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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
|---|
签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案. ]; \+ z5 }0 Q% E: [0 |% f0 E
1.5请参照本章例题,编写一个C程序,输出以下信息:( S" Q. v3 ^: y/ j
main()
) Q p9 J4 p: M% W{
6 j# d' j+ v9 b" v" W( z( K3 tprintf(" ************ \n");
% E& S4 Z# u- D' T1 ]$ T4 L& Cprintf("\n");
7 ` \' D6 U8 }% e( l, Uprintf(" Very Good! \n");) G$ C3 M& |7 y% `
printf("\n");
) J0 M1 e% `+ ?7 F9 `- eprintf(" ************\n");, r9 `- C) B2 @
}: F3 F5 t* d5 Y5 Q9 L1 Z% N
1.6编写一个程序,输入a b c三个值,输出其中最大者。
; @+ i) J h7 U# t解:main()- Q5 r6 O# G9 b- a7 l
{int a,b,c,max;- t! ]9 G# e% y8 J# F
printf("请输入三个数a,b,c:\n");
8 y! P9 j& H: F& s; I8 d) nscanf("%d,%d,%d",&a,&b,&c);7 n) F8 @9 p. n
max=a;
4 F) Z9 O/ Y2 d& gif(maxmax=b;
* w! c+ F" Z) z+ ^1 A; Dif(maxmax=c;
$ Z! g, ^5 L% S* @ r7 |+ oprintf("最大数为:%d",max); I* J+ R+ W& z, e- |
}
! h) W) [" X3 C第三章7 s' g2 q' [# Z) K# w& j
3.3 请将下面各数用八进制数和十六进制数表示:
( `$ X- b) L1 ]8 t: ^7 N(1)10 (2)32 (3)75 (4)-617
8 q, g `8 T8 n, p(5)-111 (6)2483 (7)-28654 (8)21003; ?& [% g2 @' d D% ~6 o) g9 b
解:十 八 十六
; q# v5 F* P8 m+ @+ {. g (10)=(12)=(a)
9 k/ b& X9 }8 q" z (32)=(40)=20
3 ~% A# h ~5 o' D (75)=(113)=4b/ [$ ?5 s: |, P) |
(-617)=(176627)=fd970 x+ P g6 E6 o
-111=177621=ff91
4 x# ~( V, Y0 Z: @' W, j 2483=4663=963
0 c9 |6 B+ D- U1 y4 Y -28654=110022=9012; Z* ~+ C1 \0 d& f6 Z6 n0 K* z6 u
21003=51013=520b
8 ] Q& `% P& \* v3.5字符常量与字符串常量有什么区别?7 h4 C, B! h. g9 | k/ f
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
U+ ^' c# e0 a- J: J" j$ u7 h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'.
/ d, h" Q- z0 S3 c8 k8 s6 D- M- u3.6写出以下程序的运行结果:% [* z& m5 M7 I
#include( b, G# G( |9 k( K5 ]+ H
void main(). u1 @8 {0 J# `2 a& t) h
{) p7 n; c" d. R( `
char c1='a',c2='b',c3='c',c4='\101',c5='\116';: A+ ^! \: `6 h
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);
) I) q$ u1 I6 j+ w$ X& O& Nprintf("\t\b%c %c\n",c4,c5);0 R4 y% w! F& c( j6 |6 b- F) a
解:程序的运行结果为:
$ Q' }) }" N+ J" x( C, Q9 taabb cc abc0 v% O# J& H+ R9 ~
A N
1 }7 z- g# \* `- x5 \7 O) L) a3.7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,
d4 l1 y. u5 A& h! ]6 {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".; Q* n/ J( J4 K, ]1 |0 G. G. j5 [
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* Q; c; M. u) \% N4 c! b
输出.
2 t( _: B( ~% {6 s+ M, B# d6 Tmain()9 w/ h* R: x1 {8 I$ z' ]- b$ U
{char c1="C",c2="h",c3="i",c4='n',c5='a';
' b9 n$ f9 _; Z1 {( xc1+=4;
8 l5 M, y, Y# a" t% o3 jc2+=4;% P! l) K& U& L: A' @: h: i5 H% H
c3+=4;
; H: W' A5 S6 _% Pc4+=4;
' ^3 k) N6 @5 @2 U" f( v" D' ^! bc5+=4;1 d) f8 D, @+ p! i3 \/ c" x
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);
2 o' r; ~; m1 R5 t7 J6 r}+ V+ m# j# X6 F& F; M1 o
3.8例3.6能否改成如下:2 X% Z2 Z0 o2 u" L9 ~/ W
#include* n O6 B9 D0 w
void main()
4 I; Q. s& n: N( B" `{: X0 E% y5 |" E. Y: p
int c1,c2;(原为 char c1,c2)
* I5 Y7 y# d# ^8 |# rc1=97;
0 s5 y$ I: x5 h! g) U( m" m% jc2=98;" X2 B5 Y$ B" @; _) N5 C
printf("%c%c\n",c1,c2);
) K0 e+ `: x2 q/ sprintf("%d%d\n",c1,c2);' x+ @" B$ {0 `- W6 s
}
1 t0 P) {2 l& Z. ^& ~% l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.
( Z1 T3 x6 I8 l/ U5 p/ y3.9求下面算术表达式的值." x* o0 @- c- T& Z) U2 J
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
9 f1 D7 B9 i9 o. k, w' _8 g(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
# e' D- w0 u% f, h; j R4 {3.10写出下面程序的运行结果:
' y. @2 t# C7 }6 O. h- \/ H#include/ B6 | O6 V4 Z9 x8 u9 ~" ]
void main()
; T3 L" C9 j- i{
5 {; `, I6 p" B0 e2 x' Kint i,j,m,n;
- m0 D: w. M8 H" r$ Ei=8;
# U. F3 \+ `4 p$ i' [. b4 zj=10;
E9 a" \9 U6 }; m8 x3 M a+ {m=++i;: u4 k; u8 D# ?. |: c+ V' T
n=j++;
4 K& j, E4 F4 A. d Z8 m; n# Lprintf("%d,%d,%d,%d\n",i,j,m,n);
7 E( L9 M" y/ n" G}
v# T/ A r; g! ~% ~4 R) Q解:结果: 9,11,9,10
# R( a# g y0 X2 J1 N, m第4章
' I! U+ g* T2 [" B4.4.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; [# B3 [& \/ e8 S' z. h 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:
V# H9 I% F6 F/ [% Ja= 3 b= 4 c= 50 p3 q+ z! g/ R* n
x=1.200000,y=2.400000,z=-3.600000
% A: J) Z/ _- Rx+y= 3.60 y+z=-1.20 z+x=-2.40
. U0 i2 L6 T8 f0 d1 x. _u= 51274 n= 1287652 ]4 |: n1 N% K; r8 [% _- k- Z+ z
c1='a' or 97(ASCII)
. E' T& X: P& H) b* k( Wc2='B' or 98(ASCII)
0 v8 r: r/ S4 N, q) h解:! T" b3 y @7 t V! P
main()
+ Q1 c% c2 O+ J( [% i7 t: I{
: F1 |% m, p( A% q2 nint a,b,c;; r% y; G: Z( T$ }& N+ w
long int u,n;5 \8 c" i/ y! p7 V. q
float x,y,z;% { b- k k( q$ P( b
char c1,c2;
3 F% k) |2 U1 W$ ?a=3;b=4;c=5;/ c( z6 ^, |9 b! B
x=1.2;y=2.4;z=-3.6;
& {; `) s; m# c( J8 _$ gu=51274;n=128765;& M9 f# I3 C, Z5 z% b4 [$ F. V, B5 T
c1='a';c2='b';2 l% l3 o! {! l/ ^0 I
printf("\n"); `$ G0 }) a' Z, y& P, G
printf("a=%2d b=%2d c=%2d\n",a,b,c);. K4 m* E5 m" B0 v$ e
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
4 J$ c0 _8 X2 z" _( N- _* mprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);
1 K& ~5 l1 A4 r) {0 M* ^+ d V' pprintf("u=%6ld n=%9ld\n",u,n);7 A+ h! j2 G' {
printf("c1='%c' or %d(ASCII)\n",c1,c2);
9 @6 d |: {8 fprintf("c2='%c' or %d(ASCII)\n",c2,c2);& b7 L% G# \5 a' s
}
; j+ v3 |3 V4 W4.5请写出下面程序的输出结果.
4 ?, x$ y$ x: r' o& m) o* z0 U结果:+ A4 L, L: @+ f3 b
57. L+ g6 ]8 w( }
5 7
: R8 s( y- I( b0 V0 ~; u67.856400,-789.1239623 N9 C, X8 P7 `; l- z5 R
67.856400 ,-789.1239620 y/ @) E9 L8 }7 g) s
67.86,-789.12,67.856400,-789.123962,67.856400,-789.1239629 X8 \4 C* Z0 `
6.785640e+001,-7.89e+002; p7 l8 `/ F/ i9 _' m' Z$ x& E
A,65,101,41% v/ I8 ? r& ]4 r! f1 b
1234567,4553207,d687
: v* [( o3 W" f1 @65535,17777,ffff,-1
' k6 n1 _7 T; Z! N6 C/ {2 y DCOMPUTER, COM
; A1 Q6 R# z5 R, O+ W" @1 y6 E4.6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',
& i: U+ G4 Z" x* z! @4 C问在键盘上如何输入?
2 ^' b, L8 ]" _main()
% l7 F Q+ k- ^4 P6 N{9 w! D: j' p7 m3 K4 b$ ?4 J
int a,b;
2 E, B' T8 h5 e. q! efloat x,y;
: Z5 n9 p& v. I9 pchar c1,c2;
6 b6 d2 U+ @# T4 j, m, lscanf("a=%d b=%d,&a,&b);
7 Y6 t7 }% z" d% Iscanf(" x=%f y=%e",&x,&y);
3 o' n' G B3 b! d/ H: oscanf(" c1=%c c2=%c",&c1,&c2);1 n2 K; s5 x# i w! a+ k
}: `7 x9 X3 w: k0 S
解:可按如下方式在键盘上输入:' V6 b2 R' v2 K! I( O q' n6 C* A
a=3 b=7& R* [* {3 B3 j) v* }# a
x=8.5 y=71.82
, q3 w u6 O$ g2 W& y: @c1=A c2=a3 y' j) Y6 }& m4 _- ^5 B
说明:在边疆使用一个或多个scnaf函数时,第一个输入行末尾输入的"回车"被第二
6 B6 A0 l) U2 |1 S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
" W2 B0 f' C! s0 [. H- l0 P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
$ z% q9 d7 e6 }初步概念即可,以后再进一步深入理解.
: O$ n' G3 j* ?: N$ k4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-
6 O# E" Z( j( y5 P8 s, }# U0 S3.75,z=57.8,请问
, }& e' j& t6 x7 ^在键盘上如何输入数据?
/ A( n% D M: |) [) lscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);8 ^1 ]% P9 w9 l' I% N" O* f
解:4 q4 N, J- K' P
main()9 A# d( `5 _4 d( K) C9 I
{
; x/ I) F- M9 l2 R( Yint a,b;5 ?7 ?9 D8 \5 `' Z
float x,y,z;) K! T5 | b# T, ]$ P; L
char c1,c2;
0 s+ e# p' x: ^" y' G/ lscanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);* C- W0 h- d( S0 m. U0 _
}5 m6 z; d1 Q! n
运行时输入:
' x' l( Q3 J) g. ^' R, J10 20Aa1.5 -3.75 +1.5,67.8) \/ D: }& Z/ T* x$ E3 n( s
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
+ \4 X1 f, W6 O4 D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4 C4 w M- n' O; x$ E3 }# [
会赋给任何变量。
) v) b& z$ B* e" K! f0 u7 [) x, L6 q j4.8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,
0 m' J; H2 J9 S" w1 ~0 M: O4 P/ Z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! B" _/ g3 c- v7 O: G
程.
+ O9 ] e ]3 l& Z, T$ ~解:main()
- V& _( |3 p, ?" D7 |{, Z1 I4 L5 h$ O2 j
float pi,h,r,l,s,sq,vq,vz;/ d/ h- S! U/ \: z& f+ ]! k' F
pi=3.1415926;4 ?0 ^! i, _: z' f
printf("请输入圆半径r圆柱高h:\n");
# s9 U: @0 f1 z$ E( Nscanf("%f,%f",&r,&h);, o$ _2 Z# d( v- E2 k
l=2*pi*r;
( j( X! K/ j0 ]# [8 m, Us=r*r*pi;3 O* c7 C9 s0 @& {4 h$ }
sq=4*pi*r*r;
) `1 f$ {' A$ k9 C3 B: rvq=4.0/3.0*pi*r*r*r;
9 A* O. n# h' cvz=pi*r*r*h;
; X3 D/ O R+ A: l' vprintf("圆周长为: =%6.2f\n",l);
; r% R7 v. D" c1 f1 yprintf("圆面积为: =%6.2f\n",s);
" Z0 N, V" r Y0 N5 ]5 ]printf("圆球表面积为: =%6.2f\n",sq);
2 |* y) w J* G! X6 Z+ L/ Lprintf("圆球体积为: =%6.2f\n",vz);
, J0 u' h) @6 N# i* e( l$ P}
7 n& f( g( x2 V( ]- p5 F$ D1 s9 d4.9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,% X J0 y! [' _7 T- N1 M$ {
取两位小数.
! v# S# e6 t+ j, ~+ e7 j- w# b解: main()4 y) C* Z) K! L) J8 N9 o- D
{: Q" W# Y' m; B5 H5 m
float c,f;: M$ _. f6 ^: D' Z! ]4 W) i
printf("请输入一个华氏温度:\n");% h0 l W. r& `
scanf("%f",&f); k3 A/ S4 I1 _3 G
c=(5.0/9.0)*(f-32);
- l3 n3 h% n( e7 ]) y+ y nprintf("摄氏温度为:%5.2f\n",c);
* S) m% h) J% W}9 H% q' u* ]& W& H9 p
第五章 逻辑运算和判断选取结构; a# Z5 n' G8 a& c' o7 @/ K/ S
5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.& p3 D; X" d9 R! h7 M
main()8 \8 \3 D! a1 m$ X1 e( ~) C
{
! ~! {4 `1 }. @& N) ?9 @ hint a,b,c;
9 w) v; Y* K. {5 |% V6 E( Y1 ]% mprintf("请输入三个数:");8 z7 {# b" b) H/ a: s, T
scanf("%d,%d,%d",&a,&b,&c);9 Z; Y$ c0 X9 L1 ~9 ?" A
if(a if(b printf("max=%d\n",c);9 K& o* Z* G5 H* L
else1 J' H' P9 O- H- u8 {
printf("max=%d\n",b);+ M/ {1 H4 @) O5 L
else if(a printf("max=%d\n",c);& C' f5 K) c4 z; j$ I6 S
else
( _* \- h; @4 O& g, g3 m( H printf("max-%d\n",a);
7 b; ~ [9 c1 u0 i7 s: c}
$ L0 X0 U1 b' T8 R方法2:使用条件表达式.
' \% {" K% Q3 q: G2 i+ {& e' I. y% Wmain()
7 O0 R* C% ~) [ \! X( O' O7 \5 X{int a,b,c,termp,max;
- J9 \, F7 B! y0 u- X: P& O+ }/ U printf(" 请输入 A,B,C: ");
9 y" u8 D. I) j, [ scanf("%d,%d,%d",&a,&b,&c);+ `- u6 [/ `( D
printf("A=%d,B=%d,C=%d\n",a,b,c);
# m4 E }# h+ S4 a( @6 e temp=(a>b)?a:b;: o5 ^( N- K3 b1 ?$ n! g. y
max=(temp>c)? temp:c;; Q, m1 g, q. B a
printf(" A,B,C中最大数是%d,",max);
$ }6 T" v) L% D: g. A: Y}! [" |& Z- b6 B4 o7 z8 T
5.5 main()# v% U7 _8 Y/ ~! F
{int x,y;0 T' f( Y8 T$ l) C' u
printf("输入x:");9 e% j' `( U) r: [9 E! e7 x% t
scanf("%d",&x);
$ U/ C# d' M& Oif(x<1)7 s1 N% g9 C$ X: ^2 w F+ M
{y=x;
( a. [3 f, C( M+ L5 x# Y2 O1 D. b printf("X-%d,Y=X=%d \n",x,y);
! K/ _* @6 a6 \1 `* p, H }
7 |9 ~% g% ?' E, P: B/ Velse if(x<10)& I5 \# p& d5 b# _2 z
{y=2*x-1;
+ [# q" }! C( U1 r' O% V printf(" X=%d, Y=2*X-1=%d\n",x,y);
5 f/ o" L/ }& n9 G/ u }
" [$ e# T3 }5 jelse
) ^" S5 T- E i2 A1 T1 z$ i {y=3*x-11;" h' j; y" V8 L+ }
printf("X=5d, Y=3*x-11=%d \n",x,y);2 A+ s; D, F, |4 b2 F
}
& R7 X8 r' i2 `4 b/ R. N7 e1 [5 \}
) l9 s" j# @* O$ H(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! ) t' {. U1 I3 Z ^8 D
void main()
# h/ e* _* R5 B- z/ O) X. }) \{" V; W; o4 [" w X1 J% f5 v
float s,i;
5 g8 g: _' y4 k+ Lchar a;& q: o8 |+ }4 _/ ?
scanf("%f",&s);9 G# X* B" J; |$ P! F& t7 v$ B
while(s>100||s<0) v* c! i: a- k0 D
{
0 V* ?# N* X* q# eprintf("输入错误!error!");/ s1 n) l* N; ~
scanf("%f",&s);
1 r5 n" [5 j' h/ R( w! ], E/ Q+ J}% W/ O( ~3 F# j7 M; c$ I
i=s/10;
" n |8 C2 p+ z. w5 ~ cswitch((int)i)
" a! K5 Y% j6 y) B: n/ U( ]{$ ~9 n- A5 s5 P& Z0 |1 ]
case 10:% l/ k2 b9 o8 q; _
case 9: a='A';break;
$ R, O3 t1 L; B; Fcase 8: a='B';break;4 f4 ~6 ~, U; ?) N
case 7: a='C';break;% r" C$ S( ]9 ?, L3 q
case 6: a='D';break;
7 ]. i! Q- J* \% F Z, D0 o" }2 _case 5:
# ]8 t& V" V, u: O9 R& E$ _ Icase 4:
# u& ~/ a+ w! T; j+ ?: Dcase 2:2 R, g9 d$ s' }* q$ u) D! o
case 1:+ Z, L8 c* }# v! d
case 0: a='E';) l' F( I T' N) t) y
}
' G) D C1 |* m7 X) r7 L9 Yprintf("%c",a);
5 [8 Y( ~) Y, V# u7 V, b0 {}2 }# D: A& N+ \1 m2 b' `
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.. M# S y* E$ F1 o( w
按逆序打印出各位数字.例如原数为321,应输出123.
