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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
9 |- T0 F n! V; u0 O1.5请参照本章例题,编写一个C程序,输出以下信息:
& m6 {, l" y: Y7 M" V2 J/ Lmain()
: O; m0 Y5 }. U7 S- q' ]{* k* K. |, k; h0 `
printf(" ************ \n");
5 @2 `: x; D7 V: Q. X5 q" `printf("\n");2 [+ |, b( n% j5 }# ~
printf(" Very Good! \n");" N8 y$ o8 G+ Z) p9 \
printf("\n");! @$ p& i: L& r) z
printf(" ************\n");
) T0 Y, a0 ?# S% B$ L}
: D: |5 v7 ~% a' _ O8 J4 E1 f1.6编写一个程序,输入a b c三个值,输出其中最大者。" W, }4 k- W! x1 ^3 W. u
解:main()
( M/ _$ A, z; c6 ^( y# D{int a,b,c,max;; Q' o# S" B- C: O. ?
printf("请输入三个数a,b,c:\n");
! `8 E2 _! J9 _/ a% E! iscanf("%d,%d,%d",&a,&b,&c);' J4 q& d1 g6 d" }4 u
max=a;
1 w2 `% Z# N" |if(maxmax=b;
1 J' ~7 ^' U5 v' w/ |6 F4 k6 [if(maxmax=c;
9 N% P" o# Y: x! K8 A+ T$ B+ aprintf("最大数为:%d",max);+ b' H% |& k. a
}
- a0 N* F0 S" ?$ K, h第三章
9 B# r" P) v: l1 Y& e) Y# x: u* c3.3 请将下面各数用八进制数和十六进制数表示:
1 M+ f0 t/ W, G9 t* p# t" W# T8 L(1)10 (2)32 (3)75 (4)-617) ? h( y F* I9 { Q
(5)-111 (6)2483 (7)-28654 (8)21003
+ |. s1 y3 x* j解:十 八 十六$ w3 W/ ^# R: T
(10)=(12)=(a): f' ~' Y3 n& N3 |& n( Q
(32)=(40)=20: A; G: {, R% Y
(75)=(113)=4b
) X$ u& l @; y+ u8 B% n/ S (-617)=(176627)=fd97. K0 `& [9 Q8 g: E, k
-111=177621=ff91) i0 W, U7 Z3 S
2483=4663=963- @( j( s5 l" x: b0 D( S
-28654=110022=9012
8 U K0 `5 ]4 C1 N# D! j, p: q 21003=51013=520b
+ m1 b* f/ L+ m! C4 Q) d6 G3.5字符常量与字符串常量有什么区别?
# n8 u6 G/ h! o7 C" }解:字符常量是一个字符,用单引号括起来。字符串常量是由0个或若干个字符
! L1 q: f1 Y2 D7 ^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'.
2 P. y/ J# f% }0 W3.6写出以下程序的运行结果:
: _6 e. i0 _0 l7 m) o6 O7 s#include, ^: p* t( K E# L
void main()
4 b, W2 {( n' d{
4 v# X4 d4 Y) r. Echar c1='a',c2='b',c3='c',c4='\101',c5='\116'; E& {- v7 _" q9 ^' s
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);- l. f/ A0 O/ ^
printf("\t\b%c %c\n",c4,c5);6 o% m( i: l- A4 }2 w4 p0 s2 M
解:程序的运行结果为:0 N5 M# a% f0 @
aabb cc abc; b2 S) |- Z2 c/ s
A N6 J) C. L9 x3 W" z) N6 H
3.7将"China"译成密码.密码规律:用原来的字母后面第4个字母代替原来的字母,
+ ?) i, N: d$ `, `. ~, z+ a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".
4 R% u, b: q& H# _; C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( S4 L* G* `' e/ {' d3 M/ P
输出.
# m' X1 H# ?4 \ m- v6 Smain()* @* j' |- I" ^6 N$ `
{char c1="C",c2="h",c3="i",c4='n',c5='a';
# \* z, |9 u$ y K8 m* W& n( u- xc1+=4;6 ^, C U6 t% S0 s- b( `& c" H
c2+=4;
, }# ?4 d% I; C& M9 S+ g& |c3+=4;" w1 p# L2 @7 l& N+ }5 k( `( x
c4+=4;
W% w8 [& e1 X; i" z8 v$ Xc5+=4;3 z6 X/ I H" W, |4 ~
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);1 Q4 B. x* J8 B) C% ?2 H
}
2 P0 X: j) C9 k2 i/ O3.8例3.6能否改成如下:
+ U! J' U! p8 i6 x9 E7 P. i' ^#include. ~3 B8 v, l% N1 m9 x
void main()
+ X* i0 G- U! m( I1 ^/ W3 V{7 ]5 h6 p$ g+ n% L) X- Q# j2 y" a A
int c1,c2;(原为 char c1,c2)$ y% ~4 ?5 h) x+ i& w2 ]) i/ X1 N- D
c1=97; [, S% B) i# K Y9 T
c2=98;
0 h6 D& L, O# G1 J. aprintf("%c%c\n",c1,c2);
" c9 J0 p, h+ z0 Jprintf("%d%d\n",c1,c2);
2 u8 [- }& P% R& Z3 a' |5 e}+ a( L9 C5 J, w
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.
2 f. m( q; g( W# p& |( k' B3.9求下面算术表达式的值.
1 a- B4 ^4 j$ _; F(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)8 s! m# D9 r+ P! k4 w9 ]8 Z
(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)2 d$ _- w% F! }' }
3.10写出下面程序的运行结果:
. L+ s6 S& o( m1 I- k: i#include
( u4 d+ R. d3 Uvoid main(), g$ D E- F3 ?$ h" r& X* y
{! o3 b. N; O! Y2 M: l' J" R: R
int i,j,m,n;
6 U' H$ b! s j8 W1 _: ti=8;3 N& u0 \; T9 b* K q' R5 H
j=10;2 u) c2 D+ l1 ]% T5 N2 l+ I
m=++i;' m9 a8 ~0 P$ q# Z( p" i
n=j++;
/ [( G. Y9 r6 o {* w+ R8 Tprintf("%d,%d,%d,%d\n",i,j,m,n);8 j, K$ U; P1 z5 C! s/ Z F
}
% {2 M2 n9 s ]6 M% G解:结果: 9,11,9,10( K/ x7 @9 B H3 P2 n) G% @- R
第4章2 J; h( b' e4 f
4.4.a=3,b=4,c=5,x=1.2,y=2.4,z=-3.6,u=51274,n=128765,c1='a',c2='b'.想得$ O" Q; d& P N* v2 ~. a
到以下的输出格式和结果,请写出程序要求输出的结果如下:
4 {) A+ _: S4 v/ ^a= 3 b= 4 c= 5
, C6 y3 Z8 i! S$ Q$ M( r. L- M- \# D7 \x=1.200000,y=2.400000,z=-3.600000
/ ?7 b5 |' R+ o* d& _% a: N- {9 Ex+y= 3.60 y+z=-1.20 z+x=-2.40, ]! C4 P" Q5 D, w$ }
u= 51274 n= 128765
1 f S3 C5 d$ U, kc1='a' or 97(ASCII)
- S* O9 W9 g% fc2='B' or 98(ASCII)
# |; \- D1 a; y( `9 Y; ?6 R [ X4 x解:
$ p( I8 u; n% K n2 a8 t7 h3 R* u3 fmain()5 L& \' D" W c0 V
{
* @' `: l0 m" M( |7 uint a,b,c;
8 {- B! d+ t9 K0 O# Mlong int u,n;) Z9 @1 C" d& a/ K
float x,y,z;9 e# o- p9 Y- u: A3 q( H& T& P
char c1,c2;- R8 x4 v$ K$ ^4 e" ]
a=3;b=4;c=5;* z3 p" N6 J1 I% v" H; P
x=1.2;y=2.4;z=-3.6;6 d" ^3 Q- W$ D" h/ Y
u=51274;n=128765;
/ x0 j0 n4 g% a* M6 _" Mc1='a';c2='b';* o2 C" k. h! I
printf("\n");
' Y3 i* @, N$ {$ M- ]0 yprintf("a=%2d b=%2d c=%2d\n",a,b,c);, g- C; C3 A# L9 f! a9 i+ l5 K
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
2 V# ]7 h! o5 Vprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x); O$ A) y, R8 o# E: s4 N
printf("u=%6ld n=%9ld\n",u,n);: e4 T0 z; u. b0 F5 b5 h! [4 `+ Q
printf("c1='%c' or %d(ASCII)\n",c1,c2);
6 c- h: ?' S/ A: ]$ Vprintf("c2='%c' or %d(ASCII)\n",c2,c2);2 u: f7 O! }! H0 l7 [
}
! z/ c- n. |: Q+ f N8 W, Y" \4.5请写出下面程序的输出结果.
, g4 p4 `+ P9 u: r _5 F结果:& q8 {6 V3 u6 _6 n$ o
57
, _% `2 S* S" R, `8 Y 5 7( e# s! L8 G$ R7 q" b+ I& ~. M
67.856400,-789.123962
) G4 S8 g: b* H" Y# t0 `67.856400 ,-789.123962
( d E6 Z$ V) ? 67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962% D% E. i. X1 Z# z
6.785640e+001,-7.89e+002
3 l4 u8 E* B; G: QA,65,101,41
8 s4 Y8 u$ ~# q$ F& j1234567,4553207,d6873 b/ `9 q9 E; k9 e) U, j
65535,17777,ffff,-1
, u+ t) V: N5 ~' }COMPUTER, COM
5 d4 q3 x- K' u( z, m# N4.6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',- m. d; U( C; I7 D+ ?
问在键盘上如何输入?
) a6 x% M' R. r+ K' \main()6 ]6 T1 [& }" e- F( T
{* g+ N/ h( A. a- I4 P) a) ^
int a,b;9 s: Q6 F4 N2 `2 M+ X2 I; B
float x,y;8 }$ y$ y! ~+ m4 q$ p1 R
char c1,c2;0 N& R" v% L7 I' \7 n- l
scanf("a=%d b=%d,&a,&b);5 g0 N/ f) m4 R
scanf(" x=%f y=%e",&x,&y);# A$ y' D) Q% @- ~: Z6 j
scanf(" c1=%c c2=%c",&c1,&c2); }0 ~' s+ g/ ~' q' Y% o
}
( I6 @1 W( l& `2 ^+ R+ a5 I% p解:可按如下方式在键盘上输入:4 l% a) I: j0 y+ y- Q ?7 O& g; Q# O, G
a=3 b=7
$ m& Q4 N+ \4 T/ [! C2 S4 A7 w' {7 X6 `& Vx=8.5 y=71.82
2 i! L) S6 e( P6 N0 X/ i+ rc1=A c2=a2 g, n6 j0 h1 z
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
+ M5 x, N" z, k* v L z( t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8 q* y! P, j9 H
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: O. `; c, V$ V) O: L+ m
初步概念即可,以后再进一步深入理解.
% G C/ y, ~" h4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-: ` L8 \8 p, r# ^, E' L
3.75,z=57.8,请问- t, {; ?& R6 S9 m% p$ w7 \$ c
在键盘上如何输入数据?
0 f: l0 I& k5 y4 G7 iscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z); m n/ y8 [& D6 _. e
解:
. I1 o1 B& g4 ~+ Bmain()% d f- s+ U1 i% [8 x7 d+ g6 ?
{
- Y% i/ |% N% S, P% a+ q# y5 Cint a,b;! F% P$ m" r \, h& f
float x,y,z;
* @# f) G6 _: Bchar c1,c2;, I8 j/ H/ _+ ^. d8 b+ B5 r4 n* m* r
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
. F B- g7 P# i7 D}
7 r8 O6 ~9 B3 I( g运行时输入:: G* X% e1 Z) A( o
10 20Aa1.5 -3.75 +1.5,67.8' ^4 r+ U# y& \9 t
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
3 k6 }% y# l5 _) E4 d$ ~5 E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9 ^) v$ B( O3 y k8 R3 M4 a
会赋给任何变量。; \2 _# ~6 j; f; \
4.8设圆半径r=1.5,圆柱高h=3,求圆周长,圆面积,圆球表面积,圆球体积,圆柱体积,
/ y1 o! l% e! k6 g9 |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
5 D$ u/ f/ ?: z程.
7 }' C, u6 F' s( B解:main()! f! @# c+ K9 M! ~- m1 j9 d9 Y
{3 w) {; l N8 T
float pi,h,r,l,s,sq,vq,vz;7 d) w8 m9 t& `$ A! W
pi=3.1415926;
# ]& D* w- k2 l9 S r3 M) Iprintf("请输入圆半径r圆柱高h:\n");
3 [9 Z9 @+ ?) A7 m; mscanf("%f,%f",&r,&h);, x" @9 B+ u9 a0 o6 X4 r) ?
l=2*pi*r;
F% Y9 y3 q& o+ g) g9 As=r*r*pi;
6 c6 f# {# T D" f' csq=4*pi*r*r;
4 M: Y5 ~( n7 e$ dvq=4.0/3.0*pi*r*r*r;" E: Q/ l7 Q, k" k- I% D
vz=pi*r*r*h;# u0 m$ I& F" G6 B6 n
printf("圆周长为: =%6.2f\n",l);4 B0 S7 ]/ J% }1 b' _/ ^
printf("圆面积为: =%6.2f\n",s);
" a* v9 T, X. J4 D8 J8 q/ k! Hprintf("圆球表面积为: =%6.2f\n",sq);
% _$ s# y0 Y7 a& r5 Gprintf("圆球体积为: =%6.2f\n",vz);
* L+ j$ s/ R: k( {* ~4 P8 T/ Z}
8 a- p* t2 X4 H) j5 }4.9输入一个华氏温度,要求输出摄氏温度,公式为C=5/9(F-32),输出要有文字说明,8 @% h) X) t4 c. b
取两位小数.6 X: ]& N* \* @
解: main()8 v& ]4 t( V, T3 } ~, c; Y+ s7 G
{
' s- c9 ~5 ^8 ^, qfloat c,f;
- v5 S4 m; W. n. ]! p `printf("请输入一个华氏温度:\n");
) f. }6 R; v" o. |; L* M) j- hscanf("%f",&f);
1 D) d8 W# o7 l/ Q5 D6 n) Qc=(5.0/9.0)*(f-32);
- k/ G' o! V- yprintf("摄氏温度为:%5.2f\n",c);: @4 Y/ L' U% O
}" U9 q! `4 x( _! w( @
第五章 逻辑运算和判断选取结构
& F d" ~; S v% Q4 e" l0 C7 K5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
! x' s( v2 m1 Q# o; j h- p. f6 dmain()9 W( D" N' R; t% a6 z1 f! C
{" \% C b8 V1 Z7 C3 Q9 s1 U+ o
int a,b,c;: l. E8 O9 ` S" l
printf("请输入三个数:");
/ j8 O8 c& g( I/ ]: V+ H0 F, r5 @scanf("%d,%d,%d",&a,&b,&c);. A6 b; g7 i v6 v( S
if(a if(b printf("max=%d\n",c);5 J) j/ z3 V* `! t% i1 l! m9 o
else
) [& c! Y$ ?" S' x: c( O. q printf("max=%d\n",b);* U7 Y7 _2 k, X7 v+ S* u- D/ ~
else if(a printf("max=%d\n",c);1 J% w6 E+ D, k7 W `' p9 b. _$ J
else
) F8 U0 l! Z1 u0 P printf("max-%d\n",a);
& p; F: \9 f9 J. b7 t0 W8 U}" L q$ h; E7 `% `& _
方法2:使用条件表达式.
/ ~2 A f0 A6 `# e# ]+ x) q, [/ Qmain()
* {6 C' c6 I/ O7 T+ S& [$ T{int a,b,c,termp,max;
6 L9 s$ d% l' l1 N, {3 D6 Z7 Q printf(" 请输入 A,B,C: ");
& k" x4 m9 g# F, V8 o3 x4 v scanf("%d,%d,%d",&a,&b,&c);
6 Q" F0 t0 w- T4 H printf("A=%d,B=%d,C=%d\n",a,b,c);8 d/ w$ H9 j* y1 n! Y
temp=(a>b)?a:b;
# M; Y4 z' d. O( v max=(temp>c)? temp:c;) y2 G1 B' j( X# ^( P D, F' Y/ A
printf(" A,B,C中最大数是%d,",max);4 p7 f: _9 t( F, h p& N6 g) _
}
( w# U6 U( t/ {' |7 w/ P5.5 main()
7 j) ?8 G r( u{int x,y;
* a8 Q- R( V- ]+ r# k( {1 \printf("输入x:");7 k s" X R. y0 R; W5 O
scanf("%d",&x);
, e* \ u1 g# Wif(x<1): x' B$ Y# u ~! j; v) `- X
{y=x;4 m' @3 S/ M ?6 b; X$ x5 Q
printf("X-%d,Y=X=%d \n",x,y);& ]8 l* i5 g! {. s& y
}# [1 ]7 E+ [7 g- Y, Z% g
else if(x<10)
# a5 F3 S. c' k4 G- M* W& p {y=2*x-1;8 u, k& Z) T$ E9 Q7 K7 ]
printf(" X=%d, Y=2*X-1=%d\n",x,y);
# G4 l* u9 W# ^$ @ g7 ?' Q }
$ Y4 [% z3 T1 x' r4 N/ b- X' `else, b' \6 B+ y. Z5 p& o
{y=3*x-11;9 Z3 v h& D4 @8 l8 F* e6 k
printf("X=5d, Y=3*x-11=%d \n",x,y);+ M3 D1 B5 a6 O/ {) _) K- f7 j
}% v: ~2 ]* |* D1 a* S
}3 f! L" I7 f' L. ?3 c! A) e
(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!
, C: Y, ~1 r6 @* [9 w: A! o( Pvoid main()
o# L, ?# R# l{
1 g# c) H$ x' _6 `3 |float s,i;; z: x8 R# m% h0 W1 s
char a;- s- V! u7 r! L* w3 b7 r
scanf("%f",&s);( b0 B/ q9 R5 O1 e9 i/ i L
while(s>100||s<0)
Z, I$ @5 v" z( P4 [{
3 n: Q8 T. G7 T# E- @2 Xprintf("输入错误!error!");8 F3 M5 t& S% o) ^' N/ l$ _
scanf("%f",&s);6 j( [+ F4 Z# ]( t& h$ x
}7 y5 f$ L, o- x. R& v7 _8 h
i=s/10;
/ X9 {6 e! [ M+ E H* Tswitch((int)i)
0 D9 s) ~2 u7 e. ^0 L9 |. m{
5 m) y& ~" e. A8 Jcase 10:
2 s/ `+ t" F* S$ e+ Xcase 9: a='A';break;2 V" |8 f: r* Q/ ^$ W% k
case 8: a='B';break;9 f5 l {- O& Q+ e
case 7: a='C';break;
% C/ D4 U/ ?) @. A- lcase 6: a='D';break;
% r* u: e& S1 Qcase 5:
/ o+ F' q S, ^1 X1 K, tcase 4:
+ m: G" M& i7 D1 E9 q, A& ncase 2:4 U. f" @& g( E5 z( k4 l3 W6 s
case 1:9 M" H" R% r6 v2 H) i4 A! P
case 0: a='E';
7 m* H2 R l% t- M, |. e}
" c7 @' v% C3 N( T& Q, {printf("%c",a);; |/ Y7 }" Q$ \% \ }
}5 r8 r/ j. R+ P( v
5.7给一个不多于5位的正整数,要求:1.求它是几位数2.分别打印出每一位数字3.
