- 在线时间
- 28 小时
- 最后登录
- 2016-10-15
- 注册时间
- 2010-2-26
- 听众数
- 5
- 收听数
- 0
- 能力
- 0 分
- 体力
- 2905 点
- 威望
- 0 点
- 阅读权限
- 50
- 积分
- 912
- 相册
- 3
- 日志
- 11
- 记录
- 13
- 帖子
- 56
- 主题
- 5
- 精华
- 0
- 分享
- 27
- 好友
- 16
升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
|---|
签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案
8 b8 C( H: r1 X% A& l% n& M2 Z& G7 y1.5请参照本章例题,编写一个C程序,输出以下信息:8 j" H8 @. c& Z1 [- @& F$ {$ {
main()0 O) w# Y5 z: C1 ^; A$ j
{
5 n! ]; \' b. N e4 E7 A* T6 z" F7 \! Qprintf(" ************ \n");
4 C3 `* ~) O- P' j" e! }printf("\n");* t1 Q, o( x8 c0 F1 L# e, h" a
printf(" Very Good! \n");
' h$ L* P! g+ T" r0 Z+ E) gprintf("\n");
0 C( R: l& D$ f; q+ \" k( aprintf(" ************\n");
9 w) a' Q. R8 }}
' S) m, [# s$ O2 r1.6编写一个程序,输入a b c三个值,输出其中最大者。
" h* }, v6 A. D5 ]/ g解:main()
: g) E4 L6 y; |; X9 y{int a,b,c,max;; G4 p0 i, O( t! @
printf("请输入三个数a,b,c:\n");
5 e" ~/ ~; q/ ^* d$ V4 s* t9 N7 n% pscanf("%d,%d,%d",&a,&b,&c);
7 \ F, w& K$ Hmax=a;
! Q2 Z5 l4 {( z$ n- _if(maxmax=b;2 V* Y3 K) t1 o2 e
if(maxmax=c;
* h* k' `8 U' R8 o" J3 j+ b# z+ Oprintf("最大数为:%d",max);
% a. c7 {. o% h1 I5 P3 ^}* l# D$ s! g# u3 H1 m
第三章+ t1 w$ m; D7 T3 J
3.3 请将下面各数用八进制数和十六进制数表示:
" q J2 ^1 B* Z- m6 O# V(1)10 (2)32 (3)75 (4)-617
0 `, \2 G! ?$ m5 b6 Z5 Z& Z3 J(5)-111 (6)2483 (7)-28654 (8)21003 w3 y6 ~; {; x% ?" Q8 D
解:十 八 十六
/ y1 w1 k6 `: l( I% D( ~9 v+ f (10)=(12)=(a)* }/ W% ?( ~0 a5 ~: Z1 P
(32)=(40)=20
9 q+ [ f/ Y( H" H _3 g (75)=(113)=4b8 n" ^# @, I/ L
(-617)=(176627)=fd97% e. @! z3 K- ~5 w6 t- W- E5 q2 B
-111=177621=ff911 T- K9 J. }2 q4 v4 W/ Z& t
2483=4663=963& U5 d3 x# B# g0 b; V3 a
-28654=110022=9012
9 X1 [( Q0 q% U) `4 [ 21003=51013=520b
9 u: f( R5 f" H/ n3 \3.5字符常量与字符串常量有什么区别?' r6 h0 Z# b4 A' z4 i
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! H& _) t- n# s# }
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'.
8 Y. J# k* n0 t8 h8 v! b* }: u3.6写出以下程序的运行结果:
9 z8 T" X" X7 l! ]: G#include
) S# {8 I1 t+ L) }) f. ^) u$ Uvoid main()
4 @5 Q; b1 Q& y& {5 i6 z{2 d! L& ~2 E9 a' h
char c1='a',c2='b',c3='c',c4='\101',c5='\116';
- ^2 l. y! a- B/ t! b9 bprintf("a%c b%c\tc%c\tabc\n",c1,c2,c3);
( L9 c3 e4 a. P" \! zprintf("\t\b%c %c\n",c4,c5);
4 K2 \) l# M: `* p: ^) s- j解:程序的运行结果为:4 V& s: V! r, i7 j6 a2 Y
aabb cc abc' i8 g+ r1 A2 i3 m4 m. {0 j2 |( _' r
A N* O+ S9 J- X# t, r. z0 f& W0 S/ K
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,* T( i h$ k0 n2 k& w
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". U" t( {- v7 i' y
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) ~" V5 R5 ^0 a1 r o3 _) c
输出. n' I ?: b5 N2 \, z4 t- y
main()+ C$ X x; d1 f+ f
{char c1="C",c2="h",c3="i",c4='n',c5='a';- u6 n% m) k! N& I2 L4 g- ~- E" z
c1+=4;
# G$ t+ ?. _- Vc2+=4;1 \% w7 B* z* l! i+ ]8 F# x5 s7 g
c3+=4;
8 C( C- n, e+ L7 \; x- P' [c4+=4;
2 H% ~- I0 K: X! j& vc5+=4;/ e0 O) M8 E m2 I7 x) x
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);& s% ^9 s6 Z1 e+ W7 Q H. e' _
}& H7 W6 V- Y; |
3.8例3.6能否改成如下:9 Q# y9 K) P x( r0 H4 W, Q
#include8 X B6 N# x! T5 O; ^0 G" p
void main()
' n9 \1 N" ]* w2 B2 W{
* o8 J! {+ z: ^) m0 mint c1,c2;(原为 char c1,c2)
1 T' Z. o# g# F3 a- Gc1=97;+ J# L8 L7 j7 O
c2=98;
2 Y4 B/ V7 [; Vprintf("%c%c\n",c1,c2);
2 e% y! @. d5 |0 `8 K/ T2 Y& sprintf("%d%d\n",c1,c2);
5 {# v0 H; ^9 [# y}" |: l3 T7 S P A
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.
, J x, e& o l+ t l3.9求下面算术表达式的值.3 w1 }$ L) l9 s# w* y
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
& r* Z' o% j3 O, q% i: i4 ?4 B(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
# `2 e# D5 J5 S- f S- r8 A0 {7 r3.10写出下面程序的运行结果:* ^9 b( [. V [4 ]) n! a
#include
: e6 G+ w" C& T) B; ^! ovoid main()" H( R1 f3 z0 e3 z' F% d
{
- e4 ~, g5 x$ k h" dint i,j,m,n;
5 D; p4 E- O. z3 @' u5 Ri=8;
0 k% H. N) ^+ M8 P, tj=10;; Q$ B9 h4 d9 ~( q8 M/ N4 j- i
m=++i;4 f6 m1 E, h/ _- |% y3 h3 [
n=j++;
) `9 N7 O7 F) e& }0 S8 ]printf("%d,%d,%d,%d\n",i,j,m,n);" R* E8 m7 L2 p' X# {) U# t
}
( I+ M& u* j/ p* f解:结果: 9,11,9,10
/ d1 @( N) C+ c, Y5 J K第4章7 g8 Q) G% f5 L+ m) z8 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
# n$ i W8 \0 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:
2 L$ P! x* G6 H5 g* Q: }2 U. A& ya= 3 b= 4 c= 5" T7 G5 e- p4 P
x=1.200000,y=2.400000,z=-3.600000# A# Q+ Z/ f# ]7 o
x+y= 3.60 y+z=-1.20 z+x=-2.40
5 K: g+ J0 Z2 p5 {( h* gu= 51274 n= 128765( _6 ^! L0 h) ?# k- D- D
c1='a' or 97(ASCII) }2 m/ f3 L5 G* H
c2='B' or 98(ASCII)% I- P' D( k x4 `5 s* Y# @1 N
解:+ ^: F/ W5 q2 z9 `
main()
F1 M/ A! h9 @9 z{
) A3 k! E1 b; x3 o. k7 `& i; [, Aint a,b,c;, V! |2 H* h7 @) s- d
long int u,n;# T$ Z: L1 [" m" G0 ]: t9 n
float x,y,z;. r/ w3 @: r" \. `, _- W
char c1,c2;/ q, D" x! R: C* B9 {
a=3;b=4;c=5;" e# s, I; p, P9 Z
x=1.2;y=2.4;z=-3.6;
. b) D% w3 J, W! Iu=51274;n=128765;, b% q1 v8 b5 a( ^- Y
c1='a';c2='b';
7 U' P x( Z8 }9 Yprintf("\n");/ k9 T, s x1 { g' a% y
printf("a=%2d b=%2d c=%2d\n",a,b,c);; X# A4 F% G5 m, y
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);! L: c. u u( y
printf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);9 b$ d% I5 l; G1 X
printf("u=%6ld n=%9ld\n",u,n);, s9 [6 p* U0 E [+ b0 O
printf("c1='%c' or %d(ASCII)\n",c1,c2);
1 j- j3 u- O0 ~% c6 Q+ G% i: Zprintf("c2='%c' or %d(ASCII)\n",c2,c2);* l/ n* v% d4 q F9 e( m0 C6 m
}' @* I; O2 {( X
4.5请写出下面程序的输出结果.
4 g4 N5 O4 j3 H& V2 Z y7 ^. X) U结果:3 A: Q& o' B1 X7 a- r6 b
57
4 f3 N! b9 L. S5 A 5 7
M( {: U0 {" g, A# T67.856400,-789.1239623 V' F) M8 C# R' V* X
67.856400 ,-789.123962: x& @5 Y; o7 b. E: p
67.86,-789.12,67.856400,-789.123962,67.856400,-789.1239623 o* o( T5 _ y' m
6.785640e+001,-7.89e+002
8 ~& ]+ I' q9 E% n/ V5 w2 v MA,65,101,41; q Y0 k D* R
1234567,4553207,d687
# a' B0 t2 O$ x, h% r$ h65535,17777,ffff,-1
) w, _5 s7 g& k" R& y( aCOMPUTER, COM# q1 ^6 A) a8 u) I1 M& q9 |, C
4.6用下面的scanf函数输入数据,使a=3,b=7,x=8.5,y=71.82,c1='A',c2='a',
4 G4 y( v$ o) ~' `; y2 q$ e问在键盘上如何输入?- Y2 y3 G) C3 \4 ]) B# i n7 b
main()
' A' G* B" j$ ?" F' w6 C$ {{
! K$ l! }2 P+ E) _9 Wint a,b;6 Q4 W5 P* W7 g( q% S2 Z9 c
float x,y;
( d! |! a) ]# B9 b$ O$ j9 ichar c1,c2;
" S* j3 _' M/ w% s8 E( mscanf("a=%d b=%d,&a,&b);7 N4 W1 B8 i+ }9 w& C+ G( k
scanf(" x=%f y=%e",&x,&y);
7 d) b# E! l. O" u- @scanf(" c1=%c c2=%c",&c1,&c2);2 M7 K) X' I2 ?) a' V! o( U; I
}2 W9 Z4 m# ~1 G% I3 |
解:可按如下方式在键盘上输入:6 `' l: c5 R5 v9 ~% @ V2 r% B
a=3 b=76 _, u8 O$ S' y, M% J) x
x=8.5 y=71.82) F0 ^0 `% I5 |, Z. [* r! S
c1=A c2=a
" A8 V& Y2 F4 _, C( e说明:在边疆使用一个或多个scnaf函数时,第一个输入行末尾输入的"回车"被第二
6 R& B O% E1 l% h- @% X# `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3 l4 x/ {5 a9 O: ^$ B8 q4 a/ m
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; {5 ~0 K) P4 X
初步概念即可,以后再进一步深入理解.
7 r; I. b" h3 r/ U* C5 V0 h8 A4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-$ Z' u4 K* Z6 _
3.75,z=57.8,请问
7 M$ }8 Y8 N3 @在键盘上如何输入数据?
" S/ x5 h0 o, Yscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);3 o2 q, g5 ~+ }* n
解:5 A/ b; `7 _4 ^0 U/ K
main()
4 ?9 \3 r3 y: a; B: N# o* ~2 l{+ p1 N, }4 n. V
int a,b;1 Q. I8 i% R: P3 x; j
float x,y,z;
4 E" `, S4 K5 D3 g% f+ Nchar c1,c2;
+ B% \0 w+ d" M4 X6 S& v4 p! B! v& n0 nscanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
5 W1 ~' |+ _! u- T/ F5 z/ Q}+ s k0 S* I4 Y- a9 [0 z
运行时输入:2 B8 S! y0 V: l: d
10 20Aa1.5 -3.75 +1.5,67.8
8 `7 g, z1 U3 g% p4 S注解:按%5d格式的要求输入a与b时,要先键入三个空格,而后再打入10与20。%*f$ c% ~! ^! x5 R7 C! o+ S
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8 L% V; u2 ]% g& D4 P2 v0 J
会赋给任何变量。
+ L) @! a7 k( o; d# q4.8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,
t$ Q: `) m: T) N+ P' l3 s5 v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
, X( t8 B6 x( o; R. @% p; X程. r# I$ q( }# g' {4 e7 m
解:main()
2 x' r% `* g; ?# }, S k{
- w4 R/ Q* ]) }# K3 c+ Mfloat pi,h,r,l,s,sq,vq,vz;
0 E! g% n2 L, b* Opi=3.1415926;3 _0 l. e) B, w+ b$ u G+ o2 Q, s# t
printf("请输入圆半径r圆柱高h:\n");+ e% S8 s" y2 j* M) n
scanf("%f,%f",&r,&h);. i3 F( C: A9 s" U3 ~4 ?! B
l=2*pi*r;# J) ^5 r9 p- O, M& C6 R' T
s=r*r*pi;2 a1 Q6 a0 {: R8 i' D U& j) {
sq=4*pi*r*r;
- s! U+ I! h/ o3 [0 \vq=4.0/3.0*pi*r*r*r;
2 [0 |+ z. H) z2 l0 [ y% ]vz=pi*r*r*h;
9 r$ p% _ n* X6 w6 pprintf("圆周长为: =%6.2f\n",l);
8 b# h% y Q9 d6 Xprintf("圆面积为: =%6.2f\n",s);
6 [, u: E$ K! ?printf("圆球表面积为: =%6.2f\n",sq);9 k$ w: z4 h8 c- v
printf("圆球体积为: =%6.2f\n",vz);# t9 X% z5 U, M B4 K6 v
}
& }9 s% R/ e Q6 a3 E4.9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,
' j1 o9 t4 u, o) }* m# z( @: w取两位小数.