+ o- N: W* [ N) Lmain()
! n6 @7 d- j$ k' j7 H3 } {: @- k8 m% c. t T
long int num;# C+ h; O& u! T/ L- M
int indiv,ten,hundred,housand,tenthousand,place;
6 l- W& Q6 I9 R( @7 r printf("请输入一个整数(0-99999):");
6 L! l+ W8 c2 R8 F' L9 o7 T$ x scanf("%ld",&num);/ I$ o. Y: O& L$ j* r: Z1 x
if(num>9999)* W/ f* E0 V4 s1 V
place=5;
+ ^5 L7 W7 q# S/ ^else if(num>999)
; L5 Q, j: ?8 n1 a7 g7 m5 u- ` place=4;! `8 D; h" t4 [, s; b" \2 [
else if(num>99)
# i& _6 m# j1 Q( H+ ?; v place=3;
% s- O8 h9 j4 y8 e* felse if(num>9)
* m, v5 Q& t- E" X0 F place=2; K" W" S# m* R6 z* `
else place=1;
( j# x1 y9 G) tprintf("place=%d\n",place);
5 L8 s& s, g& J" eprintf("每位数字为:");+ F9 M) K" G/ e( Y, v9 e
ten_thousand=num/10000;
* v7 t" E( _: w) K5 v* n1 Tthousand=(num-tenthousand*10000)/1000;
# K; ]" R% W ~% m/ c; Uhundred=(num-tenthousand*10000-thousand*1000)/100;1 o8 c6 x- O: y
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;$ F/ C1 o- [2 x4 u. h
indiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;
% R: F$ I. K# H! W, I6 L% R' Nswitch(place)0 I: s3 b6 q. M2 L- x
{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);$ t z v, ]" i# G# o$ ~
printf("\n反序数字为:");9 ] r. ?! ]) z8 G, U+ a! c
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);
7 G6 I# f- J) i$ {9 B+ n$ {7 c break;
3 f6 c: S, X% G( Qcase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);6 Q A0 W6 Y* g: R: p. Y
printf("\n反序数字为:");- Q! z+ ]4 S9 A# a" K' t% t
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);) |9 X+ w2 w8 s3 ^" O
break;: J5 c6 b# h' [9 x
case 3:printf("%d,%d,%d\n",hundred,ten,indiv); M& h: ]- `4 s: P& i+ ? a" z: v$ D
printf("\n反序数字为:");
( j c; Q( S" i* F8 n printf("%d%d%d\n",indiv,ten,hundred);4 ?, W$ K. _: q- C" \( a
case 2:printf("%d,%d\n",ten,indiv);
/ r: i; R+ j% F* q printf("\n反序数字为:");
% p! R y" S0 v( d1 ?) m# x printf("%d%d\n",indiv,ten);% M: q7 I, Y! T/ Y( y1 a" F9 Y6 }' c% @
case 1:printf("%d\n",indiv);3 Y! s, O7 x8 J" w: V b
printf("\n反序数字为:");
1 V# @7 @8 e0 Q( P printf("%d\n",indiv);$ z8 F. ~/ [/ ]2 |3 W
}
1 o- i' V) M( G3 q2 `5 ~# e}
, s" l/ c8 K5 S+ @4 M# p; T8 h8 x' \5.87 j0 t4 g2 P G/ V; E
1.if语句, {; @8 m; i: n5 |
main()
* }: t$ {: a8 `/ k/ M P1 o{long i;
7 G/ L6 C" E* w) | float bonus,bon1,bon2,bon4,bon6,bon10;( [- a/ m H: G/ {% x8 K" K
bon1=100000*0.1;
: u, p+ R4 X! k& x/ x bon2=bon1+100000*0.075;/ |# q2 d& ~/ X6 w4 [/ z
bon4=bon2+200000*0.05;
, [0 {2 p Y/ B! m4 L6 p" ], f bon6=bon4+200000*0.03;
; P( P- P' g/ T) j bon10=bon6+400000*0.015;
# i% Y7 ^) P: h: \/ ^" S scanf("%ld",&i);
; ]8 l/ ?" g a% Z6 t/ F7 V: Y8 g if(i<=1e5)bonus=i*0.1;
' B( ?7 H w( D; M/ `8 ~' M. Z l else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
0 c* x6 c. g* T& H. P else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
3 }+ Y; j- i) ~" A+ [, `- v6 h else if(i<=6e5)bonus=bon4+(i-400000)*0.03;9 [ T$ g" n0 G3 ?" V
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
& W3 w2 m- B- @) O else bonus=bon10+(i-1000000)*0.01;% C1 I8 u3 J9 N8 o" a) _5 s) B
printf("bonus=%10.2f",bonus);0 b4 u% ~8 A4 D. ^) I8 @- Y" L& j: ?
}
1 T7 i0 h8 b! A' c3 R用switch语句编程序
/ v, h; c) e4 J* e3 emain(), J/ `$ ? B1 _; L
{long i;
1 B- J1 [/ Y! P: Z. I' @4 M+ Q' o float bonus,bon1,bon2,bon4,bon6,bon10;/ {5 q/ K8 x1 `( O0 u6 n
int branch;* Z4 I. _3 {8 V7 i) I/ ^% O
bon1=100000*0.1;! ]* D( \# B# f4 j
bon2=bon1+100000*0.075;6 I; U/ k8 \$ s+ ?6 T! p
bon4=bon2+200000*0.05;
& w1 O$ @3 D& M/ k" V" G bon6=bon4+200000*0.03;
9 i# A! ^6 O# `- [' j bon10=bon6+400000*0.015;
: {3 m: {* |$ { scanf("%ld",&i);
$ g$ J# l' ~' i3 }2 r1 I branch=i/100000;5 S2 t3 i! c, I& P. H& G
if(branch>10)branch=10;
! X* Y" s- W7 n6 \ switch(branch)) x5 M4 K& R6 J% Y/ n* d
{case 0:bonus=i*0.1;break;
& d& W: u! f" T3 P/ O$ t case 1:bonus=bon1+(i-100000)*0.075;break;. U4 S- v. @& z
case 2:* M' i. {5 y3 u
case 3:bonus=bon2+(i-200000)*0.05;break;5 g n0 N3 [/ W/ R$ B
case 4:( |3 q! H3 o9 T( R; v- W3 N" E
case 5:bonus=bon4+(i-400000)*0.03;break;
* q+ z" ~0 A+ B; q. l& [* ~ case 6:
& ~' E& y0 R8 V2 | case 7/ ^: [" g( l. w. U8 P, C( u
case 8:% ~$ L+ D3 }; d2 p# S
case 9:bonus=bon6+(i-600000)*0.015;break;( {6 J2 J& r1 ^. E0 ~1 {
case 10:bonus=bon10+(i-1000000)*0.01;) ?/ }) ^. s t9 a4 z
}
! N5 ]9 o7 d9 T* [. A9 h9 |: l printf("bonus=%10.2f",bonus);. y# q+ t5 S7 T5 g2 O1 m: ]* u( Z
} 1 T* C3 ^8 o7 T) a* k3 U
5.9 输入四个整数,按大小顺序输出.
5 p k( G: O) e7 W7 |main(), z8 m( s. [, {2 q; `. r6 f
{int t,a,b,c,d;
3 X, }" O; T' W$ B- z printf("请输入四个数:");4 D: G. j+ f+ A7 C) h: u$ {7 K) p
scanf("%d,%d,%d,%d",&a,&b,&c,&d);+ r R/ Z- C$ K: J: `3 ?
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
A9 T9 m) F* ^0 V if(a>b)
4 Q$ c; b* ~6 V$ S {t=a;a=b;b=t;}
$ K1 g2 R) @" N+ Q5 q- } if(a>c). K+ ^2 t9 m5 _6 j- ]- r! r
{t=a;a=c;c=t;}# v5 h3 o" e/ t& |
if(a>d) O8 M s. X6 F9 B; _
{t=a;a=d;d=t;}
0 W, y2 u6 R {# r" u if(b>c), q" u# ~1 O) T9 U
{t=b;b=c;c=t;}
1 I2 o; ~2 ?. t5 _ if(b>d)
6 B7 ~0 S i8 _# u$ w {t=b;b=d;d=t;}
. S; c+ A. \5 b7 J4 i7 ^ if(c>d)
3 R6 A0 \0 i1 {3 P/ J8 O2 ~$ O. h {t=c;c=d;d=t;}
9 y' { p" a9 W+ I5 |5 r; Dprintf("\n 排序结果如下: \n");- y# \; I6 s5 T( b9 i! `
printf(" %d %d %d %d \n",a,b,c,d);% C. l$ g/ a4 y# x
}
" M3 x, ]8 w9 p: W$ A5.10塔
* ^" l$ X8 L! R. P' R8 }) B h- _main()
! L2 s2 ?& M- a2 s' J{
. T X$ l) M- J' Iint h=10;
) q: c) W1 B+ afloat x,y,x0=2,y0=2,d1,d2,d3,d4;% U! b! E1 F7 m6 V# v' K4 g C
printf("请输入一个点(x,y):");
7 W. f1 b6 f3 ?3 P0 Nscanf("%f,%f",&x,&y);
1 m8 O' y) h: k+ ~* K* f- `, Zd1=(x-x0)*(x-x0)+(y-y0)(y-y0);
) h1 @5 u5 {+ x7 y4 V i( E5 k7 a/ t' [d2=(x-x0)*(x-x0)+(y+y0)(y+y0);1 w6 v1 k: Q- j. l) y
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
/ w7 z( Y: Z4 S: b Dd4=(x+x0)*(x+x0)+(y+y0)*(y+y0);. I) t2 N, p3 S: H
if(d1>1 && d2>1 && d3>1 && d4>1)/ H ^& B1 B1 q; w
h=0;# J: t; s4 B# K6 z1 @% X1 G; w
printf("该点高度为%d",h);! ~8 y- g" ^8 a1 i4 k
}
! t4 g( _2 C1 U. ~第六章 循环语句- [4 q2 u6 [( a4 t; {
6.1输入两个正数,求最大公约数最小公倍数.
/ O, Z4 K0 { R% u. C3 I, Umain()) _% }. W" N- { v
{6 U5 |6 x# k7 l5 U# S/ o! T
int a,b,num1,num2,temp;
- a$ v; ~. _: }/ u) A u! Jprintf("请输入两个正整数:\n");* z2 u/ N' D1 y/ C" m' z$ v
scanf("%d,%d",&num1,&num2);
, U9 V& ~; F7 x* V& a& L) [& oif(num1{' ?# k. Q7 F1 g2 f# _/ x
temp=num1;. I3 l$ \9 j+ G3 s' v) v
num1=num2;
) s$ W7 e) a8 E" Fnum2=temp;
- ?7 e0 L% Z8 p& i* f ~1 {: k}
; m0 t2 s% J. \" Z8 n/ `a=num1,b=num2;
$ x0 W; i1 U/ F3 y7 Kwhile(b!=0)
/ ?5 ^ w+ ?$ f2 o7 k1 T5 |. r C; ` {
" {* E B. m- f) |' r2 d" J6 E' f temp=a%b;( p4 [' ?8 u7 c6 B" `$ [
a=b;# f I1 I7 N# s+ {. z9 B( Q
b=temp;- z5 K) j1 r5 }" Y+ A/ f
}
. s# ]( x: @) C+ jprintf("它们的最大公约数为:%d\n",a);
7 B( N p; U' B! F3 Nprintf("它们的最小公倍数为:%d\n",num1*num2/2);
; \4 w9 q0 A& M}
3 E5 P: n- K. j. q8 o9 o6.2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.
) h" Z* A. l5 Q6 P解:/ F0 c0 c6 X S
#include < >! O5 D5 K/ H4 @9 z6 r
main()
6 m5 O7 B4 c. E8 m{! {/ G9 T( W+ I. N- z: A; [- {
char c; A% h& n; K) h6 q; r3 N
int letters=0,space=0,degit=0,other=0;
8 T' Z$ ]) O# I3 z: L z% Cprintf("请输入一行字符:\n");
- h" C/ @4 q/ H1 [+ r% Gscanf("%c",&c);
* T5 h8 @3 F9 U# p8 d$ qwhile((c=getchar())!='\n')8 V* V2 T9 F( H) L$ s( H: h; @
{3 o. h) ~) y# m h/ C2 t# W+ J
if(c>='a'&&c<='z'||c>'A'&&c<='Z')( U6 a1 X2 f8 x* Z& @+ a/ R
letters++;% D, D: D7 [/ Y& Y7 Q# F
else if(c==' ')/ F; T- E+ J$ i) x. S% ]4 c
space++;5 Z4 `: N7 Y+ |* `
else if(c>='0'&&c<='9')' @6 W# n1 { p9 Y$ t3 M
digit++;
6 |9 J% a' [# Kelse
5 Y+ |4 t! x# G2 i- _6 mother++;3 R! }& v+ m$ x9 G" L# N L
}
' C0 b: r% K% j) Zprintf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
- n/ n9 b" c, @! u }4 i5 Q( ud\n",letters,space," P* ?- @6 w [7 \- V; p. H; w
digit,other);
9 B/ O8 ]7 |/ _}7 b5 |# g" n# T9 S
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.+ s; |8 W8 }7 r5 d# ?0 A
解:+ [* u% {; v6 O& ?
main(). M* [: I- D2 C* u( y1 e
{
; c9 i- P- W9 q; Q5 G0 q {int a,n,count=1,sn=0,tn=0;- m/ `" y2 T6 i- Q. F
printf("请输入a和n的值:\n");) S# Z3 f: G$ O5 O9 a7 P" h
scanf("%d,%d",&a,&n);$ Z( E; ?0 p) _( P! ~' `4 b, i8 @
printf("a=%d n=%d \n",a,n);5 M+ K/ a7 ~! ?! a* c7 n: {
while(count<=n)
% x: ^4 P9 z1 ^: t" z6 W$ Q{
5 f! n* j/ F0 b( n2 Q2 ztn=tn+a;
( @' Z/ {5 w8 B- M9 lsn=sn+tn;
9 N; W. p# F0 c Oa=a*10;" B# E0 d6 W1 u: e0 ^0 F& w
++count;% H7 M6 @( t3 P7 S F* R4 ~
}. D; M+ Q, A) B5 @7 V
printf("a+aa+aaa+…=%d\n",sn);+ {! v5 L# e3 f/ K) u
}
" |% T% E% i% t$ X6.4 求1+2!+3!+4!+…+20!.
/ | u6 \: C/ }7 b9 fmain()
@8 J3 q; m h3 ^3 a. j{
" n6 E4 H# h- qfloat n,s=0,t=1;) D7 b2 e4 |9 b% X+ d b. K
for(n=1;n<=20;n++)! H+ }6 m6 L4 ?" N6 P* L
{
- ?# [6 q* V! i% E6 [* @t=t*n;
$ M% o/ K$ ?: }' \4 w. Hs=s+t;% R/ I* j" _2 d3 G I, J
} S; n3 K4 Z) `2 u- r6 H
printf("1!+2!+…+20!=%e\n",s);; s* ?: e3 H% }7 o4 n) L+ {
}
$ }+ C! H9 S' O2 ~! T6.5 main()
6 g& ~2 H: K: S% `+ }{# w& y- I% ~8 ^6 K( Y
int N1=100,N2=50,N3=10;+ m, Z$ J7 ^. u" O
float k;9 `, T* l) c: B& h/ y; N8 j
float s1=0,s2=0,s3=0;
$ S5 R! {; c( ]5 R) |4 ?for(k=1;k<=N1;k++)
+ R" A; K" c8 c# E6 P0 F0 | ^% U{
) r s3 p9 J! hs1=s1+k;$ R9 z. A/ A0 f; B
}
0 O; v$ d: R8 w; R; |. R" hfor(k=1;k<=N2;k++)( i9 m N5 n1 t1 C7 B
{" t* }; d" _( H5 g: }, i
s2=s2+k*k;# r+ C) X J6 L( z2 D7 {
}
5 Z; F% `" Z( x7 N4 M1 ^$ {) `for(k=1;k<=N3;k++)+ M7 x4 B5 C, ?3 D" {5 z
{; c' J; \# w. V
s3=s3+1/k;" k$ }9 i! a. g2 y& d& Q/ G( E \
}
+ `" R" o) h& _printf("总和=%8.2f\n",s1+s2+s3);/ K) k2 s: ^0 v C; s4 r8 z
}
: h+ W& F( Y- `6.6水仙开花
: ]; n' T" S" A4 h* P. ~# Wmain()/ ?# C% ^: }, Q' k, d/ s2 j: R1 x
{
. J# k2 N+ r3 r C( {" F1 jint i,j,k,n;8 W7 o9 H1 m3 P# p; Z0 c, A; B
printf(" '水仙花'数是:");
- J# O. v& |' }5 R( Lfor(n=100;n<1000;n++)& H* [9 Y( A3 k0 C |$ K
{# v0 n; O7 r/ [9 s. `9 o# K @
i=n/100;, t8 |# \" ~1 [
j=n/10-i*10;
5 q: E( z+ i0 Hk=n%10;- y( |) ?0 G+ Q) V# @3 g
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
1 s( |7 f4 I; T" V$ H" ]/ r& J{
; m, T+ Q2 k" S+ fprintf("%d",n);' @' Y: i: i& K: F
}" M2 F1 l. J P' F4 A
}) X. o! U9 J6 c+ H
printf("\n");8 H7 G$ {, U: U( w
}& U1 H/ ^# ~( u) i. F% }4 `" f
6.7完数3 w. s2 E+ u, |7 E Z
main()
1 E( u+ C* Y9 f; f% f4 v( C#include M 1000+ M+ I$ G0 S5 w% }3 J4 s. S
main(). E9 g$ b7 C, ?- A% ]; s" V
{
" R% N0 k) A$ w+ O. ^5 x: d- nint k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;, n0 w: r2 i& B: n0 L; m
int i,j,n,s;
: U/ B& y3 Y, p7 {8 pfor(j=2;j<=M;j++); v+ l7 ]; T8 @, @' C' q! }5 f; a
{4 Z! D- ]4 y4 s6 c+ q( I/ J
n=0;
- n/ c" J' w' ?; a# x0 ^s=j;
, x) d8 ]: \5 [% j, N2 }% xfor(i=1;i {. b N. [% T# l6 W, ^% n& L
if((j%i)==0)
$ z' F( L9 @4 }' F {
3 S; {' C5 q1 K3 F if((j%i)==0)3 y! \" }5 ]+ j: O
{
0 |3 z! V/ q' C n++;
# O. }" N( y* D% l" n6 f s=s-i;
2 t, h" Z* ?- g* y8 U- d switch(n)
1 u# g& u# E6 w' N3 J {( ]! e! p4 ^; {7 W) q: a# k
case 1:2 X/ W( f( ?% Y: A1 [
k0=i;
) d( b: k. Q b, A! x* r( J break;
7 f; q/ @' Z9 }/ V7 A" v; i case 2: p* P+ M) b2 S5 K
k1=i;
. c! w. t5 y. O. o+ o9 N5 x+ n break;
6 Y. Y# E7 c. ^' l r+ I case 3:/ S' w8 I* N# A. T0 E! A7 o
k2=i;, f0 O; J9 j% l7 }+ v
break;
S6 W4 e0 \# A! o- e! ~ case 4:1 _4 A) K2 ^! D" G! y$ T$ _5 y. B% P
k3=i;
7 }1 k% Z' V( q' U/ g. h break;6 P) ^) ~% i! { b( S! t
case 5:
) \$ E3 T9 ]0 G% x; C+ f4 X% i k4=i;
, f. M! R# c7 h4 U0 d5 r$ J break;5 e9 r8 u* h6 \2 }5 j
case 6:1 y z, i6 @; a' `6 J
k5=i;
& K% z1 g) o! Q6 L- W break;
0 V3 i- |7 ]4 |2 b0 q0 N3 H/ l case 7:# J) z4 U3 e- @4 L
k6=i;
6 u# d6 u4 m' v+ j' h/ b break;3 M! Y" Q: J: t! }) n
case 8:
( t! G) \+ H9 g. _8 G k7=i;
, |) r+ l- ?$ w& Y. p; d. M break;
% t0 Y$ y: D! f Q9 O% G- E. U case 9:5 i% l6 u5 ^, K
k8=i;
~4 F8 P4 s+ X ]5 x) o. A; w# g break;
' L. F( r: X# E1 U( j case 10:+ K: w* B# u& Q* ^3 T
k9=i;
" V7 ?1 z4 n9 U break;. w9 ]7 |: M9 |+ h+ [
}5 ~, Z$ P7 D7 k, s( k2 q" s% M/ N& Z
}
0 Y+ ~6 f; q. V' Y5 L$ B }