8 Y& T; {( ?- c+ w按逆序打印出各位数字.例如原数为321,应输出123.
+ [' j2 [7 \9 Q# J0 D J9 M+ {8 vmain()
7 V# S/ W. R5 s, r" A" T4 i { o" ^5 s' ?$ N N# e
long int num;
: E* y, ]( z3 v3 i% `8 v int indiv,ten,hundred,housand,tenthousand,place;9 a4 p. E) L+ B1 D
printf("请输入一个整数(0-99999):");8 l, w. o& o% {! D. d
scanf("%ld",&num); }! p! b8 z: z, @. ~+ q
if(num>9999)
# L$ n e* \! m7 q: T& l; \( O: m place=5;" o/ X/ _$ m) ]/ u9 c3 F, e
else if(num>999); ]+ e/ A" t. r3 Q5 h
place=4;
. g( C: m& g, U/ @else if(num>99)
8 j% A0 c* j9 A/ {# f( G- F8 W place=3;% |' X% [( V# U. w, s0 e; z6 T% |
else if(num>9)
- W! A D- v: u2 }. |# Z( h: u place=2;5 o. a+ K# t* E6 q) ?' x
else place=1;2 U; S$ s# G; l0 J8 D
printf("place=%d\n",place);# ~/ I2 Z) E' S! D9 S& Y! Y+ U
printf("每位数字为:");' N5 @6 f: J/ C/ c6 K+ N
ten_thousand=num/10000;
& m) o T6 a @3 A, Uthousand=(num-tenthousand*10000)/1000;
' R v8 k0 ~$ k- @1 Z6 @2 ]$ m3 Qhundred=(num-tenthousand*10000-thousand*1000)/100;
5 k0 q$ M3 [% n( W) q& aten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
; @7 w6 Q$ u% v/ c7 Y2 w. Iindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;! ]+ b8 }( _7 x. `, @
switch(place)2 o( f( ]% n, E" f
{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
" R- h2 _" G3 c/ p1 a k, J8 I printf("\n反序数字为:");
5 S' ~; f( L. S# ^( j printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);
$ ^2 l1 b% B2 E- p. e% s break;% d7 l( P! g1 p# ]: D
case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
9 r* d/ i3 m2 C0 J# z7 I/ B7 D+ [ y printf("\n反序数字为:");
7 b( t: P' o' t printf("%d%d%d%d\n",indiv,ten,hundred,thousand);1 T$ ]- B& A; q7 J
break;
& x b- P- D: C1 ?5 Jcase 3:printf("%d,%d,%d\n",hundred,ten,indiv);
1 }- r2 x: F6 P6 |. i8 q$ O& p/ c printf("\n反序数字为:");
" p/ J" H& e9 l1 h$ Z+ M printf("%d%d%d\n",indiv,ten,hundred);
- d0 s5 |% i' ]; f) q7 ^case 2:printf("%d,%d\n",ten,indiv);
4 J2 k. k7 s+ X8 N7 l# Q printf("\n反序数字为:");+ a) O6 e0 T1 U4 y! E& Y
printf("%d%d\n",indiv,ten);! R8 i! ?/ ~9 {. [
case 1:printf("%d\n",indiv);
" o, A, r3 R0 } y5 n printf("\n反序数字为:");6 i; s% f+ Y. V) B& ?5 v
printf("%d\n",indiv);
( X: S2 p* Q5 N" }3 l2 f3 [ }4 [$ W ?, I9 d6 ~, N
}& f9 M- @! b, j/ }
5.8
; K% c) f1 @5 @% s& k A' K8 t+ b1.if语句
1 t& `+ c7 V- f/ f* `: a. amain()
! {0 r' Y1 q5 Y, C& U" x{long i;
) d- H, N& U1 d0 S/ x( b1 z float bonus,bon1,bon2,bon4,bon6,bon10;( H* B: w! E3 C7 z" ` ~
bon1=100000*0.1;
$ y# I+ Q4 J4 R. @1 O/ u bon2=bon1+100000*0.075;
' j+ G) h% ~0 {' H5 l9 P bon4=bon2+200000*0.05;
- b {* T L) o. m1 I bon6=bon4+200000*0.03;
4 ^/ z: r' W+ N' z/ j! Y bon10=bon6+400000*0.015;
* }; M0 K/ `1 F( X% J q1 ` scanf("%ld",&i);" Z) c& z* \4 z4 U3 M2 ?3 n
if(i<=1e5)bonus=i*0.1;$ |( G; S4 L1 Q1 J. b
else if(i<=2e5)bonus=bon1+(i-100000)*0.075;1 p! u( F+ C# q; M) \
else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
) e/ v4 l- _2 V O: u else if(i<=6e5)bonus=bon4+(i-400000)*0.03;
: T. D8 M( d+ W; [5 ~& l. E6 i0 O/ ~ else if(i<=1e6)bonus=bon6+(i-600000)*0.015;. H6 G8 Z* q% R# K5 k
else bonus=bon10+(i-1000000)*0.01;
8 d( d- A6 @8 D8 x [" V printf("bonus=%10.2f",bonus);5 a5 {, L' E! K. B
}
' o; z! t* u; j! C) v. B0 v3 {用switch语句编程序! n# v3 }7 Q* v% [* ]( ~
main()
8 ~' h" Y# b; V: ? Z" p z{long i;, d& n5 \( F; d( q4 J: g) @
float bonus,bon1,bon2,bon4,bon6,bon10;/ P8 G* ] b+ S$ o, q, Y, S3 J# l
int branch;
# w: V7 ?' I4 f% C- o bon1=100000*0.1;+ T4 k! J @+ U$ [* \2 @3 n
bon2=bon1+100000*0.075;
Q2 {! U* ]1 s* B$ u7 @ bon4=bon2+200000*0.05;# N8 Z1 f8 j% z* ?% l& H6 b
bon6=bon4+200000*0.03;" P% i/ V0 ?6 e X6 [4 R! i+ Y/ }
bon10=bon6+400000*0.015;
& c z* p& U0 D( u* N0 h) @0 p+ M scanf("%ld",&i);
8 Y" V0 N6 }" @- Q; y branch=i/100000; ~& g: `9 X" ~, e4 n' N9 c2 C
if(branch>10)branch=10;
- Z" I6 [- }. W- J, ` switch(branch)5 V$ Q ^) v: x4 n! ~( C% s6 f
{case 0:bonus=i*0.1;break;' T8 Z9 G$ T( `+ d# J0 S2 q) d
case 1:bonus=bon1+(i-100000)*0.075;break;
9 @) F! p! f/ {" q* X case 2:
$ e2 W2 Q+ i3 v; ~/ S9 T case 3:bonus=bon2+(i-200000)*0.05;break;
$ s2 I& o/ j7 Z" ~! Z E) x case 4:7 M% k& x3 h; b
case 5:bonus=bon4+(i-400000)*0.03;break; Z7 Y2 H' N7 A
case 6:
1 k/ X& `7 y6 N8 x case 7
( ]3 b4 c. u0 i% y6 G/ G' T case 8:
, J' q; E! W" y case 9:bonus=bon6+(i-600000)*0.015;break;; E8 n2 ^7 [& v& c* I9 l
case 10:bonus=bon10+(i-1000000)*0.01;3 _6 d/ A+ U) U0 M! Q
}7 |8 R% a8 D O. _- Q) \
printf("bonus=%10.2f",bonus);% b. f: r( J# k1 [' D- n1 N) ?
} 1 b+ `) O* y5 ?1 ]# i, h
5.9 输入四个整数,按大小顺序输出.$ F: o+ T# P7 y4 ]
main()
/ @ K8 X" z# {8 D {int t,a,b,c,d;
0 M) E. e. X1 i9 P6 w printf("请输入四个数:");
5 }; S5 z _4 p2 J scanf("%d,%d,%d,%d",&a,&b,&c,&d);0 j( p& Z# n- S% \/ {
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);1 l! }# I3 S. m6 d3 W9 E4 n
if(a>b)- L9 Y5 Y% |& [8 D: ^) ~
{t=a;a=b;b=t;}$ b8 g8 E- E$ p0 r+ u* B
if(a>c)
$ ^9 p: d4 l% G9 G( d {t=a;a=c;c=t;}
% r/ c6 T: v" N( \3 L4 Z if(a>d)
9 x" ?( U O% q8 y0 X9 ] {t=a;a=d;d=t;}8 J, u N% Q$ U. B6 ~
if(b>c)
3 |: @+ H9 B2 k; t0 G9 P {t=b;b=c;c=t;}3 |7 u# F3 J4 Y1 E7 Q7 e5 o
if(b>d)/ M% h; m9 ]7 o& T3 t( F! I. I( |/ |
{t=b;b=d;d=t;}- W$ s8 B8 Z: d u) b" J+ H. g
if(c>d)
. f' d" @& w) D/ V {t=c;c=d;d=t;}: l: J3 C- f7 d- q! |; t+ y
printf("\n 排序结果如下: \n");
5 s# `7 F( v% q9 q$ o8 i6 L2 X/ Xprintf(" %d %d %d %d \n",a,b,c,d);( n: J, n/ s- i3 A
}+ S! w1 }1 F: k: S5 p* K
5.10塔+ E7 ~ G. o' ]. e7 U! |/ f( x
main()
: B/ p) N S# L3 H* }: D{: ]" g% s5 }5 h8 C, z
int h=10;/ {* U3 j' T1 V s/ P8 {% }
float x,y,x0=2,y0=2,d1,d2,d3,d4;
2 j0 I$ a/ g& gprintf("请输入一个点(x,y):");. f4 ^9 @$ y' d3 U6 r: H+ |8 @4 r
scanf("%f,%f",&x,&y);
4 g4 K4 g% [$ Jd1=(x-x0)*(x-x0)+(y-y0)(y-y0);& H/ c$ Z3 r4 ]* n$ |' D' O' G% V
d2=(x-x0)*(x-x0)+(y+y0)(y+y0);
: M! ~+ W* b2 d4 e" [d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);2 [# f p: P, S+ q; s4 X2 P
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);: M0 ^' R" J0 f' t; |2 \+ l, a
if(d1>1 && d2>1 && d3>1 && d4>1)5 @: L7 P. A5 D$ }5 H
h=0;
* q. c5 r1 h2 O, {+ W! ?* a: F: aprintf("该点高度为%d",h);
1 S9 r* s6 |5 B8 L: @6 m% b; k" I' D}* T# ?3 j- t9 i6 h& ?
第六章 循环语句* o9 \0 g: U) ~! f5 g, C, `
6.1输入两个正数,求最大公约数最小公倍数.
* B6 I9 G% `1 p! C- l' Zmain()( T; r4 M" R! o" h5 q4 K- [
{
8 @4 l. a% s, Y, O0 s5 T' L0 tint a,b,num1,num2,temp;
! z3 w/ e& l( D. m( ] y8 O( l+ T" {! l- zprintf("请输入两个正整数:\n");0 d* k/ u. C; Y: {0 n# t! g
scanf("%d,%d",&num1,&num2);
y. w d( e/ G. Jif(num1{
r1 K2 [* X3 O V) Ltemp=num1;
6 ^( ]1 t8 E4 E! I3 r( D6 snum1=num2;: b+ u z& B H5 m: c2 B
num2=temp;
7 {5 B: U* w7 m; n3 E" Y}
/ A! U5 T; _9 x& O* T5 x, ^4 J0 Aa=num1,b=num2;
! l9 `+ \- s+ N$ A6 |4 g! g# d8 q" Iwhile(b!=0)) [) C5 M. C; i( u) U
{
! c: Q/ E2 r. P% j& p temp=a%b;; X9 Q) |3 P C% R! i
a=b;; U0 N' K' w; C, U
b=temp;
7 V& ? I2 a( ~ b }2 k" m* ?9 z, J, v
printf("它们的最大公约数为:%d\n",a);. l w3 j8 d5 v4 Q- G. h
printf("它们的最小公倍数为:%d\n",num1*num2/2);6 n$ N- m( E* t( G* H8 _% m
}2 d7 l# _+ ]/ t$ h, X. c
6.2输入一行字符,分别统计出其中英文字母,空格,数字和其它字符的个数.
" U5 o+ n" z5 g; ]- B R) A: P解:
5 X" D* l; @. \: g1 p U#include < >( P/ a! i5 w: n* H8 `/ W
main()4 C- \6 v7 c# \
{ T% u, t* ~9 h h/ p& L) r$ R7 |+ ~' Z
char c;
1 k9 U* d! S+ {- I) bint letters=0,space=0,degit=0,other=0;/ Z- [0 S2 R! q; q3 R2 _
printf("请输入一行字符:\n");/ w8 Y1 u. ~) W3 u$ P
scanf("%c",&c);
! e, I; S/ \ E# F- Twhile((c=getchar())!='\n')) `4 `4 `7 y' [6 w
{
. @2 g7 t% u, [8 h! Y/ Qif(c>='a'&&c<='z'||c>'A'&&c<='Z')
. C9 n2 D* ?5 i3 }- o+ Oletters++;
y5 X2 J; Q# }6 U; w% ~) @0 welse if(c==' ')8 \/ ?7 S2 \( u c% Q+ w, m7 N4 h
space++;
6 \8 U5 O v3 lelse if(c>='0'&&c<='9')
8 N0 g a& R, \* h+ R2 j$ D& Ddigit++;
o! J, [4 \+ ?5 D5 belse
; L4 H3 G" a, hother++;
/ W' s4 U* m: n. v6 y}3 v# o3 t# }9 J$ D8 `% F5 c
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
w( r$ ]. V: s5 }: ]. \% dd\n",letters,space,
' e9 @0 t/ ?) [digit,other);* o3 q* e8 M* s; X1 M3 H- I
}
! i+ y6 Y' _; Y" }. x/ T; I# r |6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.
4 ^' V: v& s" m: R: ?3 q解:
0 Z: E' t5 H7 l4 w$ u+ q+ Xmain()
% {3 v* z# j+ y* y{
0 z; A) X# z0 o R5 B7 K2 \: d7 jint a,n,count=1,sn=0,tn=0;
% N/ q4 T$ _6 k3 L/ K+ Y8 Nprintf("请输入a和n的值:\n");
5 ~4 \* A* n7 u4 {& v9 N& Lscanf("%d,%d",&a,&n);3 M3 t9 Q: V7 X" {, g* W# O
printf("a=%d n=%d \n",a,n);- v" O y; ]0 P6 t2 o( i Z
while(count<=n)
: d, P: D3 {* S+ E) B- e{ X m5 J5 J/ B0 z5 S2 }
tn=tn+a;! J# `& i) P2 {2 d
sn=sn+tn;" N6 K, ]$ P$ s6 c
a=a*10;
8 s% R* N' N n$ t1 {& G5 @++count;( o( ~' s* n2 O9 C) S
}) B0 l1 i4 j3 j/ L
printf("a+aa+aaa+…=%d\n",sn);+ J' B C: E( h% b
}
; f: ]8 o# Y- G- ^, V. Q. E! I6.4 求1+2!+3!+4!+…+20!.; r0 k( D2 Y( i( ?) @
main()
g0 h( v0 V. a# L- x6 D7 _{1 Y$ v: ]6 @0 s; n j2 Q
float n,s=0,t=1;
( Q4 c* ^+ ^1 i* y0 N' `( x* efor(n=1;n<=20;n++)4 O+ L5 [1 y# J
{
! d. p# j$ @. ?. l7 \- A; w3 N) Rt=t*n;
5 H* T8 f) p" Xs=s+t;0 M2 G9 ^* T" K/ Y. ~4 Y
}9 d( I: q" G3 x7 N$ f$ W$ o( C
printf("1!+2!+…+20!=%e\n",s);
4 t+ q$ }2 a9 M U3 s3 m, i}
% S1 ?: w" e- C+ D6.5 main()( w& S$ U, _9 Y: P! D; ^
{
; a* _2 {/ C s4 _5 Rint N1=100,N2=50,N3=10;
. {, Y: ^# `7 p1 Bfloat k;- f z' e$ J0 T% }9 K* ]& k l
float s1=0,s2=0,s3=0;% Y$ o% Y0 E( ~2 p8 X9 x
for(k=1;k<=N1;k++)1 p5 |8 {3 x" z* M- R* R4 L
{3 S+ u9 g; _/ ~9 _9 C# G
s1=s1+k;# H& T K! \2 o& w: I
}
5 U" `9 @6 W& d/ Ofor(k=1;k<=N2;k++)
/ x- D7 e; t& s{/ B; m& V! p( u ]- y
s2=s2+k*k;, H1 j% |6 `, [1 \# c
}
) K4 p/ X/ S9 W2 L* M ^for(k=1;k<=N3;k++). @9 k2 }( T. e9 q: V
{
: u5 W! r8 s$ X+ P6 h% Ns3=s3+1/k;
$ u' s. V" s7 X4 ~3 U( g0 o u}
& ]! D5 ?2 d1 Dprintf("总和=%8.2f\n",s1+s2+s3);" V6 X# h* g" t! e
}/ V# t2 E: \- Z9 Z# C1 b8 v
6.6水仙开花# y: {' _+ B4 A/ g$ ]. S
main()5 p* c+ C2 u9 R) Y! W
{
4 c( A. ^9 @- L$ V* s$ E- rint i,j,k,n;+ k$ K# D* u. j! M) r! c0 m0 F$ D
printf(" '水仙花'数是:");
# g x/ ^' k& }. |. Efor(n=100;n<1000;n++)" o+ @2 [5 p* z
{
% B! I3 q9 _% p( i5 N9 X9 Ji=n/100;
$ q- t9 D J! M( A! Mj=n/10-i*10;
; v# H+ S5 S4 J7 W ?1 @k=n%10;4 J# W0 T$ f# J3 ?" h1 S8 {! K
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
9 W6 f5 q7 j. ~- ~- N2 Y. q{
3 l0 ~; t* s+ N2 Kprintf("%d",n);& X" [: }8 c: e* X5 a
}6 t2 Q6 j3 V2 T+ q, c
}
4 z) q% e& G* c8 F. aprintf("\n");& W2 u1 V! w; j
}3 y: O, W# O( e
6.7完数
0 D4 }' t, ]9 p. bmain()8 d$ W# Z, e: }$ ]4 I' n, c& t8 k
#include M 1000
0 N: y+ J% H5 L2 Ymain()
9 W' m# o& h+ t, S# C/ F/ t" n{& \6 I0 c$ g( j7 C
int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;7 s# O5 w2 r. F5 b+ b- i
int i,j,n,s;
* O) b/ {+ ?3 J7 Yfor(j=2;j<=M;j++)
/ S$ W5 A* J3 x{
+ K( B S& R7 ~9 Y7 y5 H7 I' jn=0;
" `4 |; o* I' S8 ss=j;
8 z' p' h% B( L/ ` |& m7 Wfor(i=1;i {% I& u# q+ s2 H& w* o8 i, _
if((j%i)==0)& D" X) D7 c2 K
{+ ^ ?& {- I6 V0 p9 ]: e
if((j%i)==0): b8 X# X e4 H" r2 m7 V
{
3 f# N$ B2 k3 F R$ w5 a2 S& ~ n++;* p- G! W5 Y& r( b# H, i- ~
s=s-i;8 u. T* \! J; A6 l* V/ Y1 L
switch(n)
: d( q+ [) Z" o( ^* I/ G+ d {' l% V3 C& ?7 T; e6 s4 _
case 1:7 q2 G4 V! a3 N
k0=i;, S0 w/ d, x" E% r
break;5 R/ y+ e& K+ {: z" @" k* V( x
case 2:
0 a r; K0 g. Z! i' u k1=i;
; v R& @1 i" U! O break;
5 I( }" m5 k; _! @ case 3:
7 }+ y2 d( V! X7 A: ?9 P6 n( c k2=i;) V( W' _9 G! t
break;' s! f2 Z8 U) X' p: [6 g
case 4:
6 f* i3 N- b' _9 U# {; v% w8 y k3=i;; N0 V2 V8 w; Z$ N" K5 y) ?% j+ [
break;2 S% b9 W6 v% H% X2 u) l
case 5:
9 G" ? c1 Y1 I9 U* L k4=i;
& L+ E, J" S8 }6 {% I8 I break;; O: l7 r1 W; |5 K. Y
case 6:
. t" K0 q# K2 \8 @1 a k5=i;
% s& \1 m* b, p" W" j {$ I break;4 k; D& e9 P2 E( V1 s: M) i( g
case 7:
( w0 b- {' \. o9 }9 E k6=i;
' I' I3 r' c/ O! K break;
# B8 x( k! D4 ?3 M/ N case 8:2 r5 c9 t1 t8 e5 j( w
k7=i;( {$ ?4 { Q( y; ^4 h
break;2 S* p; Y& H: U
case 9:7 n: _& z5 F4 T5 j1 z
k8=i;, H( ~7 k! j; C$ `# {7 ]1 K
break;; u0 ?! ]/ @" ]# y9 Y" ]% v4 `/ `
case 10:
3 v X) Z* D* i% ^ k9=i;
' ^% O+ e/ x# V9 R break;
- f2 N1 s5 F/ ]# N' C- k$ { Y! i# L+ O }. \+ ]; T2 B: E0 Q! A/ G
}
m$ F( p+ @4 u) w( X5 N }2 Y3 n, C2 x0 g0 C" a
if(s==0)! ]4 }0 k' d' c4 B, f
{, Q3 G: ]0 o2 G! d1 `; v( ?
printf("%d是一个‘完数’,它的因子是",j);
3 J* W, l) `. W. p6 mif(n>1)- B" `- S. T+ l' s
printf("%d,%d",k0,k1);- r# h W. B7 j R
if(n>2)4 K( g" V2 x; Q
printf(",%d",k2);- B @+ z# ~1 K: P* ]: l
if(n>3)
2 Y( ^8 M4 z6 L9 C! a k4 Z( s printf(",%d",k3);6 @; P, H( X- ]8 w1 J+ X
if(n>4)7 ~( {" H5 P+ s6 A# t8 R1 I
printf(",%d",k4);
& Q9 t6 y: ~* P& ?3 c$ v# s1 l9 mif(n>5)
) Z; ^$ b% C6 o6 I3 Q' n I printf(",%d",k5);3 t" J& \. a9 R
if(n>6)
7 M* W: L. F. R3 a5 q printf(",%d",k6);
/ [' v' j4 W& _" l4 k( W0 Dif(n>7)
& E# ~$ M) N( R8 l" { i5 u printf(",%d",k7);& q( ?8 n$ A% p n; |% `
if(n>8)
6 E* H7 w3 {3 ` printf(",%d",k8);- W0 O {. F" I' _1 {: W* n
if(n>9)$ C8 D$ I9 p6 }4 ?- `1 j
printf(",%d",k9);% ~0 ~- `; A$ w- a
printf("\n");
7 m1 z# R+ R/ _& d }, p2 r) w |1 G& c2 ~% B' z( m# y0 b
}
, C; d0 L" ~0 Q8 X3 x5 U方法二:此题用数组方法更为简单.