: H q) C% u$ B/ c: j解: main()
+ b: v) h! k4 j! n' y7 x{
( v. U1 c6 l- z; Gfloat c,f;
0 `$ y0 R( e. h8 X' A. Lprintf("请输入一个华氏温度:\n");; i. e) [2 d+ u9 h
scanf("%f",&f);, `! `) D( _+ C
c=(5.0/9.0)*(f-32);
) G, j- I' P E, Y& t- Hprintf("摄氏温度为:%5.2f\n",c);* z6 f! `$ n: y
}
3 |3 `9 R C. ?( V+ g第五章 逻辑运算和判断选取结构3 Z% l3 ^# O( k& |8 _: o. R7 K
5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
( K% V/ F) l# tmain()
4 A" ~; |1 m# ] w. }{% W% W; J+ e* m1 {- [: Q: L
int a,b,c;+ c6 w" d8 T2 z9 Q
printf("请输入三个数:");
1 } \4 b+ D+ w" k3 _8 Escanf("%d,%d,%d",&a,&b,&c);
% v1 L- A6 C) `if(a if(b printf("max=%d\n",c);
& d Q0 X7 D6 J else
! v* ]7 a2 \$ e printf("max=%d\n",b);+ c* |3 D) _' ~4 b
else if(a printf("max=%d\n",c);# K# N( P8 W1 v
else
# v$ n+ s9 o1 `5 ^9 q printf("max-%d\n",a);
8 P# E9 p# `" }0 H5 D}
' D3 m. n& a, }$ x( |方法2:使用条件表达式.
2 i& ~6 q# i; c3 Y; hmain()* B0 f' S# }! x7 b
{int a,b,c,termp,max;
9 U M' K8 c( Q printf(" 请输入 A,B,C: ");( G5 c: t8 c5 {+ G& V- b
scanf("%d,%d,%d",&a,&b,&c);
( j- g& V0 N m) C$ i printf("A=%d,B=%d,C=%d\n",a,b,c);7 X4 R) y% W0 f- d$ L8 Q
temp=(a>b)?a:b;
& h5 n. z8 ^; k0 X. ?3 ~- _" C max=(temp>c)? temp:c;( U) B8 M# F' V1 Z
printf(" A,B,C中最大数是%d,",max);$ }# \7 U" F& k @3 B; B" t
}
! }! W& ~# w9 u1 x( w2 c5.5 main()0 ?$ K6 i9 a+ z6 \' ]
{int x,y;
) {% {: |6 ^/ Rprintf("输入x:");
, ~& x9 C5 p$ H9 x: M; Dscanf("%d",&x);
+ D4 V8 t# t- G/ uif(x<1)1 W0 D5 K3 m) s! ]
{y=x;% Q+ W' b7 C9 O& ~1 q6 Z- K5 ]; f$ p
printf("X-%d,Y=X=%d \n",x,y);) P: G3 D, o, F( z$ P$ @( R# M
}& b. U/ \( ]% r7 a
else if(x<10) F" N, s4 G6 |' r/ c, k
{y=2*x-1;
3 R4 }$ W% a: n$ h4 H printf(" X=%d, Y=2*X-1=%d\n",x,y);; X5 g! f+ k4 E$ O; k( a
}
$ j! L) s2 A/ w' f. M1 a" celse
# ~+ x$ d" [; |2 ], f { {y=3*x-11;
* I0 z: f/ c) o- O0 @5 X4 d printf("X=5d, Y=3*x-11=%d \n",x,y);6 q% ?' l @0 G$ @+ z- P6 q
}
0 y" j; z9 ~7 C2 E* p}) r- B4 l k- s9 w: X* k6 r
(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!
% m- g; \2 D3 g* }! v- Cvoid main()- l# `* y; ?! |# ^( }7 j$ C
{6 q; R7 ~7 W! y$ e
float s,i;
8 k. p# ?4 z3 ?9 P# b. E' o8 i+ Achar a;
5 N( _" @% V) F# |4 r* Z: |scanf("%f",&s);
2 U" T' U( D& f- S; ~while(s>100||s<0)
# G Q6 t6 M" \7 F2 y{
w( _# ?- d* X% lprintf("输入错误!error!");
4 ^* m7 Z2 ]) M( \% |: s bscanf("%f",&s);- X% C8 k* r+ G5 {$ g1 a" K% u
}
% e3 ?) ~- K, {7 Bi=s/10;. W( t; m, h2 H# ?2 q/ l' ]
switch((int)i)
7 j2 s$ P @ h- s4 r) r$ ~{3 k) T( u0 E/ E7 c" ?1 [
case 10:( |. ?; P( u, w
case 9: a='A';break;1 M, O3 h& c2 @; o1 E$ G
case 8: a='B';break;
! n) y, {+ M( W+ U, x9 v0 J2 _case 7: a='C';break;
7 z: a" o9 s( v. S6 z% Icase 6: a='D';break;1 I: G1 h2 {' V. `
case 5:
+ k& m/ @- [. n; Pcase 4:
$ `) }2 y* q" P+ V R, E* ocase 2:
! G) i: V' _' _+ Dcase 1:
; c( X2 h6 z" T& Hcase 0: a='E';! s9 }- N4 e! m r) K, N
}# g2 y4 d% [7 t' x( f
printf("%c",a);# \0 ]- l6 u& U* ^. S; n" e
}8 `$ W6 Z$ I( P; ^; G+ @' A3 c3 M
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.& G I8 S6 S% D3 ^# T3 u
按逆序打印出各位数字.例如原数为321,应输出123.
# Y" S2 R* N/ J: ?main()
8 _' G$ F+ g, z3 O1 e; L7 I {
8 [( l6 C7 }8 b' R5 P long int num;
1 }* j" q3 ?9 s" u int indiv,ten,hundred,housand,tenthousand,place;% @+ x8 ^& ]0 f2 o3 g7 l2 M, ]
printf("请输入一个整数(0-99999):");
; O+ O3 x4 q' n2 y- e7 g- T) A! ^ scanf("%ld",&num);6 w% ]( V0 g% N4 K6 ]
if(num>9999); G/ M( h2 E8 D2 l4 {# v6 V3 A9 m
place=5;! w0 F. d4 L, x* R3 ]
else if(num>999)
: w$ J$ O3 d2 b v( g place=4;
8 J! s; ]1 ?0 t1 [2 {else if(num>99)
* n: Y$ V" f; v' N5 d- E3 {& k# O place=3;) S t, z& S2 O. V8 T& d# U* y
else if(num>9)
2 O' |) N0 a3 i place=2;
) |8 D$ w. _7 ~$ m" ]else place=1;
, a% d0 |7 K/ V0 v3 xprintf("place=%d\n",place);- h9 L2 b6 j6 \# n
printf("每位数字为:");. P- c& J5 A$ D+ W0 J, Z) V+ ?; Y) y5 N! s
ten_thousand=num/10000;
. a. |, _9 [9 S6 o+ N6 Q: ]thousand=(num-tenthousand*10000)/1000;
8 d3 r! U/ C; ?, Q; z+ J1 y* a8 Thundred=(num-tenthousand*10000-thousand*1000)/100;' p4 E. B( z, l; f
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
( Q3 U. K7 {6 b( ]/ ^indiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;' p( ? y6 R* v& Z! i
switch(place)
9 q0 Y" F( ^/ i1 [5 h( q9 d* b{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);, r/ x' c6 L* O' g, H7 A
printf("\n反序数字为:");" B' Z5 _ b9 a4 \+ L
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);- Z- F6 m/ O @$ i; l. q8 H/ X
break;% v! p; s5 p1 `: e+ N
case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);9 x; n! C- A6 K* H. x8 A$ E' }8 t+ m
printf("\n反序数字为:");- n6 M2 Z: ~! X7 {5 q5 e$ T
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);5 c7 ] `9 d; Y2 j& [; V$ U6 I
break;5 F3 ?1 t% P( ^: l
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);4 w3 m% A% X2 k4 K) \8 B/ Z3 |
printf("\n反序数字为:");) m) Y8 i: g# T$ `9 f$ a
printf("%d%d%d\n",indiv,ten,hundred);: M' X/ m( ~* A/ I( _- A
case 2:printf("%d,%d\n",ten,indiv);( `8 @7 x. t, e9 u/ o
printf("\n反序数字为:");) H, `- o- C8 V* F% \/ E
printf("%d%d\n",indiv,ten);
# s; p G& G. }7 c! scase 1:printf("%d\n",indiv);% J$ E J A( [+ y
printf("\n反序数字为:");; z8 S+ b6 Z+ U9 G. s
printf("%d\n",indiv);5 _7 h$ l" @; E; O
}* T" ]$ H( k4 y2 F6 H
}
/ }0 i% M% S! \: |6 b3 I G5 _% ~5.8 b9 s: l# p. M% _* g4 v. G# ]) ^
1.if语句) j# }5 Y6 X4 N; k1 q* E( e" j5 T7 A
main()/ R* h9 V/ ]$ [; L- x6 y
{long i;* V$ ^3 w- a7 C% n% S
float bonus,bon1,bon2,bon4,bon6,bon10;
8 H( q) j5 Z! q( F3 ~ Z5 b bon1=100000*0.1;, {% i- j; _8 I2 W1 y! U+ i/ T5 P
bon2=bon1+100000*0.075;
Y0 v% K- b; g8 O' `* \ bon4=bon2+200000*0.05;2 i8 p3 N. w" Q2 D' n
bon6=bon4+200000*0.03;/ Y/ _! m( Q% x, v
bon10=bon6+400000*0.015;1 P r! ^5 [6 G5 X( f
scanf("%ld",&i);
$ h- {$ S9 D0 `& S) e& ] if(i<=1e5)bonus=i*0.1;- }' L( G0 t+ y$ [
else if(i<=2e5)bonus=bon1+(i-100000)*0.075; H0 p( L8 O4 ]5 c$ \+ H9 @
else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
+ o; o" T* l$ v( E7 _7 F else if(i<=6e5)bonus=bon4+(i-400000)*0.03;, F; d. w2 G4 Y$ ?+ v6 Q
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
& I @2 ^* b1 d) Y7 Q* Y else bonus=bon10+(i-1000000)*0.01;/ ~! ]; }, I2 R7 i$ M% {% c5 o; E
printf("bonus=%10.2f",bonus);* {% ?: U, G6 Z$ v8 [6 R4 B
}
0 j# k! S! o# h- r. a) ]2 Q# w8 h用switch语句编程序
" G5 ]1 k+ k) @, amain()
$ {8 r# v9 @1 v- v8 d{long i; ?- j" u; D) O* Y6 h7 {6 H
float bonus,bon1,bon2,bon4,bon6,bon10;
0 F( _* N' {) }8 @2 P* {. @8 w int branch;4 P0 K9 }% x! f' j
bon1=100000*0.1;2 {. B P' |9 f$ _+ c9 `* _
bon2=bon1+100000*0.075;3 @& Y u( Q+ D! y8 g+ k
bon4=bon2+200000*0.05;
3 L) ]5 E+ ~, @! E7 Y bon6=bon4+200000*0.03;' U# P# z) R6 ^; p
bon10=bon6+400000*0.015;& T1 s! v5 X: o7 }0 O, ^" E9 I. q, z
scanf("%ld",&i);. Z* T8 ]- a0 f7 `
branch=i/100000;
& a% q3 t; Z8 j9 l if(branch>10)branch=10;! g% s. \- f% [2 R
switch(branch)( S# [! _( C/ O/ T9 [
{case 0:bonus=i*0.1;break;
$ f* i8 N" K! m case 1:bonus=bon1+(i-100000)*0.075;break;) P) s4 T' Z- r1 m" ?% w; g
case 2:
* i1 |7 l- X6 F5 C0 G case 3:bonus=bon2+(i-200000)*0.05;break;& q5 E% R# }% h( U
case 4:2 p6 J5 c" ?% a7 ?: H
case 5:bonus=bon4+(i-400000)*0.03;break;
4 H/ g* @2 _ i& X case 6:+ ~# g. \3 ?/ x2 p1 C, A& S
case 7
: B6 {/ i! k+ S+ s* Z case 8:1 \7 m9 k7 d ]& A* G r( l* D! V
case 9:bonus=bon6+(i-600000)*0.015;break;, i6 ]" [; r& e7 V! W y
case 10:bonus=bon10+(i-1000000)*0.01;$ B/ t# {1 F+ Z: I% ]6 h
}
. ]% _. d* e \1 }; z printf("bonus=%10.2f",bonus);
; `! x( h7 g- X! H& s1 ]}
. M, E2 b- L5 N5.9 输入四个整数,按大小顺序输出.