+ Q- x- @& Q1 G5 j) tif(s==0)
% L1 b R9 C7 H" Y; a {* L3 i9 `6 }7 h7 |, j/ V L
printf("%d是一个‘完数’,它的因子是",j);0 z/ @5 j) Q3 u
if(n>1)
- w: }6 j4 b1 f2 V& A7 F printf("%d,%d",k0,k1);( u; G& Q3 A1 c1 K, S! a: M
if(n>2)
5 j. e6 |/ u. B' ^0 p0 X printf(",%d",k2);7 f! C/ T. N0 u% f* \8 Z
if(n>3)
" H5 J6 b! B' O6 H, x L, L& } printf(",%d",k3);
& `/ l/ \ e j H" Wif(n>4)9 W' H1 i- v3 v" p
printf(",%d",k4);$ Y+ z1 k' V! ^: H
if(n>5): {/ Z" } m! y$ d5 B; A; i
printf(",%d",k5);
2 A. q+ f" u! I1 [0 }0 U& b- _4 P7 yif(n>6)% d$ X& i0 o9 B2 r. p# P: w
printf(",%d",k6);
8 c' X7 {8 L' V- w$ m- Dif(n>7)
$ c1 M- j3 q' \& b printf(",%d",k7);% W: u( F5 i) D, \) @6 H
if(n>8)
9 Z6 p3 k( E+ n& @ printf(",%d",k8);
1 V7 l0 ]8 Y4 U# i S' R' e1 j Qif(n>9); J, {7 J3 k% I- G
printf(",%d",k9);
T3 m* ?* e7 o$ nprintf("\n");
: j' X' w- k. l: V& S! G4 C }
5 u# f7 g/ o. w' n) A# G0 V}, C0 @7 Q: N( ~
方法二:此题用数组方法更为简单.; T; _0 o$ `3 F% u, ?* a8 B0 ^
main()
* T- A$ ]& i3 c$ R& f: F{
. U2 }5 V" t$ G! u; K" Ostatic int k[10];+ J( I/ R, t) P6 O# t$ r5 P4 f
int i,j,n,s;0 V$ A$ o7 B0 E/ O; @$ |
for(j=2;j<=1000;j++)3 K: t, U, z7 T1 G
{# p, M, `# Q* E9 S. n& J! K
n=-1;, b+ F7 K" o1 R) o3 U
s=j;
; B/ ]3 f# R: h( w: C- d% Nfor(i=1;i{& d9 Q' j) E6 l
if((j%i)==0)
- o/ t8 X0 i; b* ^2 D{
2 h2 S4 _- s) _. J/ K* n7 C; a; Yn++;5 O2 f& O8 T+ B3 E
s=s-i;$ Q- j5 _- F3 u/ s9 l
k[n]=i;; L; t; {/ D' \% f7 o# V
} k2 p' c y% \* E% Q
}
O# W8 D0 e9 _$ \! a% X2 eif(s==0)
# b+ `; `8 Y ^+ ?" q7 i1 o{
* E# h" q, N7 u/ ~printf("%d是一个完数,它的因子是:",j);) b. y2 {0 J6 ~, h$ y+ h
for(i=0;iprintf("%d,",k[i]);* S4 A" N9 f3 t9 [9 G# a* X ~6 X
printf("%d\n",k[n]);' J! H R) A# I! L' h q! }
}/ z. q6 X3 F( c+ I" x6 R$ a
}. m% }$ L% i7 Z# f q
6.8 有一个分数序列:2/1,3/2,5/3,8/5……求出这个数列的前20项之和. ^4 Z6 }: e5 t& R0 x
解: main()
. H6 _- o& l+ T, m6 x# F{2 {) g- c2 H' Y- k% U' e, J6 E* E
int n,t,number=20;$ g, U' W U. h
float a=2,b=1,s=0;# a# V, m- R. J [) {, `
for(n=1;n<=number;n++)% \; A$ e6 w7 O% l
{
0 Z/ X3 l- `5 R7 m: Zs=s+a/b;% T [* o3 B1 ^ o# L
t=a,a=a+b,b=t;
% e( L d, a8 V# F t7 v}
# {" X& {* c a7 o4 vprintf("总和=%9.6f\n",s);. U% t2 P# L5 M
}
% G- \7 c( v' z6.9球反弹问题3 l6 x3 _! U2 H9 g' z) R5 k$ r
main()1 ]- p1 m' V" p
{7 C6 |" W; ]1 h. Q4 c, w
float sn=100.0,hn=sn/2;4 D* n, B" d$ O C3 m
int n;
, ~4 e* A- M; ]4 S, ofor(n=2;n<=10;n++)) h# E- u$ u: U+ m
{6 x9 C+ R6 f& V K4 U
sn=sn+2*hn;
3 j: i2 w$ ]2 v+ `* M5 R: Z7 Yhn=hn/2;: z3 J& G: P4 [" g; f3 l. j+ _
}
1 c; I% w! m- p( r9 A5 H9 _printf("第10次落地时共经过%f米 \n",sn);: K/ P7 s0 s9 z1 e* D. I# g. c: J
printf("第10次反弹%f米.\n",hn);
9 d4 r& \/ ]' `( D, P |% ?}+ @ _( } D% c. C' K1 j( M
6.10猴子吃桃
3 s2 p: s8 W0 Ymain()5 l3 W: {3 x0 _; ?/ V* P ]' y
{
% u1 h. W* o f) l2 Y* Q* [/ _( w% Rint day,x1,x2;3 S# T6 u* ~2 e4 B v! g5 z
day=9;( j' c3 B) E6 G; s6 ~
x2=1;
/ D0 S9 p1 }2 gwhile(day>0)" m S% U( |! w( E
{
3 M' U4 v( N' O; k$ ox1=(x2+1)*2;
" ~ \( k( t5 G: }$ dx2=x1;* E. `# ~! U$ d9 l/ b, C9 ]
day--;
9 U( _' t. v" O: f- T) T& l}' G1 s2 c4 j% a0 b! }
printf("桃子总数=%d\n",x1);8 l! R/ s3 B& W; P
}6 S. `! W- L8 J7 S- n
# G, F J) F+ S9 ~& @4 D. e6.12, i( i$ U6 \% V, i
#include"math.h"
$ z1 _- A2 I; b! Ymain()
/ T3 j- y( p! U, e" C, s" w{float x,x0,f,f1;
6 P8 [6 `' x8 p x=1.5;
5 w8 W/ v$ Q1 [! ~6 q( I3 d7 o do
' |! l% i& e( d' E) N: j6 c {x0=x;. }' a9 i2 H( s J6 Q9 q! `, I, h1 s
f=((2*x0-4)*x0+3)*x0-6;
* _: E; n: p0 \) l f1=(6*x0-8)*x0+3;
5 t! E K; Y. v2 i x=x0-f/f1;
6 m- Y. P% ^: m; {! x& ?& ^ }9 h5 f: I( K2 }! ?: v
while(fabs(x-x0)>=1e-5);
/ m6 Q3 w* V4 a0 u! ]0 | printf("x=%6.2f\n",x);
! O z% }, Y' ~ }$ u}+ b2 K4 P. a: H" R" C) K+ @4 ^
% |# l# J+ B7 V2 _
6.13
e" H3 R. [. @9 x! b#include"math.h"9 Y/ ?8 [4 M' ^5 f
main()
' a( E1 t2 P, i- ~7 _% n! m; }4 G{float x0,x1,x2,fx0,fx1,fx2;- O( e+ E4 Y6 `* ?% c
do
, H/ J9 G/ r: I' ]8 o& e$ O& M# E9 T {scanf("%f,%f",&x1,&x2);+ C2 E: a/ J! @. D- y, ~5 e `; L
fx1=x1*((2*x1-4)*x1+3)-6;
4 y, o0 F3 q- f5 p& b0 p fx2=x2*((2*x2-4)*x2+3)-6;
+ X! \2 ?8 y% ^) m- U }
# l, K( y {1 O while(fx1*fx2>0);
% y8 c* \! v& s* W do+ z" a' h' @+ x- q x4 s+ y% V" H
{x0=(x1+x2)/2;
0 U' e/ k+ V; M) Z6 k6 z1 r$ s fx0=x0*((2*x0-4)*x0+3)-6;+ E X' E0 h q( X, U
if((fx0*fx1)<0)5 J8 }+ F, ^+ {7 c* p9 Q. L
{x2=x0;
7 M: l4 T3 _8 z9 \& c. V. J fx2=fx0;" {5 l) k" C* h# ~$ r
}% C4 g3 f% q; Y+ t4 X
else
/ A: x. N/ g; w* k0 a" M+ `( y {x1=x0;
' @3 P* P1 t: d% H) {/ r- d fx1=fx0;
& [* `8 z0 c) _0 B( V3 U }
) N& |) u% t! O" ~" C }
6 M6 C7 x( C4 y& l' c while(fabs(fx0)>=1e-5);& Y% x3 D T* i" P7 j
printf("x0=%6.2f\n",x0);$ G& U, b7 B# z/ ^
}' Z2 c% B: x; J- R" \
6.14打印图案; u7 _* E* h1 Y- q: @
main()( o7 O N% n0 ~+ j) v, J5 g# r
{int i,j,k;! ~) Y; t. W6 @* g$ [; ~* j7 Y2 J( N$ N
for(i=0;i<=3;i++)) u8 K! W4 G0 v7 V$ w* W; c' v4 Q* S
{for(j=0;j<=2-i;j++)
& Z* w" g5 G, ~+ e# x y6 K& ^2 A printf(" ");2 {. z ~8 O% e$ ^7 t
for(k=0;k<=2*i;k++)
0 V* [5 d! c, G* C/ r& r$ x/ } printf("*");
# E. e) K: c4 a: E" @, J2 A printf("\n");8 u9 H7 ~( J/ u9 u& A6 o) ~; p
}
; F; R# V' r6 V ` for(i=0;i<=2;i++)* K/ z+ `8 J" m Y
{for(j=0;j<=i;j++)
- W! M3 V G A% q printf(" ");
1 g; l3 F7 d* U+ @( R for(k=0;k<=4-2*i;k++): E' ? m& o, @5 d. u! [1 T0 \
printf("*");
4 p3 }# V. v/ B( N1 y printf("\n");
9 K5 N* |7 P7 u9 g6 H* G! s }
" K' ?7 z2 y! V}
8 J1 C; L7 G$ w# |, {' C9 o( M, |6.15乒乓比赛' i2 J# k$ Q1 k1 Q: f7 h
main()) c' {+ D8 e7 X2 d* f$ Q9 F
{) f/ q. q' |9 M+ f y# x1 W
char i,j,k;
( x' q% T5 h, \: [% P8 i" Mfor(i='x';i<='z';i++)
# o5 o r( W" ~& L" Ffor(j='x';j<='z';j++)
7 ~7 ?- x3 |* T/ Z {
5 G& {6 Q8 S- Xif(i!=j)
/ t5 k9 B' [- z1 p% Tfor(k='x';k<='z';k++)
% V# e" U0 l6 f3 q. |! j4 x4 }4 } {$ t4 F7 m+ {7 O' T7 ?
if(i!=k&&j!=k)! R* x' ]9 v; l1 l/ T2 k
{if(i!='x' && k!='x' && k! ='z')
! ]6 m2 v' n! K, K* Hprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);
6 K% `9 ]. M1 y6 W5 V }9 J5 M- r/ @1 J2 y* Z8 h
}
( y1 ~6 [8 S# a% R }
7 X) y" @) I' s [) G& j}
3 l2 m4 @4 k0 F+ L1 x, k: PC语言设计谭浩强第三版的课后习题答案
: V' ~" k0 a+ ~3 ]' s3 s' m9 O. X7.1用筛选法求100之内的素数.
/ N# B8 c; a! V) v/ P" j#include
6 ]) \8 a6 f( U#define N 1017 ~: T' A6 o% n9 s. @% H
main()
' Y% a9 k7 r' R, \2 K{int i,j,line,a[N];( I# g8 `$ H8 T/ w, l# U* C! H
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
: l, H9 E9 K a) J if(a[j]%a[i]==0); s2 P0 u2 t2 t4 v; D
a[j]=0;
$ }' M0 ^, V. j; d$ @7 [printf("\n");
( G5 ]6 H+ H0 j% U! l) r/ b* xfor(i=2,line=0;i{ if(a[i]!=0)2 o& ~: G7 F, C& E4 X7 }. w
{printf("%5d",a[i]);% j$ L; e& U+ ]5 C+ N1 @5 t
line++;
" P9 K- N5 |( p* e( Z6 Z% u if(line==10)
- Q" X' f+ K# A {printf("\n");
) `2 G9 ~ w3 d/ { line=0;}
9 }/ a x: U$ Y& H }+ ^! c) L7 W: G: \. }9 }2 i
}# k4 w# [7 @; P9 p
7.2用选择法对10个数排序.- e% Q1 J' h9 C: [3 g
#define N 10
8 N( f3 ^- j& l$ Imain()% w7 ]3 N( e% L/ ^; R, P
{ int i,j,min,temp,a[N];
( f$ J; }# H5 R. t* ^; M4 u% {3 j& Aprintf("请输入十个数:\n");
2 Q0 ?6 r& u& T9 p5 u$ [' f& R( Ufor (i=0;i{ printf("a[%d]=",i);
$ A& P' _% S! c! w% J8 h5 x3 { scanf("%d",&a[i]);
# s6 \) p0 \" [& F4 _7 D% _1 x}1 P5 o: h8 ?! R& X
printf("\n");
9 }$ N& r( _5 ~: O$ Y w; gfor(i=0;i printf("%5d",a[i]);
) T/ H+ v) c; q; q; @& ^printf("\n");. Y/ z) y* ]+ _4 v
for (i=0;i{ min=i;
4 [. u( @: J! q for(j=i+1;j if(a[min]>a[j]) min=j; D: u! P" u2 |7 g5 F+ `& R; w& L
temp=a[i];
" I. a# ] G, e% k a[i]=a[min];
9 K. R* J: B' B a[min]=temp;6 @' G. H( A$ Q! D* i& L" ^6 E
}
/ K7 {6 {" C, N+ {printf("\n排序结果如下:\n");
& K8 L6 r7 d0 e+ Ffor(i=0;iprintf("%5d",a[i]);
* ]/ a& J d' ~2 c7 e" v% _! V}0 M9 D/ G% C% F4 n. L
7.3对角线和:/ x# r/ V( k+ k* `( Z
main(): Z% A% q. [( ?' x; u
{2 A4 |# x5 W: U3 T1 W8 g, N
float a[3][3],sum=0;
S* d8 M# B* Aint i,j;
8 w7 m5 l. f% B+ tprintf("请输入矩阵元素:\n");% k P7 D7 P2 O* O
for(i=0;i<3;i++)
# B% a( {; |7 q4 ^8 y4 s* N6 Q) e for(j=0;j<3;j++)! C5 q: p9 C' r
scanf("%f",&a[i][j]);/ P4 q* E9 F1 `% e* o
for(i=0;i<3;i++)$ g- r+ s0 @% A5 K
sum=sum+a[i][i];
3 g$ Y1 V9 C* o# K, ^- G3 I printf("对角元素之和=6.2f",sum);
- ~" p( z! N& h. {8 ^3 y}# R; A$ x8 G% {
7.4插入数据到数组+ s( h2 V2 y; Q+ q5 U
main()- H! U1 ?; s5 Z" p' t5 L/ c
{int a[11]={1,4,6,9,13,16,19,28,40,100};
) A+ v4 }. [$ k" H' @+ c8 \1 Tint temp1,temp2,number,end,i,j;
. Z& E" a5 S) U6 Z& A4 O* Mprintf("初始数组如下:");, n% X: A8 Z: j9 w: l
for (i=0;i<10;i++)
& i4 D% S! ~( k) U; ^8 P! Gprintf("%5d",a[i]);
; b7 W6 h: y5 [9 Nprintf("\n");5 }& ~! p: H4 P7 i/ _0 n4 z
printf("输入插入数据:");, i7 m( F" D) o& a2 e" E
scanf("%d",&number);
" P" e: v* X) D+ _2 v) \; Eend=a[9];4 w3 o' J( a E4 {' c7 e
if(number>end)8 \0 C: y0 p0 B: I4 V- ]3 Z
a[10]=number;' S0 Q q+ k) ^. D) O' t6 f6 R# w
else
$ ^0 e% ]5 z( I* }$ L+ d5 R) t0 [5 I2 _ {for(i=0;i<10;i++)
. L+ Y f5 t( d# t# }7 J { if(a[i]>number)
5 v0 Z" \2 ]: I Z( t; x {temp1=a[i];
. }& z$ t- M; \, a9 _ a[i]=number;
3 f1 k- d* T% ?: O, t1 h for(j=i+1;j<11;j++)+ p3 w: Z; n, a% S1 g* _
{temp2=a[j];
8 \' _0 Z5 n6 A1 d O a[j]=temp1;2 z2 U& G4 G7 u4 V1 H0 E, s
temp1=temp2;
V& v8 j+ s, v" M- F# A }! J# v" N: I8 e5 I: W
break;! w& \! `1 y4 ]1 ?# x# H% b5 f3 [, N
}
! v/ b" I; ^7 L( A- X% y6 k( I$ Y }7 P. P3 t9 g; g g% s
}
! V7 S# ^( f+ o( o7 ]0 x8 n for(i=0;j<11;i++)
* r; R. r; R$ m% c printf("a%6d",a[i]);
. L6 @8 N8 L {}
0 I( g- K" r+ R6 X* Z4 S% T7.5将一个数组逆序存放。- d* \! [$ v) |3 P
#define N 5/ W. c$ L! {' c% Z$ u. W( s) P/ f
main()9 m+ s+ n' d3 M }8 e
{ int a[N]={8,6,5,4,1},i,temp;, ~! S! b$ _( T% [: F" U
printf("\n 初始数组:\n");
& M. r4 ]$ E/ I2 afor(i=0;iprintf("%4d",a[i]);6 p- s4 n$ ` U. M) ?8 s- d6 |
for(i=0;i{ temp=a[i];
0 U: Y6 R" M. ~! C a[i]=a[N-i-1];5 ~9 t' V2 t+ _+ i; v$ f
a[N-i-1]=temp;
) a' w$ e$ B% q! ]& S}5 z: w. E; B+ u5 @" N
printf("\n 交换后的数组:\n");
5 O' ^9 z' _: ^+ f/ c, }7 U! V( Pfor(i=0;i printf("%4d",a[i]);9 h m, q1 g, Y6 ~
}
5 N8 v2 `8 ^4 m! L7.6杨辉三角7 r4 z) h4 a! o9 }
#define N 11
" L4 k: H& y k' T2 a# v! X. N6 smain()
) o# L5 W1 |" N& O7 m8 [, T4 W u1 O: j8 J{ int i,j,a[N][N];
0 B% e. ~2 I7 |8 Y) z for(i=1;i {a[i][i]=1;: @5 M& {" [7 z. ?0 m' O
a[i][1]=1;( T, }+ t9 I, j- y+ O7 V& [; r
}
( U# k4 D- s: C# n. ~ S8 W$ E for(i=3;i for(j=2;j<=i-1;j++)
! d0 t# g6 P) [$ K/ J+ P) f/ F a[i][j]=a[i01][j-1]+a[i-1][j];
- b4 W1 y: U6 n0 q4 J" @0 M ~4 ~ for(i=1;i { for(j=1;j<=i;j++)
* W& ]8 S, Q& c; q p- X5 z" L printf("%6d",a[i][j];
, \! E, O1 d: m5 t& f9 O printf("\n");0 G5 S8 h- C# O; t8 M: N
}
4 j; y4 ~9 y. s' B7 {: l printf("\n");
2 P) W% w7 N k0 C* ?}) y9 \( i# {1 q2 ]
7.8鞍点
( S/ i& ?5 o3 k% v#define N 10
2 d: E j1 Q! y' q; E0 N#define M 10( G' q8 j3 T! F) r. }, \ x
main()5 F l& c$ t* a9 y5 C7 ^/ z/ x
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;3 h$ p% Q/ m; d4 x4 h' d
printf("\n输入行数n:");
6 V* c" z+ k. T/ G0 N scanf("%d",&n);
* j! p8 b' i# M printf("\n输入列数m:");* M- [% `. G; I R# m( E; y
scanf("%d",&m); T1 ~% G# ^' j2 a5 \% u' Q4 [
1 C1 @$ F/ t7 b. |5 M for(i=0;i { printf("第%d行?\n",i);% k! Y: B8 _5 k) N
for(j=0;j scanf("%d",&a[i][j];
/ `6 s4 k w- P }) I5 X# w' d1 B1 v: c
for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
3 Q' W0 t: a, w7 o8 V pritf("\n");
# i; } Q8 Q/ @2 Q- v+ | }( A+ G2 E% e0 o6 [* ]* {
flag2=0;
' k/ \# _2 x9 [5 n$ | for(i=0;i { max=a[i][0];
. a$ ~6 w! z) \4 i: S; P! i for(j=0;j if(a[i][j]>max)
1 K$ z$ r4 M; E8 W' e" e% I5 z { max=a[i][j];* r1 G+ N) }( a) G% `( E5 F7 E9 n
maxj=j;+ \8 D6 s+ j( C. t b9 k
}5 D E( p% B/ |+ \
for (k=0,flag1=1;k if(max>a[k][max])6 V5 v- q, V' v6 ]! S1 m
flag1=0;
( l% T* {5 [0 y G5 } if(flag1)
9 v& E A E3 ~1 ]8 q, g+ o$ Y { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);6 f% N0 ]3 c0 Y& W! T
flag2=1;: a* L+ D: X8 |1 D% l) b3 W
}. M2 o& v, \! Y9 ]
}9 U* U* C+ j3 W& _1 g5 F
if(!flag2)
7 B* Q: J7 o# v$ ~7 \9 m6 E; x printf("\n 矩阵中无鞍点! \n");
. H- E! b6 I0 K* J; w2 Z- H}
% K6 _7 X/ P) r) V. Q& K4 e$ j5 \! ]0 H
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.