9 v5 P$ c2 g6 H' l7 gmain()
# h) }/ y/ i5 M4 N& |) d5 J4 O{
6 ]6 _$ }) X$ Z% \: u* n! gstatic int k[10];# p" M8 r% v* b# w7 v0 |& O3 f) _6 K( F
int i,j,n,s;! R# i" k, \: K d ?8 {
for(j=2;j<=1000;j++)% X2 H7 @+ t7 S. W
{
$ |" x2 W- [6 [ i- N/ kn=-1;1 X9 a; u" R. q6 o1 S
s=j;0 Y' ]+ e+ S' `7 d9 j ^
for(i=1;i{9 ]# r' |! W$ F \8 q/ C
if((j%i)==0)
s3 n' \) |: @) F2 l# E{) T# L( T$ {$ x/ x4 I" C) x
n++;2 X( {2 y6 J, O
s=s-i;6 O/ ~7 Y6 M3 T% Q7 F) ]' [8 K
k[n]=i;
. T! U* K- ^3 i+ N }+ p$ w! {" l! O2 z2 E* ~+ Q) A
}. ^1 [& P) {7 R9 [# t" |
if(s==0)
4 N; b: S* S% [4 a. m{
( E, Z6 ?% C7 Jprintf("%d是一个完数,它的因子是:",j);
! F" N: M$ G% j. n; Pfor(i=0;iprintf("%d,",k[i]);
* |% D0 X c5 I; e0 O& V! nprintf("%d\n",k[n]);0 r, \1 |: i5 ^- j
}
4 L; p* W7 L% M8 U+ O}' W) }7 E7 }6 f1 ]' ~( [
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.& w0 S5 g% e4 e6 C$ r* U1 O
解: main()
( z+ C& C# B. f1 ?4 e{. o' W" w$ t" I1 _
int n,t,number=20;
- t0 ~) t. m" S6 b+ h5 Yfloat a=2,b=1,s=0;, j# |0 Z# M, x4 N
for(n=1;n<=number;n++)! X. j6 }; X" i( }6 d
{
( M5 s& G. v+ G4 ]0 ds=s+a/b;' |! E m, ]8 x# p9 G$ R+ D3 t! M) s
t=a,a=a+b,b=t;- `- B9 M7 i1 l+ Z1 n# a
}7 ~" ?4 p* t4 y$ |7 i \
printf("总和=%9.6f\n",s);
o0 M! E1 g: a! L- o' ^ ]}
: k. L1 h) @2 o6.9球反弹问题2 u: R: X9 l/ n: K
main()
* e: c7 F* a2 B{, s$ G- n8 k3 G$ N# l' ^7 E ]
float sn=100.0,hn=sn/2;8 ?9 v' f; X% A4 S- K0 F& d3 E
int n;! z- ~1 F( P. J2 _
for(n=2;n<=10;n++)
+ v" q. U2 s8 S7 s9 g* O. s. W{2 z [% w3 G7 t. ^
sn=sn+2*hn;# W" G+ n' Q! P0 J# j# M
hn=hn/2;( W/ ^+ P- w8 S% q
}
+ }+ z# ?$ N1 H9 j6 y1 J4 [printf("第10次落地时共经过%f米 \n",sn);& E: U* T$ j+ j' Y
printf("第10次反弹%f米.\n",hn);" H- f# c/ h) L2 M8 S
}
- L c- D+ R4 z8 W+ A6.10猴子吃桃- e0 U! ?: m3 n8 E4 n5 d
main()
" a4 }6 C7 x: j% c) x) |{
9 N" v$ k. v" \8 m& g rint day,x1,x2;+ f( U- s' Q0 u) g) L8 ^8 ]; u
day=9;
9 k Z Z- k% j+ j# E$ h' [: Q- Ox2=1;9 U# A4 P# y, \/ j7 j! V: f$ W ~" E
while(day>0)
, Z; v. S" v/ Z5 V{
7 V( ?3 L, n9 d9 U% H+ {x1=(x2+1)*2;$ C3 _6 w( @5 Z9 Q
x2=x1;/ q0 ]7 O) z3 Q( i F; A
day--;* F/ [% E3 z& g7 X4 D# [) {
}
' }! e \) f. _& m9 b5 M4 ]/ xprintf("桃子总数=%d\n",x1);" o% Y: ^- J; S3 r4 A( \2 x6 u3 e
}
. I6 J. l: |! w- j; ~4 Z; `2 U
8 p: j" o7 x9 @1 {6.12
( r; k0 S u8 V5 X* W! P+ y7 ~( k#include"math.h"
2 ^* R u0 c! _! Y3 J' E9 r! tmain()
4 q% D/ ~1 g% x" f' x( i1 i. G9 c{float x,x0,f,f1;
' S8 S" a% ?' {5 S) E2 a x=1.5;
" k$ W: m9 L0 z& P0 F1 v9 V do' Z$ ]3 S: N# F: R
{x0=x;
2 v8 g# P* B$ G3 g6 v( P f=((2*x0-4)*x0+3)*x0-6;* x+ @" d3 I& a
f1=(6*x0-8)*x0+3;& P2 M1 |7 \) ?- R
x=x0-f/f1;
" [; C% x% H) t3 i+ u' [ }
3 w# }! a" O" h) i while(fabs(x-x0)>=1e-5);
. t; ]+ B2 @' W7 O printf("x=%6.2f\n",x);* u3 a% A* n9 A; e
}- \& X( r& K/ H! \+ q* O1 u
+ x" u* H3 k( |9 C1 _# a r6.13
, _5 @' Y) W F9 J#include"math.h"3 v! B0 W/ X( A1 K* u% d5 Z; K5 I
main()
& h+ U- ?9 k6 K& G2 u& {# }{float x0,x1,x2,fx0,fx1,fx2;
9 G- ]5 f) v8 Z$ j* {, x+ e% b do; e: o4 T! L# |
{scanf("%f,%f",&x1,&x2);9 p% H1 n; _+ G* C
fx1=x1*((2*x1-4)*x1+3)-6;8 j# g5 W; |8 _& b( Z: j# h
fx2=x2*((2*x2-4)*x2+3)-6;
! q+ n2 I* X7 B" P# {) R } y4 c1 p& [2 p$ f
while(fx1*fx2>0);3 c) e/ R* ^$ ^4 b' K( o, B: Z1 _
do* D) M* A- D1 F
{x0=(x1+x2)/2;2 t- y) z3 W3 I0 R5 C' y
fx0=x0*((2*x0-4)*x0+3)-6;/ a) Z& k n# ^- f6 A- J0 ?
if((fx0*fx1)<0); ~/ ^5 X! [0 ]: F5 \
{x2=x0; ~ N: T: J) I' f& z
fx2=fx0;3 q, o& E6 w% A
}* ]$ l2 c; v) f$ S9 O* j6 f
else0 v5 B% G- f+ x0 i) I
{x1=x0;3 }0 H! s% P% u
fx1=fx0;
6 j* t- @9 l3 w% {9 y' `& P9 x' w }% B% @& P j8 r( o0 T) f8 K
}$ F2 q i% N2 g! t+ @- [# l
while(fabs(fx0)>=1e-5);
" d/ |8 u3 E/ e, O$ k1 u( b5 ^ printf("x0=%6.2f\n",x0);9 R2 k: G; X# O
} j+ x% r* q/ j$ v# Q* z
6.14打印图案/ C! C0 J5 ^" Q
main()
$ p+ [3 h! W3 K; T& z& m7 V{int i,j,k;' g$ u/ y- Z+ C, u% f
for(i=0;i<=3;i++)
8 v5 h% h: h2 Y$ Q, y {for(j=0;j<=2-i;j++)
9 k! n4 J, e3 h8 J1 [- t: f printf(" ");
3 I+ n6 g# L& I6 N+ |6 w* K/ ^ for(k=0;k<=2*i;k++)
8 Z6 `/ S8 G* E printf("*");) J1 N f, ^% ?7 U4 D+ b- X* I
printf("\n");
# ^* g/ v$ b' R" @% L( m6 @# m }! p# S. l3 h0 Y1 X: i/ ^3 v
for(i=0;i<=2;i++)
* ?( X* d( n* o% @ {for(j=0;j<=i;j++)4 b* I9 q) h4 C$ p4 `+ O
printf(" ");
% ~ h! z! a8 x$ G* l for(k=0;k<=4-2*i;k++)9 R1 \8 v7 @% |
printf("*");8 i( Y8 a+ o7 A3 w0 \
printf("\n");
2 P+ j% G7 F7 G8 a3 `5 S; B }
7 {* F' I% c3 O" e0 s}0 {+ ?9 U f z* ~1 t% [: u( N
6.15乒乓比赛+ [: L# t$ |( n0 U% @ {! r
main()
5 y3 p2 c5 L+ U4 G{6 g8 m6 b, j( f: c7 Z
char i,j,k;
. b4 R6 G9 I; Q- G/ hfor(i='x';i<='z';i++)
7 f) ]: w0 Q9 ~# r3 l* pfor(j='x';j<='z';j++)
0 a8 c7 q9 D% V+ X7 X {
, k d8 A& S& X K/ W% c4 D8 jif(i!=j)
# G; R. N' C1 r9 T7 P; L* @: C4 j& bfor(k='x';k<='z';k++)
' C- G: s, A }$ [ {
3 v; F7 i. W, {9 Oif(i!=k&&j!=k)
8 [5 N; U) I& g9 } {if(i!='x' && k!='x' && k! ='z')
8 Z8 u M* ~" d q1 E+ lprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);
9 p R/ l4 b0 r1 J. ? }* Y3 s) [2 j: P9 q: i; e! Q- r
}
" H5 x+ N! n! Z& G: f5 f: y, m }
( y4 O/ _0 V# W. v8 E}9 |/ Z8 }8 b: L5 J2 S7 I, r- S
C语言设计谭浩强第三版的课后习题答案
, y' b5 X4 ~ `# _& u: m7.1用筛选法求100之内的素数.2 Q- R6 k# ]; W1 c0 e
#include( x3 L" A- x6 ^+ w& [
#define N 101+ K: t- F7 Z$ P# [' e+ Z/ m
main()
/ Q# R+ _7 O/ K+ s% q) r+ y7 C{int i,j,line,a[N];2 W- [6 q3 Z' |6 a' l7 i
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
5 n# k8 j3 Y$ h1 d6 G$ x$ G if(a[j]%a[i]==0)
' Z1 d) Y h- P a[j]=0;6 g2 i6 l0 C: q1 M
printf("\n");
# q1 N1 A* L1 qfor(i=2,line=0;i{ if(a[i]!=0)1 N5 ^% g) u- N$ p7 I/ u2 ^9 E
{printf("%5d",a[i]);% N% M# ~& p$ d1 J$ [9 f
line++;6 f3 m% A E, G' I, c( P: ?) \
if(line==10)2 ]) x7 G, N& I* D6 j$ m7 c7 J
{printf("\n");8 }' J% D* \ c
line=0;}
; D' W0 M0 Q0 s }
, G0 T' J9 o, Z! A( F5 C% I}
, D4 @6 U, @: Y& |* p7.2用选择法对10个数排序.
& e8 u! C; a% ~- Q#define N 10
" D+ D O& L: x/ Cmain()# q7 I% i4 z( ~
{ int i,j,min,temp,a[N];
% J) M/ L! `6 |printf("请输入十个数:\n");
/ i3 t5 g; Q ?7 r/ p- rfor (i=0;i{ printf("a[%d]=",i);
3 R" f1 F9 H, C: a3 v scanf("%d",&a[i]);
1 q3 \& X, U- u) X- e}
) j3 u5 |: c: t+ r5 x' c7 E; }! Lprintf("\n");
& C9 R$ V: m3 z3 x/ B ?0 \for(i=0;i printf("%5d",a[i]);
4 J7 U6 Q) M. c% O: g% jprintf("\n");2 R2 w3 ~' p1 U6 X; z) i2 }
for (i=0;i{ min=i;5 {. B5 @0 Z8 t% @/ H
for(j=i+1;j if(a[min]>a[j]) min=j;& ^0 N! f" p# `7 I* X- K) r; k
temp=a[i];2 t6 p- t+ A, y3 f# t$ c, W
a[i]=a[min];
" y B2 W a" V" a# k a[min]=temp;, a1 l; D/ O- W) x& s
}
6 z" ?$ e1 D! L$ n. D1 fprintf("\n排序结果如下:\n");. c( F3 x! ` e% ?6 y
for(i=0;iprintf("%5d",a[i]);
5 c% ]0 ]/ t- O3 V% N}
* |, O& q. o( c; ?0 p/ }) x7.3对角线和:1 n" F; F1 S# U, z: E
main()$ g! Y4 M1 x I# { {2 y
{
# z8 l9 m1 \( ?" x/ G& @1 M6 sfloat a[3][3],sum=0;
3 ^ X6 k6 \% @. y9 o$ g% eint i,j;3 o7 ~4 e; s, D+ R+ ^( w
printf("请输入矩阵元素:\n");4 \0 c! t, B3 }8 S* W
for(i=0;i<3;i++)
) {- j& _ I3 T# h for(j=0;j<3;j++)4 l9 @+ K# ?+ W
scanf("%f",&a[i][j]);
3 d$ h1 b2 h% V- o6 |1 ` for(i=0;i<3;i++)9 p) R6 c; }& `9 M0 m
sum=sum+a[i][i];
/ y: @0 j# e4 P printf("对角元素之和=6.2f",sum);. m. B- u9 m$ g7 t
}
; @4 l" w' s, \* j; y" T7.4插入数据到数组& P+ S2 Z' C& a% j& ?8 M
main()( ^ S H" d8 n/ T
{int a[11]={1,4,6,9,13,16,19,28,40,100};# o& K3 g3 o; q( j7 V
int temp1,temp2,number,end,i,j;( q, Y+ R) H% Q6 N! D
printf("初始数组如下:");& [0 z9 s3 Q, [& X, @9 f1 q2 w
for (i=0;i<10;i++)9 z7 ]0 g2 q: n; ~' \
printf("%5d",a[i]);$ M- G) e+ w/ i& g* r
printf("\n");
; W+ h" N2 R+ ^! ?printf("输入插入数据:");: k* {5 ?- p8 [! A4 _ L
scanf("%d",&number);
! V/ g& D6 E$ h5 p( ?0 a Z6 D0 jend=a[9];" a# `( F6 N2 d% ]* h6 H
if(number>end)) a" w) x3 W# ~7 H
a[10]=number;$ g& f: l3 h) I8 ^! K E( ~4 g
else
/ @' x+ u6 M/ z$ H5 w {for(i=0;i<10;i++)
8 i5 C/ B& }1 t' w* U4 U( [ { if(a[i]>number)) X ] p9 a. H% a' O& z/ d/ Q
{temp1=a[i];$ L% o M2 w" u9 B; ]. ]1 O
a[i]=number; h( O2 s! D& u% e. L
for(j=i+1;j<11;j++)# Z% J& b6 k# t, b! O5 s
{temp2=a[j];7 k' [$ r) L7 R+ {
a[j]=temp1;3 w0 z$ X1 G- w' B$ x' M
temp1=temp2;
1 c! }. p5 r) s* m }( b' P0 _# B. s N; B
break;
% c" w$ @$ O3 m- c }# v3 j% c( J1 J& u* u. {
}
7 m1 c3 ^* s1 x8 w3 Q2 X( [ }8 ^. D- [% y9 `& Z- c
for(i=0;j<11;i++)
; P8 d% S* s: P; t# }" n printf("a%6d",a[i]);
/ z2 j: d6 r [+ d+ h}
5 z9 A! g* G! ^$ a4 B8 i7.5将一个数组逆序存放。& D! [" W0 l$ r" |
#define N 5
) @6 A P9 L$ Pmain()7 r8 [2 l6 ^4 W4 ~$ D! n+ G
{ int a[N]={8,6,5,4,1},i,temp;
# l) P) ~" w/ S, f1 l. \, x- Tprintf("\n 初始数组:\n");8 ]- X5 j0 C' a, U3 q( x& J0 H& s: k
for(i=0;iprintf("%4d",a[i]);
7 J+ ^8 g" D3 }% M+ K0 ^for(i=0;i{ temp=a[i];, L' O& P# o7 _/ a( m5 T
a[i]=a[N-i-1];
! U! p- g5 J) ?- t a[N-i-1]=temp;! v( N. B2 b# F, G6 r: z$ i
}
3 o* Z7 k7 _7 q3 t: Eprintf("\n 交换后的数组:\n");
- a) S6 r% c0 ^; q0 ~# k7 Ofor(i=0;i printf("%4d",a[i]);4 m( _' R+ f# O
}
" }7 [* M7 x4 s9 E: k7.6杨辉三角- u$ K7 C0 n! @
#define N 11
6 @; E' U6 X0 w/ }& I4 `main()7 J7 m U3 E$ j
{ int i,j,a[N][N];
) ~. o5 T) U T; c3 u for(i=1;i {a[i][i]=1;- R6 R. @1 F! `) R* ]( m/ y3 V( X
a[i][1]=1;
9 c- M9 Y) b2 F" Z" l# n6 o% Q" ^ }
5 W+ l$ C0 B$ `; H$ H7 Y for(i=3;i for(j=2;j<=i-1;j++)
8 R) R) h, Q" S: a' H a[i][j]=a[i01][j-1]+a[i-1][j];; }* j3 q4 K) G" K# u) o& K
for(i=1;i { for(j=1;j<=i;j++)
, U7 K, L& _, q printf("%6d",a[i][j];
+ t2 C9 p- Z( r printf("\n");- f0 J7 }# a( q( P$ n$ ?8 W
}- w2 n% O; q @1 v* N* `
printf("\n");
2 ^9 A- [7 s; B# R# [}
% l& z/ v* y' M) P8 r3 _& O U4 q7.8鞍点6 }2 h% T6 i [1 ]/ J4 E& K2 A
#define N 109 q4 H1 N( O* B
#define M 10! W( V' y# r! f' b' m
main() T$ n- Q9 v# C0 ^. Q d
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;
. Q7 s3 v0 C/ z printf("\n输入行数n:"); P( g* ?! B3 y( U* e1 a
scanf("%d",&n);
! i5 f: g9 I J printf("\n输入列数m:");
, d% u) S+ B& R scanf("%d",&m);
3 `6 @" E u1 P" l- z8 X
3 F1 o4 z; g+ z8 O6 i# P \% k for(i=0;i { printf("第%d行?\n",i);
* I; K5 o& o+ j2 q for(j=0;j scanf("%d",&a[i][j];
2 h+ x, h, k9 g }
7 E* M G! ~' Q for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
: |1 V8 J3 u, q4 `, M9 P1 z9 d7 q8 [ pritf("\n");
$ I7 @% q3 m, F* t5 G }
% U' F+ Y0 N E flag2=0;7 B l% T% x( _+ a
for(i=0;i { max=a[i][0];* l7 p8 N5 D6 o( r4 Z
for(j=0;j if(a[i][j]>max)! f3 T" J/ a @6 c! K
{ max=a[i][j];
$ [8 d* x3 z# I9 A- I+ \: x3 M maxj=j;- @5 B: q; S0 ~/ d/ u) X' R3 }
}
, S: a6 o- w' _! H for (k=0,flag1=1;k if(max>a[k][max]), [0 k6 T% P, {4 R+ X# F) ]
flag1=0;
d, W* V: l8 ~+ Z, z" Z) C; } if(flag1)! K4 x! a- Y! h0 f5 _
{ printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);
8 a: {2 J, J1 ]7 l flag2=1;9 w, ^2 I- z( ~# Q, z: T5 K
}7 }, v0 x' }; W8 V/ F+ d/ Z
}
6 T8 P1 D B7 o" a- b, Q/ Eif(!flag2)4 {: Y( R: t% K
printf("\n 矩阵中无鞍点! \n");
- h9 h/ \; a" V0 H1 H/ A; {# y. V}
( o4 J" w O7 W# E$ ~
" w2 Q+ x S" N! \& o& D7.9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.* p3 J: J/ v; }3 p; |5 U+ f* W
#include
$ q/ a7 F9 O+ L+ I" N2 e; r#define N 15
+ N: U9 p% y6 m$ qmain()
$ t: K# \' [, n" Q# K5 y3 c( m{ int i,j,number,top,bott,min,loca,a[N],flag;
# l1 D9 w9 }6 d# M* L8 M' V char c;9 S6 p. U e6 k* G% L% }
printf("输入15个数(a[i]>[i-1])\n);
! o3 j1 J: K9 O3 P9 Q, o scanf("%d",&a[0]);1 f0 j- Q& |, p1 V
i=1;$ p4 A4 P! i$ m! p! Y- A
while(i { scanf("%d",&a[i]);& y3 r6 L& j6 E) }" G
if(a[i]>=a[i-1])
: ~8 n* K( Z- c* ]2 n$ M R i++;
( D, A. a1 e! ]& _% ^( C1 M esle/ n1 O2 f6 y9 q. Q( N$ r4 ^1 A
{printf("请重输入a[i]"); S+ q6 Y% U4 L' J: b
printf("必须大于%d\n",a[i-1]);
: R8 f z9 Z! W! o9 n }3 b7 x5 ?; n% B" `' W- ~7 V$ x
}
0 P' }3 B( e$ n4 O6 o$ B printf("\n");
+ `0 t* U+ [' }$ o* E4 P6 o3 q for(i=0;i printf("%4d",a[i]);# z+ r E, t0 y4 Q
printf("\n");
4 [3 e7 M, h u5 W
" |$ ?+ t W6 y flag=1;
# s. |- s, F! q _0 u: L while(flag)( {5 T: ~, _4 e+ I# Y4 r8 g+ T
{
$ d/ K( c7 B8 k/ K# ` printf("请输入查找数据:");
3 ]1 \# ~9 Q9 N8 {2 R scanf("%d",&number);
/ _+ G$ B9 V9 S) P loca=0;
% u9 D9 \, h. O4 Q6 F, C top=0;, f0 o5 g! |' y6 R; Q6 e0 }# [
bott=N-1;4 S4 V1 V* O4 `3 E
if((numbera[N-1]))- _ }0 {# C/ d: h- g' Z+ [
loca=-1;
2 p/ a4 H9 p8 B6 b# _( i while((loca==0)&&(top<=bott))
5 c* E6 A5 J- B2 Z { min=(bott+top)/2;
7 e* L+ c% m: Y* G P. A if(number==a[min])
4 M6 e7 z# R' h. L- K# k/ r { loca=min;+ c# l$ W1 A# a" M$ v- o' r6 p
printf("%d位于表中第%d个数\n",number,loca+1);
, c, w% r7 y& \8 E, N" z/ w) J+ x }/ e" a8 m* Y y: L+ Z
else if(number bott=min-1;) z8 @3 M2 N5 F4 ~1 [! M
else
/ A) q) [+ C# ~! k Q top=min+1;
& E5 n2 M0 \5 g& X" b/ S( ~ }
- ?1 f/ }( Y# A4 G# E1 i: ~# b, p if(loca==0||loca==-1)
# I, h6 @# K6 }: [# s printf("%d不在表中\n",number);
( p! w3 D8 v7 b- ]$ j6 V8 H9 g printf("是否继续查找?Y/N!\n");6 z# v- B- Z3 S2 E0 w r- ~
c=getchar();
/ P0 i5 Y n. S# G+ c/ I% X. w if(c=='N'||c=='n')' c0 U( Q- J, C4 ^0 m; f0 H
flag=0;& K! _! q9 W2 D" K) a! Z/ t0 @
}
/ \7 P# s0 G5 Q% E9 Y}
# J; s4 h0 i" S, H3 G* G
! |& k2 `8 y& x/ @7.10- P, x* i9 t' }8 c ?5 [
main()
+ S$ x7 D1 u s4 A( T! \{ int i,j,uppn,lown,dign,span,othn;/ d2 C" ^6 O. B1 K; R- B
char text[3][80];7 u' S, i1 S4 P' K6 l- {
uppn=lown=dign=span=othn=0;
0 s+ O2 D. |4 D9 x for(i=0;i<3;i++)
* ? t# ^; e1 ]5 |3 u$ C { printf("\n请输入第%d行:\n",i);- q0 J* l, @ c3 H- x! Q6 C9 |
gets(text[i]);* `$ I7 a X: d8 H
for(j=0;j<80 && text[i][j]!='\0';j++)0 H; i+ z( r8 E! r/ W
{if(text[i][j]>='A' && text[i][j]<='Z')
' u7 R1 l5 `3 |& t5 W4 ?, e uppn+=1;. W' G9 P; w! y" f
else if(text[i][j]>='a' && text[i][j]<='z')# q. K {' `0 Q. o6 b R1 `
lown+=1;
0 q5 w) V! F. s. P else if(text[i][j]>='1' && text[i][j]<='9')
) \# t: Z: x2 z5 D1 e dign+=1;
" \1 I+ ?+ ~% Z: q else if(text[i][j]=' ')
$ d9 J# F8 k' }' p: D; ~5 M, h) d span+=1;" I* e$ J8 E7 m9 D% a1 c+ ?