J4 y# n1 v2 w, vmain(), b1 c6 x% v) X. N8 a" T% r, \1 Z
{int t,a,b,c,d;
8 p- T" ^) ~# l2 Y9 t% w printf("请输入四个数:");
. A3 {$ x D, v G scanf("%d,%d,%d,%d",&a,&b,&c,&d);& A0 v& Q3 h" h1 g
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);3 v9 P0 p* o; s2 \) @: S2 ], `
if(a>b)6 n6 e/ d# V' p$ | J6 P- L/ l g
{t=a;a=b;b=t;}
! n `( j" m* B7 W1 `/ } if(a>c)
& [( ]0 X( b/ u {t=a;a=c;c=t;}
9 j& J" H. H3 h1 z5 [7 t7 E( w if(a>d)
$ f* _3 j0 }1 Y2 h) X; E+ E2 C5 E9 {" W {t=a;a=d;d=t;}
/ \( b6 L. |* I1 c- \ if(b>c)
& k2 A" g! [4 D+ U3 [% X {t=b;b=c;c=t;}, g; {& I$ K- }) R! N% D
if(b>d)
3 k2 l$ |% ?: X$ g" \7 v8 o8 s! x {t=b;b=d;d=t;}
" z2 Y x- `6 _% I7 i- ]8 P if(c>d)8 x7 s; S& Z) A- Q
{t=c;c=d;d=t;}( L4 o, R* H& [# o# C
printf("\n 排序结果如下: \n");
. A! R- L+ G0 z9 H! v; lprintf(" %d %d %d %d \n",a,b,c,d);
& Z+ G% D7 U" y& [& U}) D2 o! A+ J, h' s ~0 H8 s
5.10塔9 u5 Y# U7 V7 d
main()
. m3 m) B4 M0 q{/ i: ]$ `2 c, [/ b9 L$ B
int h=10;" s7 |' W V- R* m$ Q5 V* w
float x,y,x0=2,y0=2,d1,d2,d3,d4;
8 ` ?+ `2 y' e, g- L6 T9 \6 W2 Rprintf("请输入一个点(x,y):");
* U# l' x/ l d; ?: kscanf("%f,%f",&x,&y);( W) r: v# V5 H, _
d1=(x-x0)*(x-x0)+(y-y0)(y-y0);
, `' {7 O" p! f. }! fd2=(x-x0)*(x-x0)+(y+y0)(y+y0);3 K; K/ C" b" o: h: ]5 n* G+ m3 m
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);# o+ l0 n: U* S2 ~7 f( ^4 |
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);" O- \ [& G4 b* d/ b9 Z
if(d1>1 && d2>1 && d3>1 && d4>1)8 W6 `1 y' X# ]0 ]: p
h=0;
. u1 h- s* K2 `printf("该点高度为%d",h);) E8 v; D$ q D+ C
}
9 V+ O; W1 G* i6 q第六章 循环语句
( @4 x; n7 m6 M; m" N6.1输入两个正数,求最大公约数最小公倍数.
( a! n$ U5 c. z6 G$ W" Zmain()
3 s8 w% @ V" C9 l5 Y$ j/ v{1 \& @# R a: n# C
int a,b,num1,num2,temp;, e4 c- y9 ^6 w! V Y
printf("请输入两个正整数:\n");
& P. _/ I, m, T( k3 kscanf("%d,%d",&num1,&num2);
7 F5 L5 Z+ p" F8 B: G+ k8 }+ ^if(num1{! v' Y% X, ]" ~1 E( I
temp=num1;
: ~& y+ p9 m; Y- c! ynum1=num2;, W) C' i3 f- V
num2=temp;0 l/ h7 ]. I: B# F5 V
}
7 {; x: c4 T; O Za=num1,b=num2;
6 _" n0 s3 b E! u2 I$ {while(b!=0)2 k' ?2 `0 |) \1 h6 N9 ^' u
{
! i! S; A X+ I( W2 j- Y* \2 M ? temp=a%b;
3 N2 A7 H! m1 l1 Y! a/ ~" \7 o a=b;+ k: F' k9 [$ w- x1 B/ s. F
b=temp;
; j, J8 M! A& |3 o5 f2 B }* r9 \, u. c, o: a: X9 W$ w
printf("它们的最大公约数为:%d\n",a);
3 u [9 U/ p9 u: L& Lprintf("它们的最小公倍数为:%d\n",num1*num2/2);
- z1 x' Y, o3 e$ T: K0 g ~8 a: [6 s}: P, D w7 p& R8 N- T: b \
6.2输入一行字符,分别统计出其中英文字母,空格,数字和其它字符的个数. a `# X D6 n; S
解:
+ {5 \) i K4 j#include < >+ `$ i, N: D- K4 f! W9 B$ n' c' J
main()/ R) k0 v: c6 v& A$ u
{
l: B+ u& D9 u+ r$ f" J7 bchar c;' l7 B3 t% @4 [6 z4 I
int letters=0,space=0,degit=0,other=0;* @" R" }: Q# G1 x% C( X, C
printf("请输入一行字符:\n");
' q, J- k+ S% Cscanf("%c",&c);
' I; ^7 J$ Y( Ywhile((c=getchar())!='\n')0 G5 r1 \, _3 X) c$ g: L- |- S
{
( O6 u& T2 R0 \/ j( w+ }% r( ^* Uif(c>='a'&&c<='z'||c>'A'&&c<='Z')
* _ A2 P/ ]( v/ [# B/ Aletters++;$ Q* X0 |0 P, j* { y
else if(c==' ')4 [0 U# o! |4 ?! f; R3 ?
space++;% ]9 B* q/ |& P: M/ G; m
else if(c>='0'&&c<='9')! [- @! i% ^$ d& b7 U
digit++;/ k: x* U/ H9 z+ K- Q
else9 R8 s9 D. ]7 m& x
other++;$ k# m w/ E& w! F1 i7 a( Y
}
8 G3 @; G$ a2 a' T9 F4 }- hprintf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%& T4 G7 K% \+ ~( ` z1 { W: \4 m
d\n",letters,space,
& r/ \9 D3 a- \% |$ }digit,other);! I( S: Y+ z+ D! X' q0 j: `
}+ y5 y& y; v# l$ e' S5 b4 O! W2 n1 S
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.! N0 K) g( a0 O' I% k+ Z8 ~
解:
0 z- c+ n- V2 A7 v2 Omain()* d' M+ {! {7 C' j+ { P
{
; U2 i5 k+ C" hint a,n,count=1,sn=0,tn=0;
* v' C: x3 J: N' ^printf("请输入a和n的值:\n");
+ o6 i4 C- m+ S7 }5 O4 Z5 O4 kscanf("%d,%d",&a,&n);8 y" N0 L e5 l. p( _5 @4 _5 s
printf("a=%d n=%d \n",a,n);
N% Q3 z) w5 B9 [: d/ qwhile(count<=n)
$ }. s$ o' o" ?. C% A+ _# m{
* r: {$ x* }% L" ~$ ^3 Stn=tn+a;
+ K4 A! |- k: ?; c- S# c4 c, ysn=sn+tn;& F6 B' v" b O) x( `& s
a=a*10;
/ l; T- J, X/ p$ p1 c& Q++count;
8 c; U( k( d8 N) V9 `}0 u4 l: L3 u) h5 I
printf("a+aa+aaa+…=%d\n",sn);7 N$ R& c6 B# J' e" c
}' f& o# S; N0 S6 u; ^% p }1 a
6.4 求1+2!+3!+4!+…+20!.
9 s0 d1 K" }! T5 o! xmain()
; Z( G! ~, S! g( N/ x+ p{
& B |$ n ^4 K# T- t5 B, Gfloat n,s=0,t=1;
* w. Z2 j, D/ a7 E7 ?) Ofor(n=1;n<=20;n++)1 G1 N$ o1 R5 h. Y
{
( O0 V1 a1 n9 Lt=t*n;
0 w; K7 m5 ?/ C* v3 is=s+t;
$ f( C; \6 }% n0 h}
4 y/ \7 ~& b# ^4 k+ A' tprintf("1!+2!+…+20!=%e\n",s); d% e- P$ Y3 ~$ f
}
. c8 q T6 D; V% z6 b) Y l* C6.5 main()7 {# @. ?2 ?- F/ N# e
{
7 S/ o8 z/ f" ` p% }int N1=100,N2=50,N3=10;
" Y' d0 m' F* L6 m3 b3 A6 Y/ Xfloat k;3 V+ z: I2 b j" n, I4 Q0 M4 P
float s1=0,s2=0,s3=0;" K# p' w% B! j) _
for(k=1;k<=N1;k++)
% s+ L' B- Q% \6 p( M% s{
, p& M; I+ S( R. M1 N& J0 As1=s1+k;2 G; K W, B3 Z- [8 X* R9 K
}
5 O) V" `2 ?6 X5 [1 bfor(k=1;k<=N2;k++)" p2 a$ k' s7 r+ j
{5 D7 J3 W: n$ j+ G
s2=s2+k*k;' h6 Z! Q& I0 K2 e
}/ E! V7 H( |4 W3 ?
for(k=1;k<=N3;k++)) o" V* x f, g( t8 C
{* x( m( m5 g+ P8 ]$ ~; A
s3=s3+1/k;
4 J7 r* A% ^4 G}$ m$ J3 Y0 C T% W [2 Y N
printf("总和=%8.2f\n",s1+s2+s3);
) X: b- m! Y; I$ j% \; m3 q: S3 w# P}
4 T( V6 R, u$ ?$ G/ L6.6水仙开花! k) I+ J0 }8 q$ I; l2 z: |9 j) |
main()' n6 {0 q, v5 c7 Q2 M k! a* @
{
) x2 ?, l. X x. f1 {8 u9 R7 Nint i,j,k,n;
: q( W8 w7 V2 ^! N _5 qprintf(" '水仙花'数是:");) B) J- _" r X. u5 P/ S8 Y
for(n=100;n<1000;n++)6 @; n5 m; N+ e' ~# z I8 e
{
& q9 d6 e$ H& b. i7 |i=n/100;
. |! f% h) b* l0 F" qj=n/10-i*10;" q' J3 g3 `# ?2 z. p# }
k=n%10;7 y; {" y8 \" n( [$ P$ ` ]' T
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
1 ^$ l3 {; r( c4 h( }& V! m6 b{
) z0 R2 U1 J- R( Uprintf("%d",n);" k+ p" R3 b0 c0 N; ?5 E; c
}% l7 n! X* M& `# O- C3 n! u
}% J# Y" A1 [8 v2 |1 a) S: n
printf("\n");
1 H7 s- Z" m! z0 n+ C& ]3 M* c8 [5 Q}
$ P' L8 R$ y4 x% [( @% E9 B7 O6.7完数
: j! k2 W# {5 {8 |main()8 V. {) l6 P0 u) ^' b" N: M4 \
#include M 1000
! e# ^1 t- E" S) @8 W/ P3 g Zmain()7 I1 \ c8 X \
{; l" o; C: [. b) H' B6 I& `3 B$ I
int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;9 U) E5 U- J0 s: T$ [' y' D+ H, o
int i,j,n,s;; E E0 c' { D* g( o
for(j=2;j<=M;j++)" B& L3 e9 w; s( |% F u: X1 `1 y
{
9 A4 c& f0 P# _$ P% B+ ^n=0;
) ]8 O3 p! J2 T8 k% xs=j;
* A4 G `& X6 |' N, \for(i=1;i {% c- k8 P' m& n, b5 \* h0 j ]# G
if((j%i)==0)0 B. N' ~! Z. {& b2 X4 d5 X
{
, b m8 e9 Y- m. P$ K if((j%i)==0)7 y$ B/ q( h0 \9 b+ n2 n
{
2 n, w* t+ G- ~- y" @ n++;
: _4 W& i& ^( |4 s; |0 U. j) [ s=s-i;
5 D: f* X% m/ L! C( b, Z+ L9 Q8 D switch(n)
) P X5 p& _5 X3 a' i {
* p D# i2 ~5 @2 R; h7 C case 1:) V4 e# q1 [2 S/ H) d K+ \1 F/ ~7 L
k0=i;
3 f; k" w, b' H! P( v8 C break;
F% y, X. j U' |: E( a' }+ b case 2:
9 a! m- F( ~0 q/ J6 ~, i$ }' P k1=i;
/ f' u; C" U# Y9 ~5 h# } break;+ j. q" }4 x" K' d+ H
case 3:) d7 F+ W% T4 ~ J- z8 z& H! O; K: V
k2=i;% c8 E5 P( W! R0 V# a9 F" O0 k/ |; ]
break;
, ` t; U: l A case 4:
9 ]5 e5 j3 r) j9 S, i3 N$ d, J k3=i;7 ]) h" p2 F- w. X9 s8 W0 B7 S& a: E
break;
2 W; d: L9 k) d6 X! y case 5:; p/ f5 T3 N- \, X
k4=i;
' V7 r; ?1 h2 d$ U3 P- x- s break;# c3 L# S1 r. O
case 6:
, u, `( q; D6 \1 k+ C, f* N k5=i;
" l, B C* L; ]& a' s; z9 m7 Z- Y break;1 G$ ` X: x/ p
case 7:
4 o* s# P# |% M k6=i;
% k* Z* ^, \6 y( D break;
( _" g: h0 Y( d( J case 8:
# W+ d0 p( l4 Q8 G, l, N k7=i;
& x' ?* U' ?1 w, C2 q9 K/ Z* x* x break;! {* ~- x4 z! [/ q. M/ s! n* g
case 9:
9 Z' F2 Y/ k) `- s" ~9 V9 r k8=i;$ s+ h0 i% H4 n' D
break;
2 d$ W5 J' D# j9 ^# f3 a case 10:3 `$ H; [9 {/ z6 W
k9=i;
1 n- Z" `7 U+ ~* G1 s break;6 Y( i- o7 @6 E8 U, _' W& f' V
}4 q3 ~5 U. k8 x! q( }7 j
}9 J1 P( J0 B6 z# h( Z
}- r' P7 ]( U% i* E
if(s==0)
7 ]+ i6 ?8 X- I) a9 ^ f {
/ e4 v( q" F3 C- _" Hprintf("%d是一个‘完数’,它的因子是",j);9 }$ s2 x3 v; k$ F0 O/ a
if(n>1)- v, [( i; N& p: n" F
printf("%d,%d",k0,k1);
$ a2 O& M0 ]) Y9 sif(n>2)2 Z2 {0 Z+ C: Z! B! l
printf(",%d",k2);+ \: ~, L8 M; E4 h" t! h5 U# T
if(n>3)
6 j; N7 z! q7 d printf(",%d",k3);5 u# |; i0 @( o
if(n>4)
$ Y! F$ K9 R& y! P3 h, t# Y printf(",%d",k4);5 o( o, ]- k0 ?9 r; L9 n& e6 A
if(n>5)
3 ~5 U4 _( K0 j0 @) v/ T" k- y printf(",%d",k5);0 E* T+ @ ?' j' r+ \
if(n>6)
/ a, c3 m& z" D2 a0 s* w printf(",%d",k6);
* l. j) C- W8 fif(n>7)# U# J7 D7 ~1 q% V8 P; l0 C2 ?