+ u$ _- B3 G' _' G) F; r#include2 @' }4 G! d: N7 W- P( |! Z
#define N 15
4 [4 q: p& h5 I# o V6 Omain()
/ f7 @" O* @% w1 n& K{ int i,j,number,top,bott,min,loca,a[N],flag;
& N2 @( Z6 v! Q5 L char c;% ^+ N: Z7 U! ^) g0 f
printf("输入15个数(a[i]>[i-1])\n);
7 e+ c$ s* A4 J scanf("%d",&a[0]);' x, |. Y) M& H, W Q
i=1;
; W6 z/ G5 r6 R# w+ F/ F1 a while(i { scanf("%d",&a[i]);
; e. ^3 f+ P# J, S. D8 G2 O1 ~ if(a[i]>=a[i-1])# `) |+ y& y" x: _ c# g6 x
i++;6 ^! o( S( Y6 n
esle; t7 M* S% J! t# r+ B" a' d* r7 D
{printf("请重输入a[i]");
: u" @# ?, X+ j, {) q: z' l; c printf("必须大于%d\n",a[i-1]);
! A X+ J/ K' o2 ~$ j, v }
3 a* C. [" c: W }
3 ~8 z0 c" L- a! c7 s% |- H) T0 ?/ z printf("\n");8 i9 a) T3 S" R6 E6 D
for(i=0;i printf("%4d",a[i]);
4 w" B6 ~# X$ D ^& Y printf("\n");
$ ?% G) g% L/ ~4 j+ R 9 }3 b* Z# ?, K% w/ m; e* z
flag=1;, u% a1 K1 }0 l% ?
while(flag)
, F. E i2 e% p, J# w5 z1 l {
/ O1 P% J0 A2 C9 [5 r3 w printf("请输入查找数据:");, V5 P2 k9 t, H9 ]- m! Q/ |
scanf("%d",&number);' n0 l1 @0 t3 Y, i
loca=0;! K u8 f& s( |# E, z
top=0;1 [! u0 }6 a( H/ v$ i( a
bott=N-1;
- r r3 ?" G# y! ^8 q0 W# b4 X if((numbera[N-1]))9 b- O: M# g0 N( _" w) i
loca=-1;
h" C7 M5 P+ Z( w' g, d while((loca==0)&&(top<=bott)) t b. Y0 B5 C$ d
{ min=(bott+top)/2;
) l! G" N* L! g if(number==a[min])9 T( n" b7 Y j: ~- W3 [$ u% J
{ loca=min;# z9 E# v% }6 Z4 c) g, n5 }
printf("%d位于表中第%d个数\n",number,loca+1);' G$ E; ^6 u. w" E
}
/ ^+ i1 x8 g7 ` H. S else if(number bott=min-1;2 x( b# u4 E* o" [6 v
else
) M1 {2 f& o f- B. s! M- D top=min+1;8 q3 Q/ o w, C1 a
}
/ @" F4 A v$ o6 { if(loca==0||loca==-1)) j9 h8 \! K R0 d; _8 _6 D$ l
printf("%d不在表中\n",number);
V f* v; q: ~6 b printf("是否继续查找?Y/N!\n");
6 g/ G: ^+ v4 k. T c=getchar();& E' p2 [4 z4 R( V3 E D& e! ^: p
if(c=='N'||c=='n')- I: x7 y' a' }$ a
flag=0;$ U$ z) y! k0 A9 c4 A7 Q, n# p4 P
}$ |3 S* ?1 y* A( G, b
}( Y& j# j) @. J, j M
, Q* l5 o9 L9 z/ q d( q, I7.10
1 E9 K8 k- z; ]: p5 r" xmain()
7 ] }8 D9 B" i% m{ int i,j,uppn,lown,dign,span,othn;
# S5 V" n0 R1 x) r char text[3][80];
& Q; m! l9 K( ]2 N- o7 ^3 P; M uppn=lown=dign=span=othn=0;
+ k0 y$ S6 L# n5 A$ Z* u for(i=0;i<3;i++)" P8 L0 q) D, N2 ~# h' e( N
{ printf("\n请输入第%d行:\n",i);
0 R2 Z' H+ D" Q% i8 `0 V$ T gets(text[i]);4 A% b$ L3 d8 \3 R
for(j=0;j<80 && text[i][j]!='\0';j++)4 ^8 b' ^; ]+ u% d
{if(text[i][j]>='A' && text[i][j]<='Z'): g* P% h8 A. e+ o6 h( S
uppn+=1;
3 T) w# B3 F" M& n6 l9 o else if(text[i][j]>='a' && text[i][j]<='z')
0 [; W; D, U" `5 s: h g lown+=1;7 ], L4 j0 o3 L" u
else if(text[i][j]>='1' && text[i][j]<='9')/ a+ R9 j3 _6 P4 d
dign+=1;1 |5 F( _; ?1 b& x3 }4 P- m
else if(text[i][j]=' ')
9 f) d9 X. b3 p) X% x$ W span+=1;
, Q0 ?8 r1 B9 x7 ?3 r else
# F: z3 y$ ]; `+ d; }, c othn+=1;
+ u9 f) W, W% b$ M2 L8 G }+ x7 A( c! b9 r9 V3 y
}
6 T$ Q. H5 K6 a/ h: O for(i=0;i<3;i++)
+ \# _7 L' A+ g% ~4 q7 Z F1 E! p printf("%s=n",text[i]);
, E' Z, T" N7 K6 ]$ y, L printf("大写字母数:%d\n",uppn);
- _+ d: L$ B+ P8 y printf("小写字母数:%d\n",lown);# R! B t9 H$ x
printf("数字个数:%d\n",dign);
& D8 }! E& v1 u# q1 x# T printf("空格个数:%d\n",span);
/ G9 T" P! t& T printf("其它字符:%d\n",othn);6 G' [7 V+ J$ P, @
}% s! }" }4 c% L
3 n) A0 h8 f4 R# J4 q
! F7 Q. K! `( y
7.11
' y! P9 X- i$ ^& b/ emain()0 B& a. q3 u# f! N: E9 l' c- l: T
{static char a[5]={'*','*','*','*','*'};
. b7 E' d0 b8 G3 G4 n7 Y int i,j,k;
1 F$ ~" F2 F1 m5 \: N }' Z3 p, \ char space=' ';2 f% K) H- U0 ]( s+ }! c' a
for(i=0;i<=5;i++)+ U- M! r e' [8 t% }* A
{printf("\n");5 Q. J8 L( S( G- F$ Q6 O
for(j=1;j<=3*i;j++)1 |& t% @7 `* Y1 C2 ~
printf("%lc",space);
# G: D c8 R) G1 c for(k=0;k<=5;k++)* R! \' _5 q. d7 x) z/ |" G# w. z
printf("%3c",a[k];: X- y1 o' W& [7 y2 C, C& d
}
4 E* z$ V8 T! }0 g7 ]. F}
5 r: E6 b! `" ]1 a$ r7.12
# [6 E1 }* J- d; c+ \% Z/ C6 K) ?#include! L5 N d3 C/ z6 a$ h/ N
main()5 e- t! [% X7 E* }) O6 L2 [
{int i,n;1 l5 e( B8 E+ o0 z' m
char ch[80],tran[80];
' c1 F4 j3 z. f9 ^& l3 n( M printf("请输入字符:");# n$ Q. U6 S4 \' {& M4 n
gets(ch);
+ h" O5 n& @: D# Q2 J5 I, L6 c; E2 ~printf("\n密码是%c",ch);
" d7 r* ? y7 Wi=0;
* H2 C1 O* r7 q u, S2 Y/ Mwhile(ch[i]!='\0')2 M+ K( U7 ?6 F& \: Y
{if((ch[i]>='A')&&(ch[i]<='Z'))2 \ n$ E% m/ g8 l9 E% y( K
tran[i]=26+64-ch[i]+1+64;
# w- }6 d; w) B; y; Melse if((ch[i]>='a')&&(ch[i]<='z'))" a- D5 F# p7 T6 O) m5 O) p. D
tran[i]=26+96-ch[i]+1+96;
2 }& o) Z$ B& J' @else
3 M6 G* `3 i2 h. a tran[i]=ch[i];
/ v3 V0 c# O8 R i++;5 m; z( y# C1 |! g' Y1 P
}
: z7 z0 K1 B' i" D6 bn=i;
/ Z" r; r* f9 L& Vprintf("\n原文是:");+ \' L* U1 p7 D8 \7 _
for(i=0;iputchar(tran[i]);
9 ?2 H* h1 j6 K7 p}
! M3 Y3 _4 {. {0 Y! [7.137 K+ E, x. \( o) s6 X* ?! o. O
main()
2 n, {) r$ v( y" u( ], e1 u5 u* x {
% j# o- C, ]9 S char s1[80],s2[40];
3 z4 ~2 s) D" w int i=0,j=0;
2 j1 p7 r' o, Q" ? z printf("\n请输入字符串1:");
+ a0 ~- K5 h- k3 b. D% s3 F. M scanf("%s",s1);
" Z$ p6 _/ F3 \* d0 X printf("\n请输入字符串2:");
/ G+ g' M$ d2 Q* W1 i scanf("%s",s2);
" T: K0 r3 c' h0 W3 f0 t while(s1[i]!='\0')/ M0 c; y" y/ y
i++;3 j* r: m2 `, N0 [7 o
while(s2[j]!='\0')0 M7 B o4 ?* G. }
s1[i++]=s2[j++];
1 `) ^# a: f0 h) ?7 q, i" ?s1[i]='\0';( P7 W r6 A- V3 x, f" u" s
printf("\n连接后字符串为:%s",s1);
& m8 r( h' C. E/ L1 U }
; o' @% p c9 o$ }# w' `! @ T4 \0 r( _+ x
' x5 ^5 Q( d5 @' a, P7.14
" Z. z* V) q' h# z: I# X5 N# O#include
6 g8 [5 l7 a" x" y- jmain()
4 W4 Z2 q) S+ ` t( p! x{int i,resu;
7 c, T7 I$ [% W char s1[100],s2[100];
' g" l9 P0 V/ S+ q1 N# @ printf("请输入字符串1:\n");. \) g' J5 m) y
gets(s1);
3 j- p% e6 u$ G% I printf("\n 请输入字符串2:\n");
1 \1 X Q X/ ? `! o! v' I3 C gets(s2);
+ a5 o3 W5 }# ]3 R" y/ N2 M i=0;6 x" y6 u2 f( H# p: d
while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;
}5 F- Y" @, k+ D) Q if(s1[i]=='\0' && s2[i]=='\0')resu=0;' R0 }0 i( G e! g- z- n+ A) `0 l# y
else: |1 G8 @* O$ i+ v6 m3 [
resu=s1[i]-s2[i];) w! m' w& W6 H6 ^% n: ~
printf(" %s与%s比较结果是%d",s1,s2,resu);
- j1 Z: m3 l& d( \7 z& {}. p% u R6 e( s, l. V E
7.15 K+ y& \0 S/ X7 P4 T7 x
#include
( q, B9 r1 G u/ h Emain()
! |- l9 V& [" x0 n" N {
# T R: h6 f% g6 J) r, G char from[80],to[80];, e O/ x7 `1 J
int i;# \, h! l/ ?+ I3 d
printf("请输入字符串");
1 a3 y6 g5 s2 J Q; ] scanf("%s",from);1 L; x( R( {. j' _7 s# u m
for(i=0;i<=strlen(from);i++)* ?; ^% Q& k4 b7 I
to[i]=from[i];0 ^8 B7 y9 X$ D
printf("复制字符串为:%s\n",to);# [& } g9 V; w( t
}
6 ]7 s& g' S0 K3 n: Q: k" X! j6 n; C8 g3 f9 K7 r) f% G
) h6 t# {5 i% P9 J第八章 函数
% z) ~6 x8 T/ `8.1(最小公倍数=u*v/最大公约数.)( b+ r) X1 ^5 l; H
hcf(u,v)
9 j* |4 n! l# f7 @ @int u,v;
/ |% B) Q5 e H6 ?, O" s(int a,b,t,r;
' R! h; L' @" F# h7 F9 R if(u>v)
; L4 r% H9 J @% L$ s8 M {t=u;u=v;v=t;}7 E/ Q; A9 o J" S( [- t
a=u;b=v; M3 A, Q9 C( ~% l
while((r=b%a)!=0)
8 c2 O7 i9 v' k$ V- T {b=a;a=r;}
% F1 x6 z& N# r. N: l return(a);: V( i% W/ m3 K* n4 _9 C! z3 z; j
}4 O, s9 Z K2 O: [
lcd(u,v,h) k( [. M# ] ?% {/ d
int u,v,h;/ |. N: t( X! Q8 s" l6 J' e2 r
{int u,v,h,l;
1 |5 i( P+ O5 b7 t1 Z scanf("%d,%d",&u,&v);* ?1 O$ l# X2 K; V1 `
h=hcf(u,v);( ]% y+ c- t, ^$ a, }- k
printf("H.C.F=%d\n",h);
2 N; T3 }9 D0 i: ^: }0 z B l=lcd(u,v,h);
/ W% o8 p9 i! t' ]0 u- [5 O+ e9 _" ] F printf("L.C.d=%d\n",l);& ?8 _1 m' f& ~9 Y
}4 U! g! a" N, t' I9 V0 u
{return(u*v/h);}
! D) V: K5 P' s. I) \* e main()( n6 E" s: J$ O2 ?" f
{int u,v,h,l;
1 n: ? u' w. D# M; M4 ]* I scanf("%d,%d",&u,&v);" f* f/ C2 e( {7 M2 L# U
h=hcf(u,v);
, R- }8 P8 y' ]2 n, R7 E. G printf("H.C.F=%d\n",h);- }3 v$ K4 Z# v C0 E: G" m$ i
l=lcd(u,v,h);3 m/ ?- r; L8 i* I0 e s9 W
printf("L.C.D=%d\n",l);0 l; `* S! Z9 r1 I
}, O; b J. D/ p2 g
/ r; P b; B* L; Q! r) T
% U( M r& N$ [1 H- o
2 ?2 U; `, N* m& w8.2求方程根
8 R2 k& }, U! ?#include8 t& N3 c( i( c% t4 `6 t
float x1,x2,disc,p,q;7 r9 Q* w! ~- ?2 _! k3 g
greater_than_zero(a,b)- j9 F. \' M6 J7 w2 J
float a,b;6 H; L$ Z/ I) b" C$ h7 S
{' i h; }# d! K5 }" ~
x1=(-b+sqrt(disc))/(2*a);! ^6 C4 _/ i1 _1 y
x2=(-b-sqrt(disc))/(2*a); n* [0 U0 X U8 n
}) z) A( @ G! ]% a1 X
equal_to_zero(a,b)) V, E* h3 q& |& J8 t+ F' f3 q
float a,b;# ?( l* o. T& @/ n% x) `$ b
{x1=x2=(-b)/(2*a);}
) ]- D ~+ {. Y/ C4 [7 {5 Msmaller_than_zero(a,b)
2 {/ _! n) b7 l+ Y) zfloat a,b;
& q9 l3 g( |8 C' Q4 d{p=-b/(2*a);
2 p* a) E6 ^+ v3 G `$ [q=sqrt(disc)/(2*a);
" ~8 C% p* p2 q}6 N" p$ x3 M4 d- v# C) o$ m
main()7 s8 K4 u8 v! |3 t( G
{2 E$ d+ b* i! c T2 ^4 y
float a,b,c;
; y8 c+ Y3 X: D F4 q1 K7 Uprintf("\n输入方程的系数a,b,c:\n"); q1 d" D8 I. q+ P: k
scanf("%f,%f,%f",&a,&b,&c);
5 ^3 s1 @, N4 e5 w3 E+ U: Nprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);
" f; C! F/ W- X8 L, l5 Ddisc=b*b-4*a*c;
6 C8 ?' o4 V7 I, `printf("方程的解是:\n");
1 [: m+ R& \% j# [- `if(disc>0)
& r% O6 M! R" m2 ~( S0 w9 v{great_than_zero(a,b);
5 B. U4 q3 |+ G* a; fprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
0 T# N5 L' s# P5 `}
% }( v) p- o; [& M" }else if(disc==0). x2 b. j, `& [
{% e, c; ~- z- P0 V4 Z/ |
zero(a,b);5 `# [& l6 ^' v4 Q$ ]/ V/ _7 ^" i
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);3 B7 P4 r# l/ B3 `3 r$ Q
}5 r4 s; p2 ~ \ U
else" y+ l9 p- N+ y+ v1 D& t" z8 p. G
{
; p8 h* E# D! ?% ?" H small_than_zero(a,b,c);
+ D v2 Y" k( F printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);! W6 Z/ l) Y. C; Z; O: t# g
}" Y- A9 H) _* l% i
}
. z0 ~/ J* z I8.3素数
$ \% _- @' n# r#include"math.h"
; M y% K# u4 r! Wmain()
[/ j @$ } u$ g3 n2 y{int number;
& F+ O0 F3 ~6 g, s# A: @ scanf("%d",&number);
O: i' |0 z. n) c4 I# a if(prime(number))8 q# Y- p, B1 i: m' e* j- L: M
printf("yes");$ }& \" t* _# w" Q
else
* t5 P1 N9 B/ K8 |, O: B printf("no");7 a; l+ q" B u" ~/ i0 I
}
) Q1 l* q# u9 u5 ]! kint prime(number)
( j: I. M, g* M( h }int number;
% T" f* H5 m, [9 _' N# Q6 I) u{int flag=1,n;
" k5 ~$ H4 q+ Z% e for(n=2;n if(number%n==0)1 Q* i u7 S- r7 ~; |7 R1 l+ _. m+ N
flag=0;
. q0 s& k( i& [ return(flag);
5 ?9 K9 |7 k) d1 c0 c7 g}
; J1 | @1 V% C! `6 V7 _) a' D, ]2 c7 h
& ]& H9 i/ k& S- _5 ?
# d& k0 e" n1 N/ L
8.4
/ z- O4 ^, r0 y#define N 3
% _ C2 Y/ |- E$ R! X8 A0 Mint array[N][N];5 Q, b8 b3 P" b
convert(array)
; ~# F/ G6 u/ u2 ~4 F: p: tint array[3][3];
. }. [6 ?( V$ z( z& o1 p { int i,j,t;8 A5 E$ [' J1 f4 ^3 A y* y
for(i=0;i for(j=i+1;j { t=array[i][j];
+ S1 O1 ^7 A* f# W; i' M array[i][j]=array[j][i];
; d, }3 s) L' q; p& m+ b( Q$ ~ array[j][i]=t;: U1 @# j6 o- v
}
+ v4 R5 P2 M( T% u" V) H" ]+ R- c }+ w' O _" n( ^. U Y! T2 u v- z
main()
* L- |9 D1 K* P6 O! p{
/ Y" a6 D+ ^, j0 d' n/ d int i,j;
' k) G g' H5 U$ y* v9 e printf("输入数组元素:\n");* G! H: z* h$ m/ R4 H1 K
for(i=0;i for(j=0;j scanf("%d",&array[i][j]; `4 Q" ?2 X3 E( U0 z+ z( x
printf("\n数组是:\n");
}% L( U* d- F! L1 Y" \9 s# tfor(i=0;i { for(j=0;j printf("%5d",array[i][j]);
5 a: f5 c) m5 L7 l. |6 |4 v8 N printf("\n");
6 E: J) g* \* [4 i } U; e8 j0 \ y, w4 Y% i
convert(array);! z# p( C2 R0 R, X
printf("转置数组是:\n");
& I8 }2 A9 v7 P- U for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
+ b+ e' V6 |0 X9 P7 |2 N printf("\n");
{2 }6 f2 x/ f- _- t! ^; F }
& g. T/ G1 o4 e. j# r0 A) o g}- f- E% m& y9 S n
$ H0 }! O% a# w% ]- K/ F+ j* I2 b: ^- }" h' t
6 w! p6 K* P$ g8.5
& H. p, T- K! f* a1 ?% Amain()
( j: |0 a$ X7 q, a7 n/ ^{
/ g, G$ g y+ ^# _$ |& echar str[100];- q/ x1 S# H- R# f, y
printf("输入字符串:\n");
# A5 E5 k' I8 n X9 @+ l scanf("%s",str);% N$ L( H7 W* w6 S
inverse(str);
: I5 ^7 P' w0 u7 y* B printf("转换后的字符串是: %s\n",str);
+ {; `$ [/ D* C) C3 c}
6 H# ^4 N' a* z+ j/ m2 s9 ~inverse(str)
( V1 h+ {2 X5 j2 H; T+ X0 }0 v/ @char str[];" g# f4 P# h0 O3 V
{* Y, H% K7 {- B5 j# D& \
char t;
5 ^; F. t1 \' N0 ^* D int i,j;( X+ S5 j6 F- Z
for(i=0,j=strlen(str);i {0 n! |( v7 J( [+ M: A! D/ n5 ~
t=str[i];" n `& b9 ?. c
str[i]=str[i-1];: M. t& p. X& J" W* _
str[i-1]=t;* p0 s" c9 Q4 e
}
: K4 g1 }! \5 Q) o}
5 U& Y1 g/ N/ A5 A/ e" _+ J2 @" }8 ^9 S* G! y1 b9 [4 X D1 l
( y M& Q# W* i- [
+ [4 J" {6 y' |0 a, |; x
8.6- p" v7 S& p% E7 C
char concatenate(string1,string2,string);
( D( S7 J5 i, h9 T8 v; lchar string1[],string2[],string[];
; J/ V8 T U9 t+ d4 G+ A0 L: i$ Q{6 j7 u% V6 {# [- J# p+ l
int i,j;
& t8 w" z2 ?; j4 K! z- ^) }- g. Ufor(i=0;string1[i]!='\0';i++)
. }/ @: u2 z7 l& s! e+ Q$ y: f2 | string[i]=string1[i];
# S. ]( k, z/ Pfor(j=0;string2[j]!='\0';j++). F9 d q& Z+ y% D, c8 F
string[i+j]=string2[j];
' r) V! x4 ?$ F3 L) L string[i+j]='\0';
& Y1 x( c" A; V# a( `% v+ r3 E}4 Q* W; D2 w7 I
main()- B9 I8 r: I2 ]% S( `3 D3 K
{4 }0 M# J% J3 S
char s1[100],s2[100],s[100];) i$ t Z, r2 Q1 P' X
printf("\n输入字符串1:\n");
5 ^% W4 y, L: t$ J2 m5 V7 L! | scanf("%s",s1);2 p% h/ j, Y$ E) R! m
printf("输入字符串2:\n");1 k& x$ q% O/ i; V F
scanf("%s",s2);( N J y& {! z: ^
concatenate(s1,s2,s);1 F* ~: |. U7 D. b* @5 T
printf("连接后的字符串:%s\n",s);
8 t& n& h1 g6 j7 h3 N* E( {+ H}
" l7 D6 l% a, J+ M i# M b
U2 u' B- |6 N* r
a& Z3 d: f3 ^8.88 n u# k% ~. K4 t! m
main()
% ?4 L9 I% I. i- S6 H{
# H+ T: A. S _# z! p+ n char str[80];
- c8 G8 c3 E' g" T2 e5 `/ | printf("请输入含有四个数字的字符串:\n");6 \: T4 P: k$ f& P+ X; X$ L$ m
scanf("%s",str);
' q5 F1 U' w3 q3 k insert(str);
9 g4 f) {+ g; R, k# q; E. @}1 x l9 V3 O2 y* R: s _* h% u
insert(str)* P& s9 h+ q+ [
char str[];4 l4 o1 m* r, A: i* c+ [( E
{ _& f( g. J4 q0 |: W1 C0 G
int i;+ O' G: D7 g+ Q, O* \8 z
for(i=strlen(str);i>0;i--)/ i* i" _ x- y
{ str[2*i]=str[i];* R9 S+ P) A' y- w1 g# ^+ O- K0 x
str[2*i-1]=' '; c3 U$ Y C! s& \' [. t8 [
}
; Y+ a3 d' y' m+ ]+ n printf("\n 结果是:\n %s",str);1 T+ F+ F7 A. A
}/ W$ ?" u- o+ N
9 X' {0 x/ {$ H# X
: d/ ]: @- y2 ^% ?/ _- K2 J# a" R% \: n. v3 ~. ?