else3 }( l. [2 v. V
othn+=1;% V7 V7 s& z# S
}1 l; K+ d! F8 F" x- z4 Y5 ?. w1 T
}
% H% n$ ^8 X# z5 w- \6 C1 b for(i=0;i<3;i++)/ T& c' }0 C; q. q5 K
printf("%s=n",text[i]);
: Y6 o; S9 _6 Z/ o printf("大写字母数:%d\n",uppn);
$ o* |2 L# t* l$ I5 z$ j printf("小写字母数:%d\n",lown);: P9 C T& w f: J8 q
printf("数字个数:%d\n",dign);1 e7 e8 e. C! s, O4 g8 N' k
printf("空格个数:%d\n",span);& ^! t% Q4 R [, K( v. [
printf("其它字符:%d\n",othn);3 r0 q9 z5 t' K" M
} w* d; m& t2 j+ s
. A: b+ U* t' g0 _
9 L$ E- v! B' Q3 \+ k8 u, H
7.11
% C: f; i4 o$ @' H& x) C( k: vmain()
* w2 e+ T( l2 x# s& D- q {static char a[5]={'*','*','*','*','*'};
7 z* }/ f' j1 j8 `% U int i,j,k;' m( P+ p; X# i$ J! B! f
char space=' ';9 N, F- r* {! R4 a3 m4 U+ ]
for(i=0;i<=5;i++)5 H+ o5 O8 @' n# o8 r
{printf("\n");
. a/ \3 C6 Z- f; y. R! t+ X' _ for(j=1;j<=3*i;j++)1 @- h$ D' q, G4 S) a3 ^
printf("%lc",space);
0 M# _% ^: ~8 h0 e for(k=0;k<=5;k++)
3 ^2 B# ?( p2 Q7 I6 i printf("%3c",a[k];5 U: N I9 `8 v7 D$ s
}+ Y/ p$ e7 z, D) H' I* u. V1 R
}- R0 Q* k: R3 y) a: @, a0 F$ e; K+ K
7.125 K& K; A7 Z* V" m- v8 s) K/ L
#include
0 b) O8 d4 u6 A+ h$ Mmain()
5 ?) f7 t7 G6 Y7 o2 [3 k) }7 g6 P# w6 q{int i,n;
5 {7 `2 S0 M7 B7 }9 u- P char ch[80],tran[80];
& a4 t6 \$ ~& Y' I" S: r* }/ A printf("请输入字符:");
& G* k' h1 V+ j+ S' e7 D9 v) \# S gets(ch);; J9 T" S; M# U. x; Y
printf("\n密码是%c",ch);
: n% v9 D7 b$ zi=0; `7 ^& c! s9 m; f& r
while(ch[i]!='\0')
. J* |$ S& d3 g7 E# C7 h/ Y{if((ch[i]>='A')&&(ch[i]<='Z'))* @) L3 G1 X7 F5 w8 \# P
tran[i]=26+64-ch[i]+1+64;& N n6 B R7 F" N$ {& E
else if((ch[i]>='a')&&(ch[i]<='z'))
( r+ |. S" ^7 d% q tran[i]=26+96-ch[i]+1+96;6 c6 r B" x& r
else* Y/ q- |5 N9 { T5 C
tran[i]=ch[i];
h0 R4 Y. c+ v: X3 R7 C# N, _/ l i++;- a. i: Z' z- L5 I% A/ u5 l: `
}
- K" A. I/ ^ Q: `9 R0 Fn=i;
1 o# \- i' `9 j1 F- Vprintf("\n原文是:");! v; i& ^; Z3 v3 }* t0 B# i1 O
for(i=0;iputchar(tran[i]);9 u( ]1 r' Z- L
}
n+ [* L4 \- U! L" R7.135 {" T1 @. A) h& R7 f+ Q# k0 K8 F
main(): @; f( {/ f L! `* o
{
. }" H9 ?3 C- W8 C# l5 G char s1[80],s2[40];9 i! ~1 n; B. i) |! N. r) s
int i=0,j=0;
% d) m1 ~/ a" Y# z, ]- D/ | printf("\n请输入字符串1:");! `3 _. B9 z8 X. N
scanf("%s",s1);
$ e) u9 A; P2 o. A& c printf("\n请输入字符串2:");% G+ g2 i2 H, l6 W8 t8 l$ E& R
scanf("%s",s2);! ?7 q9 H7 T/ S5 _4 n- }8 P
while(s1[i]!='\0')7 f) q5 n5 @9 y, x: @ w9 c
i++;
$ w# L# p5 j# t' T+ y4 c! w! Y7 Qwhile(s2[j]!='\0'). E1 d! G1 W1 Z1 t
s1[i++]=s2[j++];
6 _3 S, y& o: K! c) x1 c9 A3 ws1[i]='\0';
& U2 S% Z" \8 ^* z9 Kprintf("\n连接后字符串为:%s",s1);: [6 P* a3 \2 L) Y4 ]. G
}! g) E- q4 s$ l( D/ ]9 x
# w4 I M8 s6 W6 i0 S/ v. V
( |+ U: i z8 M) p, e1 `) X7.14
: Y$ n8 F- t+ a, j2 E#include
/ S! }( z5 I# ] _) Y! gmain() p/ @) c9 Z2 S/ K" p5 J/ K
{int i,resu;8 c- K# \ [0 L, v: [; b6 _
char s1[100],s2[100];8 W" d S U- M7 Y+ U; B
printf("请输入字符串1:\n");- O' J. | ^2 S8 Q9 A3 m
gets(s1);/ f" Q5 W8 m" r' r5 J ^0 r
printf("\n 请输入字符串2:\n");& d7 b$ X8 q3 x8 v1 j
gets(s2);
7 ^# u; s; T- d8 [ i=0;
" `; X; {4 ? m. s! S while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;* D" ~5 Y Y2 o$ l6 x w- ?
if(s1[i]=='\0' && s2[i]=='\0')resu=0;
b1 O' f4 h5 y) l7 {8 p& R/ w( c2 U else
" w) z( a; L2 i, m) H resu=s1[i]-s2[i];
0 U4 A) g9 S) v$ _' n2 T printf(" %s与%s比较结果是%d",s1,s2,resu);
' L$ s2 T3 N& j}
2 h. ^8 {" A% {7.156 h! l# p3 P: d K/ z) I: Z8 M
#include; ~7 y- [+ a! x$ M ?
main()
) ? e$ L x" S7 y* I! c {
+ [! w. O& b7 w; P: U6 W char from[80],to[80];9 ?& r# P$ c2 o+ b: _* l
int i;# f; t' ^& C: \& k( ^5 y) w' i
printf("请输入字符串");1 m8 U9 S# a4 j: ^8 s- C
scanf("%s",from);& l T2 q6 W3 ^/ n( c! A
for(i=0;i<=strlen(from);i++)
4 Q: Z0 g/ A, o. s8 l to[i]=from[i];: T# z! G, V5 F
printf("复制字符串为:%s\n",to);& N' u; T& X8 D; M* L0 _( W3 {9 {
}
6 a5 P9 Q% |* k# h2 O; ]* q m/ X+ Z7 \& }8 D9 \
/ [1 N0 B( s+ J7 m+ k. W, [
第八章 函数) |# [+ E! `' {' R; b6 \$ {( N. U. j
8.1(最小公倍数=u*v/最大公约数.)2 v4 V$ Y, k) k" z6 t: [- |
hcf(u,v)1 v. U; m! ~" D0 v& v( Y+ n( x
int u,v;: K M6 L$ s8 I h! o4 {
(int a,b,t,r;0 m4 e; x! T# y( W5 j- p
if(u>v)
9 T0 W1 m4 B- o0 D: H6 L, ~. ~ {t=u;u=v;v=t;}
H) \& P m" C) j Z; Z8 ` a=u;b=v;# W% f$ f1 T' [* D8 r% m
while((r=b%a)!=0)
) y" g3 w* ~0 u. W# F$ D {b=a;a=r;}! p' Q! c, ?) n7 Y7 L/ W6 d. j+ f( d- c. P
return(a);3 K. u1 q2 R6 I H9 h7 v0 Z& x
}
5 t" ^/ ^* u( `2 ~0 C- A+ R* t lcd(u,v,h)) a% c& `; F3 ~& {
int u,v,h;
/ Z2 B8 _1 Y7 q. r) V5 n" s {int u,v,h,l;
& k3 k* w/ T& R) V7 Z7 p scanf("%d,%d",&u,&v);4 q! h6 K$ x7 w- D
h=hcf(u,v);. B( G3 a! v5 U4 _* ~ r9 i3 ^
printf("H.C.F=%d\n",h);
: M/ V! X3 _8 Q7 w" v l=lcd(u,v,h);+ Q# s0 a2 p7 U; c: x; O4 b" R( f
printf("L.C.d=%d\n",l);
: K$ R+ a, \3 q" s0 t/ t }& I+ I$ o7 s2 V. |( Q" _ m
{return(u*v/h);}* _9 g0 G1 N! c9 k2 K* o3 E
main()& Z p/ [- ^8 V; B) X0 ^+ D
{int u,v,h,l;. r; O0 {7 [! N' }) m1 W- J$ x
scanf("%d,%d",&u,&v);- t! c; d2 _: B- U1 F4 L+ J
h=hcf(u,v);# X: N2 [9 H3 r+ C- W* _
printf("H.C.F=%d\n",h);
" }! o; r( O" d l=lcd(u,v,h);5 ]0 n0 f1 p7 s @' F3 W1 }6 o
printf("L.C.D=%d\n",l);
! l) r% c; ?5 _* g }
, d! {" P. \* Y; O; M: V0 ` w+ [' S0 Z5 v$ K6 ^# t3 [# @8 k) e3 e
' e8 o6 e# t0 h" N0 v/ J
, e9 h3 U8 X/ c' i- z8 ~8.2求方程根
, v: I9 J, g/ s: I9 i#include
2 U3 y Y* B5 Z; Q! y- r& B" ~: t$ \float x1,x2,disc,p,q;/ A! g5 I0 q r% b+ L8 Q$ S
greater_than_zero(a,b)
, }6 J$ C2 H, {/ o$ p, S0 Q: h; |float a,b;) `' r. E( Z+ C5 l6 d. `6 j) ?
{
9 z' ^6 i: e' ?+ }x1=(-b+sqrt(disc))/(2*a);
/ l, r7 N" r$ B# G5 Wx2=(-b-sqrt(disc))/(2*a);
& t7 j. _& L8 A) g% |/ ^. H& l2 s}# I, K Z, d3 O" u) U1 `7 u
equal_to_zero(a,b)- G) r" L# }/ Q1 v0 e( e
float a,b;+ y) }! @8 I0 J: Z/ y: w
{x1=x2=(-b)/(2*a);}
9 D8 z1 Z8 e# t3 ]/ q. esmaller_than_zero(a,b)) w: K( L" a& U
float a,b;% M+ J& z/ J& K: v
{p=-b/(2*a);
% P. R3 D7 V3 u: @q=sqrt(disc)/(2*a);. O3 u7 e% v% k
}
8 a( i; S' E1 @main()
* g+ w2 t7 Q5 q/ Y{
1 R+ V; S1 S' u: |* X' S' k( ^float a,b,c;
8 W- V6 A7 F' }printf("\n输入方程的系数a,b,c:\n");% N+ i: ]: q) w; U1 m4 @& a- V
scanf("%f,%f,%f",&a,&b,&c);5 c; [! a7 p9 |# C/ {) f/ [
printf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);6 `2 V* H- \. h5 b- j f5 Q
disc=b*b-4*a*c;
# e5 _, ?% ^3 c/ l1 ~! u" k6 qprintf("方程的解是:\n"); s1 g- V. d" ^: h2 D" j1 X
if(disc>0). F3 P, E" l9 C2 E) v7 ]" M
{great_than_zero(a,b);' O! o$ i* c @
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
\) y; S9 E9 u2 ~9 P: O+ p; h}$ y; ?- ^. D: A, [. `5 g. `
else if(disc==0): M- L* ~6 {& n
{& v( D8 P" p) M5 U: l. M' P
zero(a,b);
( S5 P& j. c3 G1 P7 Y5 v1 oprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
u( x, Z' r: P. P }
" U# ^* `9 U ^else: r& x6 [1 s* k& }" V
{- m( i/ I4 ~% o8 c X
small_than_zero(a,b,c);
, B# i0 o/ M( ?7 ], ^* x# O printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);
|0 X* A. }! h5 V$ @% ]- b# W7 O }& Q8 X2 ]& p7 `$ N$ ~- i* c
}4 s0 X; H: T' {. ~5 i- _$ t
8.3素数
7 v5 s+ [$ M; u4 n* l* N#include"math.h"
9 j2 b: S1 O$ E, _8 y* i @main()
* Z/ n5 R$ h F/ H{int number;1 S. C6 K2 K& Q' i
scanf("%d",&number);
0 Z2 y, M* K4 E& c if(prime(number))! g4 z( U8 \. Y
printf("yes");
- C9 K+ C* S4 W, u else: A* p+ z, }$ W) j$ v8 a
printf("no");/ l2 l" x. q( f* {5 x: T
}* `0 R; y s1 c! `5 D9 w
int prime(number)+ P/ ~0 y, K* C' \
int number;# ?1 Y! d0 W2 E; e0 k3 b
{int flag=1,n;
$ G) }- c8 w6 I. c. O for(n=2;n if(number%n==0)
! F5 x( ~- G9 @6 w- I# Q flag=0;/ s" O, N, S7 B( C% N
return(flag);
3 s7 `- b' d0 E. w# o# t* `. i}
/ b% d3 n7 ^7 G
2 J, T2 v" W) [ o& q* @( [5 \6 r) r& G( O9 e9 |
' g% M5 l0 }1 D* O" Z. i' j9 h' G8.4
. `# d4 Y) O+ o2 C3 v#define N 36 L- q. G; U6 M$ j/ J) ?- Y
int array[N][N];0 u! @) V" K) U" u/ T
convert(array)
( _4 D J7 Y9 e' r9 A, h9 Oint array[3][3];
9 y, Q4 P6 X: b, W& O# G { int i,j,t;
. o4 E. L0 D4 A for(i=0;i for(j=i+1;j { t=array[i][j];# v* `9 F; @3 c
array[i][j]=array[j][i];
6 L u1 \3 P/ Z O8 K% e, f array[j][i]=t;
1 H6 T+ Q! b$ m. D- g5 Z1 O/ L }$ m* f8 P/ o% @4 @2 ^
}
9 N5 W) `/ |6 ?# U; {) a, w* ^! a+ Ymain() c* `: n9 o; M( l8 S
{/ k9 g- f" S% o% F( C
int i,j;3 I+ `. ?; k6 E) \
printf("输入数组元素:\n");& z2 E1 g0 s U0 j# V' a
for(i=0;i for(j=0;j scanf("%d",&array[i][j];0 |5 S8 O4 h7 x) x; u$ }( ]; x
printf("\n数组是:\n");0 U0 y% l* ~9 G
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);+ I R* A% v: e" K. e
printf("\n");
/ C; |+ z$ I% B7 o1 j }8 [ ~" A5 v" ]
convert(array);1 C! O+ h- n% u9 B: c% c" F
printf("转置数组是:\n");
5 t6 G( |& L; g& g' W( _: T for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
2 D0 C0 x/ u( t, n printf("\n");
. ^' w8 }- v; f }# I! b9 `! I' ?' a* ^$ D( \
}+ L/ q0 ^9 l9 @& X) ~' p
. T1 e4 C1 T; T0 g) j6 K
$ Y% [! `1 S& ?' u+ b" h: p" O$ Q* a d/ M% V
8.51 [7 b, q/ T6 J" q% j a* R
main()7 V; q4 O9 I1 W( @/ @0 J7 _
{
' {" W2 x) a& _5 H. j* i" ?char str[100];1 i! @" @8 K2 W
printf("输入字符串:\n");/ l L- m$ q Y$ E
scanf("%s",str);
) A2 b% A2 f5 \ inverse(str);, K5 J+ Z7 }, a$ g7 R! R3 W& U$ C" O1 p
printf("转换后的字符串是: %s\n",str);
9 }% O" u$ m$ l0 E4 n}0 D9 v; E& V: H
inverse(str): @* H- l/ h7 A: h }8 i; i
char str[];7 j; o G( u) p' e7 ~" l5 ?, t+ W- f
{
3 `4 \, ^2 O# t0 O% U char t;
0 c( W+ E; F$ N# ?5 k. H% W$ U* v int i,j;
1 J ~/ w9 ~; z for(i=0,j=strlen(str);i {
9 r9 V6 \; V6 T- _: w t=str[i];
" u5 ~5 C6 j9 e str[i]=str[i-1];
0 Q' D0 e3 S: }7 k5 s2 W str[i-1]=t;8 p; g# G8 i9 S
}* Z& R- W% R% H0 h% J/ h) w& s
}# `4 _1 d& R* b7 C% H) b9 M" `3 ]
2 M& s! t, O1 Z; m8 R
6 U0 N/ _' K3 B6 n \5 B
+ H- W/ C' m. j- {8.6
, d0 x7 T6 L' [& G _) X% Xchar concatenate(string1,string2,string);
8 h: ~( m2 U1 M0 u; D' h wchar string1[],string2[],string[];
]0 ~1 Z9 K" L/ n. O{
. n+ ]6 G; s: z. x# @int i,j;
5 _4 U. d# B+ \$ Y6 k# n- W8 Bfor(i=0;string1[i]!='\0';i++)
/ X3 [- Q3 f% b4 D3 Z8 E string[i]=string1[i];
$ f/ T1 I4 t, t* y/ ufor(j=0;string2[j]!='\0';j++)6 M- d5 g; B& T+ L
string[i+j]=string2[j];6 x- {$ w, q, c5 n' F
string[i+j]='\0';5 ]* l& y, U9 M0 D, s h
}
" y8 F2 T' L7 F9 r. Qmain()
* F% ^! y; n& ^/ H4 L, a{' ~$ i- Q) p4 C y) V: Z- L) ^
char s1[100],s2[100],s[100];% i6 L. x, S, \
printf("\n输入字符串1:\n");
% ?% ^- Q! N. c. z7 n4 w- f; O scanf("%s",s1);3 w# M5 Z6 K7 }* l, ?# ?8 Y
printf("输入字符串2:\n");
$ b* D n! z; i9 K( Y% j8 | scanf("%s",s2);
3 d) r* m$ E; l/ e( B+ A; i& G concatenate(s1,s2,s);$ i7 J; B1 j- H p3 T" N1 j' ?$ P
printf("连接后的字符串:%s\n",s); o( k' _# V" `4 a% {& B4 D
}: W: n' p9 S6 D. I/ v- c1 c+ e
& X5 q: H$ [, ^0 a8 b) \
) [1 ?8 n) U" c; B @8.8
. \- d) q+ z% `% W/ W7 w/ s1 Y( qmain()
$ Z s, v% W) a) h! B{1 H4 H1 I; |' M) ~7 \) F* }' b
char str[80];
" n4 |7 d; h/ }4 t! L0 i) ]$ o1 N6 S printf("请输入含有四个数字的字符串:\n");
. o6 G! M7 }" E- K scanf("%s",str);
8 _; Y) b- d u- y" H2 Y& i insert(str);
: j2 X# Q1 H, }. m7 y}
% d2 T5 P% E4 x/ [) M( Q0 ~insert(str)& z! ~0 r! @2 U' F! m' l7 l( u7 Q& Y1 i
char str[];
4 ]. A3 V x: w. { L{8 t" ^9 D, Q2 b% r* F- Y- w9 N
int i;, j) i0 C/ N/ D* x7 e C$ Q
for(i=strlen(str);i>0;i--)) F {; d" Z' g2 w6 F
{ str[2*i]=str[i];
# o$ S S) Q5 `1 l str[2*i-1]=' ';
9 t8 z4 h# I0 Q2 E: w }
4 j' ~, G1 j% ] printf("\n 结果是:\n %s",str);( U' r8 r v$ ]+ K5 z. n7 y* T/ v