printf(",%d",k7);
6 Z. s: F Q. a( {" C5 dif(n>8)# O" O( \$ X! ~# Q* I8 a
printf(",%d",k8);
1 r; V! P3 b( _, X$ [if(n>9)
% ~8 i: z! X1 [% s+ v printf(",%d",k9); {- S6 H( r! v4 i2 ]; W' t' s8 l
printf("\n");
7 P$ g g& u( l# K$ L. X4 z' ^+ r }
5 n6 c1 K% {9 N$ ~4 k; c" T}2 D% t% p5 \2 K; `
方法二:此题用数组方法更为简单. ?) r5 L- }4 E+ W, n
main()
! m5 o6 k% }# u$ j5 [3 a$ P6 |( L{% A- K* F2 s5 o
static int k[10];
" R0 B# _, h3 K* J, K' Z7 z2 Aint i,j,n,s;! P( [0 V. G/ J/ e' t% G( X7 u: n
for(j=2;j<=1000;j++)' t/ m X- Z E. M& I
{
0 \ y! s: ]# r) C- L5 un=-1;
3 }& Y* w7 R% S; e) w, ]s=j;, K7 o6 x9 d d" q$ O
for(i=1;i{% A4 q, Z/ m' s% k% O
if((j%i)==0)
; {. {, L1 A7 v! I5 `{9 ^$ a8 V3 ]1 t$ U- d
n++;/ x% I" K) N) t. p* m n1 N- N
s=s-i;
# {( ^. }8 H. a1 {k[n]=i;
4 s0 f# |8 T6 Y1 j( A4 d. e- ?5 E% r }
6 M ^0 _$ Y* I/ f9 ]5 s8 Q c# u1 G }
) \" N+ F/ {! Hif(s==0)6 p! t( f [9 u5 E: Y
{4 ]6 ~+ U+ a+ c8 I2 @1 i
printf("%d是一个完数,它的因子是:",j);1 `! p+ g" ]% u1 S7 r1 f" X6 x
for(i=0;iprintf("%d,",k[i]);
* y' U: F [! l" o- I' {printf("%d\n",k[n]);+ [( d8 k9 Y6 {. _3 a
}! G$ i/ m! [6 U' E
}6 |; b# b1 @* r" ]7 B
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.
2 k$ m$ @. a1 X" F解: main()+ G9 H, }, s0 g+ C0 r V
{
. u& t; H3 S5 eint n,t,number=20;
; [6 Y5 \2 V( [- yfloat a=2,b=1,s=0;; K1 O% p' Z! k0 C' J! I, h; J3 A
for(n=1;n<=number;n++)
; R. `+ S& C2 |* r% ~1 w{
b( r2 I1 y; l0 ws=s+a/b;/ t0 u; V+ G1 U( g7 s
t=a,a=a+b,b=t;6 n. N6 I- U$ g* u+ h1 ~) r
}4 ]6 `0 c7 \9 v$ g8 G! p
printf("总和=%9.6f\n",s);
8 G' n$ ^: U8 S$ U& c" O( {}
% s l2 j4 G# _6.9球反弹问题
* W }, d; e- J0 qmain()
) b, n. T. L; j6 l{
4 U$ y5 d; |3 N% d- w( A6 j5 n: w0 ofloat sn=100.0,hn=sn/2;7 _. s1 b; m9 n, x/ Y2 }
int n;) r/ v8 e( z. D. p( O2 T
for(n=2;n<=10;n++)' g- Y$ b2 [6 x! V9 h
{( X/ p5 g) u' g2 J
sn=sn+2*hn;
H4 J. |2 V# O: S6 y, N( V i/ z7 ehn=hn/2;
~$ {& F: x( Y* S4 m+ `} b; _7 y4 |4 R4 r' I; A( s% ?
printf("第10次落地时共经过%f米 \n",sn);3 n. ]* _8 L' a/ [# d% k* x
printf("第10次反弹%f米.\n",hn);
5 l1 i, A6 t2 }}
: e4 g9 | \ s9 }* W7 Q# ?9 y6.10猴子吃桃
3 @9 d, a( t- {6 zmain()
; ^, i+ t q' ]7 }{
1 `0 B, B+ x/ t0 h: x" T: o3 W3 Bint day,x1,x2;7 E! C1 ?) h7 X8 p3 g
day=9;4 ?! T) Z2 y" F+ b9 L, P. m2 W; {. _
x2=1;) A1 Q4 k( I# `0 T3 t! E
while(day>0)1 L- v& O8 V( Q: \+ H: C( c
{
$ O- \( j2 C4 U& f, u9 F" H5 h, ex1=(x2+1)*2;5 r2 j V! h3 n( N
x2=x1;
& A9 U2 T1 Y$ C! v, U# Q/ \- tday--;
& [1 S. z9 ^1 P5 n' ]}
( A. Y, a n G1 ?5 q) }& ^printf("桃子总数=%d\n",x1);
: l8 r% k% ^+ u}
, G9 o: Q0 s# A- G2 Y, Z* ?) U. Z$ S1 z+ w7 X* m
6.12
: o9 {8 O) R& ~0 j#include"math.h"
8 N: F6 c$ L5 G9 Q8 X3 B5 Qmain()
# Y7 i, e! [1 N& }{float x,x0,f,f1;1 s* u+ K4 z! N/ O/ O8 T5 G
x=1.5;& h# |+ w+ C$ J" }& L+ G+ D, V
do
3 Y5 k; x# v l; F- ]6 f {x0=x;; |7 {1 a8 ?: M' q
f=((2*x0-4)*x0+3)*x0-6;
/ _/ ~( A4 W M8 T' U1 h$ S+ z f1=(6*x0-8)*x0+3;2 X, e' w9 [& G, E8 |- c
x=x0-f/f1;; _& r5 S S [2 ]
}0 |, \5 e' N( c2 k
while(fabs(x-x0)>=1e-5);& {! a& Q0 `: N9 G$ {( s
printf("x=%6.2f\n",x);- T5 i3 q$ E) k6 ^9 y; E, r
}
& W' j. a) }! l; C! W
$ x7 I% q$ W8 p6 U4 V6.13. }0 T' Q3 r m8 B
#include"math.h"
3 \+ v& {7 Y- x4 O- Ymain()
: P; k2 x* T% D4 `. y7 \. l{float x0,x1,x2,fx0,fx1,fx2;
& m( F. x. _8 Q. v3 Y do S4 K4 @7 ]8 m' w
{scanf("%f,%f",&x1,&x2);; }; X" V* q+ v( d4 Z
fx1=x1*((2*x1-4)*x1+3)-6;% A. _) p2 H- Y6 P$ _( g: ]7 h2 e) l
fx2=x2*((2*x2-4)*x2+3)-6;
& Z, H" f1 C) l( P# |2 O0 u* { }
1 \+ ~+ R8 p/ A+ C3 [ while(fx1*fx2>0);; k- G) N2 i# b( @" Z
do; }+ z; h! s7 b$ g3 c/ a, [2 ?
{x0=(x1+x2)/2;! R0 x" i& \3 M
fx0=x0*((2*x0-4)*x0+3)-6;. ~2 Q. Y) M% j: \# f1 S- K5 x
if((fx0*fx1)<0)
4 X V8 w8 M! Q8 t. t7 Q {x2=x0;
9 V; A9 Y1 @- a/ Y0 I fx2=fx0;1 _* s. a' x1 O. `# p1 S
}
1 v6 s& F5 r* N, D+ B* A! O else
8 l0 d2 o* H, |/ a0 e1 ~3 b3 h7 C {x1=x0;$ M0 D% d; u( H; D
fx1=fx0;! A) B% y, b ~; h* T1 w3 A( Z
}0 p% Y7 a. T6 I' S1 B; C& j% Z. z% ?
}
$ @) G1 Y# {# C while(fabs(fx0)>=1e-5);
- o: [5 b/ q. I# Q8 ` printf("x0=%6.2f\n",x0);: Q; ^& z* C/ K' T% z$ z
}5 R3 p# {6 d% B
6.14打印图案9 Y2 ^) b |+ W
main()* k9 Q) O# k" g2 L
{int i,j,k;
9 h# z" t o# i for(i=0;i<=3;i++)
- Q) C! [# x( l- F8 J3 z7 L {for(j=0;j<=2-i;j++)# t2 N: T3 [6 E, Q, c8 N/ V
printf(" ");0 X6 s3 w& \0 a' A9 ?
for(k=0;k<=2*i;k++)
v' H0 y! m( ~ printf("*");0 `1 [+ S" L/ {) u. V0 i5 C" [" ]
printf("\n");3 F/ c( q, y9 r+ [2 L# L
}& h& U1 e3 T! |% ]
for(i=0;i<=2;i++)
. o8 ~& h7 v4 M {for(j=0;j<=i;j++)
; [, p9 L+ N" b7 W5 n; P2 \" w- X printf(" ");
. l1 L$ y7 x+ D) U9 Z/ Z for(k=0;k<=4-2*i;k++)
0 I' e. M' l/ }( f6 z/ X: f8 f printf("*");
( d% q/ J3 K, D7 s* k- M. Q z printf("\n");8 y( q) ~4 X' c
}
5 Y# A% g* _$ Z- P}
+ V; S) A, Y) d! T& `, W q; z$ T6.15乒乓比赛3 F! Z2 M& t" i: {
main()/ C/ F0 i C, f* ?" v
{0 t! L7 c2 A/ ~+ k) ^1 k) X
char i,j,k;* U$ ^ y3 Z. r; k& u" M3 b" U
for(i='x';i<='z';i++)
# V# S" N& u i/ V) Dfor(j='x';j<='z';j++)+ u" u) A1 f$ k8 d
{
8 t; a+ a7 V8 O1 O# l5 Dif(i!=j)
0 N# ?% y! s( G! {for(k='x';k<='z';k++)
' d0 D- u$ R8 Y' A {
1 n% s1 S& M: {0 x: V) x+ Q% Eif(i!=k&&j!=k)& L" [! B3 w! s9 P
{if(i!='x' && k!='x' && k! ='z')
6 S/ `3 L* D# U/ d" J: Uprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);
+ {* p3 d4 w- M0 \ }; p7 u! O2 q$ e$ x: }1 ?3 S
}
: x% b+ M, h2 x# {: T& I }
- n' a- H' ?4 M, s/ D}
" M3 ~: J6 h# ~) [9 d" A, F+ AC语言设计谭浩强第三版的课后习题答案
, N1 p8 I6 n7 l- E7.1用筛选法求100之内的素数.