8.9 X1 l; E# j$ S
#include"math.h"" l4 Q( I3 `1 L% S( R) A
int alph,digit,space,others;
* L9 ^% I5 l2 a* b' Mmain()
$ V/ a2 O `; F{char text[80];
9 A3 u# v7 Y- r& Q/ h0 ?0 M+ u# } gets(text);
2 ~+ S; u6 e+ Q4 x alph=0,digit=0,space=0,others=0;3 z0 b: B3 O; w4 |6 t* K7 c; V
count(text);
6 D3 B5 z& c2 P* B }! E$ q- f printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);
+ q8 a- G% l$ [) s$ H' _}
$ M0 j! \- h8 ocount(str)
. U/ S. H) V+ E9 ^6 Vchar str[];4 m- u6 r7 `7 Q/ M- q# \, A
{int i;# g/ d9 t4 x. E" K
for(i=0;str[i]!='\0';i++)
# Q4 Z( I9 s1 G/ _2 p- B if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
V; e/ |& z* t4 z alph++;& g+ k0 \! ]# \* l
else if(str[i]>='0'&&str[i]<='9')4 c) d2 B; X* ?. q. J% V1 d
digit++;
5 t& W+ `6 S# x% g else if(strcmp(str[i],' ')==0)5 D" ~- G- _ V2 t
space++;, V1 D/ T( [( Y4 {0 p6 P- Y, W
else
; M4 P2 e/ u% }% X; K& n X6 m4 f others++;8 q) J: i ^* M ^! I+ r* j
}
0 N! W' ?2 H$ }) u" Z/ G. G
/ U8 ~+ ~* Z& F6 W" W0 P, m$ v9 S/ \. f* ]
8.10
, Z9 f) H8 U8 n# ~0 tint alphabetic(c);
; e7 K4 W' m6 K: }3 m, wchar c;7 W/ u' H/ w8 s/ l& }
{
- J" E% m5 g# e if((c>='a' && c<='z'||(c>='A' && c<='Z'))0 I" R, Z- [! Y; B
return(1);
( w1 D* M3 [- y- z4 J! j else
; [$ m1 G* [' ]0 W return(0);6 @$ V* q0 S3 ^ J6 @ u
}0 V6 p+ k, t3 K
: {0 {/ n0 H4 v c/ q5 ^
int longest (string)
% R8 L R: ^: Xchar string[];9 A# Z/ X6 [0 W; |$ m& o1 V
{, x/ v9 Y. Q' Y, R8 z! {
int len=0,i,length=0,flag=1,place,point;
$ J* c v7 }6 k2 Q" [ V8 ` for(i=0;i<=strlen(string);i++)! |& n4 |. @' A, s
if(alphabctic(string[i]))5 i" W$ S" W7 e3 V
if(flag)' c; T( Q4 B; c
{
( j+ {2 |& I- B- c$ g$ a point=i;
" ~0 J3 y+ [% L& Z flag=0;
3 ~' S( ^' b* X }
) t, z: g7 _3 b& `. ` else5 L$ S$ p- n; ^2 Y: c' P- S/ Q
len++;2 N0 ]# T5 U$ i; H) U- k- ~5 @( ~
else& x8 _$ k; H( B& }% i [
{ flag=1;( ]- e( Y& E2 m/ z) ]
if len>length)- @8 V; ~* Q3 q4 S/ @' ? J2 q
{length=len;+ M- H# W: r5 l# v. X3 U
place=point;
+ f7 d% @; N1 W! ~- l+ \% g+ r* G% Y len=0;
% v1 { P1 O( t3 w2 C) d! d, }, W: [ }
8 w7 I: G0 I! O) A9 R" ` }
9 [* h% A' H* ?8 e- V$ s return(place);
' Z. G6 [! ]9 y2 n8 w: c }
7 v, z1 z9 ?7 R+ q/ Umain()* A% G& ]( O0 J& |6 s
{0 K1 n: h7 A; n d2 q8 i
int i;
% y8 |& w2 A& V% n. C6 gchar line[100];5 \/ x3 Z8 { P$ U6 K
printf("输入一行文本\n");
- H* D7 b$ ?8 l& K4 P0 o* pgets(line); [0 a& _: b8 G9 C
printf("\n最长的单词是:");
. `9 _7 L8 J- z1 |6 }0 q* Y+ Ufor(i=longest(line);alphabctic(line[i]);i++)2 `# d! J" ?( v9 B" U
printf("%c",line[i];6 |- u( F; B) F1 n" a1 d2 _
printf("\n");
1 B1 H0 X# J3 l3 j( h* G+ X}+ Y4 f! t# j1 {, _4 [3 u
# \: c( ^+ c- W% E1 j3 B* j3 }
! T, n( q% @5 X. d+ d; Y+ @8 A2 a8 l; o& {3 I8 y( K J
8.11
`& P. W! `. R6 d: ~' Y#include
' k4 |& e K/ n% N V8 X( @: V4 @% l z- p7 X' A6 B/ g# e3 s
#define N 10
7 \9 q u$ w( c; ~: K) |, h" [char str[N];
3 T( f4 X0 c8 Z) cmain()4 a, a2 M' K8 |9 a6 f7 Q; j( s1 T
{7 l( X2 Z' p- X0 y, Q; N, W
int i,flag;" v5 l j# K4 K, `' Y% @- Q# X
for(flag=1;flag==1;)
C4 k. l& d8 W8 ]{* t" `5 D3 V; f- h
printf("\n输入字符串,长度为10:\n");1 s( M2 U& y: P, T' M T( g. p$ \
scanf("%s",&str);
8 j3 ]( ^8 l5 s" n4 j" A6 E if(strlen(str)>N)
, n3 i# E7 C3 W6 i/ G printf("超过长度,请重输!");
0 E/ u) q5 D& C) D else
o; D7 Z( ~* r( o flag=0;
; s" {) L: d2 }0 L}' e8 i7 q* \5 [: S+ t m* _/ y! O
sort(str);9 p5 g9 y/ z- V' b/ V% l9 p
printf("\n 排序结果:");$ E9 q2 _) \( f
for(i=0;i printf("%c",str[i]);! C4 I* y* H* n2 k
}
: l [0 y4 `" vsort(str)
. _2 v, C) i/ {$ {* o3 z. vchar str[N];
3 P6 E5 `: L& K @4 ]0 m7 \{
/ T4 I- A1 b* N9 c) M4 q2 ?4 mint i,j;# K" t& H3 t! t6 `" q4 l
char t;
6 U, S* U$ W5 H! S# Hfor(j=1;j for(i=0;(i if(str[i]>str[i+1])
. \1 v+ [4 S! z: J7 p* a { t=str[i];
5 {1 n' P" F, V4 X4 | str[i]=str[i+1];
( ^$ {2 C, u4 E5 t8 z$ D, X str[i+1]=t;
2 \2 ], _" e$ p- Y; P }
/ j7 j# l/ z. I# N}
" m7 w/ F/ e0 m9 Z8.122 a0 ~9 z* X% |9 E- y. L
#include
/ `% q3 v/ ~2 N#include
h( [4 H& i+ a5 mfloat solut(a,b,c,d)) a b$ z6 A# j. m* y
float a,b,c,d;% \6 Z" l2 A5 z5 k
{float x=1,x0,f,f1;7 {3 H( I9 u7 O) y; X# Z7 q# v$ u
do( K# o1 I+ ]! E% L5 q- x4 N% H( X
{x0=x;
+ ^8 c% s# R+ @; l f=((a*x0+b)*x0+c)*x0+d;6 n5 Y. W! R3 t, B
f1=(3*a*x0+2*b)*x0+c;
2 I3 y! ~, M2 K( r x=x0-f/f1;
, U3 c: p( w8 T: }* w. l! H }
, d$ E" A) M% ~6 h0 ]" e while(fabs(x-x0)>=1e-5);5 y8 [4 Q, i. @! F& e2 [# {
return(x);
j3 r0 W9 U$ O}
# J) ]! c$ T6 ]- i4 q0 B, `main(): }0 w ?0 q# F9 K
{float a,b,c,d;/ [- I) l& S% n
scanf("%f,%f,%f,%f",&a,&b,&c,&d); V: |: v$ y$ Q& _; h5 c1 H/ f
printf("x=%10.7f\n",solut(a,b,c,d));
" h9 @$ i4 j B3 s. s9 c}7 E* K% \' x9 H! e4 R
8.133 N4 M; A5 U% l) |# p; D& ]
#include
; R$ V# [$ f" j$ Y; @& k" ~main()% c% h+ f; T! r7 i. ?( B/ F
{int x,n;* ]. z5 Y* u/ B `8 n- ^
float p();: e' K d0 q3 c2 F8 Y
scanf("%d,%d",&n,&x);; `& y& }8 ?# @3 a; |5 g6 K
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));3 [- N6 C+ X: }6 l; j/ i- `
}9 L6 Q/ m+ a* `3 F4 N4 P; D
float p(tn,tx)
b1 K& e3 V+ M- I$ Bint tn,tx;
, x0 N! n+ Z$ M( x3 J; C{if(tn==0): ^1 \* W/ Q; p2 W9 b1 ^
return(1);
! a0 J" A9 Q" v- x# i" q else if(tn==1)
+ _, F) \) U( v- B return(tx);
& G; {& s! D L else
/ {3 C7 n' m0 g, w& c return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
0 M# n. Y) A8 S) @$ P0 k}% g, P2 [% p- H% _! e( D
8.14, l* z4 }5 S1 m R0 {
#include "stdio.h") j. O9 }6 D4 W' r1 x; i7 p
#define N 10
6 v0 L) ]1 l: N0 }) X& Q#define M 5
6 a# b+ U$ I5 _! q- }, t5 Cfloat score[N][M];$ _) B2 v7 }7 C2 k9 G5 F! |1 v6 o* h
float a_stu[N],a_cor[M];
: I/ y4 n J# `3 H0 ]) Omain()
5 j1 h+ l- U7 Z* e{int i,j,r,c;, w0 B. z5 i7 ?
float h;# b( ]7 s. M; o1 s, l
float s_diff();
3 i- {( g$ {* T: p) E4 L5 ? float highest();
& N, ^. N" T8 q r=0;
8 s$ Q7 p6 @2 \! U c=1;# M G+ \2 T% F9 T0 p/ g2 Q
input_stu();
# t( A3 O4 ]& j0 P8 i! M& F" @# ? avr_stu();. c; v" w" g) m* h, H$ S8 u7 j
avr_cor();& [+ V5 s3 E7 f W$ M- G: A
printf("\n number class 1 2 3 4 5 avr");
- C1 S0 }0 d% ~/ F# y+ z for(i=0;i {printf("\nNO%2d",i+1);
/ J. r7 J, [: K3 A/ m0 P for(j=0;j printf("%8.2f",score[i][j]);
; E) n. p; o9 |& L5 h, m2 r printf("%8.2f",a_stu[i]);
, V3 y& ?% B/ N1 L }
1 A' z6 w2 R2 V0 m printf("\nclassavr");! ^" X& i) w ^, C% O$ X
for(j=0;j printf("%8.2f",a_cor[j]);
6 @4 M. {; i" j0 n5 X* Y h=highest(&r,&c);; @ M: e; q. x/ m! Z& V
printf("\n\n%8.2f %d %d\n",h,r,c);7 n; G0 }* ^, R( }+ `5 z
printf("\n %8.2f\n",s_diff());
2 w0 k4 j' ~& Y# |}
, x+ M+ D+ S/ T6 d4 kinput_stu()3 N0 w7 y$ X/ X, T/ B
{int i,j;9 {3 X; {8 B) ^
float x; s; N9 M2 b. ]7 ?
for(i=0;i {for(j=0;j {scanf("%f",&x);
9 j) P9 o c4 \$ r+ u; e# j: J score[i][j]=x;1 P3 @: O+ }/ {) l* t
}
. g, C1 O" k9 V; R o }6 I$ O( J3 H2 F; [6 f/ v) j
}
- I. Z6 y5 D$ x7 Iavr_stu()" P/ |7 H* g4 A' D' O6 R
{int i,j;
; e7 d, n$ O: y# \4 T float s;4 a* U0 L% d5 R) U: S5 J0 @; N
for(i=0;i {for(j=0,s=0;j s+=score[i][j];
/ ~ b% u- d }4 w( q8 c0 L a_stu[i]=s/5.0;
. b: z* s+ @0 {$ l6 T' L }$ Y4 d1 g% F3 d
}
; g( M( J) j/ u9 u+ h! Y6 favr_cor()
) E, z( D8 C+ Q* s$ g' W{int i,j;3 o4 b3 Q& }! z% [( ~
float s;3 ^% R7 Z% t. i
for(j=0;j {for(i=0,s=0;i s+=score[i][j];
8 V& R* C+ Z; h; {1 d9 f a_cor[j]=s/(float)N;
% ^+ [( e" `$ f }- c! d" ^& x, R! N5 h* a1 x" u! x# @; a
}# M" R$ l3 N: q% a) h/ V5 e
float highest(r,c)/ s8 i/ Z' q5 |$ W. o7 R
int *r,*c;" }( j6 `. d: B) Q4 t# V
{float high;
. j# W4 j, J2 C6 R int i,j;
0 ^0 g% w. s- U7 [ high=score[0][0];
9 _- s" f% D4 H" |/ Q& e6 m for(i=0;i for(j=0;j if(score[i][j]>high)( i+ p N8 `; C- u1 ?/ q9 i
{high=score[i][j];
( P" X+ k3 r1 v, h; `% S *r=i+1;7 S( }3 ]6 c$ q/ P- o: ]5 E4 G) C
*c=j+1;
! g- ?0 k: T+ r. z }2 d6 D8 T6 j0 s1 k
return(high);1 m0 {: b$ D: d% Z- v
} O/ Q3 K% E0 T! w) @, C; ]
float s_diff()
, T0 b2 r, q& F' c- X4 Z( A& T' E/ R{int i,j;) l( }+ D: O# H" U
float sumx=0.0,sumxn=0.0;
* T9 k" B2 z3 g0 {. w/ F( ` for(i=0;i {sumx+=a_stu[i]*a_stu[i];
& Y' _' d" [0 M& f sumxn+=a_stu[i];
6 ]: B. y4 r+ `) Z }
2 r8 ^% ?) g& ^/ X return(sumx/N-(sumxn/N)*(sumxn/N));! n& F0 B6 y; H9 T& G+ K' y
}
3 x1 O( `% h0 ^3 D; o" W4 v8.15, o% [4 }7 G+ \8 @, E' u
#include& v8 }" O! r( j
#define N 10
) m$ s* u, d- k# A$ a$ X$ \ Ovoid input_e(num,name)( `- V% I* k2 O
int num[];. j0 J4 X1 }# K' o+ z# q6 e
char name[N][8];
1 j9 c0 e' I! j5 o6 s" q9 c{int i;0 o0 r; z) i0 l2 t
for(i=0;i {scanf("%d",&num[i]);) E6 L# B2 H( c
gets(name[i]);
! O/ v! s( [7 b$ B3 L6 H |) E q }
8 G2 l4 \' C0 Z0 E+ a& @}
# p6 n$ n5 f! N7 B9 z( q, [0 @3 Rvoid sort(num,name)
: Y- P! n, `' K, h' p3 o& I+ i3 n9 Sint num[];
5 S ]4 J$ u$ q& X1 V Lchar name[N][8];# T& k/ l) ^1 L) V% w2 k, E$ F
{int i,j,min,temp1;6 Z2 v7 ~5 W: l1 m
char temp2[8]; |, w4 d8 D% Z. u1 J
for(i=0;i {min=i;) p+ G$ e# y* T5 d
for(j=i;j if(num[min]>num[j])min=j;% H" l" ?0 C3 X# |: U1 H" R. J
temp1=num[i];
! ~/ q( \' b4 P8 o9 b& U2 g; j) n3 M num[i]=num[min];) O9 t9 Q i7 {5 q4 C- `& E
num[min]=temp1;
. }, R( g& Z7 ^7 y strcpy(temp2,name[i]);
" S( r: b5 ^* {4 `) }) ?' ^# w strcpy(name[i],name[min]);# E: T8 j. F) Q9 b6 \- u9 @0 ^
strcpy(name[min],temp2);
/ n- P3 `4 F3 n! ], y }$ y& l7 V# T* |: x7 S
for(i=0;i printf("\n%5d%10s",num[i],name[i]);
( U0 y3 E+ t) s' ?6 S# j}
}: U+ X+ i, d$ ]( I3 Xvoid search(n,num,name)# _' _2 }) r6 r! a! e
int n,num[];
3 f% O2 |( f6 {$ ~char name[N][8];/ O; D N2 ]3 }
{int top,bott,min,loca;# ~0 F- P, X% H+ W. | a
loca=0;
1 d! |& r" |! f: `; @; m top=0;
# u0 D# p. k" h3 c. ?1 ^2 S bott=N-1;" q+ c. m) {3 ?
if((nnum[N-1]))
- L' h) f6 g7 ^ loca=-1;
, H/ K/ z* d. X* Q while((loca==0)&&(top<=bott))
0 s: Q- _2 B3 Y8 M8 K, Q {min=(bott+top)/2;
' ~: m8 y! x3 l0 j/ [' Y if(n==num[min])2 a& f, s0 G- {) ^) O8 ?' ~
{loca=min;6 F6 z/ u# M2 E8 u9 Z! j
printf("number=%d,name=%s\n",n,name[loca]);& m" K6 ^ R7 j$ i4 M9 M/ E1 R6 V R
}1 E4 H, [! Q& V, B
else if(n bott=min-1;, ~, y( r- I V/ m- h" Z/ @5 [- v
else
! k) F4 m; w4 |4 C2 L( W! U* m top=min+1;
2 U4 `3 N& d3 [6 y }
5 R! I( |" e, [9 z4 m: L if(loca==0||loca==-1)1 [* S! {$ W3 L; i
printf("number=%d is not in table\n",n);
+ v0 U4 Q+ N$ U( s% v; B}/ \& l* U, f K! {
main(); R: G& o. [& q5 ?/ Y
{int num[N],number,flag,c,n;
& f) \6 _- n0 \) D char name[N][8];: x S: a! i/ E! {8 X6 _
input_e(num,name);
. W2 m" C/ K# y3 Y) U% @4 w sort(num,name);
! q8 p+ \9 c+ t% \# L for(flag=1;flag;)
7 P* f) v! j9 X$ K {scanf("%d",&number);* `. M9 J0 N4 O& R. ~, Y$ D
search(number,num,name);3 Q3 b: q* `/ M# @/ a
printf("continue?Y/N!");3 D6 I, L+ q0 q; I
c=getchar();
: V; E a) H5 {9 O4 k3 e Z% l if(c=='N'||c=='n')5 s: W8 Z2 ~! y! U8 s D* w& G
flag=0;. Q& @8 H. m( @
}
3 n3 I7 W! D; j! Z, @+ V}
5 S, X- s$ _6 [5 K* N: N8 x/ k4 c& k$ P5 F+ ^% H
8.161 C \! _. N! F/ S
#include% L' ?' |. ~+ g$ T
#define MAX 1000
* z, u4 K1 n' k& ^8 ]& i) Jmain()) t$ a; |0 I3 n2 Y+ c, V
{ int c,i,flag,flag1;
$ g* w, b+ b( S9 s! ~ char t[MAX];
7 @: L) E X" | i=0;
/ ?8 A7 g. D' n& g5 R" V flag=0;7 c+ w4 k" }- Y, v" L. F
flag1=1;6 ?. G6 w, \5 i2 i( n+ i
printf("\n输入十六进制数:");, I9 H' N, S% v
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
4 h8 g0 s# Z3 a2 k* z7 Q6 u {flag=1;, v2 g" ~# A$ h# a4 X2 M# }+ U
t[i++]=c;
" r& E4 x& |8 W7 P- Q; h' H9 W }! o/ |! ?) U) P/ L M1 U8 q
else if(flag)/ S9 D5 y6 K( P7 A; k
{
( M. u7 R+ r# U2 ?1 L t[i]='\0';9 j% I% k+ U: J% \: M
printf("\n 十进制数%d\n",htoi(t));( c4 g* U. z9 ~& _. p. }9 S; K
printf("继续吗?");
^7 ^& P$ s) k3 {1 Z# t c=getchar();+ ]; d. y7 [/ M+ A3 Z6 ~3 l
if(c=='N'||c=='n')3 G' \9 |$ I3 A+ j0 J! q+ R
flag1=0;0 y0 D0 J% n/ N" K
else
; f* o+ I( k: v# r. @ {flag=0;. j: y* F- H* s- B" l) T/ _0 l
i=0;
) h0 x2 `1 h7 f printf("\n 输入十六进制数:");* H3 Z6 q1 n' e: E) }
}
) c h& n+ n) |1 T1 W' \}& K/ F" o' b9 K
}4 i" M" |! K( C! n$ L0 [) J
}
& e0 B6 q9 H7 Q& M1 xhtoi(s)
: S# ]$ M4 {( W1 G- }* F; j9 dchar s[];
) z) ]8 k% e: z{ int i,n;7 v( l4 _( o8 n `4 q
n=0;
' f/ g9 R& a Q1 i1 ~ for(i=0;s[i]!='\0';i++)
' n1 U: J4 y6 } {if(s[i]>='0'&&s[i]<='9')
* r0 N. q4 h( l0 _9 m n=n*16+s[i]-'0';
' r& c* E: f( W( j! }1 J# r9 a if(s[i]>='a'&&s[i]<='f')% `, E( o# g& m
n=n*16+s[i]-'a'+10;( e o0 q4 v# V: B
if(s[i]>='A'&&s[i]<='F')* i. A3 p2 T, g3 b: C
n=n*16+s[i]-'A'+10;: y5 X( S2 p3 i( N
}1 t, H# l4 I: P( W; y
return(n);0 t. T) k( t, d# Q: P# z
}
% o6 n7 _! q5 B1 z, U0 ^; e: g' O5 X' [! [: z S
' m& |2 P3 k: E: |6 \2 O, s* W. f g! ^; Y# e
8.17
7 @; ?/ Z- l) p& Z+ n( |3 [#include" g5 k- Q1 D! `& l
void counvert(n)
7 O; O8 J/ R9 H) H) p+ M/ P3 A( Xint n;7 f" J$ W" ]% I( H! O% V
{ int i;
/ {9 m, ?/ U$ K7 k" Y4 F, e9 o- w if((i=n/10)!=0)) S' O7 O0 a6 a) R0 s3 ?; w1 Z! U
convert(i);1 n1 Z, c- F$ A6 r" ?8 l$ q) C, ?