}
! g/ P- H$ W1 S# ? j$ m
! x! ?; W* e0 h6 C' s! O, `) E8 J: d; ?+ }/ D
& D$ T. [; u: @( [4 j$ b8.9
" y* X" Z6 {& _* Z& D- z#include"math.h"' m8 m% `' G9 Y; L2 I6 k
int alph,digit,space,others;" \ q6 R% v. V+ E$ d0 T U* q1 w
main()
' D6 l+ k8 W# A+ Q( K( s{char text[80];* L4 K- N) I; {8 C6 @3 _1 s
gets(text);) K# a7 b5 j O( x* `; I
alph=0,digit=0,space=0,others=0;: x- ?/ a+ Z) M; h2 s
count(text);% o/ v8 | ` m' K K2 f
printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);% S7 w a ]' W% u/ y9 c
}- Q5 o2 \2 q) S) k3 p. t. C! z
count(str)+ @) n0 x) Y$ R, R
char str[];
, w1 Q# {/ U- t) U: @% p" M{int i;
8 @- ~" @ L' z* a$ |1 j for(i=0;str[i]!='\0';i++)
" u# |0 S0 t2 D/ {, d7 n2 u4 Q if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
: h4 s* f8 O; r1 \1 k alph++;
/ m N0 \( }% [! l, L else if(str[i]>='0'&&str[i]<='9')
' X- X) |, q7 d4 h7 N digit++;. v. b$ K' t# f
else if(strcmp(str[i],' ')==0)
2 t8 h( v9 d. W6 e space++;
* d ? f( P+ q, M+ R* _+ N else
9 C/ L2 h$ l, {9 s others++;
: d) Z: _0 g# |) f' x; y}
6 V3 E& Z' V i8 R) s1 w
% ~5 `9 {* l' H. C, ?5 `5 p1 U) x( o+ W1 G; Y; d# i' [
8.10$ C& B2 \' \! i! W* l+ \
int alphabetic(c);4 V! R' v$ E' w$ u4 ]& P" N! o
char c;/ F( ]) }1 E7 h; N* r7 C9 S) w
{
! c$ r% {% k% ]2 u if((c>='a' && c<='z'||(c>='A' && c<='Z')). m3 V1 ]) i _ C/ P" Y. G
return(1);
4 X2 Z# ^: w3 G& F else" j, \: p( }2 Q* Z$ v7 }2 X
return(0);* _% O& V0 j& N* U- g2 w2 Z/ D
}
2 c2 T; K% D9 q1 z4 g% r/ V1 R/ n( x' d3 p }4 z; r
int longest (string): s2 q* j% k0 U# R @2 v' e3 r
char string[];
6 ~5 O( {. z) N. M$ a$ A{
& D( R* C& ^/ U j9 F. _6 P% n int len=0,i,length=0,flag=1,place,point;
5 Z: f; J2 v0 ? J/ |" } for(i=0;i<=strlen(string);i++)
% i" G/ {* S2 O& w# j if(alphabctic(string[i]))
& Q9 p" M9 _8 x# u if(flag)
9 Z3 l9 e$ }# j, D! H {
) ?% Z `, c+ ], ]8 e point=i;
* Y; z/ [2 B' r flag=0;4 X0 c0 ^$ L6 N2 c
}6 C! \8 K# ?% z
else
9 c5 C# E. e5 q, I4 r7 x! } len++;
+ n4 i% v/ u# y, R: |$ a+ I1 G& ~- q else' i' S$ P& @$ N7 T
{ flag=1;% t! _+ z0 w8 |. h: b5 k
if len>length)1 B; k2 R# \1 Y# C# ^
{length=len;$ X3 g5 o- g3 D' w* U) ~
place=point;
8 O$ b* m: {9 M: g len=0;2 ^5 q q0 {) a' y+ `
}
" r) Y( A, A. F }
$ n6 y% V+ |# r) B4 U$ V$ }( I return(place);4 c& S8 M; s" D4 E; m+ b5 x6 O! R
}# a/ P, X5 U% @" U
main()
3 X$ D& X( l" {* Y' `8 O' `/ [$ z{
6 v% d* |# W' O V- g! c+ `- u* Lint i;
) ?* `# x2 f2 ]6 J- [7 Zchar line[100];
9 X" }' }. q C+ W9 [printf("输入一行文本\n");
7 q5 t# R* n4 y1 q, f" m1 {, k9 [gets(line);
( {: a: f; `- ^* ], [, R# R0 `printf("\n最长的单词是:");
# y2 P/ u8 F$ k: E. ifor(i=longest(line);alphabctic(line[i]);i++)
9 t! f- i( S2 F0 h$ f- _ printf("%c",line[i];% G2 f- h2 ]8 p+ T
printf("\n");7 I4 A) D+ H1 r# h6 l, s
}2 J0 I2 j3 u% g1 Q) M, V/ ]
6 m% T3 s0 S6 O3 K
8 C0 H; a A, {; S# M3 ?7 C) t V% |/ S
8.118 ` ?, ^& E, ^4 M' b" O1 o9 L
#include3 o$ f# k) a$ y+ [0 G
+ Y5 w4 |) A: g" K; K#define N 10
- g" M, Z, g! F: `8 w+ Cchar str[N];% [/ u% V$ ~, e) x0 I5 i* m/ Q- V P
main()4 d* v$ p' c. A1 L- W
{
7 |: h% ~$ `$ A& p, mint i,flag;( i% a: Q% Q/ D4 [
for(flag=1;flag==1;)
2 O: a' Q# a9 ?; m8 f! a6 N{
6 k: J( U) z O3 ] printf("\n输入字符串,长度为10:\n");7 L! X6 j1 }8 R4 J/ Q5 Q$ w
scanf("%s",&str); l: z: [, f; W: g5 o! v6 M
if(strlen(str)>N): t! P% T, E6 {6 t
printf("超过长度,请重输!");
1 U' @1 w# b* @( O0 E else4 \3 w% ~. }7 X5 a7 o& g
flag=0;
' m- G7 @/ t7 B+ j; O7 q' }}! V m; c9 a( a9 n+ g0 \) m
sort(str);
( n9 h* P: w8 U. _1 [; @% f" zprintf("\n 排序结果:");! M1 ^" j! o( M+ [" p2 c
for(i=0;i printf("%c",str[i]);2 W' j t$ q. g; ^4 D5 G" @4 A, R+ b
}6 }4 ^5 ?) Z5 L
sort(str)
X8 k& z3 h- g& C+ Echar str[N];1 |% h0 i |3 u7 U3 W$ M
{
) t8 K5 ?! a+ R/ b7 b2 Wint i,j;
4 \% N! H: u3 h/ h! A7 fchar t;, z7 |: v* G4 @: j+ m+ q
for(j=1;j for(i=0;(i if(str[i]>str[i+1])5 T& y8 c& B! e# e* T) o
{ t=str[i];
6 A8 j# L7 l9 U+ W$ O# f. v1 b str[i]=str[i+1];
* x2 M: E% P, q+ c" s str[i+1]=t;- S' [, g- d6 J
}8 O$ e- N, u1 q3 Y- _& L
}
1 H0 {3 S4 P( K& ~0 K/ ]8.12& j2 G1 b( x F
#include! v& `& i# N. P1 H. E
#include
V8 r. a$ K6 B( R1 mfloat solut(a,b,c,d)
1 G0 P) n/ P% A: l: rfloat a,b,c,d;: h1 z! Q+ a3 |8 `3 S# o) d' f' F
{float x=1,x0,f,f1;
3 _! w) G9 R, |+ @! ^/ j do7 W) p7 e$ ^; V9 d! |7 ^# O
{x0=x;1 j6 P: s5 } y1 w3 f
f=((a*x0+b)*x0+c)*x0+d;
2 i }' z+ U" C0 W8 J6 a7 I+ Y f1=(3*a*x0+2*b)*x0+c;1 o+ M5 j* C0 {* M t* J
x=x0-f/f1;
; s* r7 _9 }6 q/ ~, c }
- _6 w. Q' Z* ]7 {& I/ @ while(fabs(x-x0)>=1e-5);, A. _* Z% M' y) \& l0 p: c- t: E
return(x);
9 y0 f; L/ _, N0 y/ k}
- i5 |9 Q7 M) \' M$ b+ ]' S {& \main()5 j4 \6 }, r5 v0 v7 m
{float a,b,c,d; R5 b- P& T3 F' w- A* D
scanf("%f,%f,%f,%f",&a,&b,&c,&d);8 J: H2 [( c N" }
printf("x=%10.7f\n",solut(a,b,c,d));
/ Q3 p4 W% V; W- T' q5 L3 w3 d}4 s, L( t- X, Z+ b5 @
8.136 w& B$ K; c. v& X3 n# m
#include1 F7 L/ V" G. W5 H" |$ {" x
main()
! r7 `/ z; p) ^: h6 V{int x,n;
4 r& u8 R# b+ G; X% A( D float p();4 y0 i0 A6 k3 e5 Q* c/ E1 c$ i- E' j
scanf("%d,%d",&n,&x);
8 X. D9 L& ~: o/ a8 O printf("P%d(%d)=%10.2f\n",n,x,p(n,x)); p8 w2 x3 h. y2 A
}# q) ~# |% z3 g+ S: K
float p(tn,tx). \2 S2 ]5 ? t& |
int tn,tx;1 o0 Z0 \* @9 T; y$ V$ I, X8 H
{if(tn==0)
2 `1 l8 p. i0 g* o0 m& ]: b return(1);
; Y8 ~! f0 b+ s* z else if(tn==1)$ z. {0 o9 B+ d1 H( D" f6 S8 b
return(tx);2 E4 J% B$ _% T7 D! k5 @
else3 {6 ]# H* \6 d' v. f
return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);4 L4 b- l! N3 B% H# L2 n$ m
}
4 ~$ Y3 P6 C6 I8.14
8 ?6 @; ~+ u& e( Q. o4 X#include "stdio.h"* D( W4 ]! P% R1 y2 I5 R
#define N 10
_ s8 t. l+ H C' X) x#define M 57 S8 D$ s8 K6 D3 V: `% [" L3 _: @1 N
float score[N][M];9 B* A2 _; H" N5 @6 S* O" n
float a_stu[N],a_cor[M];
$ o. @1 A8 Z, ^1 I0 j2 `main()
0 k8 O- X+ f2 @8 [{int i,j,r,c;
n( P1 {' \7 Q/ Y0 U# R% V float h;
0 D; A$ F0 C' p7 O4 A float s_diff();- @2 V9 h+ _* w4 O9 }/ e
float highest();
, @2 B6 ^- a; C& S# ]. J# g r=0;4 p- G6 l0 A2 C$ A& T
c=1; o0 W0 q, A9 A5 ]" U
input_stu();( u% [$ j4 a0 K! n5 Z3 M" X* B1 h
avr_stu();
$ n) y2 K. w$ @9 i: P) Z, I) o avr_cor();$ N5 ~& R4 P6 T f ]" l
printf("\n number class 1 2 3 4 5 avr");, d1 m- W6 s6 W* E- s
for(i=0;i {printf("\nNO%2d",i+1);3 v, P! v0 X9 U
for(j=0;j printf("%8.2f",score[i][j]);
+ l" I3 a" i6 R) s printf("%8.2f",a_stu[i]);) M, [/ ^: b U+ s
}% V( p) B1 P; B1 Z
printf("\nclassavr");
' f0 ]9 Q) z- W+ f for(j=0;j printf("%8.2f",a_cor[j]);
1 k; ~. T. B9 z& U h=highest(&r,&c);5 @. X, f5 s) S
printf("\n\n%8.2f %d %d\n",h,r,c);6 A% m/ p* }" B4 n+ x: H
printf("\n %8.2f\n",s_diff());
) F; C- |0 v0 U: K2 l}, O: O0 H( [3 ]( j4 @
input_stu()
* Q+ r0 W- C5 ^0 K3 {{int i,j;
. I7 Q* X- D5 w& o float x;
* h+ a1 K7 C* L1 H, q2 t5 |, c for(i=0;i {for(j=0;j {scanf("%f",&x);: r; ^0 ~8 J+ w5 \ \( l
score[i][j]=x;
' [. a9 j7 L: {4 K8 I }
: q9 `( d. o1 s' Y }
& G" H6 x$ D$ C7 l. S- T9 h. L}! P" X$ P1 M$ I( `: `
avr_stu()
5 r2 N( j! b3 H4 }2 Z{int i,j;
9 S8 f% i- d9 g& } float s;
2 K1 B1 s, M1 I' @* Y for(i=0;i {for(j=0,s=0;j s+=score[i][j];
3 {; m+ D; H0 k) L3 R a_stu[i]=s/5.0;
9 _( f1 x4 F0 X, K- ]8 n" f" U }
, c; E" m# K7 y& ~0 t, v}
+ z2 M; [! w; s( `! Davr_cor()) |$ W% d' h, ^
{int i,j;3 k- k$ h3 ?/ c# p/ p( m
float s;
7 U7 I. I; A" v' a1 n# y for(j=0;j {for(i=0,s=0;i s+=score[i][j];
L8 q& K' l; C* { a_cor[j]=s/(float)N;
7 f9 S/ f/ O3 l3 G }
& b5 V$ M5 M- w& P}
+ t9 e. A) b5 u$ j$ x$ |float highest(r,c)6 [# }- T2 G, N2 T4 t
int *r,*c;7 d3 s" j9 q0 L. X; t. \! h: r) F
{float high;
; g3 Z) c" T* `8 x$ x int i,j;
9 D4 Q1 n" f. x$ ? O9 d" G& g high=score[0][0];
0 \6 ^' {: h6 y for(i=0;i for(j=0;j if(score[i][j]>high)
& P! J+ u3 A) B; {9 K {high=score[i][j];5 i- u4 T" Y {' [
*r=i+1;9 |6 F5 p) F$ V6 B
*c=j+1;, z/ H( t/ K' y) z
}
4 `+ [* U! _3 a5 o9 p return(high);9 O- [' B& {; N
}. M, r9 V2 O4 E/ W1 P, @
float s_diff()
8 \; \( w0 B' |) K- f{int i,j;; C4 I0 c! u2 I* p6 s, H V- a6 C
float sumx=0.0,sumxn=0.0;/ q. x$ _, T I& X& N& e
for(i=0;i {sumx+=a_stu[i]*a_stu[i];
" z/ s5 B/ @1 d: _; e2 y$ h- Z# B sumxn+=a_stu[i];/ w0 m) F ]$ W' S! E
}
) |3 B5 M! F O- E return(sumx/N-(sumxn/N)*(sumxn/N));
# u" l6 q' R* ^}4 ^1 b/ I# M( o3 ~# q: u
8.15
( N1 r% k! J* u9 o9 B#include
& z( S7 n8 ` K2 n4 \* C#define N 10
! F v- A' g% m3 Yvoid input_e(num,name)
& K% ~$ D, i4 V7 X/ V) }9 l" Pint num[];# ]1 \9 s1 L0 O1 U. F- ?5 {& M8 m* [
char name[N][8];$ A3 y: |! a8 h6 _4 v
{int i; H4 Y: U! U$ z4 S" w) ?; z/ r
for(i=0;i {scanf("%d",&num[i]);
V- _ z' M4 {$ i4 x gets(name[i]);. q7 I- Z3 x* B+ b% t
}
6 L/ r3 }; T% P$ h& o t8 h; T}: F, y8 G3 n2 D: b4 E
void sort(num,name): L# m; W+ _7 Z. @% r1 v0 Z: D
int num[];
! G, Y6 \7 ?1 J8 Tchar name[N][8];
( c% v2 F4 T/ I1 c- W1 H) {{int i,j,min,temp1;* P8 V8 K7 U# Z5 ?# F& Q5 V
char temp2[8]; ~6 Y; i/ p$ w7 {. m7 J
for(i=0;i {min=i;
" f+ j3 L* n- E* L6 |# z for(j=i;j if(num[min]>num[j])min=j;
" h8 a v* G2 z temp1=num[i];
! d& N+ D% C( j num[i]=num[min];5 N: D1 ^8 j) o0 W; v9 H
num[min]=temp1;- [ H' m& l% M! X
strcpy(temp2,name[i]);
+ [% C! _5 e' l1 Z+ n/ h strcpy(name[i],name[min]);3 u0 n: _. J& `
strcpy(name[min],temp2);: S9 m; F: e7 _- @4 F
}
' M( ?& Y/ I* _6 X; Z) I) b3 a for(i=0;i printf("\n%5d%10s",num[i],name[i]);' J& N3 O7 @. U Q1 j) i1 {( Q
}; j5 _$ q! u5 h3 j0 P
void search(n,num,name)
( b) Q0 `- v, |- j- W' ^int n,num[];
8 ~" }. [6 T- l% f4 {. ichar name[N][8]; C# n2 i1 T5 [. k
{int top,bott,min,loca;6 `0 @* z7 |: D% d5 c
loca=0;% _. S7 U& [4 e+ d+ l) r, M) |
top=0;
& U# j! X' c! T bott=N-1;. D. }" X1 l I4 {; @$ B6 g, Q; _9 O
if((nnum[N-1]))
9 h7 O7 O7 s) f- {: q! t7 Z loca=-1;! d5 Q) Q9 T" D. K6 B$ @
while((loca==0)&&(top<=bott))
# a6 n( L! K% w {min=(bott+top)/2;5 `9 n7 z; {( h0 {
if(n==num[min])# e" ]* e; c. b1 U( k, {% g |
{loca=min;
- A* i b3 o6 p5 x; }3 e printf("number=%d,name=%s\n",n,name[loca]);/ I# ?8 S. n3 s) y
}
' t: g- n2 @% H5 ^+ b7 C5 T& d# j1 P else if(n bott=min-1;" n% a. P3 W# F% t
else
, h. K" N( h, {6 y* u# A" {$ \, l top=min+1;6 j( r+ v, \. i7 C6 ]
}
! N. {4 O* |1 H) i9 j; c- B if(loca==0||loca==-1)
$ F5 Q2 J# F z2 e* ? printf("number=%d is not in table\n",n);0 T) C# {* s" G# U/ T' B
}
' D- M% o2 _+ i- {! |; [$ y+ Q& ^$ [main()
& A5 O M- E& B8 G! r/ t, o& ~9 W{int num[N],number,flag,c,n;
" k9 K4 V0 ^) g. T9 p4 d M char name[N][8];
. |7 b" l6 z( M$ k; B2 v9 q# \' [ input_e(num,name);: v2 }8 j& q% R( O
sort(num,name);
% Z& A6 }3 D2 S6 O7 l for(flag=1;flag;)8 j* C" {# P4 R
{scanf("%d",&number);5 {1 w9 e8 C" [
search(number,num,name); }- B4 P- E. g# B
printf("continue?Y/N!");
9 O" N4 f, ^3 g ]! H$ i! g c=getchar();' X, y% a$ e+ H' E* J
if(c=='N'||c=='n')
4 ?: {1 o" U* q7 ` flag=0;5 D4 O1 s. l0 {( x' ^' B& G: ]9 U2 h
}
, L8 Z8 k8 E$ e0 K# i7 s/ C1 w}' V3 c" T& H. e6 w
, `* [ M; I1 y% O1 A
8.160 s$ u, i' P: A
#include( O' N$ O& v% m0 G
#define MAX 1000
! S. t6 d: \8 L9 nmain()2 |& Z, W: W q% N9 s' h
{ int c,i,flag,flag1; |+ |( J3 M: ^8 t- s8 p" }
char t[MAX];
" Z5 \$ \3 q; n0 {, F, T9 B i=0;0 p) ^5 q: p9 y
flag=0;5 n# t, l i/ B9 |2 v( ?