6 b. ~0 V+ O4 l8 V2 y#include
" I% ~% C% w/ O4 E) C6 ]) G#define N 101
0 Z. `3 Z! B; O: {0 I# {main(); K/ {5 F& C4 A
{int i,j,line,a[N];5 X4 G8 y! T2 `1 F; B1 N
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
& |+ F. h! o) N if(a[j]%a[i]==0)# L0 n$ A+ y- r: c5 q' p' N2 ~
a[j]=0;
1 @8 m' v! O% Sprintf("\n");
' p6 }9 G* J! M. dfor(i=2,line=0;i{ if(a[i]!=0)8 B+ ^' t0 } B- e
{printf("%5d",a[i]);, }+ z0 c$ p3 {5 S/ e3 l
line++;
5 x9 W+ p6 w6 |' I: g6 e" H if(line==10)
; v W% E8 f. Y: w: ?+ S {printf("\n");
3 ?; L# _. w W; X* I/ N/ `/ O line=0;} d9 G4 X( u" b
}1 i* I! H, q" |( r1 d
}: { x1 x ^% o6 m' L |$ ~
7.2用选择法对10个数排序. X4 i! L4 G8 C7 C$ Q! Y
#define N 10
, N' m# v; n. X$ Q' q9 p9 _# O" Qmain()
1 \' q w) S3 g& m{ int i,j,min,temp,a[N];8 y7 o9 R* M( R- w! q: g) Y
printf("请输入十个数:\n");" `4 H, N$ c" @! N u5 \' ^' i0 t
for (i=0;i{ printf("a[%d]=",i);" t( R- M6 \7 d$ L4 `
scanf("%d",&a[i]);: \ S4 y& @& W6 t
}3 Q5 A M& n& a& E j
printf("\n");2 p- U# S' X5 X7 Q b; _; M
for(i=0;i printf("%5d",a[i]);/ E( m7 Q1 n6 N# q1 a5 h' M
printf("\n");! c9 ]% {' M4 I6 k: ?8 `
for (i=0;i{ min=i;9 P) V5 @' |2 O% Q4 G2 H3 W* D, ?
for(j=i+1;j if(a[min]>a[j]) min=j;
, b7 U/ w$ R, p$ b" t& I temp=a[i];# R& a0 z* j2 z5 |
a[i]=a[min];3 m1 D* X! W% B, B, a. N
a[min]=temp;2 V7 d; ^' g& C E! i6 ^9 W$ M
}4 t; }! q/ F2 g8 x* r
printf("\n排序结果如下:\n");& E) H& O3 _+ S+ P s' q5 h0 \8 L
for(i=0;iprintf("%5d",a[i]);
! ~- a3 }6 p2 \3 J2 i}0 \; d) `+ p% u! p
7.3对角线和:/ V# O/ @* @! h6 t( |* M7 P8 c% r) R5 N
main()
2 v) |- n J, O7 s{! w/ y8 j( L. j
float a[3][3],sum=0;
, X2 E. h# W: S t. ?int i,j;7 i2 l5 j' C9 j# C j' o0 K ~$ @
printf("请输入矩阵元素:\n");; h! u4 F: `- p& d
for(i=0;i<3;i++)
" \: r8 @. B% q. r0 |5 e for(j=0;j<3;j++)
5 o `- ]% v8 c$ S5 _5 \$ F scanf("%f",&a[i][j]);( ~2 }' A: i0 g: Y( Y
for(i=0;i<3;i++)
8 U$ a+ e, _) w1 C+ q$ d5 ` sum=sum+a[i][i];
6 E q+ b8 [# J+ h printf("对角元素之和=6.2f",sum);/ z9 c T8 p# Y5 G+ G- t
}( M; j- V, c; @' |
7.4插入数据到数组
1 H# A$ W8 s* P1 Smain()
2 `# Y, E: e* }# y7 o( h{int a[11]={1,4,6,9,13,16,19,28,40,100};6 {. L7 B9 U1 ~2 Q6 E8 q
int temp1,temp2,number,end,i,j;
/ }& @# |% q$ t% `' Pprintf("初始数组如下:");
, k8 N& g ?8 t; t* Afor (i=0;i<10;i++)- _% }. q& S7 ^/ S7 n v5 e
printf("%5d",a[i]);
8 T( O- s+ s4 e8 P/ S5 Gprintf("\n");
; `$ e/ [+ g4 X! L# Nprintf("输入插入数据:");
, Y* q# n, t( h4 ~% n! X2 `scanf("%d",&number);# ]+ m: Z0 j1 \ x: w
end=a[9];
9 ~. U1 l. a% f- x+ jif(number>end)
) M Y4 K- w; z- @% n- Ua[10]=number;
& E2 Y7 [. `6 Yelse' H+ [* f7 E( q. h z7 g
{for(i=0;i<10;i++)
% M* H) Y1 G% i- Y/ v2 e$ E { if(a[i]>number)- \( F5 ^! W! d5 B% Q: c6 ~) w! }5 v) _
{temp1=a[i];
; m" W+ u( m& E$ M( x3 Z% ?. y5 c a[i]=number;8 F, {4 U S) S$ ~9 k
for(j=i+1;j<11;j++)
2 B. P" M0 z, p8 Y {temp2=a[j];0 J0 N3 j# o+ o2 S' h; N2 X! p7 v& I/ [
a[j]=temp1;
% S! s6 W5 Q/ S& b0 {' V temp1=temp2;! K& Z ~( w( ?# \6 U" T
}
) C8 O, G A) D/ s break;
" t4 R) j6 t- }. z; c }
3 f4 {! P2 H. a6 e: u$ R) C }( [4 ]' S) m2 X% [
}5 J! q- J0 G; m2 W# x! J7 n7 [4 u! m
for(i=0;j<11;i++)9 S! k1 G2 X, Y
printf("a%6d",a[i]);
2 ? }7 H! p0 R}
! d- W" v. J$ J8 }3 }& c7.5将一个数组逆序存放。$ ~% {1 g4 W$ P/ Z/ E) c Y; K1 {
#define N 51 ?. v; O1 E) b# P$ d
main()
9 k- ]" `+ T1 m1 m3 R; Z{ int a[N]={8,6,5,4,1},i,temp;% u4 ^7 N% P% P6 M. a: {1 {5 H- M
printf("\n 初始数组:\n");
! Z7 r4 Y/ X& n+ @# n- r3 Pfor(i=0;iprintf("%4d",a[i]);1 u; A& V; `: u" W. [
for(i=0;i{ temp=a[i];% i8 C3 h L7 _ {) I1 l+ m3 X8 p( Z
a[i]=a[N-i-1];
9 a6 m6 l1 B2 V4 J$ N# m' x a[N-i-1]=temp;5 C- G' i z5 Q% [, y
}
5 _" M8 ]( p. U( T5 \6 W% ?9 Wprintf("\n 交换后的数组:\n");
, P4 y4 g3 [" H% [1 x/ Jfor(i=0;i printf("%4d",a[i]);
" m6 H0 L* \. g' M}. t/ Q7 c) o* [" A( f/ _: L8 }0 ^
7.6杨辉三角
$ }, J; y2 \% x: W1 n+ g) t' m#define N 11
1 W' Y$ d- H- j( S; p+ cmain()
' q( ~4 h) T; [$ \{ int i,j,a[N][N];$ i$ ?: f ^% @6 i8 O; M7 ?" j
for(i=1;i {a[i][i]=1;
7 f& Z/ u3 ~1 L4 }2 d a[i][1]=1;! k( N6 ~9 x `
}/ |4 S3 h6 f, v- h9 U' Z
for(i=3;i for(j=2;j<=i-1;j++)
2 Z+ J4 c5 _! |' d- t' K a[i][j]=a[i01][j-1]+a[i-1][j];
( k9 A3 ^* \/ S" O! a for(i=1;i { for(j=1;j<=i;j++)7 S: @& v# ^( {1 @& N+ Y0 v' a. Q/ p
printf("%6d",a[i][j];
" M' A% z1 |& r7 }% ~$ n printf("\n");# ?- I. h, v- T
}
' x, }* J, F( f- o+ L printf("\n");& E6 {$ [" I0 G/ m
}
- J3 j0 k7 E7 j% b+ ]7.8鞍点6 w4 G: H( u; G P0 z/ z6 x
#define N 10
) G( x x+ s0 A2 z2 b I#define M 10
. j) N+ B1 O X! vmain()+ h" k% }1 N$ ?& ~
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;( U7 ]- n( u" P4 S
printf("\n输入行数n:");: y8 w; e) I& J4 M
scanf("%d",&n);
0 b' r5 @6 T% | printf("\n输入列数m:");
/ O$ o+ z: L3 F8 q& E% l scanf("%d",&m);1 j! _9 V" x& ]& r# [/ ?
, N- @- ]8 x1 l; I3 v for(i=0;i { printf("第%d行?\n",i);
( e$ \; }+ v2 l1 _* { for(j=0;j scanf("%d",&a[i][j];$ S; j1 p* h3 c* n) x
}
& n8 c( ~: \% B: c+ ?1 X for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
: H) r: ?# |2 ~' A pritf("\n");
7 l9 N2 M4 O$ | G; `& Y: i6 y }
' G/ u5 z5 e" w% ?( O flag2=0;% Z i, o9 C0 Z; e2 L0 F4 O# @
for(i=0;i { max=a[i][0];
R0 D6 O3 K' x for(j=0;j if(a[i][j]>max)7 `, Z. x( Q6 g
{ max=a[i][j];
4 x$ `* e9 T7 a$ k9 z7 ? f maxj=j;: @6 z9 s2 {+ y
}" }9 T: Q6 b- L( W9 F
for (k=0,flag1=1;k if(max>a[k][max])
5 j" _* G9 |" s* Y, Y% G3 W flag1=0;
3 A! y, J8 e4 I3 ?) h) C if(flag1)5 {8 p) m4 r7 O' j/ U+ V
{ printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);
, \/ X$ Q' i5 x* W2 q: ` flag2=1;! W$ W" U; ?' T% c- L
}
1 i6 b( y2 Y/ T6 k}
) p }* U0 m& e. d* i& {6 Yif(!flag2)
F" j. \+ w6 B) n; f printf("\n 矩阵中无鞍点! \n");
! h* s$ R( k% a* O}
- n& ?6 \5 g2 U7 l- R; e# |( t6 Y: `! V2 Z7 I- Q6 I* N
7.9变量说明:top,bott:查找区间两端点的下标;loca:查找成功与否的开关变量.
1 [3 G6 K: v$ G! p B#include# L5 O. e6 b( w/ _2 F
#define N 15
: ^7 h) U4 r! h X) Gmain()
. E' I, Z1 Q# ]: K8 M+ ]$ @) n{ int i,j,number,top,bott,min,loca,a[N],flag;
2 y3 f$ a& {2 O2 N' ~ char c;. z: s; p- `3 ]# J, f7 Y; \) D8 I' r
printf("输入15个数(a[i]>[i-1])\n);# g4 n7 V0 S" n$ p" L& |" M
scanf("%d",&a[0]);
/ k) D9 ]$ A- L* m i=1;& D( c, K" E$ O0 |$ `
while(i { scanf("%d",&a[i]); |' r) `, M, C
if(a[i]>=a[i-1])
& u( Z q* v3 u' |8 g9 X2 l i++;
4 b+ F( i( S3 e3 W esle8 N7 c6 y2 V9 ~0 H& _% K' g) W3 o
{printf("请重输入a[i]");9 |& z6 B- t2 ?& l/ i6 ?
printf("必须大于%d\n",a[i-1]);
- ^: H, C" q. U$ Y }8 ?6 J' d+ z o1 F
}
, h2 E4 i* f5 x5 U6 X printf("\n");
; B6 j( x- y& K. c for(i=0;i printf("%4d",a[i]);9 `) v' L6 C* `: U' n4 h
printf("\n");
, n1 m" Z. g/ m" L; ^/ w! [ 3 ?, Q7 S- A. |1 L
flag=1;9 E7 h) c1 M8 W. [
while(flag), T$ \( D' `$ `# M. j
{# R+ `" O1 E- d/ m! V/ S1 U
printf("请输入查找数据:");% a4 `2 r% N4 J
scanf("%d",&number);
) b/ J. _: Q( T loca=0;
. }; V! s# V; G top=0;
_) Z! F- m$ e, E- b4 l bott=N-1;* Y4 u; v4 J" `, L. x
if((numbera[N-1]))9 I1 v8 j, @+ `3 c& _5 t
loca=-1;2 f$ w1 T/ \4 m8 u7 _" J2 w+ F
while((loca==0)&&(top<=bott))% F" S$ \! @) ~: [+ [4 d
{ min=(bott+top)/2; c7 t9 s: }0 J6 e
if(number==a[min])
) R- w s! F) l3 v { loca=min;2 t! |/ d) W F; D
printf("%d位于表中第%d个数\n",number,loca+1);
* w6 b" D' I# X* P* H7 a% W }7 X8 r3 f/ E# c L
else if(number bott=min-1;( ]8 x3 _) F' z9 w% u0 u. T
else
" G& s# `5 D1 Y! W top=min+1;& ^0 B. O: N" j. A0 Z6 ?
