[13]={"illeagl","January","February","March",
4 g4 ]+ ~3 r' i# \# k, S2 I6 w "April","May","June","July","August","September","October",
6 b" Q) |" L9 c% X "November","December"};# K- T- e h% O
int n;
+ ]3 h g. U7 O. K; D1 G scanf("%d",&n);
' l0 C3 ^/ O0 E1 H if((n>=1)&&(n<=12))6 _7 X" r: e! l0 K" h
printf("%s\n",*(mname+n));! T7 N8 O/ A1 Q9 \* z9 }0 I% t; K9 A
else
5 S7 S1 l" @, e, t: e/ _ printf("error");* ^+ h. l% H/ a+ P
}
1 V0 Z' a5 T" d' b, H2 z$ m3 o10.20
4 I, o) R4 |) Q. rmain()" _/ ^9 E* D4 h& }/ ]. h2 N
{int i;5 z) v& V0 K d. }1 e+ G* y8 |
char **p,*pstr[5],str[5][10];
# g1 ]" x5 X% Q5 `2 U# Q. [ for(i=0;i<5;i++)
+ v F' I3 Q$ C- H+ V' y pstr[i]=str[i];
# \2 P( K9 z" e4 a& ~5 H for(i=0;i<5;i++)2 y$ n1 w( h- P2 M5 n! H1 O
scanf("%s",pstr[i]);
_0 F; P, |/ D: ]/ [' M p=pstr;$ L% I Y8 P/ k9 f3 x- M
sort(p);
4 @/ l8 C) @8 S9 s9 N for(i=0;i<5;i++)
4 g+ V* n- o& S% T/ g- m0 p/ H, ^ printf("%s\n",pstr[i]);! n1 h6 ^1 s7 D' t& E; O1 {
}
( l3 X4 M% V& d5 n% j( e+ y* msort(p)6 g5 X- |/ Q2 p9 c* \
char **P; O+ b, p& h, N6 {% Q
{int i,j;% B7 s# W: k4 v
char *pchange;& M q9 r2 | H
for(i=0;i<5;i++)' X9 S* p, Y3 L3 ] Y! i5 k4 |2 y
{for(j=i+1;j<5;j++)9 d. l' C& [0 t; Z0 B
{if(strcmp(*(p+i),*(p+j))>0)
5 J% g3 S* A ^. [ {pchange=*(p+i);
* X7 }! R' P2 o *(p+i)=*(p+j);
- K- T; q0 \- b* \' O( t *(p+j)=pchange;
. J7 O& Y9 ~ F% i) R4 I% W }' g% R( @7 V$ z+ S6 u
}
* ~! D9 v; D: K }
2 Z/ l! ]$ n/ g9 y}6 S& H* A% d% H
10.21/ w" n7 {* d- E2 s' U9 g0 C' s3 ^
main()) L8 s. k) S& w# w( z# y
{int i,n,digit[20],**p,*pstr[20];
5 B; E6 @" x* F# g( V a6 F scanf("%d",&n);
% ~" o/ E% s7 ~& \9 c9 z! h, ^8 o6 P for(i=0;i pstr[i]=&digit[i]; Y% ]1 k" M4 a" a
for(i=0;i scanf("%d",pstr[i]);
4 [* K: K# A# p* k B, \, ^ p=pstr;6 k/ M2 b, r# j0 ?0 @# D
sort(p,n);
% M; P3 Y+ V( M2 ]% U for(i=0;i printf("%d ",*pstr[i]);
. E8 Z( \& O8 b# s; Q8 w}3 Y- n1 m+ p9 \( H
sort(p,n)! F/ v" t* Y3 i- a7 T
int **p,n;9 y8 o+ }8 m3 ^2 t
{int i,j,*pchange;4 F, R7 p7 x) ?& V4 p/ Y
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))$ s# D( O$ W [! w1 F
{pchange=*(p+i);/ A+ R* D' `/ e) N
*(p+i)=*(p+j);
' H8 g% f0 }! P: O: h" I$ V *(p+j)=pchange;
- R& I; ]+ S5 I }
! ]/ V. N, o0 m9 s' ` }
/ A5 g/ l- n5 @: _0 w }4 y. M }/ H/ N9 D C+ Y
}4 B" b; K) C b6 L9 A! A q1 h! s- i
第十一章 结构体与共用体
N7 N4 ?! l- O @0 E11.1% j7 w" `( d& Z# H
struct
0 D+ W" G2 b' p" y; ~* N9 j0 ` {int year;- H& K2 \- b- n3 K- \ o
int month;5 f7 y/ ~8 n5 v' P0 H9 c
int day;% _: Z0 ^8 }3 m" n+ e: Y
}date;
8 u$ `! M# v' Y. E |8 n" @main()* V3 a: S+ |9 J( Y2 q
{int days; m9 s- ^3 |" O1 w
scanf("%d,%d,%d",&date.year,&date.month,&date.day);2 k, Q5 ] j1 H4 z+ I