flag1=1;) Z7 y+ G h, K! y; b0 H) j5 L
printf("\n输入十六进制数:");' `+ }) @( Y7 T2 B6 ?9 g
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')& L! q% a, u. n% S
{flag=1;
" H( {1 G& C+ p, @2 n8 z& s7 t2 D t[i++]=c;. D0 F/ ]& v1 h' W8 P, m6 m: E
}
4 t+ t/ G; u* U0 V6 B+ T3 m else if(flag): P1 x. O. t4 I) p, P
{5 P: T: a4 l) Q3 T) Z
t[i]='\0';
$ ]; f4 i- V) y6 c/ L printf("\n 十进制数%d\n",htoi(t));4 y! y( e$ F/ I7 B, `
printf("继续吗?");0 ]. F! F; R+ m
c=getchar();2 W! [* U. i$ t7 w1 }( W
if(c=='N'||c=='n'): H+ F( [1 Y# g) J" L
flag1=0;
* O/ g- _( W4 y) Q7 U) [ else
: |- h+ F6 x% c N: H6 q1 o {flag=0;- w, z2 m( Q4 y3 i1 C- e$ x
i=0;! P4 T& L! a) q( o7 J; L
printf("\n 输入十六进制数:");- o( i) i& C3 R# C- A
}
7 F" p3 U' z; P9 |3 f}
5 P! c# |1 r' F) J}
# s% k9 `: a; Q' L8 ?( q}) G! I; d8 h# Y- y/ o
htoi(s)
0 i2 ]8 G; F1 N, F9 [4 A achar s[];
; K, o- z7 R" e# _9 y# M' p, f{ int i,n;4 B' W8 _) `. }: [) Q6 R3 A
n=0;9 b- h8 Z6 ^( w6 _; |# X7 `0 k
for(i=0;s[i]!='\0';i++)
: H' K( C5 M* T+ ` {if(s[i]>='0'&&s[i]<='9'); ^6 c; [; ?/ o
n=n*16+s[i]-'0';
- H! R8 V/ g# `1 x if(s[i]>='a'&&s[i]<='f')( v! w2 J2 G# ?2 ~
n=n*16+s[i]-'a'+10;
* t% _0 h3 d$ O. P7 H, ]3 G( g if(s[i]>='A'&&s[i]<='F')
7 M2 F& I7 s& f* F n=n*16+s[i]-'A'+10;* t9 I8 h$ J1 H' p
}
; q" L& q9 x1 E5 O( u3 H) \ return(n);, ~3 w2 d# N- }
}9 |, i9 w" D0 i r% l) n
) S4 \ ^3 ?7 X+ A' O( z7 }% c1 p+ D' w
5 U5 }* Z) G& N- E
8.17' M0 o: A/ [ W% b
#include
1 f% V% n) c3 g4 \: O& nvoid counvert(n)7 e8 n2 v, \/ E: a- u
int n;" a3 _3 E1 K6 a" n
{ int i;
, {/ k) m5 f+ t5 x if((i=n/10)!=0)7 q2 w* r7 v, D7 @2 x- C. f
convert(i);
) P& z i6 T: i h4 K7 _ putchar(n%10+'0');
( q/ K+ \( }4 S5 H& [% J& o+ \}
$ z) ?- K0 u( m! d0 x5 R& z _main()
( v' J4 Z8 Y0 P6 d( P" f{ int number;9 \$ @' ^1 m! a, ~8 @8 O( k
printf("\n 输入整数:");) ~$ N% `1 R+ T# @+ M
scanf("%d",&number);
4 H7 u3 x* A K( g X: c O printf("\n 输出是: ");
. H$ e: h! n$ C- K- p: k7 n if(number<0)
9 {! k! I& w3 G) c) i# p+ c4 f { putchar('-');
, i5 K) ~; b. M$ m% y) D5 \/ m number=-number;& T' x" I/ Q9 c" C5 p( g
}
( i5 A9 u1 Q5 C5 gconvert(number);% ^# e* g. u3 Q( ]) l7 O6 b. H
}9 P/ k+ i R" q4 a$ u w6 K0 i
# p: d" c9 h$ Z" X2 @; J( Y7 U. k
, H+ j4 U! `+ ~: R' s& x" W9 s, `$ v0 \' Q) e8 y4 f: f1 j k
8.18: L. k& q+ U& w" z
main()3 B8 L" C' ~) Z2 G/ g
{8 }; b8 r, ]0 c4 @" O2 p
int year,month,day;
$ k, E6 p+ H/ l$ u- }- i" s! A; U/ ] int days;0 j2 O- I! {$ s- ^& ?2 _/ D
printf("\n 请输入日期(年,月,日)\n");/ j/ b d: Y1 ]
scanf("%d,%d,%d",&year,&month,&day);/ i6 @4 } \0 h+ |, `
printf("\n %d年%d月%d日",year,month,day);
, w" H% ]7 H" W days=sum_day(month,day);
4 _8 K0 v5 t$ b, v( G if(leap(year)&&month>=3), \/ C9 ?3 u. r3 t$ \
days=days+1;+ A' H/ B( q- {9 P
printf("是该年的%d天.\n",days);1 @! \% N* I9 \3 ~
}
5 q( i4 {$ ]8 R+ l, e% l; p static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
Q% Z6 p& ]% y$ n! B int(sum_day(month,day)
# J8 S v9 t, I* z int month,day;0 Y; i- c( f/ E1 Y8 O4 `% h: V
{7 X6 N4 q. W9 W* y0 S
int i;
# f. O* }: W7 ^8 c for(i=1;i day+=day_tab[i];
. i' R A3 `/ L8 [( u9 A return(day);, C' g: m- ^/ z2 ~/ ~) C9 g1 S
}" c& c) N: V0 a R
int leap(year)
. Q2 e; T& f0 B int year;
5 ~" @' w! B5 i" n+ o {7 v2 Q5 T. v/ q
int leap;
* s2 g/ M) [$ C6 q+ b leap=year%4==0&&year%100!=0||year%400==0;2 B2 w0 K+ J- o. n, R
return(leap);
. m" t8 p3 r, b: E' z }9 }& }. G- N; e K6 J& z
第九章 编译预处理5 q" k# d. e( F& @! Z
9.1
7 Q6 H3 t% T% j2 y/ W. ?; i#define SWAP(a,b) t=b;b=a;a=t+ b# }' X; v3 A0 ^
main()1 w; b0 F: T! J5 a/ E3 Y2 Q
{
. S7 P& L' z$ Y% i8 @int a,b,t;
$ Y) O$ a) M% _1 G7 Qprintf("请输入两个整数 a,b:");
$ V/ Q b% j* D$ d$ P' H# z; }9 Iscanf("%d,%d",&a,&b);
0 r- D( F- I( X+ N- DSWAP(a,b);
) J: a4 _: _0 _- `/ d& oprintf("交换结果为:a=%d,b=%d\n",a,b);; B5 T/ K) Z% G5 |7 Q! J8 K8 u
}
+ O/ [8 u2 M" e3 V4 i( } * E6 A% D, t z7 _
1 V- D' F7 `5 c, X3 R) Q9.2
% R3 P v& H1 \, a$ H" L* i# c4 E#define SURPLUS(a,b) ((a)%(b))
$ c: o1 a% T& [* A! ^main()6 }- y! i9 d2 q
{
4 Q4 i# ^+ s0 G" c5 r# @ int a,b;
$ J& P0 Q/ z$ I! w2 ^+ S printf(" 请输入两个整数 a,b:");5 c6 Y i- X! A6 @: X' [
scanf("%d,%d",&a,&b);
9 W) C: M$ t% Z3 f9 `printf("a,b相除的余数为:%d\n",SURPLUS(a,b));
; y! i, K# C V' N/ X- R }( z; C, q/ R- B: f. r
% t" ?% z& c; P7 J5 r o, j# v) B2 u" k( p X5 G3 ^' \! ]
9.3
3 y: }3 v3 e+ l" ^# @1 g#include
0 p3 `9 s) j0 |: v, v% t8 c! [5 M, F#defin S(a,b,c) ((a+b+c)/2)
: U, {4 A) s! f9 o9 q' u#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 q4 C! B& T7 ]- O. G3 d
c)))
! i' @5 n9 r ?) M4 F" e1 E& A% \main()
" b$ `/ ]9 ]; T: W& v* A9 W {
- p9 y3 x7 |8 D$ | float a,b,c;
$ Z1 c* ?5 y4 ]4 \. w( k printf("请输入三角形的三条边:");" p: r* s# C' S# A
scanf("%f,%f,%f",&a,&b,&c);) A& Q* n, G) o/ n& l
if(a+b>c && a+c>b && b+c>a)! Q; a. B+ [' J; B3 p, h6 x
printf("其面积为:%8.2f.\n",AREA(a,b,c));: W# w; ?. o9 C! K ^" s
else
+ M; C0 e( ], q9 a. k, z3 Z9 a8 _ printf("不能构成三角形!");
$ g1 J: q& i& l }
+ ]/ g7 }: H3 g/ w
: ~" k: ` M6 W' e' b, A" d; g
- v! T5 r9 x2 L: p# {8 b1 `
Y5 G8 g) x0 M& P7 V9.4 X, D& p7 R$ p2 F( C4 c; L6 T
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)
; e8 ~. w3 @5 U! ~, Rmain()
9 J1 ^) H: L5 N {
% \0 A0 R0 ]- ^7 M: d+ \ int year;
7 B- z4 M A$ m printf("\n请输入某一年:");
" `3 k; D6 @1 r3 j- ~; K" Z. G I* K scanf("%d",&year);
: Y# O9 m$ z" @) F, E7 H' H' u if(LEAP_YEAR(year))2 j6 g3 Y* _: b1 O' G4 e* Q
printf("%d 是闰年.\n",year);
9 V/ ^9 a/ Y7 M0 d* }2 A# _ else' R0 M+ ]4 F% x3 p( V$ p
printf("%d 不是闰年.\n",year);9 i" r& r# z9 g, L3 L5 |( J
}
|0 ~2 v+ u1 |7 `( ^. J& M# ]9 A G: A- b3 @
. O! J& v9 Z% i, `4 O
& i2 \5 }" f; |, n. b, ^
9.5解:展开后:
0 z5 r% g# }: u+ @printf("&#118alue=%format\t",x);3 V1 X6 Q' i+ j5 n1 j
printf("&#118alue=%format\t",x);putchar('\n');
& e( ?0 I7 z5 O6 Yprintf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');
. e5 M) u, m) J5 }9 F% q, o输出结果:
* B$ G6 O! S2 n7 K&#118alue=5.000000ormat &#118alue=5.000000ormat
8 r9 P u9 p. c8 S* l7 H9 n&#118alue=3.000000ormat &#118alue=8.000000ormat' i1 e3 Q+ U0 F' e
8 L1 z5 o) s- f: Q
) A3 B6 k% @! ^9.8
& S, B4 N" c [6 W/ pmain()
( ?3 ]0 q2 n, C* h0 \( k {* C2 n- ^0 h0 F2 _
int a,b,c;+ A' y& I1 ?/ t4 ^2 c' W
printf("请输入三个整数:");, i" P0 G7 [7 B; w8 M
scanf("%d,%d,%d",&a,&b,&c);5 v6 r% C+ q+ a( J
printf("三个之中最大值为:%d\n",max(a,b,c));
: c# \+ _( n" |, U+ N }
8 Y$ g3 V/ M! s6 y, d/ S2 ? max(x,y,z)
* Z0 a$ f6 k4 q) a int x,y,z;$ m3 n2 A6 R" J" V6 k% L
{
2 Y0 R5 U3 ^0 ~ int t;3 i* F3 _+ x+ j+ O6 w
t=(x>y? x:y);
1 P! k! C7 Y+ i( ~$ ?9 j/ d return(t>z? t:z);5 n8 R# R2 J0 S s. L3 x1 L$ l
}
8 W+ @% k6 X4 R' u1 {- w$ O9 d+ D& l& {8 P) b3 C! [8 n0 S
9 a" f: }* ?8 _. U2 p) M3 T7 R$ v' x& L7 Y
9.10
/ y0 ?, ^: M8 @% S#include
( _. h4 F2 l6 Q, o* `5 y#define MAX 80
! ], X! Z/ i- M- x# h/ V#define CHANGE 1
; H, C& g/ n+ X1 R& \main()" S3 `9 ?' {- E
{7 b$ f' p9 ^- v3 y3 U
char str[MAX];; ^: D4 l# X( Z) e+ p# b) J
int i;
: C' q* U6 M2 X+ K# U9 T2 c! N! ~ printf("请输入文本行:\n");
; I! f. E% n8 x @ scanf("%s",str);2 v% F2 N# f1 Z# l7 ?( d, ~
#if(CHANGE)- J- N1 G _/ J5 e) P+ B
{2 O Q1 _) N3 ], p
for (i=0;i {6 @7 ?8 t6 E. f7 c4 b# C
if(str[i]!='\0'
6 F6 h- R* _( \8 a' P7 S if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')6 e4 H% a7 R: u( B- P8 m1 m6 V
str[i]+=1;$ C5 E' d9 R+ {! a8 Q) e
else if(str[i]=='z' || str[i]=='Z')
* o' u V- Y8 p str[i]-=25;) ?" W; N7 }/ ~' h
}4 S9 `7 U7 ? Q6 c# T: a
}5 Z: O7 w3 d) L+ E
#endif
8 L, w f# c0 i V7 j" C# Vprintf("输出电码为:\n%s",str);/ p* G, v- Y. i/ y2 j" m5 F$ w
}4 }1 I3 w# f5 S0 r% P
第十章 指针+ P" X6 p* g, k. R
10.10 I7 y" i7 u+ l8 ?; |
main()
0 F, G. k2 H i{int n1,n2,n3;
2 }" K5 ]5 {1 C% h# H3 B int *p1,*p2,*p3;' Q3 A- o1 Y& a" r; i2 r( n1 x
scanf("%d,%d,%d",&n1,&n2,&n3);
. C r' N! t$ }! Y) l- }5 f p1=&n1;2 |6 Z& N P1 ]/ }* Z- B2 h! {: q
p2=&n2;
6 l6 a: y9 q6 y6 A5 F p3=&n3;0 E& @# |1 i% c. Q* i/ X
if(n1>n2)swap(p1,p2);& Y2 P' @* k9 w) o; y+ F
if(n1>n3)swap(p1,p3);
8 B8 `1 e8 D2 Q5 W" a if(n2>n3)swap(p2,p3);- Q$ b4 j. Q+ d& D/ ?0 P) w+ _/ I
printf("%d,%d,%d\n",n1,n2,n3);
) Y- c, Y/ f$ ]+ s% S( J}
; V2 Y* I X8 d1 pswap(p1,p2)1 L7 [0 _& N- z }7 H% \
int *p1,*p2;
; R9 s7 k4 S' e6 k/ ^{int p;
2 I8 U; ?, Y7 X p=*p1;*p1=*p2;*p2=p;
7 l6 a8 Z8 i x" \3 ~}
}! U# b% C% }10.2
3 y5 [5 ?& ?, h" R [main()
8 [% g( ^3 x: [# O t{char *str1[20],*str2[20],*str3[20];+ h3 K5 h% Z8 T
char swap();
6 w4 J1 E3 Y8 z G scanf("%s",str1);
. e' l! I; [( @3 G: q5 u4 ~ scanf("%s",str2);8 W1 ~* R5 m" x, o3 t, ^) |$ ]7 K
scanf("%s",str3);
; b1 j6 W1 f4 d; i, H) S if(strcmp(str1,str2)>0)swap(str1,str2);( D3 |/ N+ S S+ ~2 I
if(strcmp(str1,str3)>0)swap(str1,str3);
( C, f( r" P6 ^ P& ]' I if(strcmp(str2,str3)>0)swap(str2,str3);8 n9 f* u: x) P0 o
printf("%s\n%s\n%s\n",str1,str2,str3);4 L0 T8 [! ^! X) F$ R: J
}
7 @7 {+ a' u- D. b& w( x+ uchar swap(p1,p2)
& z7 O/ I/ b9 R* c# schar *p1,*p2;
5 h$ A- [4 n( g- ^: N( Y{char *p[20];$ H; E) L( B" j( Q- W
strcpy(p,p1);& B7 k+ L% x/ Q+ ]
strcpy(p1,p2);/ o9 r. M6 V `8 t
strcpy(p2,p);) V$ [: T8 }! m6 r3 F' Y/ W, O1 O: b
}
$ K. i) _% ^+ F. a7 v' v6 T2 t10.3
4 F7 B# c" F+ T& P; I7 jmain()- G0 X4 \/ |9 q7 x
{int number[10];" S6 \4 m% p* M/ P5 i4 O$ W# ~- X4 g6 K
input(number);
# S7 E1 h" w& [7 V0 u4 t" F X max_min_&#118alue(number);: Z# o# A; k" f' y, X
output(number);5 m8 w# I& k/ G3 T; o* x
}9 {9 f' A, {' p+ s8 r- e, @
input(number)
; R1 a3 @! b' ~! d7 ^/ X5 fint number[10];# n% M% J) j: G# Z/ D" W* S
{int i;
: E4 L0 l) D) Q# f) R for(i=0;i<10;i++)7 I: {8 _; e: v0 t8 g) b* f9 |- K
scanf("%d",&number[i]);+ _. y0 }) p' B; a
}
$ ?- D* X" n& z* ]- hmax_min_&#118alue(number)
) B- R# c3 C6 r+ Jint number[10];
+ A4 @# }$ K3 x# Z8 ^$ n{int *max,*min;% b* r1 ~3 @ g5 ~
int *p,*end;: z) C& R' k1 \2 W
end=number+10;
) ~+ q! N& m' w max=min=number;
4 R6 G' n% A! O) }9 {% i for(p=number+1;p if(*p>*max)max=p;
- w; F% {8 Y& ~6 e3 h else if(*p<*min)min=p;8 d- I; {8 c* d" }( S
*p=number[0];
' d0 }: O; N5 J! R; A' l number[0]=*min;( o* `, N1 V' s2 X7 t# N! `
*min=*p;
( A5 e' [6 ~( _# M0 L/ H *p=number[9];
7 O+ G8 H1 L7 A5 P, N" \ number[9]=*max;
3 f8 G @' n9 P/ F$ A: B2 `( z *max=*p;
; U9 Y$ L, p: \4 K+ \; @ return;1 g' `8 p" J. `0 P6 D4 c2 h) e
}
$ M8 x. n- n' s$ t0 e7 B/ T$ joutput(number)
" `% G9 A; L3 z6 X& Hint number[10];
4 v4 m3 v: d- Q5 Q' M1 d" S( |6 b{int *p;
/ p. m$ t* \8 D# ~3 i7 t for(p=number;p printf("%d,",*p);& c' S. A z# G; J- Q
printf("%d\n",*p);
; O+ |/ ^5 L5 i$ s$ o P- y}
& V- y; m2 v1 y* `10.4
% c1 y* J. m' {$ V3 X2 q' Qmain()
+ B& f% ]% M% _* W, @, { B# [{int number[20],n,m,i;4 Q7 |% b* l {
scanf("%d",&n);
) M1 z& k( x5 A) R0 f scanf("%d",&m);
! k. S' k% _* E) E2 Y for(i=0;i scanf("%d",&number[i]);; o( f8 x3 O; M
move(number,n,m);
: U, l* U! v: k0 q( v- k for(i=0;i printf("%8d",number[i]);: t& b/ L; n* \8 {
}
. Z' |7 M' f" y v8 Ymove(array,n,m)
7 J; x& F1 q, |( @/ O1 @int array[20],n,m;
# d3 D% h$ T& E{int *p,end;
0 d+ f; Y4 T9 ^. G end=*(array+n-1);
; c3 M6 C. i ^& \ for(p=array+n-1;p>array;p--)
a" M5 d$ b4 _+ l9 Q' B *p=*(p-1);
0 i. n) k* d8 ~5 J- H) N* k# v- h *array=end;: l9 r9 [. u+ J; E3 Q0 W
m--;4 E# ]- F: A/ p+ M2 T! |: x
if(m>0)move(array,n,m);( T& V; W9 \3 E; e
}
2 _% j; F) `! L0 F1 i8 n, p+ Z10.5
; n2 {$ G9 s' o! D5 K#define nmax 508 {1 ?$ R8 ?$ h( q6 t+ R2 V/ P5 ?