}7 N, `6 |# A7 A, M- f5 F" ^* x" W; v
if(loca==0||loca==-1)
, V6 |& c6 S! p+ j) h. D2 ? printf("%d不在表中\n",number);
- X8 d1 N3 _2 a printf("是否继续查找?Y/N!\n");) }8 ]9 Z* |$ w
c=getchar(); k$ k) k0 R$ D$ }7 v, }, P
if(c=='N'||c=='n')$ U# r9 E1 n9 N9 U9 X2 |5 S0 q
flag=0;7 O! W5 F0 M& \# t
}, z$ G8 X, _/ K, W; ^+ @
}0 O- p# ^$ b- A% p
$ C/ t [3 [0 S1 m: @" n, z
7.10+ N: S+ R0 p2 A7 v, C
main()
; z4 v- t! L0 g0 m l P7 ]{ int i,j,uppn,lown,dign,span,othn;
7 i" S# {, G# u5 n* d+ [6 C char text[3][80];" [* a: F$ ^3 L" X! M
uppn=lown=dign=span=othn=0;
) d: k5 J: K( A3 \* O8 K4 s) A for(i=0;i<3;i++)
% S9 `" n7 A3 \& N$ w2 \! E) q: U { printf("\n请输入第%d行:\n",i);
- y& ~3 L5 o3 J: Y: [5 n gets(text[i]);5 S' {* ]9 s7 s1 F1 Z
for(j=0;j<80 && text[i][j]!='\0';j++)
1 R; i3 `4 L8 |% r! n {if(text[i][j]>='A' && text[i][j]<='Z')7 c. I$ B5 F5 p2 ^9 K$ W; W
uppn+=1;+ o6 c8 j, D5 s$ V% A
else if(text[i][j]>='a' && text[i][j]<='z')# A& ?" F- {0 `1 y- z
lown+=1;( `2 x+ @" O1 u# w) u. P! @
else if(text[i][j]>='1' && text[i][j]<='9')% ]( z8 O) T1 d/ \
dign+=1;4 g% g Z L6 r, C
else if(text[i][j]=' ')3 c/ D5 I6 O8 E$ E8 a& T3 E8 I5 y2 |
span+=1;1 @ r( F j0 b) f
else
4 a/ ` ^6 c# M othn+=1;$ [( n$ y1 g. g
}
: s* @5 e% }& C4 n( x }& Z; r7 L: ~% z% l, ^
for(i=0;i<3;i++)
7 k5 e8 h, X' T7 T! T/ o* T5 _3 X% E printf("%s=n",text[i]);
5 u& U# y& k% N' v) k, ~ printf("大写字母数:%d\n",uppn);; r& `) o d: ^2 V& [
printf("小写字母数:%d\n",lown);( ]' M, E3 |9 h2 @+ w3 \
printf("数字个数:%d\n",dign);
9 Q: ~; v8 w& @6 y2 t printf("空格个数:%d\n",span);
& G- x e9 } {2 x printf("其它字符:%d\n",othn);
; T& C! v3 J2 A3 }}
/ S9 Q8 S7 m9 n# f. r p7 s ^* S+ p' h9 b6 d
/ R1 D2 I3 {7 x4 j6 O7.11
$ U G. u1 _ w! ?main(): ?0 u- h! M" I- I3 Y
{static char a[5]={'*','*','*','*','*'};
$ d, ^5 h+ x* {7 I; H int i,j,k;3 k1 } s8 u9 e) c9 U
char space=' ';
; A5 Y5 X3 h& o for(i=0;i<=5;i++)
3 n5 ~8 o% M! ]8 g: K- I5 k: f7 ? {printf("\n");
) e" y [" k+ E( X' Z+ m4 f for(j=1;j<=3*i;j++)8 V& ^2 a1 F4 s4 B3 d8 c
printf("%lc",space);5 [5 v/ O$ s) [) N7 p# A1 I0 u% N
for(k=0;k<=5;k++)
8 y5 o! ~+ E. g7 a3 w. M printf("%3c",a[k];6 {+ d, b, \2 x6 Z3 d) U, [" n
}
! R: w9 [' p5 K( B. F/ b& Z5 Y: b8 p}
3 ?- n. M# V2 w* u l& a' u l; ]7.12) G! J; r% @( R; L
#include$ d! d4 r* ?9 W" n
main()5 o( X6 H! L" r9 ^7 n% i% r
{int i,n;
- }* Q: I+ T$ d char ch[80],tran[80];
; w8 Q% a% J3 n; y, J printf("请输入字符:");* B* O4 J$ ]1 w. R
gets(ch);+ \* Z- G" q; Q" r8 Q% j5 c% u
printf("\n密码是%c",ch);5 P0 ^7 l! i: L5 S1 b J3 ~/ g, z
i=0;
7 s7 @( O4 ]) j3 K& M7 r1 r, p9 {while(ch[i]!='\0')
2 J4 T3 z8 l1 G{if((ch[i]>='A')&&(ch[i]<='Z'))/ t% [ C7 |8 z
tran[i]=26+64-ch[i]+1+64;7 p8 B' n \7 B+ r/ J& B; }
else if((ch[i]>='a')&&(ch[i]<='z'))
0 r( J$ y8 }1 J h/ s1 M$ q tran[i]=26+96-ch[i]+1+96;
5 n) v& O" h& s* {0 @else* @" i; ~: E3 ^- V' @
tran[i]=ch[i];: s7 o9 s( Y: v. N6 I
i++;/ X H E& f7 ?3 j/ h6 H+ o
}3 Y7 W( a s1 {# L
n=i;+ Z N7 m/ n, z# W9 E( V8 w- \
printf("\n原文是:");' t( Q4 o$ o$ Q6 Q0 k( u
for(i=0;iputchar(tran[i]);
7 H6 `2 b9 U% I: ^- v}$ }, {; R2 N0 m
7.13, C" A) ^- u7 d6 p& Q/ w/ a) D% s
main()( M. A. N, P7 i# _
{
$ @3 K, H) o- f X. n char s1[80],s2[40];
" L, D- \. S& J2 b! ~! e/ ~8 z1 r int i=0,j=0;
# s% }$ X) k5 S. m; K. y; J printf("\n请输入字符串1:");8 n* C- f9 T1 s7 u' m; q+ T
scanf("%s",s1);
+ X% L5 \$ _- Z+ x( c4 I- q3 e6 u printf("\n请输入字符串2:");2 m6 r# S- p) S' g" r
scanf("%s",s2);
. \7 w* f: L- a9 C while(s1[i]!='\0') y. m/ [4 h# N% s
i++;
0 e; A8 d$ ]) p( _' R9 qwhile(s2[j]!='\0')
+ ?% n9 N+ W# S K! r s1[i++]=s2[j++];
# U+ q, g+ t& ~! W" P9 ^s1[i]='\0';& J; z- d- M2 Z- l
printf("\n连接后字符串为:%s",s1);% J3 j& M$ a4 v
}
% z) W3 l7 }' U8 Y) m6 f% T
& v3 t/ X* H+ Q
7 x) @7 i) B. }/ Q7.14
- r5 X4 H2 b" T! H6 o. k6 h! u#include
2 B$ A& x$ s" A- U$ P5 Y" gmain()
/ n N( t0 Z5 \6 x* `1 R" c2 q{int i,resu;
( y( J, Y8 ~$ S6 P char s1[100],s2[100];
2 q5 l' v* N$ t3 h* B: G printf("请输入字符串1:\n");
) Q- v+ Y3 T2 {" W7 g! G) w gets(s1);
; Z4 }& y. L& A$ Z, |0 o* X" ^ printf("\n 请输入字符串2:\n");
+ l7 A8 [, N& q! m gets(s2);6 G* c) F- O# q( D) q
i=0;' ^+ _5 T( J; _9 A' u
while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;
3 L' z* v4 q/ F$ d8 Y8 X if(s1[i]=='\0' && s2[i]=='\0')resu=0;
7 h! `- X) R8 L$ } else
! x) X. j* a! u# m resu=s1[i]-s2[i];
1 N& m1 H% c+ m1 G3 i* {" W printf(" %s与%s比较结果是%d",s1,s2,resu);( W4 @3 u. R% K8 g
}
* O- M2 f3 n" [! I6 G( Y7.15$ p/ A" v" n3 ]1 D0 x7 k+ r9 ~
#include
( e: H. y3 h* [5 f% Rmain()
5 |" f7 m! ~7 |2 c$ Q {/ h2 z: }: v C; X2 @4 J
char from[80],to[80];2 m. F) w) G: f2 ?
int i;
3 b1 E0 x9 e4 S3 }7 j7 |( X printf("请输入字符串");
# r# r: C: }- C& b$ K) j3 L scanf("%s",from);
) |5 R; O, M. h% n# P4 g for(i=0;i<=strlen(from);i++)# {9 z/ D5 |5 @! K' X Z9 A7 c
to[i]=from[i];. n* ^( V- {: b
printf("复制字符串为:%s\n",to);
/ o, ^; W) Q5 K" t }
% H, x& P" d/ U3 D; m
+ l* B( K- D$ Q
% w; q; a" p( [* P第八章 函数
- p: a2 N- ~5 B+ D4 o, U8.1(最小公倍数=u*v/最大公约数.)) p: }) {9 f5 I- ]/ e$ j
hcf(u,v)
4 t4 g! t7 H7 P! i- G" m5 Cint u,v;( S* v% Z6 s6 e0 e# O0 g8 E8 q. w
(int a,b,t,r;, X$ E7 X$ ~" ~
if(u>v)
& E- J$ k& Q/ n/ B5 s5 E {t=u;u=v;v=t;}
$ p6 l1 l/ Y p9 W) B% a$ j7 h a=u;b=v;
D6 X; V: @/ m! ?) K( X) k while((r=b%a)!=0)' g3 \% V, W. }8 r4 S
{b=a;a=r;}$ K3 e/ |; g$ q/ ]( l6 v1 d
return(a);2 N% s3 z! n( C; L
}! S6 }- Z: s; M+ l+ l
lcd(u,v,h)! }& a# a+ b' ~8 t$ q. Q
int u,v,h;
$ P* y* x/ ^2 Q8 F {int u,v,h,l;3 F( _$ P- n }: e
scanf("%d,%d",&u,&v);+ k" N( S6 }/ L3 u8 j1 c
h=hcf(u,v);
. D1 u1 Y& P6 D$ J9 S printf("H.C.F=%d\n",h);
$ N( A! i2 _( g4 @$ I& D5 M& m; Y l=lcd(u,v,h);: |0 |9 V8 J$ J" V1 d c
printf("L.C.d=%d\n",l);; V& S' q$ w& e; i/ x: E! l0 R8 g
}: ]7 P( u, G, F
{return(u*v/h);}
- }/ O! E; D+ D, l6 ~ main()/ j* {; d' U" |0 H0 w
{int u,v,h,l;
( p1 G8 S$ y: `# `3 k7 M scanf("%d,%d",&u,&v);
) Y! C' _7 M5 K# i, W h=hcf(u,v);. w1 _* S1 r# W+ u
printf("H.C.F=%d\n",h);8 a5 s$ A4 |, N8 [0 n' G8 W T$ D
l=lcd(u,v,h);
+ b; ^! E2 J$ [3 U3 I, W printf("L.C.D=%d\n",l);2 x T+ G8 s# ~! V; V5 ^
}
+ C( z+ C, l5 J" m
' c, W+ ]/ ]* J1 H! F
# n: a; o' }: T; T4 `+ w3 l( i$ `$ R; a
8.2求方程根
2 P- C6 I" `' W: o6 i5 m& ^, z#include2 T5 G9 P; z; o/ m
float x1,x2,disc,p,q;
9 s3 l4 L9 @3 G6 Ggreater_than_zero(a,b)
8 u# u. I+ e% _8 H4 rfloat a,b;
{* `1 V$ `8 Q$ a- y$ @{
# }% o/ ?% n- w) I( n& v' l! Kx1=(-b+sqrt(disc))/(2*a);! h- n6 [" g: P% ]! d8 I) M+ |
x2=(-b-sqrt(disc))/(2*a);
) b. R0 K/ x/ f: k- d}, X* c) O4 w* p% k
equal_to_zero(a,b)
% T" `) s# v5 F& E ^" Nfloat a,b; V \5 ^# F ~( O% [: r4 k
{x1=x2=(-b)/(2*a);}, ~5 P; O- K$ y$ _* S
smaller_than_zero(a,b)0 S. H" I T# O& ^* c
float a,b;
: ?' G" S* j$ o% _0 X6 q0 V{p=-b/(2*a);
3 P5 \ V4 n2 o2 \, B4 [/ _q=sqrt(disc)/(2*a);1 X7 {5 ~- Z9 c8 g. a' w
}( J4 G3 ^4 z+ c) n0 O4 Q
main()0 j2 n$ Y6 S, m0 ~
{, A/ G v& V1 E
float a,b,c;: s* j; a" A: ^- Y3 N1 F
printf("\n输入方程的系数a,b,c:\n");
% Z* J0 G c2 pscanf("%f,%f,%f",&a,&b,&c);
. ~( H6 ]6 ?! x9 _. T4 Cprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c); p8 p1 `1 M" k3 @! Z
disc=b*b-4*a*c;8 Y0 S, E9 n* M
printf("方程的解是:\n");( l% I& j5 ]$ o. v! l$ h
if(disc>0). v: Q; B0 E& E* y* |( c$ s6 t
{great_than_zero(a,b);
! l! j9 F2 G: S! m0 iprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);* Z, A/ u) D: d4 d, F2 }- M
}
) {) ^( t! ]; Delse if(disc==0)
# q2 A) [1 y, Q1 q5 y) K; c {
0 ~" p, b- q( C, Azero(a,b);* e2 l& e d0 ?' l! L
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
9 U- s- E9 V% g& H, @4 Z% k4 |, L }- ]1 l( V* J) ^$ r+ s7 \
else
) `3 H; e' \5 u/ Y" ]! k( s {
. C, y! Z, K3 L7 [# ?! h small_than_zero(a,b,c);
$ U. j3 b4 W8 o+ x printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);2 m2 V+ U3 ?2 _& C: {' L
}6 f. v" H0 o& R9 Q. f9 e/ B% v