switch(date.month)7 P0 P6 T2 l2 A% P3 S5 T0 a% D5 n
{case 1:days=date.day;break;
; B( r# j9 I" D+ U, m) o case 2:days=date.day+31;break;3 K$ K, Q; |1 u
case 3:days=date.day+59;break;$ P. _" b6 e: a# L5 _
case 4:days=date.day+90;break;6 m2 r ^) w: s6 U: G# C+ b
case 5:days=date.day+120;break;
! \3 u% ?& n4 C7 V( e* T case 6:days=date.day+151;break;
0 ~# k* r0 L( z1 k. t5 ^- v case 7:days=date.day+181;break;: i2 w! K, q4 F, f* |* h
case 8:days=date.day+212;break;6 f. l6 ~6 z# x- ~7 l
case 9:days=date.day+243;break;
0 M5 A& H, R( {5 O case 10:days=date.day+273;break;
' [3 l: E! Q$ N _/ O: u9 C case 11:days=date.day+304;break;4 N8 v$ E; v2 L$ @
case 12:days=date.day+334;break;
0 Y h9 L7 @9 Z; T }" f5 x7 O4 E3 g
if((date.year%4==0&&date.year%100!=0||date.year%400==0)3 H; b/ T. ?8 G# P7 G5 z/ L; C H
&&date.month>=3)5 H- h1 Y2 x- A
days+=1;! n! S( [$ \ N$ n0 [
printf("days=%d\n",days);
! W, N( _$ B0 q1 x; N& V}
) u( d1 F" G" h7 B9 ^* z, Y11.2
% Q8 t: b. \+ k& {# [; c% n0 Vstruct dt5 k. a' d! U/ M/ c2 { Z
{int year;5 {% d3 A" Z- i- d+ a; v
int month;! e1 b) r6 c( S- e9 g6 Z
int day;/ u, i# s7 s4 i" _& g
}date;5 M8 y" e7 x% Q. i; A2 C
main()
, q( K: N. L) d$ m0 q0 g: f2 ]{
) R% ~& w+ Y b1 I0 q scanf("%d,%d,%d",&date.year,&date.month,&date.day);0 H5 f3 g2 H5 H0 |$ C
printf("\n%d\n",days(date.year,date.month,date.day));% d; F! M* \ d! F. T) J! p& H
}
6 x( L2 y" E' B! S I7 ddays(year,month,day); H( e* K- C! t! M ^
int year,month,day;
$ J0 h% {0 S: h{int daysum=0,i;' W; ~: a' |4 U8 k% u* X
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}* {3 _' z7 E7 P/ Q! H7 A. w$ E e
for(i=1;i daysum+=daytab[i];. Q8 q" e6 h) ]% B+ p" y
daysum+=day;0 X2 f, ^, `- E* D- B5 k6 O
if((year%4==0&&year%100!=0||year%400==0)&&month>=3)
1 {* e- z" ^- ?& ^% L daysum+=1;4 N2 n/ V* e! A) p, F7 n; _. r
return(daysum);0 t; p8 U: k1 i
}
# k9 q; `, A! n2 W2 ]1 m+ m4 C11.3% T; I8 J* `+ Z2 l) k( a
11.4# R! Z u1 r8 Q( M& K
#define N 5% I) `, b& W, c4 M; u' W
struct student; | X3 Z4 B) i' q
{char num[6];5 ^% S f& l/ U' V, V# J* }3 s
char name[8];6 i$ E+ y' k: H6 k! e0 e0 Y5 T
int score[4];
9 d# `& [1 r- e f- d% p! o9 |1 _ }stu[N];0 q/ k' t5 R/ n7 Y
main()
4 G- T: P0 X1 h4 }{. U, }4 Y' @3 T
input(stu);
3 k1 B: r! d" }, D print(stu);4 u+ `4 o1 Z+ p5 ^3 w' n1 B
}% w1 l. ]: i: R( y4 ~! U k
input(stu)
# N; Z3 Y* N/ F j7 V, p1 Istruct student stu[];
- ~$ P" A2 {& `1 g" B) X2 Z{int i,j;* M2 l7 H0 ~) Y
for(i=0;i {printf("number");
V3 T( O" [+ s/ k, x" ^: } scanf("%s",stu[i].num);
3 _/ X2 ^" C& \3 Y$ w2 a7 y$ X printf("name");5 N7 g/ c* O3 o) g' x+ V
scanf("%s",stu[i].name);