main()
j* L k. m+ A- d2 V{int i,k,m,n,num[nmax],*p;* r& Z! w1 [" _4 S, X. {, v
scanf("%d",&n);
1 y3 S4 _+ M5 X1 [. [ p=num;
+ |9 V8 u# e# E0 m1 \; S3 O+ }) ` for(i=0;i *(p+i)=i+1;
, r, Z( ]+ b8 f8 y i=k=m=0; u+ V3 s; @0 F+ b# ?0 g4 t9 `+ B
while(m {if(*(p+i)!=0)k++;
# Y( ~% p$ _& X+ r; n if(k==3)
8 C0 S. r0 N4 O/ o9 I1 L {*(p+i)=0;
. j2 \' \( [4 B) ^ k=0;% L7 _; ]3 O: s0 U$ B* |7 C* a4 k! G
m++;
+ y7 M3 y6 k$ j- ]* @ } ~) M7 _/ N) Q. D5 e2 j* T; T
i++;! z, i$ ]# b4 ]% h
if(i==n)i=0;+ @8 \1 p- b2 X
}& d! J) n2 h* x- Y# E! n" P
while(*p==0)p++;
% J$ t, }8 ?! V! k* h+ S; K printf("%d",*p);. |3 ~# \) p1 d0 i1 G3 s
}8 \/ I9 ^' R2 j W. H, ]7 E
10.66 P' w4 Y; E7 k6 A
main(). c4 D! Q7 p& D' K" V
{int len;
! f) ~+ N e5 b) ] char *str[20];
$ v! X/ ]/ B7 j j4 R I scanf("%s",str);- M& ?* O6 F% ?" m2 s& z
len=length(str);* D, G; B* o- O% ]0 o# V
printf("\nlen=%d\n",len);
, d7 J! A5 Y8 D}8 {' @# Y+ T% k7 } J; i
length(p)
9 S$ U' P6 w: y- i7 i+ L8 uchar *p;
: G- l& ?2 E/ n{int n=0;
) _0 l. g- Q2 t3 t9 O while(*p!='\0')/ E% `! b/ G2 b
{n++;p++;}
, U" o0 D- g% v/ T* e5 f return(n);
6 k& @1 c: L" H2 b}9 B# | _) L$ X$ b& w
10.70 h+ n* [; _9 h
main()7 O2 O Q( f* }4 G/ @" R2 j
{int m;& I3 r7 Q6 {9 x8 h3 L @ a0 k
char *str1[20],*str2[20];5 }: ]: W, E k3 u9 X5 h9 {
scanf("%s",str1);
8 g H+ X+ ?' v; y1 m: Z scanf("%d",&m);
+ f7 z1 p/ q+ d, ?$ W if(strlen(str1) printf("error");
. T w7 P1 V8 F4 Z else/ G' r) n* t) t9 T/ z1 U
{copystr(str1,str2,m);1 L8 @3 @: ?3 Y" `$ B5 `
printf("%s",str2);
* D. ?1 S7 y: t" K; o }. t% j9 k0 Y2 \6 ^& Z
}
x6 b0 o* s$ E2 Zcopystr(p1,p2,m)
0 x9 k1 ^" H6 v- r/ N2 schar *p1,*p2;
+ [! ]3 U% t+ C% w0 }int m;5 n$ L. ~% N+ W8 n
{int n=0;: C, y7 B3 \( ` P7 F
while(n {n++;p1++;}1 y% y6 J7 b3 ]1 n& G: |& \: `" z. ?
while(*p1!='\0')8 m+ A$ m: C; ?" z o" _) S9 ]
{*p2=*p1;
: F/ r$ n0 z$ V7 r" D9 s* _ p1++;8 N* \) N1 ~( k4 l- n, g8 [9 L
p2++;0 z4 P- L7 ^. Y+ g
}! S6 F- ]: |( k4 @
*p2='\0';( ?9 ?9 J0 ~+ ~# |
}
3 W/ P6 a. r- `% I w2 I1 n10.8
, b, b. \" h5 Z1 U% n#include"stdio.h"/ c8 y0 B5 H/ ?6 }( s
main()( _% C8 G$ G+ W! }8 [: n5 e
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
5 E- d6 o* X; H" Y1 e$ U' t char *p,s[20];
/ n0 _7 m# @; @$ ?5 H v for(i=0;i<20;i++)s[i]=0;, }" n7 e/ d* m( H4 \! v
i=0;% Q4 g& ]2 T7 U
while((s[i]=getchar())!='\n')i++;
. f( B4 l r- a1 H5 p H k. u p=s;. b2 a$ b' k9 n6 A i; m' ~
while(*p!='\n')* C0 N9 I- Z* M7 f
{if(*p>='a'&&*p<='z')
8 U3 g6 B C6 f# H ++sle;
9 g2 f' b* }# s7 l$ J( M else if(*p>='A'&&*p<='Z')
. q; o( N) x2 P# o, t: w& A5 ] ++cle;
' @9 m; D. e$ A2 K+ L3 ~1 R- e else if(*p==' '). o6 H) ^3 x; b3 \: i4 R. z
++wsp;9 M* k9 \ M; |/ S
else if(*p>='0'&&*p<='9')
, b5 O- p; `$ B8 |9 H$ D( g+ r ++di; r0 K4 @0 ]2 |' h* ]/ G6 S5 x
else
% X/ I4 q' O( a0 D# w' Y0 | ++ot;
( f" t3 R! d; R" x" m# e* q6 e( J p++;
: @' o" H9 f" _2 b }
7 R; A9 X, b' ] printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
$ T( s1 ]) d& X \5 }+ K4 g}% @- g! `8 _% h. T
10.9
) {" R; }! A$ O+ Xmain()/ R/ J) r7 a$ A+ Y- a. M
{int a[3][3],*p,i;4 z4 [! S' N. K8 z p9 K
for(i=0;i<3;i++)
# T/ D& g, `/ G" R0 }3 V' L* n scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);# D/ ~. K r2 p1 n& {3 y
p=a;8 [1 n) v* k# P5 M8 s6 H+ k
move(p);2 S/ I" z. I+ m- f W
for(i=0;i<3;i++)& c& K# c6 s: S1 O: L$ _& Z
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);' t0 Z2 |6 x4 X% Q. G" D, d: ?- I
}
. V1 A9 C* p0 D" D2 Z ]: fmove(pointer)
" v0 [- e' k$ b$ L8 Sint *pointer;+ X) m K9 h2 b1 X5 ^6 m: F
{int i,j,t;
' m; m/ A' b; \5 l' y2 z" q2 \5 \ for(i=0;i<2;i++)% [# r# w3 i* e5 U1 R$ B
for(j=i+1;j<3;j++)
% V+ W9 x. s& O; k* I5 i {t=*(pointer+3*i+j);
' D4 [; \7 K# l *(pointer+3*i+j)=*(pointer+3*j+i);: U |& i9 Y. @# ]) Z0 z
*(pointer+3*j+i)=t; p" Z9 d+ j3 E5 }3 B
} M( Q0 d% p9 x2 \: Q, S6 X- p
}: \9 E$ Y3 w6 r2 }" t) ]- s
10.10' n6 g; @' {) ?2 F
main()$ g' w) t$ K7 `
{int a[5][5],*p,i,j;5 r( Z* r4 x% P+ l/ G8 S. ^
for(i=0;i<5;i++)) [2 N$ L* m& S- J, V' X4 r
for(j=0;j<5;j++)
' b9 S+ O' v! \& r) ~4 a3 { f scanf("%d",&a[i][j]);
+ D0 R. ?9 x; G p=a;
6 ?7 G) R* \. f) Z0 K change(p);
: }: H! Y9 v; I( Q% R/ ?0 m for(i=0;i<5;i++)% h$ q% S: @3 B% B) p! {
{printf("\n");
0 j& ]/ {5 s: i- d( [& {$ s for(j=0;j<5;j++)
" e5 t t' q# k; T printf("%8d",a[i][j]);* K: _& J6 J9 Z- j4 M. T
}
4 C1 H# D: H' i% q! v6 e E}/ a* ~% Q* _* R' y7 t
change(p)
7 Q0 v l2 n- J* h9 q9 T) e% P7 U% U# Yint *p;6 ^8 `7 V1 W' W: D# ^
{int i,j,change;
& b" P+ o1 L. g4 X int *pmax,*pmin;$ j! J' ~. u; w
pmax=p;% k% O; E9 G- J* t: b
pmin=p;
& w/ {% }% o6 G: k0 | for(i=0;i<5;i++)
7 D2 @, ?/ M y2 \' e for(j=0;j<5;j++)7 s' L- n9 [+ b7 R
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;
# q" j$ G! V; q0 [. L6 M1 z; K* } if(*pmin>*(p+5*i+j))pmin=p+5*i+j;
$ x) Z, [5 E" h0 h3 b5 o } c5 t7 f6 ~! V
change=*(p+12);
6 ^- X. D2 P' Z, K *(p+12)=*pmax;# e R- Y% |9 i2 ?
*pmax=change;
2 u9 t7 G5 x! C$ e, r2 l0 l& B change=*p;
/ E% Q2 w& X/ a. ?8 f *p=*pmin;) I p: d8 _% H) n2 p5 Z; D
*pmin=change;7 f2 V$ o0 c5 f `4 ~2 \3 J- c
pmin=p+1;
# m* |( u# @4 s6 A! _ for(i=0;i<5;i++)
. V8 P- [. l% I) ^' g8 Z for(j=0;j<5;j++)+ B& ^8 j+ c2 N
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;5 _# S5 Z! B/ G0 b- M7 e' p
change=*(p+4);, |% I7 e/ I+ B& ?3 D
*(p+4)=*pmin;6 V; [+ N8 {- d6 B
*pmin=change;
) f/ M& U6 L- S/ T" } pmin=p+1;: C- i6 N. B: o; d9 T$ n
for(i=0;i<5;i++)6 F. S, \: b7 y, h, d- p' Q
for(j=0;j<5;j++); K/ c' L o0 H# k- l( y' }+ M
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
' D+ D$ x2 T, k- r* |" T! {$ @ pmin=p+5*i+j;
2 `2 C/ J9 _* v- k# g" r change=*(p+20);
1 A( e/ H Y @. P: g2 U2 w *(p+20)=*pmin;
( w C+ W2 H/ l" u; E1 d1 ] *pmin=change;- Z9 l9 j. y$ h% J4 v, u# o
pmin=p+1;# K2 q3 n! ~5 S- ]4 ?+ K
for(i=0;i<5;i++): S) P2 u6 x7 w( u, ~# {
for(j=0;j<5;j++)
|! r; l; Q0 ]- T; m1 a9 H if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
9 R2 i8 q D1 o$ ?$ I- A& J &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;2 t; h0 [0 j/ G
change=*(p+24);
3 G. T9 X# N% I" e7 Y$ X *(p+24)=*pmin;( k4 |) y' R& }. k
*pmin=change;
: ^1 Z/ U! h. b. Y, j}
1 v/ _1 V k" q( s3 @" Y6 M10.115 w6 I" z0 F4 `8 W+ F+ k3 s! o
main()
% r W, x$ P u7 [) `{int i;9 D H* f3 y2 h3 x6 V' a
char *p,str[10][10];! }5 g- A( ^! c& C ^
for(i=0;i<10;i++)- ~2 `7 I8 P, {
scanf("%s",str[i]);) s1 Z3 U& P, v2 T
p=str;' ^' }. P3 E* Q3 x8 D. @# Z. o
sort(p);
0 x3 M; i/ m# n2 M& N4 q for(i=0;i<10;i++), r" B" h0 _* C9 d G
printf("%s\n",str[i]);
" z1 [& P3 s+ N: H- y}7 y- _7 l& t% p; Z4 X
sort(p)
7 G* X3 i# H& bchar *p;
0 D7 u# b5 T: O: k1 S) @2 } i{int i,j;
6 P" w0 q# @4 u char s[10],*smax,*smin;3 D# U+ C# p ~. A+ e
for(i=0;i<10;i++): o0 G* {. |& L4 {' k D3 A
{smax=p+10*i;
+ d- \" ~0 w3 H! i, { for(j=i+1;j<10;j++)+ N5 {0 }2 Y/ ]3 Q: {
{smin=p+10*j;; f. N: A4 J* t7 g# G9 U6 i8 ]
if(strcmp(smax,smin)>0)
( h$ M+ F8 c7 X6 L1 ]! L; p+ a0 j V {strcpy(s,smin);1 y& i: Y0 p6 k- Y/ S
strcpy(smin,smax);
6 t: W, `: U1 @0 r0 b! i7 S strcpy(smax,s);$ p Z# a1 }( m
}' `/ Z0 R2 l4 g' D& Y
}
" V) G4 i! c' P* Y# Y6 @2 w }+ g: `8 S7 x# g
}
6 J: e! l$ I3 p2 w+ Z$ C10.12' ^; o' @* x5 D
#define MAX 201 F1 ?1 A0 k5 Z( x$ P+ h' e
main()
u+ T2 P5 A$ ]7 }) U5 `+ t{int i;
. N6 u3 U7 y* S0 a% n char *pstr[10],str[10][MAX];
0 t" N! c* b U! r for(i=0;i<10;i++)
8 L* q4 }: R) J pstr[i]=str[i];
! N1 U$ ], n7 a9 r' U& O5 a for(i=0;i<10;i++)9 f, W2 P1 [. B4 Y" m
scanf("%s",pstr[i]);
- X7 _! q( M1 |3 } sort(pstr);: o0 O2 C- ~" h/ d( u& k( H
for(i=0;i<10;i++)2 ]$ C: a- ^( H) X$ }( w& s2 y
printf("%s\n",pstr[i]);
2 h# K7 Y+ h* B. `& `( Z7 ~6 J; {}1 X- g, U# {3 Y8 \9 V. u
sort(pstr), G8 k: X- J+ s1 C4 B' p
char *pstr[10];# S; {% F0 C. c* @) b _2 {( y
{int i,j;$ ~* T' \% k- O! p
char *p;: U" g8 Z' @: r7 P1 S, x
for(i=0;i<10;i++)
1 ]& T+ g( G" \) O4 n* u* x# L' @$ W {for(j=i+1;j<10;j++)
: ~3 I; _% X4 {# k/ z0 P {if(strcmp(*(pstr+i),*(pstr+j))>0)
4 k5 ?* w' T* z" i# ]. Q0 O {p=*(pstr+i);: q V ?+ c; @; `
*(pstr+i)=*(pstr+j);
- ]/ c$ v7 F. @# M5 p *(pstr+j)=p;* R; s0 C6 j9 ~; _& O$ C! V
}9 F U" j6 A/ R0 i( M1 L0 X |2 `
}. u. s0 F/ Y3 t8 { }0 j. y
}! E4 s6 M4 S" H7 S' ?# X1 `
}( p% C& ]( H, {7 t4 p; E) }
10.13* O" d R. T* Y1 I8 G1 X, H" V
#include"math.h"6 d4 Z3 c" D( b7 F( \: S( ~6 o
main()
8 [; S9 V1 T2 P{int n=20;. f/ d$ G! j$ U4 t+ d
float a,b,a1,b1,a2,b2,c,(*p)(),jiff();/ f# q4 \3 Z2 x& g0 c) V) e% P$ S
scanf("%f,%f",&a,&b);& E4 D6 Q$ h2 _+ c. f. h
scanf("%f,%f",&a1,&b1);
7 P* y @% @& j0 R: C scanf("%f,%f",&a2,&b2);' h$ ^* m$ d z+ u% V2 b& P& h
p=sin;
. @. A( B5 b. R c=jiff(a,b,n,p);
. x' Q$ |$ {/ I. ? printf("sin=%f\n",c);7 c! l6 K' I& Q: M: J
p=cos;# ]: N v# S9 f/ J" N; N, T
c=jiff(a1,b1,n,p);
7 s! b3 @& Q: M. X printf("cos=%f\n",c);; Z( o7 R4 r8 ]* T% n
p=exp;
1 W+ @8 ~! C/ D c=jiff(a2,b2,n,p);0 G9 c( W# U8 e7 R: b
printf("exp=%f\n",c);
. l+ W' f7 B7 L `6 K}
0 ?1 Z! y- [" e0 H. l; N+ Q$ Hfloat jiff(a,b,n,p)
- m$ A% f' Z" z! Kfloat a,b,(*p)();
7 d# O3 U+ f) i& k& hint n;
$ h# F3 L$ C ^ M) N2 L{int i;
" _% k9 Q' ^' |# j- n float x,f,h,area;
* Z, N) S1 Z7 ?1 p ?# A h=(b-a)/n;
: e+ q. Y! U4 U/ s3 F$ y x=a;
2 Z. Y; K1 _4 ^& Q# x1 j: q! D: | area=0;
5 g- Y! Q. O! A' L. V for(i=1;i<=n;i++)! _- i' \2 { f, G* r
{x=x+h;
# ~: `) }- j/ J1 }3 g area=area+(*p)(x)*h;
4 U) ~4 Q' m) ]: I }. P1 n v% X% \: r
return(area);
& o0 J. h2 R* a9 @- R$ W# L5 R* T}( i; P, R3 B& V! P
10.14/ |4 a+ N- `9 R; \" r8 t7 }# B
main()4 W5 h4 Y8 i6 Q0 {& Y" P) T
{int i,n,num[20];( R* n0 Y y- e; U* S. I
char *p;2 E' {% O& k$ R( P, u* ~
scanf("%d",&n);- O- f6 O( X; z$ u% }( V# l
for(i=0;i scanf("%d",&num[i]);" z* U9 W P& N) ?2 K
p=num;
9 }/ B+ G4 D8 `# C. _ sort(p,n);: a' d+ }: G: R( Z! ~! K4 z; n
for(i=0;i printf("%8d",num[i]);$ F. `* V) h0 f
}% ^2 `/ P8 [$ y; m4 O @
sort(p,m)
. K7 o6 O9 q3 E6 \- v2 kchar *p;
3 c2 [! {) w# N4 b2 Oint m;
" h7 N( A; U- {& _{int i;: w# V Q' P5 {; c' l& M5 i
char change,*p1,*p2;& i. o" \5 j3 B3 m0 b' d
for(i=0;i {p1=p+i;( t% U+ E1 \- J
p2=p+(m-1-i);- q) K! q+ T5 a2 Y1 V; O) w1 K3 D
change=*p1;
: s& y9 L% {6 Q0 T, S- _1 f4 p *p1=*p2;
2 q& P* I* d9 W V$ h4 |1 b *p2=change;0 G. `! r9 |* u, }9 B; Y% r) b
}
4 \+ d/ g. r% b% m7 ?: s- J0 T}( p& v8 _' }5 @* E6 w% `4 z
10.15
3 N2 `" O& P) _. K+ Umain()
1 }- B7 u8 @! z8 Y# |$ e{int i,j,*pnum,num[4];/ X s# z, n" Z
float score[4][5],aver[4],*psco,*pave;; _5 x; b: B* t) S6 k5 S7 L
char course[5][10],*pcou;
4 R$ t- l1 c* h2 S% a+ m pcou=course[0];* R& U, ]" j0 C6 V
for(i=0;i<5;i++)
' ^( M' B# i9 [, ` P! r& U% q scanf("%s",pcou+10*i);
) g0 p% L' o" F4 a$ { printf("number");
+ ^% i4 Z9 D0 c. s4 w0 u5 t for(i=0;i<5;i++)$ L8 e; j0 p1 G4 {" V
printf(",%s",pcou+10*i);
1 v! @! ?0 J$ E4 f% z0 w printf("\n");0 W' H" g7 i) b- m4 L
psco=score;1 m. ^9 y# v" X6 {6 ^
pnum=num;, K* E5 _4 _" O- D9 O9 C
for(i=0;i<4;i++)7 v( N9 f1 ]! K: A$ R3 f( D; c( x
{scanf("%d",pnum+i);
$ I+ ?, `# x- S% r% Z4 X' D for(j=0;j<5;j++)
5 u, @" O( R, B$ c! { scanf(",%f",psco+5*i+j);
! ~* R( q! P) c# x+ q0 z- B. s }
3 ]9 p3 x0 x. W: E pave=aver;
& F: }6 a6 N6 ~4 F4 _& H9 s printf("\n");
7 U7 J. r$ g# Q9 E9 V9 C' E avsco(psco,pave);! `( r% R3 T" }0 U
avcour1(pcou,psco);5 R( y$ X; I- v0 S
printf("\n");; ]$ y* @7 S& I5 c& }; W0 V1 W
fali2(pcou,pnum,psco,pave);
1 r8 M: V$ Q! x" |. U2 G6 e. c5 B printf("\n"); N; X8 ^3 k [. j5 R) s
good(pcou,pnum,psco,pave);
; X7 z# [8 G$ c# P}! S# r) z2 o1 h, f
avsco(psco,pave)
8 z; g/ u8 L- V0 P- Ifloat *psco,*pave;6 A$ ~) i, S! Z, J" E
{int i,j;: X8 l/ R# G' j
float sum,average;. b6 N6 E! C6 x1 [4 V$ i
for(i=0;i<4;i++)
& d- f& I* q# s+ K, T4 U {sum=0;. T/ Y. n0 s6 z% ]- ~
for(j=0;j<5;j+)& c5 m3 @4 X9 W8 G
sum+=(*(psco+5*i+j));
! J0 {) P1 t1 u z- d average=sum/5;
0 }4 s- h) O. j! M8 S *(pave+i)=average;
2 f8 {, S& H9 Z. h }
s" P; k) {3 ]! `5 y' `}
c1 S, O. j5 P8 s! }& Yavcour1(pcou,psco)
: H. j# b8 ^4 _+ D0 J7 Tchar *pcou;. s! r7 x( }- b- e% K' b X
float *psco;8 O+ b( K8 m: D8 f" v" I
{int i;4 N) u9 P' b H, m
float sum,average1;
$ `9 p' W% Y$ p( M/ f sum=0;3 D3 o, U5 \8 m7 b
for(i=0;i<4;i++)
! i4 U+ w8 ?0 ?. K) H$ V% n sum+=(*(psco+5*i))
, s) H" i3 ]- ]6 t: r5 M# S average1=sum/4;
p2 G) z% w9 ~ G/ T! D printf("%s %5.2f\n",pcou,average1);) y) E F2 B4 V3 x7 p
}
5 w! f$ L4 n8 ]8 F; `0 [5 @& Xfali2(pcou,pnum,psco,pave)5 s1 {- s& Y5 O! q3 W B: a l- v