putchar(n%10+'0');
/ Z7 r" D, H6 m) o}* m7 S( G5 w$ k2 f
main() N' I- v" o- Y; v( p3 ~, p& n
{ int number;
- x& k& ~ i6 d8 p0 O4 a# F printf("\n 输入整数:");0 k8 l* t: J( J& [/ y
scanf("%d",&number);3 J; U+ Y0 E6 L# e9 V' Q& f
printf("\n 输出是: ");1 j4 C5 _% [1 q/ M
if(number<0)$ r) C# [: e' S- {% N+ M
{ putchar('-');- i) L$ J- |* Z2 x
number=-number;$ t# R$ i- o. u7 R
}
$ E- J# G& q6 l& G t5 X# }convert(number);9 q0 n% L* K. k8 ~
}
! L# l0 V B; N# F8 y
# v9 e9 X/ D- v+ {# {9 |- x1 P6 p/ f+ F- D' I- k
/ z8 A4 k& T; p5 c8.18
, U* t7 Z6 z; k+ c" j9 Cmain()( |, z* p. b& @8 D, {& w0 E
{
0 H& Q8 r4 Y' u, Y [; Z+ s; P6 m int year,month,day;0 n" r3 y4 {3 l" y) D2 a
int days;
, H2 x. U; U7 G printf("\n 请输入日期(年,月,日)\n");6 o1 _: I( K' P" h& }6 A2 H
scanf("%d,%d,%d",&year,&month,&day);
) E0 g! K `' J4 j% i9 i printf("\n %d年%d月%d日",year,month,day);
0 V" G; N2 X9 E days=sum_day(month,day);, \( k6 ^& Z# D* E& D. n' ^. p
if(leap(year)&&month>=3)& X% Q6 \; N6 V6 d4 p
days=days+1;' d2 s9 z) o8 e* u" n
printf("是该年的%d天.\n",days);# _; N) O/ r9 f7 e
}
9 K0 l" z ~ y9 J# e static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}6 p0 K& n4 N1 ]. u
int(sum_day(month,day)
9 s9 N$ u6 h' h, S- e int month,day;
, {. A. n4 X3 l9 E% w& t1 D {( ]+ B1 S: z' H
int i;
) s3 @" t" s+ f+ e1 V5 n% Y( t4 {' M+ ] for(i=1;i day+=day_tab[i];
# X8 ]" L+ @( Y return(day);
/ P, d8 ]# n& l3 E: ]: B: U }) ]1 C" r! |( Z3 ]* e9 S, U5 u
int leap(year)- Z9 Z: z/ s2 U9 S! k) P
int year;
2 c' G( m) j0 D! z# Q) k* Q {
9 w/ J& E! v! Z1 N- N( [+ R. I int leap;
" M" G$ h& l9 h' _1 K leap=year%4==0&&year%100!=0||year%400==0;) ]3 ?3 g, p' k$ [/ g% F: q# I( Q2 v
return(leap);
$ T) I B0 k4 o# E }+ [4 ~6 U: ]' t1 [7 ]5 ]
第九章 编译预处理8 _. m3 ]$ p9 S8 p' h6 E
9.1
) ^+ ?5 R4 b) @: A. X#define SWAP(a,b) t=b;b=a;a=t
7 ~( x" [; x3 q6 a2 G* K% g9 Wmain()
$ s" n. \3 i6 e- c- d& F" U{
+ l4 K8 q2 O8 g& L0 xint a,b,t;" B, B z' }: X/ W' q6 H
printf("请输入两个整数 a,b:");; _/ c& |$ B D. c4 N
scanf("%d,%d",&a,&b);0 ?& z% V2 B" W! g4 n$ r8 o5 Q
SWAP(a,b);8 H H# P: z! @: k2 x: S$ i
printf("交换结果为:a=%d,b=%d\n",a,b);
; a0 x0 w/ M3 o3 t& ]. v9 b, K& O}
1 }. J. W) \( a& ` ( ?- A) |* n$ K/ K4 N X. o; X1 w6 F
- j( p- g7 `3 F: r& i2 @) T# N" I
9.2$ v9 l& r* @: V( w9 b. X
#define SURPLUS(a,b) ((a)%(b))4 f* Q; ~/ o9 c5 t, q/ x% s
main(); h) q7 L7 Z! I; @# l
{
# s# ^; Z' T: h, A ?4 f int a,b;+ }& v; _5 n6 ?8 D
printf(" 请输入两个整数 a,b:");
/ j. L) ~! f6 Z( Y scanf("%d,%d",&a,&b);" G( f X1 b$ r# m* [/ G
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));5 D" x$ p5 O+ w9 y* p4 C
}
3 f( Q* q4 i( u( E3 y. u$ b# I4 f: M& s" o3 k2 v0 k# v. E9 ]
6 k: W7 X9 b0 t2 D3 e* C
9.3
) f- S/ v: x2 \1 Q! w#include
/ g3 f! [7 W0 E2 r; _#defin S(a,b,c) ((a+b+c)/2)
% ?3 H/ g- X2 }* e: \#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-
9 e' U) L8 a; d. rc)))4 S+ O) x: n% s6 F
main()
& n4 W, q: g6 C# z; ?6 ^ {
6 N% u! t; c2 P$ Y7 \* C float a,b,c;& B @& K% w# _% O, @. z8 u: L
printf("请输入三角形的三条边:");: E* V I& }& d0 x2 s' P: O
scanf("%f,%f,%f",&a,&b,&c);
I" |' ~3 z' b- I8 h if(a+b>c && a+c>b && b+c>a)0 U q9 S" p& H8 b* C. y
printf("其面积为:%8.2f.\n",AREA(a,b,c));1 B* x6 r5 W, g% o4 r
else. e; q4 L) |: m' i3 g$ W w
printf("不能构成三角形!");
3 p% m) ]* l P2 Y }- p! x/ t: o* A! e4 z' U
8 J2 A: I4 c8 T/ J; y! y: s3 F/ i/ _
: }, k& w, ?8 ?8 h
$ `1 r* Y/ |0 B6 F: s9 I9.4
5 }+ C! p: E5 u% G#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)* U8 a7 k7 F8 O0 v& n, g P
main()
* W2 p& Q8 x# A; N: c3 M {
( k8 [; R$ o6 j int year;
' ^" F; x& Q1 p4 y- y( ]+ z& L printf("\n请输入某一年:");, k$ C% V2 [8 w
scanf("%d",&year);( a8 }6 I5 j: p K' I( a1 @( a
if(LEAP_YEAR(year))8 j' A! ~) D( a6 ?" h# D
printf("%d 是闰年.\n",year);) L% ?: a$ L# T: U
else
* ?) ?' g/ M9 n5 c6 M% q9 U* v printf("%d 不是闰年.\n",year);+ c+ L2 m+ S/ J( t. }3 g! j, y) d
}: b4 c; j+ `# n( m- T" d5 F
# `% K$ k4 j3 ~/ c
& c" S* x- ]+ w1 F0 E# l$ o# @: q0 M$ y8 X
9.5解:展开后:
9 T2 n+ E" v9 ^: o; t0 {$ r4 p+ vprintf("&#118alue=%format\t",x);
, n; ` ?4 ^/ {+ [( bprintf("&#118alue=%format\t",x);putchar('\n');6 s/ z G; I. K i
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');& T& D* E" C) H, Q+ ]
输出结果:! R9 ]' Z1 Z; ]) m$ |
&#118alue=5.000000ormat &#118alue=5.000000ormat9 D n% I! Q" n
&#118alue=3.000000ormat &#118alue=8.000000ormat* A; ]7 s3 |% r I
1 a W9 C) ~7 m' P
5 k; y [* W! `
9.8, [: U. [4 g; W6 k% w
main()
7 ~% O5 j4 q, }6 `; y {6 W O0 F: m4 q- m
int a,b,c;
( H" `$ h1 r+ h printf("请输入三个整数:");- b) D# t+ I6 r9 n
scanf("%d,%d,%d",&a,&b,&c);
8 A! I! _, n D: w: g4 } printf("三个之中最大值为:%d\n",max(a,b,c));
$ G' c5 f, U( f: q' } }! L0 U+ I- V& h7 z
max(x,y,z)
# d$ L' O+ P* {& f! ?) f int x,y,z;8 u3 Q/ q) N8 I5 g0 m; m! _' W# |
{, G9 j5 a1 Q" i. F7 |
int t;2 Q* T4 ~& v7 x% K: c
t=(x>y? x:y);
! ]& B: _/ Z0 u; L5 \3 C1 n {! g# I return(t>z? t:z);
4 A+ v, h" ?5 X& x+ G* `) Y }
& Q; U$ X$ B% Y
3 Q: i) H: ?+ f$ ^- A. \9 k* s ^- D8 H" q0 A
4 G& x4 r3 V: f5 \ ]! l9.10
6 X: M0 h: h. C( C4 N#include
$ j3 y0 F, E% {! s#define MAX 80
+ d# l) F; d" P6 d) C! x#define CHANGE 1
" }2 e/ C* o. D4 z6 ~6 Wmain()
9 ?9 G* T8 G" t! W- ]. T{
/ ^) W) w8 F1 k4 U/ O char str[MAX];
5 H" ? X3 S* }7 j int i;
3 s) T; o, D1 D: }) \3 m; q$ n printf("请输入文本行:\n");: W, F' u: [ L
scanf("%s",str);7 V/ S- M+ |% E( T
#if(CHANGE)
. [1 [* F3 u* @) w {
" I* _$ N. v0 T6 j1 v2 ]8 T7 \ for (i=0;i {) c7 s( ~5 G/ S$ M
if(str[i]!='\0'
+ u& X% Z; D7 B7 w7 \ if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')3 F Z/ `& p" c6 M+ |5 |# M( D
str[i]+=1;/ h1 \/ g3 X1 @; @' @8 [
else if(str[i]=='z' || str[i]=='Z')
8 `; i" l2 d- S( ^# H+ N; q7 z str[i]-=25;) @* f c4 l. S$ B
}
% u5 b2 k! `8 C6 R) {}
& f" j* [/ l1 E0 e9 b! e# |* ^#endif) E( }3 o) P: e' s9 B
printf("输出电码为:\n%s",str);
. c q7 f! O' ?4 C}
0 n' e3 i# W1 B: q0 j第十章 指针
' {2 y6 {( A8 }% r; U; }' U1 n10.1
* I& w7 a$ S& C+ {1 ]main()
! e( e* l% @3 q5 A; ~{int n1,n2,n3;5 M3 R2 v1 {/ O
int *p1,*p2,*p3;
8 {: w5 A2 k, c% w5 @: F scanf("%d,%d,%d",&n1,&n2,&n3);
0 J. _$ i9 s/ ]3 W+ `1 m, P9 F9 H p1=&n1;
" T! [ ]1 w8 v" B8 R) A p2=&n2;! a* R) r; B3 ]4 T4 b
p3=&n3;
- l! Q* _1 @8 g" b& E if(n1>n2)swap(p1,p2);8 {, g6 `; d3 B8 J* o' N
if(n1>n3)swap(p1,p3);! F7 d! _4 L& j$ ^+ _
if(n2>n3)swap(p2,p3);
- U. `; K9 H, L6 G' U& z6 v printf("%d,%d,%d\n",n1,n2,n3);
! [, ], `" ]& s1 G' q$ a" O* Y}
/ C3 y6 T/ d* @9 u, k5 vswap(p1,p2)* | q7 s; A4 F
int *p1,*p2;
$ v5 r9 s* v+ d{int p;
) ?& C2 Y" c* I2 k+ ^ p=*p1;*p1=*p2;*p2=p;
6 v& |5 A! T* U8 }+ _}
9 T4 z7 ?8 S1 m( c+ y, e10.2
0 d# q2 @4 o: l# K/ Y7 Ymain()3 D6 F9 N3 {6 {
{char *str1[20],*str2[20],*str3[20];4 B+ O4 a' F5 ~/ M
char swap();" Y C% f) I4 W7 H7 E/ k9 o0 [8 x
scanf("%s",str1);) Z& G& X7 M! F: g: ^1 I
scanf("%s",str2);
7 r* m% N" Z/ m; @ scanf("%s",str3);1 D# q8 w' j o; o
if(strcmp(str1,str2)>0)swap(str1,str2);0 `+ S5 w6 f. Z% U5 K' F
if(strcmp(str1,str3)>0)swap(str1,str3);
5 \* z, t1 H+ r( q$ f& c/ d) i2 G. J if(strcmp(str2,str3)>0)swap(str2,str3);7 ~0 P4 j7 S5 d, T/ W# R$ Z
printf("%s\n%s\n%s\n",str1,str2,str3);
& d4 [& c& X5 M6 x4 u8 z1 x}" P9 C/ x; T3 ~
char swap(p1,p2); v* N2 S( p( {2 ? X
char *p1,*p2;
+ b: J/ b9 E K# g/ q' Z" p{char *p[20];
! i% n1 Q# d- q1 T+ J* ] strcpy(p,p1);
' c/ o! G9 O% ` h4 e strcpy(p1,p2);
3 x6 `" K- N; \5 W& b strcpy(p2,p);# N, L5 U$ p! g
}
2 M$ V; b5 c! ?10.3
, F" E0 H, \7 ?. N* Nmain()$ R- j; d! h' h% Z9 b/ Z* W) I
{int number[10];3 v* |$ c, O5 w$ ~
input(number);( z3 v# a9 ^* L' g0 g, o W
max_min_&#118alue(number);; s7 w+ `9 p b! G# r
output(number); g$ M A! [8 J' `# B
}; f. h: E, F, ^# f0 D X! k
input(number)
8 j, {" i; Y' o7 p Z' m6 s) Fint number[10];
+ `* b0 X8 ?$ J" F( _8 w/ P$ e{int i;8 P, A+ G N8 k, W& L& z" X
for(i=0;i<10;i++)
2 K( w/ r, I7 d1 D: Y" ` scanf("%d",&number[i]);8 c$ k8 Q. ~4 J0 B) `; ~7 M
}
2 k+ m: L: f7 ~. ?max_min_&#118alue(number)
, y' @7 T) m7 R% Q5 j3 I* Q9 _int number[10];
- b, C2 H7 b: f3 s# O5 c7 H; p9 I2 g{int *max,*min;# [: v& [7 L: _: r) o+ y0 F$ W6 X
int *p,*end;0 ]: r. g( h, F! A* P
end=number+10;
& r5 O4 ~/ i! ]4 J max=min=number;
0 j5 o# x) k) o; P for(p=number+1;p if(*p>*max)max=p; h3 {7 s v) r' P8 I
else if(*p<*min)min=p;) l6 m2 M1 @/ W) c1 H. Q; k
*p=number[0];" d- [0 D7 V* @5 K1 s7 M2 M
number[0]=*min;
: M# I' Q! B4 J/ [ *min=*p;
. Q; Z; G( L o( ~ *p=number[9];
- p0 Z/ K: \& l number[9]=*max;
# o" V8 o: e9 y/ g2 A; X: Y *max=*p;
3 R& I/ \& O* d' x { return;
% u4 S! |% U1 u" x8 v1 l L}4 M8 \ h6 l n2 x& l
output(number)- z* P$ x6 p& E! @% v
int number[10];$ _6 Q' Q! W& @1 V
{int *p;
* q m, P% }6 ~7 v" E! ~ for(p=number;p printf("%d,",*p);
, L- L4 H7 z6 t& y) @0 ?3 n" [ printf("%d\n",*p);
) {& i" `; H1 l. K! n. n4 e0 n}' T' c7 H! E5 X% F5 q" U' t
10.48 ~/ u5 G. F( U) x1 l
main()5 d5 s, }8 Y& T- ?) k* g
{int number[20],n,m,i;: w% l9 d Q! \! @4 n" S4 F
scanf("%d",&n);
( u* M7 b' x0 \ scanf("%d",&m);6 g& Z+ s' p$ b, u2 c. E" Z
for(i=0;i scanf("%d",&number[i]);/ H i8 \( Z% {9 C
move(number,n,m);
. y7 O. \' k( z for(i=0;i printf("%8d",number[i]);
) O5 I9 {7 h+ q* Y1 }/ b}
; z7 L3 l- O4 C- _move(array,n,m)
# ~5 m7 g5 }1 {7 f8 I6 {int array[20],n,m;
7 m: o4 Z) r& j+ v* ^, }{int *p,end;* R" D4 O9 M' P- A, D7 F
end=*(array+n-1);1 f1 D! P+ K s& g7 }
for(p=array+n-1;p>array;p--)
- V1 I* P1 n5 w8 i; }' m *p=*(p-1);5 W2 l" e6 h S8 v. s5 X9 ?* a- [
*array=end;6 I0 \, h- k- ^; I+ {, E* u) h
m--;/ ]2 S# s% G1 N2 _, X
if(m>0)move(array,n,m);
! t6 J) k0 ]+ J* y6 |}
6 E' W) d- L9 J: b% [7 A5 p7 e' Y/ u10.5
* |# U! ~) I# o7 o8 F9 F2 }5 B#define nmax 507 b- u2 M0 [% E
main()! M% D, C& I% \$ s5 ?$ K' ^; Q
{int i,k,m,n,num[nmax],*p;2 W' Z; V, h, b9 I, U
scanf("%d",&n);9 P% z+ a6 `" J2 c4 Z! n1 e( \
p=num;+ g& l5 i) c' T3 ]4 a
for(i=0;i *(p+i)=i+1;* g9 ?& ~ L$ ]/ Z& e8 H$ f
i=k=m=0;5 q9 h8 \+ O' l% N# w+ I
while(m {if(*(p+i)!=0)k++;- t- D$ l0 F( _% R$ I1 {8 Q
if(k==3), j6 {5 {- f3 b# g/ w( i% G6 D
{*(p+i)=0;: y. _1 l7 t* t6 z& \, B( B" Y
k=0;
2 b: D1 I+ o. ^2 P# E k$ z% C: r6 i m++;! u1 u! p! x0 C& S) S
}
% S0 Q/ P5 W* D7 Y7 u3 k i++;
" K+ n& K2 x6 I I if(i==n)i=0;& u( _# U7 e/ F# _/ M0 F
}7 s7 Z& p# w) k* l' Y8 C; |/ b
while(*p==0)p++;
# N6 I* Y) J2 `, n( t; k& n; g printf("%d",*p);
- x. D4 W* L4 Y- B% w}& @3 S; W# u' C" [9 C
10.6
& q% D; b4 |5 r5 Q4 }* W6 T! H9 zmain()
; i8 R$ l9 R' ]$ K{int len;
: o6 j+ ?; B0 y8 |3 { char *str[20];
9 t* D6 V4 R J' @$ f: z$ A scanf("%s",str);9 K' C" ?$ g: w
len=length(str);6 l# n8 F( K* F# ]! s/ F {1 @
printf("\nlen=%d\n",len);- t, W- |: i0 M% D
}: [9 r2 x6 n: Z) j1 i" {, N# m
length(p)
) x. d2 f; `7 B$ wchar *p;
o: g: p$ J) h; @- {$ H{int n=0;
3 C* \- d: \ l( S* K7 f while(*p!='\0')
* d; I1 L6 p: D2 C V' @ {n++;p++;}
2 d' \4 n+ N, Q# G$ Y/ i- ^ return(n);
4 L, I8 {0 Y- Q8 s9 Y& _" D}
9 o7 w. e; ^3 p7 R* n6 F10.74 q4 H3 U; y7 p$ e
main()4 r' n+ [! f3 L( i
{int m;
7 O8 i; Z1 n1 D. k2 ?; C( {& x char *str1[20],*str2[20];
3 y' r P% z( B* @6 w scanf("%s",str1);% m0 v2 ~" V- W+ E5 G
scanf("%d",&m);5 Q% ~% G( @- F3 ?# n! k9 U0 a
if(strlen(str1) printf("error");' ~7 t6 y) @+ `! s- F7 u; T- x
else/ \& ~. a; y! _* C
{copystr(str1,str2,m);' @3 c' G( M% q+ j# a8 X; J0 L+ g
printf("%s",str2);
% D: J$ c+ X4 M$ v1 N% y4 i2 I. e7 Z }
7 z0 @8 [8 I2 z0 ^}
" W* I& d3 \/ W& N& a2 j+ s& dcopystr(p1,p2,m)
+ D W% N- D# G1 [: ]; Mchar *p1,*p2;
7 f5 P' n% G+ o7 t( |. Uint m;% ^% I# x! f8 u, H
{int n=0;
5 z. _2 x2 u" r, K7 {7 A& W- z while(n {n++;p1++;}! v6 M$ e2 ]0 _) v ^8 k. u
while(*p1!='\0')5 M% d; ~* v$ R' W$ g- j- z6 T
{*p2=*p1;! g( I/ A* M j2 s
p1++;