}9 Z3 T: o& x+ ]: m# y
8.3素数
, z. m( D# t2 u3 c9 E#include"math.h"+ h- Q6 F3 i# @4 q# L- Q* E: R/ ^
main()
2 y4 B% A5 v& ]2 E1 u{int number;
/ F- }* [1 w7 }! h* t, w scanf("%d",&number);+ D$ H7 Z% g+ b2 {2 h( O5 Y: D- T
if(prime(number))! ^" r' P8 n6 _4 c
printf("yes");% p% t7 w2 Q* C
else+ D7 @7 {/ E" I) b
printf("no");* {" h: [: K$ x2 x+ @
}% V% L2 \5 F5 `* l- t" C5 t' e4 x* y
int prime(number)$ w* w6 e+ ~8 {: q- e |
int number;
; S$ N3 J& l! o' t{int flag=1,n;
. c2 ?: g* Q/ a/ E for(n=2;n if(number%n==0)
9 I! n5 Y' p) n0 K1 w flag=0;
* \) z7 c& L0 Y return(flag);
' m( Q2 h+ }. o9 a}
9 D) [- Y. h+ R' ~7 W& ~ B t
( L6 t2 C) k6 C
8 w+ c* D% W% K- h
% ]1 r# a5 l" {: ?( y6 ]. M6 Q* L8.4! d$ o5 Y/ ], D2 E' H/ d
#define N 36 w1 E, v$ U2 u
int array[N][N];: t9 J4 u: F) H8 r
convert(array)1 w7 r3 i @5 B/ a) R0 { `8 a
int array[3][3];
# g0 n" m c2 C7 d5 o7 c { int i,j,t;2 E5 ^- T: L# |* a2 g& J! l
for(i=0;i for(j=i+1;j { t=array[i][j];) Z8 g' H: E2 \4 h; ^, \
array[i][j]=array[j][i];; {9 @) k T# P5 Y
array[j][i]=t;2 F) l( w/ P$ n! a
}
& X, Y+ ? ^+ |0 |8 _ }
: O; C5 ^0 h) N" a: S) X. mmain(); S$ a+ `1 z# a: ]+ W
{: p" g% r! r8 L
int i,j;
. t ^& V" p( K printf("输入数组元素:\n");
$ g4 a4 E" \" i" o" q0 s for(i=0;i for(j=0;j scanf("%d",&array[i][j];
$ x' K7 W5 O3 m) p( _+ j$ @8 k printf("\n数组是:\n");, @! `7 q# _9 ]. y9 W7 ]
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
& D' U: k, d! `1 w; z7 {- {/ Z printf("\n");; Z2 w# z, J1 }2 v5 [
}0 ?+ ^8 O8 Q+ z1 V6 O
convert(array);
% b7 H2 Q: H: A" C printf("转置数组是:\n");
: m+ L2 C1 `. Y for(i=0;i { for(j=0;j printf("%5d",array[i][j]);% e9 I7 e9 S X$ {4 @
printf("\n");
+ R R V% ?( e }
9 _9 D# Y0 @8 m, T, k( V}, N( [5 a+ q7 l9 n8 W
$ K" Q) u2 S8 b6 v6 p7 o: N7 K
{% m$ y# `# G( r4 ~$ ^$ e% S5 x3 P( i( j3 w2 n# X
8.5
1 P% k$ l, Q# J. r* J: nmain()
( h0 c. W4 _* G, S2 F{
: J' k; E+ S" C0 jchar str[100];0 c+ R# c- `5 T( n/ X' M# L/ i: @
printf("输入字符串:\n");( h8 a% L! f2 @( n/ A3 |
scanf("%s",str);
5 U0 {$ ?3 b9 H7 m A inverse(str);; g. d7 ~/ B+ c# j7 h# m5 N( a6 u
printf("转换后的字符串是: %s\n",str);0 q! x6 ?/ k, u9 @/ X, n- u
}
+ f; g# \! {: o ?inverse(str)
& C d7 u0 S$ l& Rchar str[];
3 \/ r8 r @% f! L9 w{6 C/ @% Z F1 I: e$ B& h
char t;0 D' N# t( B( I9 X. d) p8 y' c( {
int i,j;
' W/ X9 ?$ ?4 s* ?) R, ~5 e$ { for(i=0,j=strlen(str);i {
$ ~1 |; D( ~# X7 { t=str[i];6 Y N/ E, M5 i. O k) c; E/ ^; a
str[i]=str[i-1];4 \8 R, r4 g8 c7 m& K
str[i-1]=t;# D+ L0 q7 z( G6 |9 h- V, ^, E
}
5 i0 X' D3 N+ t}
* a) i$ B7 h' ]* ]. R
}" J, F- D* B J+ \, z) X) i- J
! m$ R, M( x+ i6 R) y4 ^- j% Y& P$ `) I* v% C5 {
8.6$ `! _2 g$ \& G# E+ z K+ ^
char concatenate(string1,string2,string);
F- m, R& Q7 Bchar string1[],string2[],string[];3 T* G! R0 M1 j. c) [3 Q0 {
{# |, Z8 i1 T5 j; {: l: j" R
int i,j;
, W9 Z9 e/ N' ^1 f# A( dfor(i=0;string1[i]!='\0';i++)
* {9 r y/ g. L; C6 G7 k& j string[i]=string1[i];
2 m; Q: c* ?; g3 ^, U* \for(j=0;string2[j]!='\0';j++)* y# t. q% Z# t1 n2 s3 z% b& H( s9 m
string[i+j]=string2[j];
- `/ F3 \6 N, D$ i) `0 p string[i+j]='\0';$ i2 B/ B3 q5 i% n( C& t3 |
}4 u$ F* A5 t$ u
main()- R8 C5 I. o3 ?7 n, u
{
$ z5 s0 T+ m6 |/ u: u$ s( b4 } char s1[100],s2[100],s[100];: i4 _; l( r& |
printf("\n输入字符串1:\n");; [) ]. F8 ~6 h5 m7 \6 H% W
scanf("%s",s1);3 T k: Y2 a- G; g1 s: U. ] N) [
printf("输入字符串2:\n");) g# M5 G1 W7 I. q
scanf("%s",s2);
7 t4 T% r- s# t concatenate(s1,s2,s);0 |$ N8 D! f% A/ b
printf("连接后的字符串:%s\n",s);) A! N* J8 H: y5 V+ g: z
}$ V7 G' G9 H6 I
. A1 _( X/ C4 n/ B
4 k: i G, b+ Q% l% `8 u
8.8) A5 v2 D" ^( |! U
main()3 v' Q/ j/ h) B1 Y; ]6 S& G" g
{
# J. p! e. t4 o: [& t char str[80];* x' U; J: O8 t# h+ G
printf("请输入含有四个数字的字符串:\n");
* m7 ~+ A; ]% [: T% L7 M, x* r5 ^ scanf("%s",str);
- i- e/ X% X, s insert(str);
9 \% N! b" j1 p}+ G% y. K2 }5 m) w. X
insert(str)
3 j8 v& I8 w8 p5 A) p6 f$ t1 W& n) J char str[];
' ]% \6 ?, C8 `& ?7 M7 P0 e{3 |1 k" h, `5 N$ S" x
int i;
, l! @/ v2 o( }) r for(i=strlen(str);i>0;i--)
" c* n2 x% b2 `6 { i1 x) C8 t { str[2*i]=str[i];
* P3 l( Y0 \0 S E' e4 D, r str[2*i-1]=' ';
8 N8 u) W6 m, q- M, j! _/ V }
+ U' f |9 S4 O: n5 W( o+ X printf("\n 结果是:\n %s",str);' F2 I8 B# t0 e3 O
}
% ]. u" Y: T! y6 h5 J" z
+ z6 g& d( }/ L. ^2 w+ q
3 ? W$ d7 |6 ^1 Y3 w @: {% E( ^0 r7 V% P
8.9
+ N$ Z y0 q9 F/ E( u#include"math.h"+ ^, ^: x; I, q7 ? ^. |' `
int alph,digit,space,others;
" t, T) y, \& f+ A* x. u2 Amain()( I+ F. h$ o2 P
{char text[80];6 w% F# h) b& |; d
gets(text);* f6 r2 G4 _, s9 R5 C `' z. U$ m
alph=0,digit=0,space=0,others=0;! S9 t4 p( T- k' @9 @
count(text);
; B) K" f- P9 {; |- ]' W4 K) X+ G# A9 O printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);* y# I) p+ P2 E" ?( i
}3 @" ?' `5 o+ @8 \
count(str)
9 E2 K' J. g1 J9 ~char str[];
( ~+ L" i1 Z$ p9 p7 ]: A9 v z) \" j{int i;
. _) y. l2 V; [. F* a6 @ for(i=0;str[i]!='\0';i++)7 e* Y- Y M) |1 X' Z0 n3 o
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))+ y- |! t, b6 R t# i, `8 G4 W
alph++;8 T# C, L0 T2 G7 G$ S8 k5 Q7 F, [
else if(str[i]>='0'&&str[i]<='9')
/ e* O( K' Y0 U3 \ digit++;
# F- a; B, b8 C$ e+ \+ M else if(strcmp(str[i],' ')==0)4 t% `9 _% b; j& n+ K. \" Q
space++;1 `1 {: q# g, e
else
- _" ^" L1 s0 S' q& d* b others++;$ X V$ v+ ^- a
}
) c8 r% _' X* n3 w9 x0 Y9 U+ k+ a3 M% s- z$ k, O* a
7 e& r7 @) D$ O/ c: L
8.10
+ V1 V8 k$ m* [2 vint alphabetic(c);
& \% J' F! e; m" e7 Ichar c;
G# H# \3 J" Y, W7 J{
8 P' c+ j z) A( O* |4 z1 o1 V* D if((c>='a' && c<='z'||(c>='A' && c<='Z')); @2 h5 G* V* A+ U/ M
return(1);9 o8 [8 Q3 g* B( O4 p
else; ?: v4 g4 T/ b4 ]1 x/ G
return(0);
, W5 R( h. \) l6 L' E}9 [# p8 C$ c1 d# t0 D+ A9 T1 d& p
: T, f r/ g' x0 v1 mint longest (string)
: E0 G9 w6 w6 h- ]- }2 e0 Z3 b" Tchar string[];
% M0 O! n1 ], g+ e* r' {0 g{
C: L2 U6 I. Z# }% s4 N6 {/ j: H! i int len=0,i,length=0,flag=1,place,point;$ k% |" ~% k$ v- i% s
for(i=0;i<=strlen(string);i++)
0 e6 \ T% s0 |9 v: v7 X6 m if(alphabctic(string[i]))
/ i0 L* C& q) J x8 a, {, X if(flag)
- r1 P3 Y: e& }( o, \# ^ {
# ]" i/ A5 D0 m7 E7 f5 J* {8 q8 L point=i;
# v1 T4 z! s3 j flag=0; P" ]+ d4 {$ i" w4 e( S$ [6 P
}2 v/ ]) ?6 w% Z) q9 e
else) H1 [" U0 @+ c2 C0 X
len++;7 `& h8 w" r5 h0 J5 s
else
; m- D) S8 d# C! Z { flag=1;
8 p: m0 A: E7 M" w' t( Z& S9 { if len>length)
& K1 g$ W( X* w: g: _ {length=len;1 }3 @/ Y, e% t, s! @8 Z8 \: Q- `8 G
place=point;
$ m* U! `1 M/ S- g8 u% k/ e len=0;- `0 U" W) [ q* c) a
}
D% Y; l3 q% P3 z( T: J }
/ g: \7 w, [- T# t; Q- X3 L5 R5 g return(place);! w# S8 F( X+ u; S5 }1 [
}
; v8 d: E' X! x6 C! ymain()9 W7 b' P: n; \6 Q# T/ a6 k
{
+ T, a- ?6 x2 k Cint i;1 U. ^- ^" S: S3 ~: M9 C. e8 B
char line[100];
4 I Z- Y* D( e {3 oprintf("输入一行文本\n");' b" C3 k9 V8 M6 j E
gets(line);8 x) ~( I! u. G/ e
printf("\n最长的单词是:"); K0 i6 W$ j5 O1 J+ H$ T
for(i=longest(line);alphabctic(line[i]);i++)6 K: j1 O) |' z7 E- n q3 Q9 [
printf("%c",line[i];7 D! k! _/ j6 [8 J& i7 \
printf("\n");& @& [. _: v, m
}& j! @2 f. Q0 M0 u* q/ d
7 _3 \' x0 X* L, v2 j
# O* @0 ~& F: X$ m z7 L, ^$ i
# G( n% D" w7 E" M N; I8.11* ?6 u" J1 D3 L+ A9 J) V ]
#include0 [! H! |; |2 w& c; v
6 m" e$ _7 j' [6 E: N: s
#define N 10
- z5 K7 S! p7 o- Vchar str[N];
. {- \# E% N" {7 L: e, S4 @% Pmain()
6 m# k: Y8 ]8 ]& ~+ e# h{
: B; }3 F, }' o" xint i,flag;4 r3 T) K1 h; T5 `
for(flag=1;flag==1;)
! A7 q0 e- x u9 j! k{( P6 F n2 H/ F: o
printf("\n输入字符串,长度为10:\n");9 g- X, B- E; _7 I4 @
scanf("%s",&str);
! w, Y! I+ ~( m' v if(strlen(str)>N): O1 y/ i% F' a5 I- O
printf("超过长度,请重输!");/ L. M( l+ o! l
else; W9 }5 z; b; b' a5 J
flag=0;
# X3 a3 ^; |/ O5 u# S' K}' Y, B0 C! ]# h; ?& C; o! t( M0 p
sort(str);
! `; `+ s8 f1 M% P/ G5 E" qprintf("\n 排序结果:");
! A5 N6 R9 q. G% b% {$ afor(i=0;i printf("%c",str[i]);