char *pcou;
6 U- u* y+ W" h& d% x4 V% Iint *pnum;
9 } t2 m/ H( e7 tfloat *psco,*pave;; p" b: Z9 ?% _ n! ]3 R
{int i,j,k,label;2 a& G3 F/ E$ R4 W- V e
printf("\nnumber\n");
& G2 ^/ L, X( u4 V1 J5 _6 | for(i=0;i<5;i++)
, N( n q; I& V5 p+ h2 u8 P7 ^ printf("%-8s",pcou+10*i);$ R2 w( I8 J0 D9 `# a
printf("\naverage\n");
/ J1 b8 |. E4 K for(i=0;i<4;i++). m6 [9 Y# J7 L, a
{label=0;
) Q9 v, Z* W- H s4 x: @ for(j=0;j<5;j++)
# |) _' w, Y. J7 U, F3 ?; w5 H& P/ k if(*(psco+5*i+j)<60.0)label++;7 M; n- U# S) x5 v) [
if(label>=2)- i" [1 I' ~* N1 \
{printf("%-8d",*(pnum+i));
3 |3 g0 G7 j$ n7 h) U7 f, S for(k=0;k<5;k++)& t2 {2 s- }3 G2 c8 B3 r
printf("%-8.2f",*(psco+5*i+k));2 \& y8 P$ C! P7 p
printf("%-8.2f",*(pave+i));
8 u* Z; ~/ R8 F5 w }
) a9 F! P& Q) ^/ O$ H! k0 m }( ?, Y9 }+ w6 b
}
+ r" F, e e/ X. sgood(pcou,pnum,psco,pave)
8 o7 G- }) s; Schar *pcou;$ M: G4 x, B: ]# R7 Y$ H
int *pnum;
5 D( |9 c% W+ b" c+ J6 Zfloat *psco,*pave;! x( T2 d" a+ l7 L3 R) D, `
{int i,j,k,label;# Y* p( T* F- ^0 w. E' y
printf("number");* Z' ~ Z' `% d' }6 q: {. _, H) h" R: E
for(i=0;i<5;i++)
4 A( m. m' `( B" U+ g& ~& m printf("%-8s",pcou+10*i);; e$ L% L, _, i; o9 L
printf("average");
?9 [; o/ r+ t4 l: i& I8 N for(i=0;i<4;i++)
3 Y# D5 x, _! m {label=0;
3 H' V- O5 k4 q+ [- r for(j=0;j<5;j++)
: [8 E5 L: n4 g" F2 W1 A if(*(psco+5*i+j)>=85.0)label++;2 k: U# q1 K% V1 v! |3 h
if((label>=5)||(*(pave+i)>=90))
9 D; X, V' x4 k# F+ @6 C4 ] {printf("%-8d",*(pnum+i));8 F! p0 N# a9 X; J
for(k=0;k<5;k++)
/ U" e2 m& h1 H printf("%-8.2f",*(psco+5*i+k));8 V" L5 n |* ?" ^( o* V3 Z
printf("%-8.2f",*(pave+i));+ o" N8 l' t6 z# Q9 p) N0 T' w
}
T2 m# f% }9 ?4 F' I/ H' U' I" [ } F" f5 S/ C. v& z4 t
}% `( q# \+ k* Y7 u
10.16
+ _1 z! B' x9 Y#include"stdio.h"+ E& |* s' z- U. p D* H& W. P" C
main()$ @( e& E. F% ], }% ?8 W2 \# B
{char str[50],*pstr;. E9 Y' w+ j4 {) P3 l; \6 b- D) G; o
int i,j,k,m,e10,digit,ndigit,a[10],*pa;0 f8 r/ M. f0 G9 h5 g- L7 J4 ]
gets(str); E! ?/ \4 K- Y0 w9 }: {
pstr=str;
, a) X! A3 f: v2 w+ W4 X) {, ]8 A& A pa=a;( f. `( }9 o0 [
ndigit=0;
! G% E- p% J+ }" @0 H% a i=j=0;9 x9 s% h9 X: {5 i- a& z1 a! J
while(*(pstr+i)!='\0')
6 W, Z/ [3 i6 a {if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))
( ^( E/ s( V3 i5 ] j++;
: F% `; Z- X* E; { n else) H, X/ Z3 R* s# S8 {2 r3 D; n: f
{if(j>0)
, T% z3 T2 X1 r% H; ^& M {digit=*(pstr+i-1)-48;: t" U, w; c/ m
k=1;
# I7 {1 X2 V, p/ l8 A' z while(k {e10=1;
! z( n' ]' d& ^$ w( T for(m=1;m<=k;m++)
% e, f6 o, R2 n# a e10=e10*10;
# h/ s3 x3 @/ A$ f. X) ? digit+=(*(pstr+i-1-k)-48)*e10;6 Q! R7 H# f# B
k++;
( n/ n: o1 Q6 }" z/ w! S- t }
, D7 J g2 v9 t5 t *pa=digit;
) A9 v* F" y) ]* Y m$ h ndigit++;
9 Z1 G6 c# G4 P pa++;9 z7 f0 ^( @) s4 _9 j
j=0;8 \5 e: {8 p, D6 q
}
@- [% R' n$ p9 t! P# u( V }6 A {7 h2 K" U) F: ~
i++;, M) m3 M1 f1 N" c
} ^: U: ]2 L9 T& c/ r& Q& \" _
if(j>0)
: h/ i. F4 c6 X2 \ {digit=*(pstr+i-1)-48;/ @2 A L8 R0 p: V( q: P7 I; W
k=1;
. B# @1 A1 F+ H C$ [* D- o while(k {e10=1;
4 d$ n& ?" E+ \) Y for(m=1;m<=k;m++)
0 V5 ~/ h3 b" V }* k. o# D( I- j e10=e10*10;
+ K* \7 c- O, U s$ W' [; k4 W digit+=(*(pstr+i-1-k)-48)*e10;8 {$ m- {* K; {6 `, B$ h: h
k++;$ }* B# ?: o0 B$ g7 K+ c$ e: J
}- p0 J0 z& k" E% m- A- x2 t
*pa=digit;
( A6 I1 H% v+ h0 d* R; V ndigit++;8 ~! Q, Q$ s1 r* U
j=0; {) T7 Z* e2 G# u
}
5 r9 t4 H+ b+ r9 E$ a* U printf("ndigit=%d\n",ndigit);
) x y& b0 f9 s, Z7 Q0 } j=0;7 `: p/ k2 U; Y' v
pa=a;
$ g, c3 h+ T) @' z$ u. u# J2 M for(j=0;j printf("%d",*(pa+j));
9 z9 T8 @) A2 `" i2 t* O* p5 a}
+ c! V5 u7 A3 B9 X10.17; `: I( W4 D& [ u @
main()1 i" C) y4 x7 z' u
{int m;
3 K5 m' V: @1 D0 w; N char str1[20],str2[20],*p1,*p2;, v+ n) J5 h% f, }" J
scanf("%s",str1);
6 |5 Z, T/ M; x- z8 H2 \6 X scanf("%s",str2);, F. |0 B. y5 s; g1 O9 j
p1=str1;( u6 O$ O# D' J; u
p2=str2;
6 [1 p4 ^$ v0 ?" C6 [+ x- w) N# Z0 } m=strcmp(p1,p2); ?1 C' c2 s9 `; X! s1 \$ l$ b
printf("%d\n",m); [# b W$ t9 ~- l1 M9 K' w* Y- O
}
, Q" H4 c' `- I% o4 jstrcmp(p1,p2)
& v7 N+ V+ v! G% hchar *p1,*p2;, ?: U& i* Q% b$ _- e# _ j
{int i=0;
0 ]7 ?( h `9 ^; A' ^+ F) |$ C while(*(p1+i)==*(p2+i))! [/ }+ _8 O8 I
if(*(p+i++)=='\0')return(0);
' Y* k! Y! M1 H return(*(p1+i)-*(p2+i));3 M j9 [" [: i4 X" e
}) S+ e5 n) z4 r
10.18
+ L/ H& k& d+ q5 G5 H3 Y9 x" C3 wmain()
5 ?5 z% P7 i/ ]0 G+ ~1 K( s% l{static char *mname[13]={"illeagl","January","February","March",
- C" B& Q6 @9 j! q8 g, N "April","May","June","July","August","September","October",/ f# I2 N! g- X* q
"November","December"};. n& M& f4 h; s) n8 L* M" P$ k `
int n;0 a7 t; x5 h9 h! s* C# g6 _9 m; ^
scanf("%d",&n);
2 U, T+ R- [ n9 Q0 P- R if((n>=1)&&(n<=12))
+ ]. w) ^# d) ~ printf("%s\n",*(mname+n));
, H6 Z) `( |! G, G' ^8 V; B else
4 L) r) \ U% h7 B, ] printf("error");
- _3 X6 i' ] s) c5 G) z}
: B/ j+ K( O2 c2 L$ b' ~10.203 W7 {. d6 K' Y
main()- ]" o9 P; U" u. p* s* J/ R
{int i;# n% w- ~. t1 [# C" I6 n6 D I; Q
char **p,*pstr[5],str[5][10];2 W! U1 x8 i: s
for(i=0;i<5;i++)
. s# T2 m0 r; E pstr[i]=str[i];2 P( w, C% x9 }1 B: F f; [
for(i=0;i<5;i++)6 m0 G* y- I/ L! n$ M, C& ]
scanf("%s",pstr[i]);
4 \, W2 @5 j5 E1 F/ C p=pstr;
+ T) B' P, L* A: x5 X sort(p);
( @9 W9 V7 N* ~4 q3 N for(i=0;i<5;i++)
3 \) r/ E- r# f* Q' F, d4 \ printf("%s\n",pstr[i]);9 s9 G. ^6 M: K6 r/ B
}+ y6 T% T* `1 R0 M; w
sort(p)
0 v: R+ T. ?9 k! `0 Rchar **P;
/ C {: k% Z" n* L& d* S{int i,j;. x: Z W" L. d4 U; Y
char *pchange;
, h Q1 l# L3 |& r' H for(i=0;i<5;i++)
4 m C( g; Q( O$ K4 h {for(j=i+1;j<5;j++)" \1 u0 S4 Z& s& B+ w) F- v
{if(strcmp(*(p+i),*(p+j))>0)
2 C/ F9 M4 [9 O9 B; i9 e$ { {pchange=*(p+i);0 z/ v0 l* Y4 B1 \, T
*(p+i)=*(p+j);
# u3 b7 R o9 H6 }7 A! H( B *(p+j)=pchange;% X* X$ N+ l y# [
}: B" N0 J. p4 F+ l- z, K
}( A; \: S$ [+ y7 c: L' \
}
/ Z: V' @8 c! \) R! x7 D" W}' t0 S- l( Q. ^$ L2 p: W) V3 s
10.21
, M5 Q& W7 `7 {9 hmain()5 R! Y/ E# s: ?8 K
{int i,n,digit[20],**p,*pstr[20];
8 _5 u( `9 V7 R' G7 n scanf("%d",&n);# L6 y1 v i$ o& r' v2 R4 N
for(i=0;i pstr[i]=&digit[i];
3 k4 B) k/ t* e3 C9 J: H for(i=0;i scanf("%d",pstr[i]);
/ \6 s( M$ K% B: {* c a p=pstr;
4 O" n N$ p! c) a sort(p,n);' y4 _8 K5 p! Z0 q& D3 N! D- _
for(i=0;i printf("%d ",*pstr[i]);
Z* i- b* v8 z( w}
# f% Y3 ]' a5 W% k0 y6 Ysort(p,n)+ R$ S# y Z' V! V1 b2 X# U- T
int **p,n;3 `" k* Z. W g+ X
{int i,j,*pchange;
T6 c% W) r7 @9 I* t0 ] for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))) w; c; U _, e
{pchange=*(p+i);
3 }7 g8 L+ q% D& o* D8 L *(p+i)=*(p+j);2 J! I M7 \# S5 q. }' H3 z
*(p+j)=pchange;3 y9 R% N& |2 t, B8 b
}2 `. S& l% f: t8 u: B$ H
}: L. R* Z) r3 j# b' `# [5 u
}
+ {& p8 `! K1 s% p- I# |}/ }; w7 ^8 H6 h7 w
第十一章 结构体与共用体
: y5 D& h1 Z' V8 F11.1+ w% y2 |6 x9 h5 u* f
struct
. ]" M( \ y( y' E) G' K u: f) ?' a4 n {int year;; }; r) k8 \7 F# [$ m
int month;
1 k) @& B& A2 F5 ]7 F% g int day;
7 \* U* [- |7 u$ D' s }date;# C! q8 Z- i5 j7 A. ?
main()3 m8 P# W" K! s/ D" v$ o4 k* H2 Q2 w
{int days;
- p6 A8 @+ N. D5 Z" @" A% B7 e scanf("%d,%d,%d",&date.year,&date.month,&date.day);' a" X2 P, f8 o l' q/ l
switch(date.month)
$ k8 ]) Z' w7 S( |0 `1 g' p6 Y- F {case 1:days=date.day;break;
. a$ z: v7 s8 w0 ]: }$ S" Z case 2:days=date.day+31;break; K' [* x7 i+ k# F) c( j( l
case 3:days=date.day+59;break;
' @ o7 e0 e' `( M- s case 4:days=date.day+90;break;
z& v. x! ]4 A- ?$ { case 5:days=date.day+120;break;
& Y+ W$ d6 N8 W. b$ u3 h3 M case 6:days=date.day+151;break;* m/ M" z% y# H- [0 k
case 7:days=date.day+181;break;9 q3 w3 Q. x6 b- v
case 8:days=date.day+212;break;
# | D3 f- C) q9 h& g! w case 9:days=date.day+243;break;* T3 A. ~3 u( c% c
case 10:days=date.day+273;break;5 r0 C5 ]: J4 y
case 11:days=date.day+304;break;
f, n! w k* D! x case 12:days=date.day+334;break;0 n K9 S2 v0 ^" t
}% Z0 v% A; a( N8 y
if((date.year%4==0&&date.year%100!=0||date.year%400==0)
7 Z" \5 S( J" B( Y5 q- K &&date.month>=3)5 [: t% R# c2 t) N
days+=1;3 o; U2 S5 X v5 m* E# ], f) L7 o
printf("days=%d\n",days);
+ F2 E9 p. @% e9 O" C6 N}
% @& h2 N X! N7 d: D6 b3 b$ a11.2
4 L( M& X) J: J0 l$ \# C8 n* tstruct dt/ @9 L/ t! w* t; w
{int year;
0 N' V9 ?4 N; t7 w0 J; `/ I% i int month;0 A1 f/ r6 j! y) _" A
int day;
! n' O# j4 t7 S* Q: k6 Y8 r* E }date;
# R; n& t/ M% i( D9 H8 [) T3 Xmain(), C' v7 j$ `7 y' C2 x9 G
{
q7 y, \" ^+ j& l" A9 Y1 l0 l scanf("%d,%d,%d",&date.year,&date.month,&date.day);3 A% g$ \0 S& K2 u7 k: E
printf("\n%d\n",days(date.year,date.month,date.day));
+ u! w$ R) h7 I. K}4 W( \0 Q9 \# A7 B
days(year,month,day)
, }9 |( j( T2 Pint year,month,day;
+ Y0 P1 s2 P# M: U2 D{int daysum=0,i;% ~ Q: v9 Q) z8 o8 S1 ^# n7 i
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}$ @; N8 C5 _* v& v
for(i=1;i daysum+=daytab[i];; k8 @0 q& K' d0 i" V
daysum+=day;
! @8 ^ o; V' o1 r if((year%4==0&&year%100!=0||year%400==0)&&month>=3)# j( K. j w0 }
daysum+=1;% l+ p, ~' \* H9 c0 k
return(daysum);+ `. S5 Q0 P+ ^8 i5 P7 B
}8 X6 l0 Y( y( A& U
11.3
( ^" }. ~. r% y( t8 k11.4. k* F4 H6 I+ ^& Y
#define N 5
. f+ N j; F) Jstruct student
) P; C" f6 D8 u" `" h" J {char num[6];
' p/ N! p; n8 i$ D# ?- g* e) s9 [3 a char name[8];3 Y/ M; O7 }3 {& { E7 K$ a% i
int score[4];
! d, m7 j1 r% u }stu[N]; X Q# y$ a4 v0 w
main(); a- d1 t5 {4 K/ g" }
{' A3 c# k- O1 N" j' c
input(stu);( W* b2 M' ^( s" @( x6 W- r
print(stu);* g( t9 L1 B J- R. j$ A5 r9 w, ^
}2 U' ]9 i( z) k% w7 S, A0 u2 L9 H
input(stu)
% v$ I2 g7 o# \- q4 xstruct student stu[];1 s* g3 p0 D* [5 A+ S
{int i,j;* X v, `% H) n. H8 @. p( Q
for(i=0;i {printf("number");
# o9 O& V9 o: Q( p$ z scanf("%s",stu[i].num);
: u8 z2 B0 T# h; [6 n printf("name");
4 @; X. D- U' L$ U scanf("%s",stu[i].name);
$ g. T" G" T3 Y* p& l7 T8 `- I5 r% x( _ for(j=0;j<3;j++)7 x6 M* B- ^ M( H
{printf("\nscore\n");
7 U) r3 y. r$ `3 F9 q scanf("%d",&stu[i].score[j]);
4 Z P) e- D( f7 i2 }- ]: n }: \" x! j" r2 ^# g) I
printf("\n");
5 |) P" {; o, x3 I }3 C' L4 ]. E" p+ ~0 z8 n2 l6 w
}
' v. R, e0 ^0 x4 i Nprint(stu)
* x$ i; P* ^* D$ k6 f) xstruct student stu[];
+ m* y% \* |( K& g{int i,j;* M. \$ P# H4 \& |1 G* P0 i
printf("\nnumber name score1 score2 score3 \n");
* B$ b! U. A4 ?1 ` for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
4 h4 {6 K& A1 `5 M3 f0 K3 C% K* o for(j=0;j<3;j++)8 M' W( u3 T% Z8 Z
printf("%7d",stu[i].score[j]);& i" V+ {4 P2 P. U( }
printf("\n");) `" |6 I7 \2 Q3 B4 L9 U" q+ C$ y
}! `- R" n1 w$ T( Q/ ^# {4 ?- O# B3 r
}
$ r$ [& [+ e& w6 ^4 m" J7 i11.5$ m8 {4 m! T8 ]7 |- ?
struct student
& t) B2 ?0 c9 {9 ? {char num[6];
( F" n4 F" V9 j0 X% b, I char name[8];
) [2 j3 i1 u6 r) ~5 W int score[4];
) I" e4 C( D8 g7 c float avr;* _& V) C5 A( `
}stu[5];' F$ p# w# Q4 T a/ H. K
main(); j) L8 o' Z- @9 _
{int i,j,max,maxi,sum;
% H7 D/ z' p9 e% Y- }; V4 u& C float average;8 z. T+ `- z( D+ l8 O
for(i=0;i<5;i++)& U5 M6 B; U. ?+ A
{printf("number");
6 e6 S* w# i/ S* ^ scanf("%s",stu[i].num); \* M: m5 Y. F# F' f9 K
printf("name");$ i G0 ^& k+ q c
scanf("%s",stu[i].name);- K- e B+ {5 C0 J
for(j=0;j<3;j++)
( R' s: U( ^5 f& z {printf("\nscore\n");
0 ?& U- |* q3 X* \+ X3 j" k scanf("%d",&stu[i].score[j]);$ [$ G8 n. Y9 D" w$ c" }
}5 p3 g8 J4 n$ D+ a+ @/ N
}# |% K J0 y( a8 u5 m/ H
average=0;# h+ O2 T5 @" _! C
max=0;
, \4 s, K7 H" j2 q maxi=0;+ S: E: A- I$ K
for(i=0;i<5;i++)
& c5 F4 s0 G/ x, }5 W8 h; Z {sum=0;
/ l4 O1 O- D% M! [; p for(j=0;j<3;j++)5 B* v( ] K- \
sum+=stu[i].score[j];3 m' p5 P# _; X+ X5 \/ ^ _$ M3 W
stu[i].avr=sum/3.0;, f+ Q0 v# S+ e Z* q+ N
average+=stu[i].avr;
. ~4 E7 G9 C c# s C& B4 ^* Q* Q if(sum>max); W3 K# P5 l! S$ ]- e
{max=sum;
3 t6 d$ W$ R) l/ C maxi=i;4 K1 T/ I$ y1 y, g6 q
} q: Z2 i$ j/ \
}3 S B! R7 T) j1 o
average/=5;! E0 K! I$ p# z3 S% O" `
printf("number name score1 score2 score3 average\n");7 a" J, S/ v6 C! Z& H' ^/ f" R
for(i=0;i<5;i++)
- G' R: ^. S9 X! }- s% ^ I {printf("%8s%10s",stu[i].num,stu[i].name);
6 h7 I, t! x* v8 ^+ i# ^ for(j=0;j<3;j++)
8 [& J, G, M1 u0 U; ~! a7 q1 b% x printf("%7d",stu[i].score[j]);' _; h: ?7 P2 @
printf("%6.2f\n",stu[i].avr);
* O g9 p4 p7 J5 |5 L$ Z }
& R8 u8 a8 F2 @8 Y0 m; d printf("average=%5.2f\n",average);
% I9 f$ F. ?. B1 R s printf("The best student is %s,sum=%d\n",stu[maxi].name,max);- e. o: t0 |/ C: H
}
: y! L$ [1 i6 I- Z# g( h9 V) {
: i' s# v6 `( r+ A5 [! p/ T: ^) Q x |
zan
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