% k+ \9 Q$ E5 X! X p2++;: f, E+ y7 }" {2 s$ e. P
}7 ^6 u* D$ I7 p" f0 _1 x/ |' W) g4 a9 w2 ^& {
*p2='\0';
% N+ ~* V3 t; H& C0 ]0 } O8 |& s}
+ I9 J5 @( H' u/ f10.8
% p( [" C6 {1 R* \9 A" a#include"stdio.h"
7 |# q2 }8 }4 U0 w, g9 amain()8 b' p. S2 ?2 V2 W4 a
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
) a1 d- `% W3 I# n7 d0 e+ z char *p,s[20];! `4 P7 {8 f p5 w
for(i=0;i<20;i++)s[i]=0;
4 C, h; M, I$ F/ N( B i=0;
6 G- U6 ]' `% K, l4 [9 l0 p: f! F5 W# p while((s[i]=getchar())!='\n')i++;" x% ~4 y" g4 U1 K! R
p=s;2 P( q) a6 ]* n6 y
while(*p!='\n')
% r C5 `% j7 \ G+ q {if(*p>='a'&&*p<='z')
- D; J9 B3 T% J2 o$ m) }) u0 H ++sle;
% x: f3 U, i H2 U) i$ R R else if(*p>='A'&&*p<='Z')! q# Y+ i K9 U6 _" w
++cle;+ R# I% U( p/ ~$ x6 T' }
else if(*p==' ')
7 d$ h, X: c, B+ f E1 U( q6 T ++wsp;
: o9 I2 F9 _. D8 [% t else if(*p>='0'&&*p<='9')
5 K3 D; q$ q$ N* O ++di;" U D% d' M4 [- g, d
else8 n9 ~8 m2 r% w! V& {' d: O0 l+ c! _
++ot;
; m: c4 D- H) `# Y p++;+ q, Y* t, _& |* w, `, O
}2 z8 [6 M1 n6 x/ y% L
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
- t$ z; N$ H# v& Q+ O( E0 G# @1 o}7 e# p6 S U6 C: {/ R
10.9
' R/ X5 z/ ]. A" @. ^9 j; n' rmain()
, S! v6 J$ n2 a f) _" F{int a[3][3],*p,i;
: u! b) q" {( b$ f& B for(i=0;i<3;i++)
2 h& z* g* k2 M8 D scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);
8 o! x! b# I- H p=a;0 X1 O( y+ \& V
move(p);" k2 I d. h3 x& j' L% b
for(i=0;i<3;i++)6 @/ O% S2 }* s( _, f7 j
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);: b2 d; k/ N# N1 y
}- Z: R8 t) Q# o+ e" A: L
move(pointer)
% i, x" q7 A4 B; x6 m3 hint *pointer;
- |4 l4 k( U% G S8 q{int i,j,t;/ V9 v0 E' N7 H, W- `
for(i=0;i<2;i++). e& \9 N8 `& h6 a) H. a6 u
for(j=i+1;j<3;j++)
2 l2 H; r, T+ i" z+ F# d/ p {t=*(pointer+3*i+j);
- P# T5 D- j; R3 T8 F1 x [ *(pointer+3*i+j)=*(pointer+3*j+i);
2 ~4 ]# u; ]# o6 N *(pointer+3*j+i)=t;
6 [. g8 m. u5 M6 b# Y. r }. w9 n! b! I' J5 J
}
$ W6 w; G2 c" f10.10
5 M% u3 T& c% ymain()3 ~$ V! m1 `/ l
{int a[5][5],*p,i,j;- ~( D- ~" s; T" y
for(i=0;i<5;i++)
/ [8 ^ x* W4 U( m for(j=0;j<5;j++)
! r. D7 X; ~! l( s scanf("%d",&a[i][j]);# G9 v( m+ s, c+ b
p=a;/ F% C8 D$ E+ [$ {$ }' [ U v
change(p);
; x, {0 A! H8 S7 B& N6 E# ? for(i=0;i<5;i++)
/ v' Q9 w! X6 M1 f7 ` {printf("\n");8 r9 d i( h2 O( @. z C# i, t
for(j=0;j<5;j++)
+ ]/ l* r! B) x8 T8 u printf("%8d",a[i][j]);
5 y* k+ X6 Q6 Z! d A }
' o9 d, r. y% w0 L1 J% G$ {}
# N: d( E( E- dchange(p)/ }' J- i3 E- M! d! a
int *p;
, O$ d, b! j, d1 v1 }& C& w{int i,j,change;
# H0 A% \- k1 k int *pmax,*pmin;
* V! @3 [2 q2 |6 r6 k% f- ` pmax=p;& m8 A! E+ I; [
pmin=p;7 e2 p: L( H, G8 z
for(i=0;i<5;i++)
- u3 ^& j1 h$ t3 j for(j=0;j<5;j++)
& n" p6 ~) q6 }: E) }* R1 j* ~ {if(*pmax<*(p+5*i+j))pmax=p+5*i+j;3 l% B) a" M! G& ?( a. c* [6 w3 u
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;2 D: z; K$ H! i& Z
}
0 R, M, o5 F6 w2 A% U8 R" I change=*(p+12);
, \5 n6 T+ L/ u *(p+12)=*pmax;! E; Z/ @ G' `
*pmax=change;
# M5 |/ o; J) h$ b: O change=*p;
0 C0 i0 a% E4 q! k# c- n% _2 }2 v *p=*pmin; d/ H( C8 a9 W, S* E
*pmin=change;
) t% U) A& z! c, t* I8 E2 M pmin=p+1;0 T$ i$ [4 Z8 z
for(i=0;i<5;i++)
4 ~, c& g' [, M% w7 y- X for(j=0;j<5;j++)
3 V i# W) m' \# k5 D- {* L if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
' q* R* Y* P# F$ c' | change=*(p+4);
7 O7 O: e$ u" ^ *(p+4)=*pmin;
# D3 K: A7 f* G4 N: D7 u, r *pmin=change;' d) y, W! [" M; ?# R
pmin=p+1;* I/ Q4 c; _0 j, @4 \
for(i=0;i<5;i++)
: F$ K q& B/ B% ]5 E8 w2 D: O for(j=0;j<5;j++)- O! x! K# j8 u p
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
& E. _ Q X( l3 A pmin=p+5*i+j;
; q1 U2 @* A# b change=*(p+20);
* O" O' G$ j( r3 o! u *(p+20)=*pmin;2 F) \& p5 }# ?- l
*pmin=change;
0 i& e# ^3 A, i+ ]2 |2 y. s# H pmin=p+1;: P( \$ a' e/ C3 g
for(i=0;i<5;i++)- h8 \; T# y. s1 I! {
for(j=0;j<5;j++)
- ?+ v, \5 G! J if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))3 u6 f4 U) K* Y' Y9 ?" ^
&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
' {$ }& ?4 ^% J: y% p change=*(p+24);
9 X1 C$ k' y6 p: j, m6 J *(p+24)=*pmin;
7 [9 n" a. P \4 b *pmin=change;# ]1 t1 }$ L9 }9 t+ Q
}: B @" Q7 v! R, V( A, R5 i
10.11
8 _7 \$ d3 d+ J' D5 Y. smain()2 r/ _ ^2 K$ H5 t# i1 j0 r3 ^
{int i;
; x# Y2 w3 }" y; r& ~ char *p,str[10][10];) n) `" y4 ? o) B: _. f' R
for(i=0;i<10;i++)
7 Z% R% L- J7 r& N) [ scanf("%s",str[i]);3 X3 ~ l; c3 _( G; j$ F
p=str;5 e( Q2 s' ^* X! L
sort(p);( s6 o1 A9 M* Z$ w4 e
for(i=0;i<10;i++)0 _; `" |$ K1 @: z
printf("%s\n",str[i]);
, W% J8 h7 w* U x+ r+ I}
- \/ \/ }8 ?/ Q% [sort(p)
% g. q0 E, G" V1 `/ Tchar *p;: K% W* I9 W; y
{int i,j;4 l) d/ Z. I7 U3 Z, ?
char s[10],*smax,*smin;
. p9 j# G" P9 u; x$ G( G for(i=0;i<10;i++)
" I' D1 i: d! s+ b! v4 R {smax=p+10*i;
# I6 O8 {6 n# C4 ?2 x for(j=i+1;j<10;j++)9 K/ J ~" ?# X
{smin=p+10*j;$ B! \9 L& q# b6 B5 u1 L
if(strcmp(smax,smin)>0)
. G) K$ {* ?: F1 i* F$ A {strcpy(s,smin);
4 h9 ^' @: H" o( u4 v strcpy(smin,smax);& r7 r8 F1 l" B; s
strcpy(smax,s);" |( _4 q$ _4 W8 U# Q# `$ q
}3 ]' v1 m3 m$ Q R3 O
}% `0 V. E+ R" t3 ]1 o$ U" f5 j0 P
}
Y$ `) L& S( W0 J2 Q}
" t& s& B- j2 a7 b10.120 F8 r4 s" j, r4 Q( {! p- c* Q1 r2 a
#define MAX 209 @4 C7 y% f3 t' H6 U
main()
+ `5 F2 O9 v7 \0 K- \( q{int i;
# M7 l; {/ R* V6 b. ~1 z char *pstr[10],str[10][MAX];' X8 q6 B2 H: X, Z" O
for(i=0;i<10;i++)! ^. U2 T: O& Z4 f1 \- R2 O
pstr[i]=str[i];" d: G& x8 U: n- x O; S
for(i=0;i<10;i++)
+ s2 ^" y- ^" _" Y3 e scanf("%s",pstr[i]);& H' r' o; R/ Q+ w9 q& j
sort(pstr);: U( G# X( f8 B- T( W! G' h# P# g
for(i=0;i<10;i++)
- P1 V- P. G2 k+ I/ F printf("%s\n",pstr[i]);
' e7 |- z1 n J @}! }, _ c9 ^$ N }3 ]
sort(pstr)7 s/ F3 T, ?. R9 p/ \& O7 B( F+ Y
char *pstr[10];- w, o' y9 I( B" W3 j
{int i,j;* l0 B8 i5 ?7 s2 i0 c6 U
char *p;3 x* W1 {4 O& c- P: n3 ?. l& ^
for(i=0;i<10;i++)
9 C! t- t. d: R6 L0 g0 X {for(j=i+1;j<10;j++)
' d d4 p9 G' n1 k/ M- d {if(strcmp(*(pstr+i),*(pstr+j))>0)
" C4 s; ~0 t i+ l1 L6 U3 _ {p=*(pstr+i);6 w' f) Y* ? B3 b: I
*(pstr+i)=*(pstr+j);
3 J& g+ \0 R& \5 G *(pstr+j)=p;7 L. k/ m: V3 _
}
. T0 X( k& M( j }6 t/ G$ G- W8 N& }
}
3 p! N% r V {( N. K3 i}+ x# ]# e# L/ _# n
10.138 E" b* Q1 ~8 N' x
#include"math.h"& @" N: G; R8 |) u0 g* W* N, D
main()3 f- S8 R5 N- ]* c
{int n=20;
; W1 V' R: ~# E% `( i S float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
! s: g& h! M$ f" O scanf("%f,%f",&a,&b);
$ U5 I' H; f1 \, Q scanf("%f,%f",&a1,&b1);
* P1 H/ s0 \1 Y$ D) Z4 C scanf("%f,%f",&a2,&b2); A+ E7 | A: K7 T9 U2 ~2 J
p=sin;) h, B! W* Z# N! l9 X2 E
c=jiff(a,b,n,p);% B. W) e1 |( j ?2 @) B% ^- m
printf("sin=%f\n",c);6 P* L7 ^. i" S
p=cos;5 I3 h4 p4 k" G, h7 v4 G
c=jiff(a1,b1,n,p);: `7 g6 n9 D6 W# W/ o9 g6 R% ? z
printf("cos=%f\n",c);
: f7 g/ B" h% @0 p- u p=exp; P) F1 H/ P* h2 C
c=jiff(a2,b2,n,p);/ k( r/ I3 a- L' W8 A; p5 c: S
printf("exp=%f\n",c);
) O/ F- d! D q1 m8 j) N}
0 z! n: O" q& G% x, q9 ^float jiff(a,b,n,p)% ~! B5 X, e* t
float a,b,(*p)();4 ]8 t( O5 d+ L/ Q: E8 ^5 p
int n;/ l7 |8 O: s2 }; l+ _, g
{int i;
$ @5 d" i$ j: r' L! ^/ n float x,f,h,area;
$ u3 M; | C/ I/ f5 Z8 L9 C7 A h=(b-a)/n;
) B! X+ W! z$ [1 M! {/ N3 v) \ x=a;
" k& A4 f" y' }1 F) p1 I* z area=0;" t# r" {: e7 X' @6 D* {
for(i=1;i<=n;i++)3 c2 P( O4 C9 b% P. s
{x=x+h;; b8 q8 ], K1 Y! A, b. V$ Z1 R
area=area+(*p)(x)*h;
2 ^! z. D5 D8 K/ A0 H$ M }
6 K$ e+ \( m& W& ]% L9 ]/ b return(area);* e" A& ]; q' N% H1 {
}
" E. @) M' `7 _6 E8 ^) i P10.14
2 H; S# i5 W2 Y+ N; l; ^main()
2 ]: b, Z# d/ x3 T7 w3 w' \{int i,n,num[20];" l% A8 S J* g/ f, o
char *p;
& I4 x+ w( w5 M2 G$ ? scanf("%d",&n);7 M! s; W& Q4 i& a& @
for(i=0;i scanf("%d",&num[i]);) A6 ]# T" H/ P$ B! l
p=num;, F6 l3 ]$ X4 }# D
sort(p,n);$ J$ P5 t0 E) V8 s
for(i=0;i printf("%8d",num[i]);. p1 ~. P$ P% f) G6 C5 r0 Y U
}
0 H8 V& k# z) f+ ^" w. n+ M- T6 V7 G" ]( ksort(p,m), z) y8 V0 W0 ^/ I2 M; j% L
char *p;) J* K' ~7 J8 s9 q5 O# X4 y+ V
int m;
; Z3 n& M. K* O. p' x9 O& \. c, e{int i;
) H$ a5 c/ m. u; f char change,*p1,*p2;4 X V- e: Z3 {: s& `5 c
for(i=0;i {p1=p+i;
8 Z, R0 \9 O$ @8 t4 } p2=p+(m-1-i); q0 P$ Y6 q6 F. ~0 J& E
change=*p1;
- P# m& w5 V6 T' O* Q! S8 U" N *p1=*p2;& ]! P: @, d Q$ `$ ?; @! e- H6 T5 @
*p2=change;' W4 z- d* r) P' Z u% @( }) R h
}
' p- I& J5 O" p, Y) B}
7 H0 s# u/ p0 x( I$ t* \. F4 g/ \; Y10.15/ m7 _' g) o+ U4 Q- p+ k- I3 e
main(); G2 N2 q+ ?, J" Y6 ^
{int i,j,*pnum,num[4];
# A* G3 b, e, o8 B2 T8 X& n float score[4][5],aver[4],*psco,*pave;1 N6 ~ H: h: E% v+ {% D5 ~
char course[5][10],*pcou;
" U, \! K0 J1 h4 c pcou=course[0];
8 T& E: ]5 W7 F& S9 C* t( R/ K3 c/ [ for(i=0;i<5;i++)$ [2 d/ t: N4 Z
scanf("%s",pcou+10*i);$ l j( N7 E! o' H
printf("number");3 P, ?0 N5 y" u% k, w
for(i=0;i<5;i++)8 A; |! w9 L# l' Y- h2 O. V5 D
printf(",%s",pcou+10*i);; t7 W8 U7 E4 p9 ^$ ~1 V/ S9 R
printf("\n");
/ M, L! A( b: J psco=score;# p7 G& k$ y( i' b: z' V
pnum=num;
, b0 _5 S! Y- @, ~1 N for(i=0;i<4;i++)
1 Q- _, C- I f/ Y) s- T {scanf("%d",pnum+i);
' [- N9 b% c2 U+ q3 ` s for(j=0;j<5;j++)7 W# w3 l: j/ M: w- F
scanf(",%f",psco+5*i+j);
8 |; X- [) D$ J" [ }
; S; b* w" q4 j7 V: y pave=aver;' r8 r9 l& @, e, F, z! Y
printf("\n");5 P0 H ]3 v' ^/ A* w
avsco(psco,pave);
& n# k- M P* L avcour1(pcou,psco);
* w; X1 o2 L) t+ T) N& D4 m( g printf("\n");$ b: m7 T& t) R
fali2(pcou,pnum,psco,pave);! Y4 B. v4 T' |. ~0 _
printf("\n");8 _9 S; T- ^' n0 S7 R. E
good(pcou,pnum,psco,pave);
6 \! V. {# V! B/ X" x1 h: j& U}' F- {* ]" [1 j: h
avsco(psco,pave)
- W# g7 v, }" R1 _5 H5 wfloat *psco,*pave;+ r* s# X/ B1 a0 U" F. ?0 l
{int i,j;" h- i2 \) M& U7 `8 T# B$ n4 N
float sum,average;
7 ^9 u" T4 u0 E8 h2 t for(i=0;i<4;i++)1 X3 _$ S5 I) R% E ` S
{sum=0;/ P" w$ d: O! `; Q7 u
for(j=0;j<5;j+)5 A- @2 X* z& y% [& B
sum+=(*(psco+5*i+j));5 w: v. S% p! d' t
average=sum/5;
. I8 c+ S, ?1 ~- Y *(pave+i)=average;6 i o; B! k) }# n$ K$ j. z. @
}
% _* `* I8 u* f; d; U! z9 F2 n}
. b% ]0 G: h9 y0 q& L cavcour1(pcou,psco)
- k3 I' R U+ p3 Achar *pcou;
# E6 q4 t$ ]1 U @9 y+ N* P" Efloat *psco;
3 D5 O5 ]0 a6 ^' Y# ~{int i;
0 i3 A1 X5 B( C1 s( }5 q4 \& D# d. \ float sum,average1;# u) j: T7 R& V& N n
sum=0;; k' y. V4 V" p) ^# t5 V
for(i=0;i<4;i++)
" G5 V3 b4 ^* ~. c+ H8 h% ^ sum+=(*(psco+5*i))6 P- f' ?- K& X t
average1=sum/4;9 A8 t: Z! t3 b1 W; i, t: o
printf("%s %5.2f\n",pcou,average1);
4 x, _3 R( i+ H- n}
+ m% n0 A) |9 Z% Sfali2(pcou,pnum,psco,pave)/ M% t4 H: w$ o4 I% w: K
char *pcou;
r- }4 z8 F0 v9 U8 ^2 E7 e) Gint *pnum;
# K! ~3 p/ ^1 i; O& u/ O5 ifloat *psco,*pave;: w- A" @& t7 z* K* E# {% _2 l7 P
{int i,j,k,label;
x6 U" |; k( e z& r$ i: \6 e printf("\nnumber\n");% V s; d0 e% a* r
for(i=0;i<5;i++)/ h: ]) V8 B- F, c! E/ C0 g: X
printf("%-8s",pcou+10*i);0 j; [% i: b. p; j, N& @1 o
printf("\naverage\n");0 ?2 v+ l7 j: T
for(i=0;i<4;i++)6 u& L% w- M' q1 P! V
{label=0;3 S$ c8 }0 C* ~: K& p( i) {1 S
for(j=0;j<5;j++)
8 {, K" K: F. B if(*(psco+5*i+j)<60.0)label++;
" d2 k5 a/ L5 i5 n if(label>=2)
( @1 C e/ B8 J/ b0 Y {printf("%-8d",*(pnum+i));
, m6 E* d6 z7 \& _" ] for(k=0;k<5;k++)
5 z( ^0 P) V' v" }$ P+ L3 Z- E printf("%-8.2f",*(psco+5*i+k));, r, I' [0 L( I; x( ~0 M4 Q
printf("%-8.2f",*(pave+i));$ M2 I1 j5 ~0 y s" ^" g, u0 l# Y% b
}; ? ]4 o4 [6 u
}: k- R3 {- S/ S: c" t* A. w
}% C: |: e. M/ `: w
good(pcou,pnum,psco,pave); ?3 M- @) z/ `- @! r! _1 n
char *pcou;
$ X7 \/ }. s( {' S( v0 q3 Yint *pnum;/ r/ H( F6 Q5 P+ b
float *psco,*pave;8 J- u7 M4 j: H: E4 l: e! g9 Z6 E
{int i,j,k,label;
2 k: N; N) B! ]" _' G- r9 d5 x printf("number");
, K3 X8 l0 q6 L! w9 g/ g" G0 N for(i=0;i<5;i++)3 P3 U0 O e& ` x6 Y6 R3 h2 j/ G) ?