8 Z! `0 o% L8 P: F( a: t for(j=0;j<3;j++)
* I. \8 J1 t. k! @ {printf("\nscore\n");+ `( v- ]0 I# @6 ~2 K2 h
scanf("%d",&stu[i].score[j]);* E4 ?) z5 r, ]3 X' u
}8 A8 k+ W5 R8 S& L: B; d7 o
printf("\n");' q9 Z6 x. Q, ], ]4 d: B
}
) O G9 e1 Z- P! r1 @( Q}6 b' q: [. b& G# g
print(stu)
; ]2 u( f" g5 H$ rstruct student stu[];: U/ ?8 }3 J/ z3 u: H9 r8 m2 R
{int i,j;
: n9 N: a! [$ U! q. M0 F; S printf("\nnumber name score1 score2 score3 \n");9 u3 ^4 L" b& s- N- u; X
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);1 I \% r' W3 ~& |
for(j=0;j<3;j++)
) c) @" ^7 A0 X* L, Y printf("%7d",stu[i].score[j]);6 M1 Z, k2 Z) q6 ~1 H) V
printf("\n");
5 A# q( y; M# @4 X5 \# A }0 G1 f0 @! j: |5 B7 c: U
}* V( P' {, { k0 r4 K# u1 h' g' J) G
11.51 e5 }" c7 e& f
struct student
; h2 T" p$ B6 B; D& F5 } {char num[6];$ k4 D' p& `: {, u
char name[8];
% b9 P* ^; v2 B& V4 P int score[4];* j2 C8 a A: x6 `5 S
float avr;# [& Y# K. s3 ~$ _
}stu[5];
; l* `& X @/ k4 d; L/ C% Imain()- u+ I9 H4 T( G* u
{int i,j,max,maxi,sum;4 b3 S9 |, H1 s. |0 `2 }
float average;
2 o. i8 I2 }7 j6 n/ j for(i=0;i<5;i++)( Q. m" {/ ]! d+ e; ?
{printf("number");
1 U/ P1 c2 D3 q' m" U$ c scanf("%s",stu[i].num);
; U, `9 r5 v. L6 `2 _* z printf("name");# A, Z. ?" M' G4 |
scanf("%s",stu[i].name); Q5 K( }& ~9 E5 n* [) `6 B! g
for(j=0;j<3;j++)
4 d/ X; [' O, U( t5 M! X {printf("\nscore\n");
' x8 h _$ z. `% X* l5 D# X scanf("%d",&stu[i].score[j]);
* G0 J+ {0 o, }, v$ r4 f8 R: M5 P }5 M9 v: S0 \! d1 |1 C
}# w V% @( f" R! y6 D: @% ~& i
average=0;* b# d0 V2 |6 }9 z) W1 k, x
max=0;, T' w4 k5 Q, f, m9 X; ?
maxi=0;
( U6 d% \$ `- `( @ for(i=0;i<5;i++)2 X( u5 ?7 X& \ Y+ i
{sum=0;# ~) P* O8 d; y; | V6 y
for(j=0;j<3;j++)7 z s$ a8 o. y
sum+=stu[i].score[j];
$ F$ L+ s; V! y9 R stu[i].avr=sum/3.0;0 h& c2 O$ d, a& v* f# ?: n1 f0 F
average+=stu[i].avr;
3 j+ `8 i( l/ E1 z& q* E if(sum>max)
* H" l) y) E$ {6 ^. |6 J {max=sum;( H7 A9 L, f0 V8 E ?9 R# k) ]+ a
maxi=i;
5 d2 M7 S- D4 \8 M/ P }. \" j7 q; D! @* \
}9 P: u V% z. Q
average/=5;
I/ r, U( t" i3 W printf("number name score1 score2 score3 average\n");
1 H) V1 e, E; S s3 y6 m5 O0 R# P for(i=0;i<5;i++)
2 h9 _& P8 i. K4 y+ r, R {printf("%8s%10s",stu[i].num,stu[i].name);6 H+ ^ Z7 n2 L p( Y; z
for(j=0;j<3;j++)
# v3 A6 A( r# A8 b% g7 y printf("%7d",stu[i].score[j]);/ s+ \; T3 x* b5 Z- R
printf("%6.2f\n",stu[i].avr);
' @& m( x7 O! N }. F/ g; |: r) r! ]; p& M0 r
printf("average=%5.2f\n",average);' \/ t1 y9 S) F
printf("The best student is %s,sum=%d\n",stu[maxi].name,max);* B$ e" w" q& z5 v
}
& M, e1 Q$ R2 q$ \1 z: X+ B; B5 M# q
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