printf("%-8s",pcou+10*i);
' a& R2 `9 U: I* k# p# @; w2 C, N printf("average");
# x* K' I$ |$ |& ^4 k- I for(i=0;i<4;i++)
4 z: [* h& v4 |0 a8 \8 {% s {label=0;
) r8 K6 X0 B& _9 A& d5 W for(j=0;j<5;j++)
6 F/ k" F# w+ h9 ^) q! W if(*(psco+5*i+j)>=85.0)label++;
- E3 C: `4 H) X0 h, O" [) g4 N" l2 ~ if((label>=5)||(*(pave+i)>=90))
2 u0 I; t2 d$ d' l {printf("%-8d",*(pnum+i));
: O, S% i2 d; V5 }0 i% z3 e9 v for(k=0;k<5;k++)/ c5 j$ Z" h; e# _, q+ S2 i
printf("%-8.2f",*(psco+5*i+k));
. w; @) o4 j4 v. s9 V5 \% L printf("%-8.2f",*(pave+i));
* |2 @' c: z# N0 N# E J1 X } ~ @8 j5 w# m$ }- B8 e8 w% v6 v9 a
}
& H7 u5 @) W$ y3 x4 c# a/ G( Q. Q1 q}' V) [1 {9 f' y7 x
10.163 q4 R2 {! z/ S" b. n( p
#include"stdio.h"3 a/ I. Y5 y2 g5 A
main()
; U" P G5 V5 T$ K7 V. i9 k{char str[50],*pstr;) u; R. Z6 R# T* a: V9 N
int i,j,k,m,e10,digit,ndigit,a[10],*pa;
; }5 ^; g- v& _3 p8 X2 p- {8 v gets(str);
; w5 u, W# u3 \- g$ w. E pstr=str;* z' }0 g+ } P3 `3 Z+ `
pa=a;
% y, T* R" W! r1 W! m4 k ndigit=0;. o, S: i) ?* U0 T4 ~5 o& E7 \
i=j=0;
: p3 _6 a; y' k3 I. U" [ while(*(pstr+i)!='\0')% [# y( t# ?8 h. p* N
{if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))
. E" s+ _# {6 j0 y- v j++;
4 Q) l- p3 J* m" n: c else4 b7 O1 N& x; P) D1 A
{if(j>0)( Q+ _9 {- c r+ |! M9 R
{digit=*(pstr+i-1)-48;
, {: ^! v- x) z& G* U$ X* ] k=1;
3 R8 O4 Z1 i# _; ~! k while(k {e10=1;
9 O2 u2 T N, R( {6 O- L3 j# f for(m=1;m<=k;m++). V" `% @/ y& g& `- z; V( s
e10=e10*10;( ?$ A+ I8 K6 t' y/ `; k! i; ?
digit+=(*(pstr+i-1-k)-48)*e10;( R" U. d/ \! W/ [- f
k++;1 T) v5 t% m; p
}
2 z) R4 L+ j1 W2 I. t8 N L$ ]% Q *pa=digit;& R2 V4 u2 e9 K, E" G
ndigit++;6 X" E8 u' M) K5 p7 t/ ?+ D
pa++;9 G8 ~' t ~6 E4 |% S( ^0 @+ R
j=0;
8 O6 Z7 h! J: C' ]1 Z) f }
- b1 L7 ?! ?# Y+ J4 O }+ x, T7 f7 D- v& }" t8 A. U/ V
i++;
# k0 Z: {9 {: ~9 b$ n1 S7 ~ }
7 D9 T) c; E; q, ~/ O( J1 Z$ n if(j>0)* z! b% U9 I1 t- ~
{digit=*(pstr+i-1)-48;( c& N2 w, u( h! s3 Q
k=1;
- w! p6 u! W7 R0 [0 Q while(k {e10=1;9 I; k2 g8 h3 n8 z ^
for(m=1;m<=k;m++)
1 e' ]6 F4 \! f" u% \- f3 H e10=e10*10;
4 j% J! M: w; A0 }1 t% R2 S digit+=(*(pstr+i-1-k)-48)*e10;
- c/ Q8 }7 e# h8 d/ _ k++;$ D& q, a- V3 d2 q
}" R0 b+ `8 \- n" y
*pa=digit;
' R& S5 V" b( v. t; z2 P% O ndigit++;0 E4 I+ y3 J6 P8 v
j=0;7 k3 n& \! x- u1 T
} . ^( Y( y' T5 q
printf("ndigit=%d\n",ndigit);
$ ^3 b' Q0 l! t+ r! x8 f R1 \& i j=0;
) {7 W" c3 W( W, [$ v5 G pa=a;
! ^7 }) @1 S- r& w5 ^' y6 b8 w for(j=0;j printf("%d",*(pa+j));$ i! U( t R% P/ f& Q' W& E+ y+ @
}, m& u+ O) B) L
10.17
! t, ` O6 N7 ]. p gmain()* d3 O8 j# J/ B. w/ B
{int m;
% x+ {3 B# \4 w+ b- @# ?: k8 i, p5 Z char str1[20],str2[20],*p1,*p2;- H) A- y/ m8 A, O6 n8 ~
scanf("%s",str1);1 F# _ [$ N, `% F( C
scanf("%s",str2);
) o; |/ ^; x. e p1=str1;
$ J( B- {$ R) I& ?: d+ m( C+ n p2=str2;
+ e+ Q6 U& F0 y2 d5 \# K( g8 J m=strcmp(p1,p2);' ^% ~2 F2 I" J3 T
printf("%d\n",m);. R0 e9 J/ V" w* g
}
- n$ O/ f! h. A* w* I8 Astrcmp(p1,p2)
- O k! s$ S1 ichar *p1,*p2;
/ y# ?( F U9 |9 q6 x% _{int i=0;0 M$ K( h. `/ h* S7 T+ a
while(*(p1+i)==*(p2+i))0 R: Q7 \1 e$ C1 \9 m
if(*(p+i++)=='\0')return(0);! N" D: ? b' b5 {5 T/ K# M
return(*(p1+i)-*(p2+i));: M1 g- n0 m8 f4 |
}/ k8 [3 [& j. l6 v
10.18
) N. T |; s* r' Mmain()
- B$ u( p" l- r- a* |( _{static char *mname[13]={"illeagl","January","February","March",+ t* U1 Z/ u, v9 Q. p! H
"April","May","June","July","August","September","October"," O, L9 z+ p( w' E2 R
"November","December"};
/ l: L+ Q7 `( b+ M# X int n;
4 P( Q. i; T# L# I: V n scanf("%d",&n);6 P* s% b$ j- }$ ^, O
if((n>=1)&&(n<=12))5 c6 {' N( i5 T* v U3 U/ U
printf("%s\n",*(mname+n));
" m! H3 }0 T4 _) o& M | else- f7 p4 e4 |: U& p( ?
printf("error");
) d* F1 y7 |9 f ]4 R' Y}
- U( A* y( b# ~8 M10.20
; k% ~# ~" S% G2 @, Bmain()
+ E+ l% e/ c) e, ]" ?7 }0 u{int i;
1 M( y4 s2 W! p, x9 |: | char **p,*pstr[5],str[5][10];6 s2 y+ R# ^4 ]. j5 t7 n
for(i=0;i<5;i++)/ `- l. X' D' _' c% ^) N8 R
pstr[i]=str[i];
/ ~6 w% [# V7 E k for(i=0;i<5;i++)
/ z6 x) @* T7 u7 O( R scanf("%s",pstr[i]);; {2 ^9 W. \+ i! _" G. H8 u) {
p=pstr;' p7 o3 K* u! ^( z
sort(p);2 {8 i& r# E2 l* b+ p
for(i=0;i<5;i++)3 s5 m- T, B2 H2 {% j
printf("%s\n",pstr[i]);2 Y; S1 L* b2 Y Q* T( A/ ]; Y6 ^0 q
}
" A# E$ C9 C( m+ }+ a9 I' {sort(p)5 x4 ~9 R6 n8 J3 t/ z+ S1 p0 n
char **P;/ q3 G. W# b+ Y6 s
{int i,j; ^8 F r( K* |) h
char *pchange;
4 V- ?9 s! l. P5 n& c5 q for(i=0;i<5;i++)! I$ s5 e+ x5 @6 D6 x/ e6 x
{for(j=i+1;j<5;j++)7 s- i# R, { U
{if(strcmp(*(p+i),*(p+j))>0)
# v/ a) {9 ]" n {pchange=*(p+i);5 Z1 f0 A* d6 R4 c4 _
*(p+i)=*(p+j);
1 @" m! W/ a) O$ G1 o5 G# ` *(p+j)=pchange;7 V$ I. ?0 C8 ?. E$ Z2 Z( r
}
& h) Q4 k. K5 \ }
5 U+ s7 ?: [4 T }6 F3 H5 E m5 s/ r
}, O+ j- x, E3 I, R) [0 s
10.21
0 \) p8 m3 t, E0 Hmain()
% n. F$ P. K9 d+ z{int i,n,digit[20],**p,*pstr[20];
: n( _9 a& \; X( C4 z& n% A scanf("%d",&n);
' j7 O2 H8 \7 c0 V( ^# [' Q for(i=0;i pstr[i]=&digit[i];
- x" b% Q4 H% F- x8 |& t- T& k for(i=0;i scanf("%d",pstr[i]);2 H9 u F' ^& Z" q' c$ y
p=pstr;
V; H1 n& {. D; {3 m sort(p,n);
7 h+ A) b( |5 ]" `2 Z for(i=0;i printf("%d ",*pstr[i]);
6 f- O6 T- \$ s0 b3 S' [: s}
: G s& z. A3 r J, osort(p,n)
, O0 B7 U" b! w" y) _6 Y& i* Gint **p,n;
1 N- T6 a! h8 y' W{int i,j,*pchange;& l- V& u! _' \" w9 ], U( R# u& J% f9 |
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))3 _5 q$ N6 n' R
{pchange=*(p+i);/ O5 j8 e% {7 U& P% Z7 C7 v
*(p+i)=*(p+j);7 Q) W0 P9 i, c. K
*(p+j)=pchange;3 t1 H; |+ z/ n0 b' \+ k& s1 C% d) V/ {
}2 s4 k- \- w) c
}
! @$ R) w, Y7 y" u, c }& x# o/ ]& ^5 a$ ?( L$ d; s
}' L2 F' H! e7 e
第十一章 结构体与共用体
9 j+ q4 S6 w( _" q' |. v. t11.1 H. R3 A: h2 A$ B# H) j
struct
- z5 N, T# }1 {6 u C4 w: R) q; C {int year;
% l4 Z7 k8 m$ z1 w int month;9 _, F, a, S3 K$ q. D
int day;/ t6 p3 F7 f) w* D6 F
}date;
1 A0 A6 C) o+ V( L) Amain()- D* ]* E& ~- E
{int days;
# L0 `0 g% u/ Q1 n! N4 R/ ]( E scanf("%d,%d,%d",&date.year,&date.month,&date.day);
* I$ E5 i% F" U6 D3 z5 b, i% F2 ?6 O) { switch(date.month)2 o: [+ ?+ Z2 o( }
{case 1:days=date.day;break;
5 m+ E9 }4 X9 J" ? case 2:days=date.day+31;break;
& |8 q# c% ]9 F6 ~, t case 3:days=date.day+59;break;: f. f3 o( x0 `9 f2 Q
case 4:days=date.day+90;break;
" B$ r; N; R0 M, T case 5:days=date.day+120;break;) w. N. z: ]$ B- T+ v
case 6:days=date.day+151;break;
7 @+ x: X3 y9 y$ q0 m) P& H case 7:days=date.day+181;break;
6 ~# c( h0 I! Q/ r* { d- F# t case 8:days=date.day+212;break;
% a% I3 V: s: o case 9:days=date.day+243;break;) @4 K* x$ _! C: v+ C
case 10:days=date.day+273;break;
' |1 K$ _1 O% v4 N case 11:days=date.day+304;break;7 E. t4 V7 d" s/ y8 _7 p/ R
case 12:days=date.day+334;break; D, P; A y& U8 X! u( h9 P& |$ p
}9 c2 u2 E' h4 J, @5 k0 b: x
if((date.year%4==0&&date.year%100!=0||date.year%400==0)
3 b/ ` _) Q7 Z4 S. S* i &&date.month>=3)6 h8 t n }( K
days+=1;9 _, B2 W' R& b% ^7 ~2 ~3 v2 k9 X$ i
printf("days=%d\n",days);
) e8 X) W d& P: \# G# J3 r}1 [- u2 p4 Z; n( v& S+ N
11.2
0 i2 y' e- l! M6 U! n4 m3 G- Lstruct dt
, C6 [, l1 ~+ a* P {int year;
0 E8 F+ a+ W( Y! n3 h4 g, a$ q) `" T int month;: z2 N6 H# ^- }
int day;4 g& L6 O( {; n( _3 s: j
}date;
( c `7 ]' X/ X+ e7 ?- d' r ?main()2 ^# r5 m4 C- e8 q. w1 q; \
{0 Z6 e( `6 m- H6 b& t# g9 o
scanf("%d,%d,%d",&date.year,&date.month,&date.day);" g* s; [' T% H8 q
printf("\n%d\n",days(date.year,date.month,date.day));' Y" X1 {7 V. M/ _& I& f: O
}
9 Z0 O9 `) P% n. ?- Qdays(year,month,day)# z! p$ j% n' Z0 t6 l2 ?
int year,month,day;& u s6 P9 }2 ]2 o
{int daysum=0,i;* `7 @% u" P$ C: o
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}6 q! w5 d) \ b5 B+ w' |' I0 A
for(i=1;i daysum+=daytab[i];
4 G- c4 m5 p1 b0 ^ daysum+=day;
( w# |$ u3 h) x( Y' e1 _ if((year%4==0&&year%100!=0||year%400==0)&&month>=3)
+ A' g2 F& M9 J daysum+=1;
2 D6 G3 C3 ~! \% l& Y return(daysum);. C, j1 B( n" B9 C! U+ U
}( N3 G$ I: O! \# D% O! N C/ o
11.3, [9 O& e3 _: _1 h
11.4* @) j- F: R' I" e- C' j. p
#define N 58 {6 f! B! D( x; g* `
struct student5 P1 c, y0 M8 u# d; N0 ] }- o& Q3 ~
{char num[6];
8 w' i+ `- @* T& V/ k char name[8];# x3 m2 Z5 R+ O7 g1 ], M7 x
int score[4];
8 b( T' I% {" }" ~! r9 F# ? }stu[N];
! @" Y0 j8 n# Cmain()
% V" @3 h2 {0 w2 p$ k- ~{4 G4 A1 o" w+ R" J; V, ~
input(stu);1 i3 t j! n/ E: v/ x6 X7 @
print(stu);
) Q1 Y9 ?, u: @1 i1 O7 g5 l}! ~/ s& Z+ @0 J
input(stu)1 y. |& v8 I2 v2 H6 o) T4 y
struct student stu[];+ K3 T: o1 K! j ?6 F2 @. o9 M
{int i,j;
0 n5 b- `) y* Z: K3 \ for(i=0;i {printf("number");
1 J: s0 W$ K' o% h5 F scanf("%s",stu[i].num);' r6 v2 j) \; p8 S l
printf("name");
9 g9 c0 U; B0 d scanf("%s",stu[i].name);/ z3 c2 `+ ?& q
for(j=0;j<3;j++)! Y+ N' F- L" P. Z& |
{printf("\nscore\n");
5 `0 r2 S$ K# { scanf("%d",&stu[i].score[j]);; y e' @4 ]/ ^$ m* ^* Y+ l
}
& i+ [" u& @) o" X% @, v9 | printf("\n");
( ~+ b4 y9 {# m4 H r7 U* R }
' j( Q$ p0 W; }}* E- u% y$ p( h, |# ^
print(stu)
5 l1 x, v8 h) i2 H; O; E4 f8 P: O% ]struct student stu[];1 E5 L1 ^7 O! o
{int i,j;
5 \/ E+ j0 ~9 T9 ], ~9 G printf("\nnumber name score1 score2 score3 \n");
\' ?: l- j0 ] for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
6 x2 _! C! Z, f2 E8 v9 X9 x for(j=0;j<3;j++)
' T& j& a+ `6 A7 K, W8 e5 ? printf("%7d",stu[i].score[j]);
8 D. L7 X9 z2 A# G1 w* I4 N* ^5 x: B! M printf("\n");
; c; g1 I& g$ q! ^9 h! o S }
- f* ?+ C% M- h- h}& @: `8 @+ J: {
11.53 J5 u# i# w& Q& l0 K, Q
struct student
' e _5 C/ g0 _ {char num[6]; E( [# m. J3 V- x- K
char name[8];& t m3 y; `( R V' M9 \3 `
int score[4];
; V% P# X+ O& \( M, E9 b( ` float avr;3 N" M/ d* c4 v& W% Q. l$ {( Y( }
}stu[5];7 D8 h: S$ c! o( u0 P
main()
; s! \0 f8 O" _* A9 u; I{int i,j,max,maxi,sum;2 ]# d0 D, ]( [6 A
float average;$ y. u4 C$ e- H. e3 n) q
for(i=0;i<5;i++)
/ ]8 ~0 J8 J; b- b0 d3 p {printf("number");
* i# K' N0 H: [- n scanf("%s",stu[i].num);
/ `% w) o; f* y* n printf("name");5 X8 J+ M: F$ ?- s( \/ ?2 l4 f
scanf("%s",stu[i].name);- X+ T4 j5 o4 o0 z% \
for(j=0;j<3;j++)
* z0 p/ \2 b( r' J: }7 z* a {printf("\nscore\n");* _, r, _" E7 U u
scanf("%d",&stu[i].score[j]);
. n! }1 h. L# L. E# }* j! q0 e, B }" U5 V: c- l& M# D5 U
}
) P9 u3 o& e/ E' ~ average=0;+ n: r. Y* R( }& m z2 X C2 U
max=0;% Y; N# E; _! E# ^; }" @6 F
maxi=0;
2 m- e; t; N7 L: g for(i=0;i<5;i++)
, n! g* D% z9 C5 f O {sum=0;
+ n6 R- ?5 L/ X- M" B, I5 Z for(j=0;j<3;j++)
. a' J2 ]4 F- v5 a# D0 ]; H! s: y sum+=stu[i].score[j];
4 t! o X3 w, }' ^0 a" {4 _, b/ m stu[i].avr=sum/3.0;' p9 t$ i& ~7 {5 t1 Q2 N5 J. Q
average+=stu[i].avr;
- _ E( t8 H% D v5 B0 T if(sum>max)0 q4 [2 |& M8 h9 J9 \
{max=sum;
' v) O2 X' F: U7 A" u2 ?1 V! t) ^ maxi=i;! Q0 E5 m, b/ p- ~4 J
}9 h& O: M+ `$ n
}2 N: {! Z7 Y. b1 x" P3 J( v
average/=5;
' y5 T" q" o) d$ h+ D printf("number name score1 score2 score3 average\n");
9 h7 L* W2 ]1 x U for(i=0;i<5;i++)
( B8 {5 V' ]/ q0 w {printf("%8s%10s",stu[i].num,stu[i].name);7 d% \, s; d1 w8 k
for(j=0;j<3;j++)* \9 T( X4 p$ ]* P, w5 [3 c
printf("%7d",stu[i].score[j]);
( b# A2 z% D& D8 |& T printf("%6.2f\n",stu[i].avr);+ A* B5 E, s8 J. }/ c5 K
}0 W9 R2 A5 n7 E/ u
printf("average=%5.2f\n",average);" i- M" b& W1 G6 g1 f) A
printf("The best student is %s,sum=%d\n",stu[maxi].name,max);
6 ]4 t( X! x0 k' S6 A" G& \}1 B4 C) Q' D: e3 L6 v7 M* ~7 x
% N% k/ y4 x6 h3 v5 h( f& e' U |
zan
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