- N* i( |& x, G& }}
, Q }( W$ i- q' T/ osort(str)
. ^5 Q5 g* l* q" Fchar str[N];1 u2 A- l! b1 O& n9 u5 |" }
{# }: p* o- }+ T4 G. Y A5 m) d
int i,j;
$ q3 b1 W. m% a/ ochar t;: J5 T8 d( d" M4 B. l4 c
for(j=1;j for(i=0;(i if(str[i]>str[i+1])
+ p4 J I R# q8 U; B { t=str[i];; l2 ]' G" G, ^! G3 L4 N# ~! b
str[i]=str[i+1];$ c B1 U4 p+ g7 b! q
str[i+1]=t;* I' q8 X, H8 K7 X0 ^2 L2 Q
}8 V6 ~+ m% _, F- T0 m" g
}8 } q+ v/ k: q* E6 V
8.12% ^2 b( L3 S4 E" ]# } `
#include+ l- |) Q7 f7 }. Y
#include
: G) [9 y4 |, j8 t3 Q: Vfloat solut(a,b,c,d)) d0 r' \' k# d. X' O- X
float a,b,c,d;1 _, o4 J# M0 u; {2 M1 O6 K/ T" }
{float x=1,x0,f,f1;
8 H! H! I7 M$ H" g$ D do
' @9 s! c3 b( Z1 b1 D" B# U0 N {x0=x;
# A$ e* a# l1 y+ q' @ f=((a*x0+b)*x0+c)*x0+d;2 i6 h6 {/ s" F
f1=(3*a*x0+2*b)*x0+c;
% L+ M" c8 }5 Z& { x=x0-f/f1;4 N7 c' f$ W- j+ V5 E6 G; |& F
}1 z) I: @% H3 \' v2 y3 Z! r) i2 \: W
while(fabs(x-x0)>=1e-5);% [$ O/ R% d" A" k' d- c
return(x);
1 F. s) V) ]* ]8 }( s2 S}* a0 b9 a/ W% E4 `
main()
, C5 n$ r: Q+ {8 k! d4 S{float a,b,c,d;9 Z7 X! L5 ?7 n. @+ s, Z
scanf("%f,%f,%f,%f",&a,&b,&c,&d);
( w- X! m4 v* u' m1 S& Y printf("x=%10.7f\n",solut(a,b,c,d));
9 S( r: a" ~0 y}
/ J% O. H* j- X5 n8.13% s: h9 b' Y5 X8 O$ Q( p
#include" y1 t1 d- ?/ \1 P
main()
. \( k2 z q0 G5 v. g{int x,n;
. o( s- G& Y4 W6 g. v9 ~. z4 b9 ^ float p();0 ?! D9 j' b8 D
scanf("%d,%d",&n,&x);9 C* b1 z7 M2 p) N( {9 h) C
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
" S0 J& l- I( m ~7 b0 s}2 x$ P6 _' d! z i* _( Z' e$ x
float p(tn,tx)8 p" H/ w/ P! W0 A: [( j* {& J
int tn,tx;8 T3 \9 j- D) S0 H8 k4 d2 ]' l
{if(tn==0)
, A: m8 z7 k M% g; c$ w9 b return(1);! ~5 Y- C4 v" k+ f: {' N/ t
else if(tn==1)3 Z: I; v f8 \/ W% `4 p! g! B
return(tx);
% u/ k. X2 N" N, X; b, B0 | else
) h9 R7 v$ ?- R) [. ?$ Q return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);* n) y0 [# P6 O: V$ E/ m* L
}
$ v$ p) r; L" _" }' a- |! I8 |* b8.14, y+ T E7 W/ b0 l, U; a. J
#include "stdio.h"
7 `5 T* e' g+ \: E* p% O#define N 10
; s6 g' [4 J H, [' r$ z$ h#define M 5
( E* d- d, r5 g, g) }# p2 |8 Jfloat score[N][M];
. T4 W0 Y. D- T* V# S+ G: yfloat a_stu[N],a_cor[M];' W+ r9 j+ U' o& X3 d. ?& o8 i
main(); n' V2 k( ?& i5 I
{int i,j,r,c;( j$ Y$ X+ @2 A' k/ g+ j
float h;
4 |2 a8 I8 y* w- Q float s_diff();3 f6 ^ e' X! W( p0 ?8 \
float highest();
7 {( { ~; F* e/ V; x! H! x( Q4 a, W r=0;: O' y1 {! H! w
c=1;
3 F) g6 Q% i5 A' j& B input_stu();
2 {6 M$ U: J- w, j# L- R avr_stu();
( U7 U# y g% b% U0 \" p8 N! g) N u avr_cor();3 K7 b* B- o3 o
printf("\n number class 1 2 3 4 5 avr");: h `9 E9 V5 \! E
for(i=0;i {printf("\nNO%2d",i+1);
3 |: b& _2 m0 V. v for(j=0;j printf("%8.2f",score[i][j]);* u7 L* v. v: }+ u) B4 j% Y
printf("%8.2f",a_stu[i]);
: w4 {; E/ g0 f5 k }- F7 S$ w' g# q" S+ F
printf("\nclassavr");; V/ g6 m) u6 s; o
for(j=0;j printf("%8.2f",a_cor[j]);/ R# a c u& C1 g' k ^& Y, z
h=highest(&r,&c);; A0 j. m6 F7 S$ l5 I
printf("\n\n%8.2f %d %d\n",h,r,c);
3 r% P# T; F1 A/ b7 g' l printf("\n %8.2f\n",s_diff()); a5 l. R" q# y0 l* |9 P7 ?
}: D% Y9 f4 f3 k4 ^6 k
input_stu()
' \9 Q9 A: @1 r- d, X{int i,j;& L& ~8 W6 k' `) r7 v. S
float x;' Q* E/ i+ `2 n6 v9 L! Z$ D* C
for(i=0;i {for(j=0;j {scanf("%f",&x);- h; h% L0 b L3 k9 L
score[i][j]=x;
4 n0 n( e2 S) [, y! x% p E. S& P }
% X3 u/ r+ r( x0 y }2 k) U" K; Z& t% ]
}' o! d3 W: U* H6 D
avr_stu()" [/ S) r( @# p6 ^, }3 o8 J
{int i,j;
! s/ E8 W& }% P2 J& e float s;0 R" w+ b9 |& ?* m# @, ^
for(i=0;i {for(j=0,s=0;j s+=score[i][j];
1 W$ P# I# ]* d% h a_stu[i]=s/5.0;
4 U& W; G, K% }' I- Y5 z }
8 f. t% x/ g9 B, M}0 b/ k E$ Y! g3 z% j* K- b; w1 f" U
avr_cor()7 Z' M' g# P% h( o1 j2 Q7 Z6 }- [) D
{int i,j;! e- |. p S1 Z% _6 m# ?/ T
float s;; }6 O3 p; R" {6 `/ }
for(j=0;j {for(i=0,s=0;i s+=score[i][j];% _5 i. Y% G: M/ G; x) @
a_cor[j]=s/(float)N;$ m$ N, k; U+ T' r* g
}
. g0 k, C+ e, |5 S, N* V' D. D}4 \# K% ], ]" @+ P8 M) k
float highest(r,c)
2 ]$ m' W7 |, v/ L' p# bint *r,*c;
8 z$ @) q' t+ o8 d: k W5 j: I2 S8 E{float high;
- T2 K; S i& E& ~) d! ] int i,j;2 s7 P3 M3 B4 X6 u' o
high=score[0][0];
" o/ L" y3 x3 A1 _5 h0 A for(i=0;i for(j=0;j if(score[i][j]>high)# `: H7 m( v" o9 F- @% b( M
{high=score[i][j];
+ c# b- M" ^9 ]. S+ h% q *r=i+1;4 r3 [! k& X; A" c& g/ _
*c=j+1;: S" w$ \" I# z+ V% A+ g
}
/ _' M5 h) @; J: O0 b { return(high);
# h* W3 g0 P9 l& W}* U, U# w, J, ~: _* u
float s_diff(): j+ p/ } c- W+ ^3 n9 T
{int i,j;% n z" h) B; ]- m/ f
float sumx=0.0,sumxn=0.0;
i- X1 M3 i- o2 L+ } for(i=0;i {sumx+=a_stu[i]*a_stu[i];( H, I+ Z8 E* h4 i" {4 @
sumxn+=a_stu[i];
u" S5 ?: Q8 Y/ Q }
$ O; Z3 k8 w- C( v9 Q3 h+ L& Z return(sumx/N-(sumxn/N)*(sumxn/N));
! A' l$ R- B0 x9 h}
% h3 O8 M5 J6 k+ l" p8.15
; F9 a; y, o. K6 J/ P( I) d8 C#include2 p: Z$ K( a5 M$ T, {( T; _
#define N 10. S# K7 j, ~ ?+ h
void input_e(num,name)$ O: L/ U9 |* c% c4 ], ]' F
int num[];
* K* ~5 M" D1 `1 r$ Mchar name[N][8];
# x* S# n% R4 D% B6 ^, L2 I# r' |6 Z{int i;. c8 y) Q! V3 V5 w% _2 j
for(i=0;i {scanf("%d",&num[i]);
# G/ K p* U" O gets(name[i]);
8 b1 Y2 A- F3 H" ` o7 v }
3 S/ r1 J: y% K}
) K- }. B- O5 r. r2 }void sort(num,name)1 d7 q6 _; F, c( L' k# p V1 R4 o5 F
int num[];* l! }/ ~0 h6 \0 f7 ~+ Z- ~ L0 c
char name[N][8];
0 O) T6 G1 a7 r6 L# b/ a{int i,j,min,temp1;
9 M" b' \8 E. s8 K+ n G char temp2[8];
" g) C! q8 Z5 k! {2 m% Q for(i=0;i {min=i;$ p5 O. a- m8 n& x8 y! ]; j0 f0 e
for(j=i;j if(num[min]>num[j])min=j;
5 T5 U% y6 a0 o# x' C1 Y$ t' b+ H temp1=num[i];; T/ `9 V' p6 F
num[i]=num[min];
. |+ r. W0 A" ]' t" G7 h) \ num[min]=temp1;
! F/ m8 n- O% j, z9 z/ S8 e strcpy(temp2,name[i]);
$ N" W1 T# {5 X' h strcpy(name[i],name[min]);& g: Y( X. a" a6 Y
strcpy(name[min],temp2);4 g% D9 N+ D1 ]+ y
} B) a0 S0 ^- Y! f* u
for(i=0;i printf("\n%5d%10s",num[i],name[i]);
9 e6 W+ j. V* D1 x P! k1 m }}
# Q9 ?# i( h4 M; [5 y r4 j+ F/ y1 Gvoid search(n,num,name)
! M8 K, W. S' Iint n,num[];7 j+ N/ q9 j5 r u2 N
char name[N][8];
4 c1 o+ L9 Y4 o/ B! b3 Z* |- J- K{int top,bott,min,loca;4 I1 k* _! {' a3 a: y
loca=0;! [. T' g3 o" R1 h; r# V1 r, {
top=0;
1 N& ~3 Y3 [( ?( Z# V bott=N-1;" n) o6 f9 @2 S" e$ {
if((nnum[N-1]))
- }' ~$ f# n( @1 Z. [ loca=-1;
- O4 P3 Z* N0 X& C" _ while((loca==0)&&(top<=bott))
0 u3 B2 k9 o) c, D: K7 V {min=(bott+top)/2;0 c, i$ d, Z0 h6 N, _- o2 G% V
if(n==num[min])$ s8 X6 x' B8 r* o7 ?: k2 v6 ~
{loca=min;) t K7 N q/ k& w, A
printf("number=%d,name=%s\n",n,name[loca]);4 l7 L: f h2 s5 v. ?
}
: [& F/ K! W# H8 u. k% F2 c# D else if(n bott=min-1;
9 x7 f8 D& K7 W9 c% s% { else( @) L# s% y' \" p
top=min+1;
) @* h% A; e% w j/ l }& L. n2 _: L- c6 }* ^8 D( c2 P
if(loca==0||loca==-1)
# n/ ?2 s" u+ r% z8 G, y printf("number=%d is not in table\n",n);
- K1 E8 E1 L: C$ v}
4 g6 H# i8 ]8 y. W+ V3 N# ]main()
0 m1 }7 O' @ L9 B9 F0 a{int num[N],number,flag,c,n;
, L2 k* L( g8 c9 v$ z char name[N][8];
! W2 J3 j# @. [; _- ?* ]4 V input_e(num,name);
~/ w6 \2 O5 f8 Z( w, o sort(num,name);
6 a' e& v: Z" S for(flag=1;flag;)
$ n3 H: r8 V" X {scanf("%d",&number);$ W+ x# }$ M) u0 L
search(number,num,name);
0 F7 L4 F. W6 ]2 c printf("continue?Y/N!");) H4 Y3 {, V) ]5 O( S/ q+ o+ {2 a
c=getchar();
1 t# A( e, J1 I: ^0 w- a if(c=='N'||c=='n')4 d5 e. }6 f! n! B7 q. s
flag=0;: s$ F+ J, P- @ _$ y
}
4 V. \7 }% B8 F" E: K* n}$ z- X. g8 ?* `4 a
" U7 u6 _0 y# ~) R1 s8.16
' P& V4 W6 B% h" y) t#include/ S N* w l6 T2 a
#define MAX 1000
/ Z+ I }. m4 c- D1 ?6 @main(); S# d" T. Y0 \8 k# y
{ int c,i,flag,flag1;
9 T1 G* Q v1 `6 w7 X0 `5 [ char t[MAX];
: v& t# X9 B- p' M; d i=0;1 E0 ]# `7 \+ i* V6 z& X- ]- j
flag=0;: K1 B9 m# Y1 f
flag1=1;3 W& d" R C4 m
printf("\n输入十六进制数:");/ T5 V6 b1 @6 f1 _
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
/ Z# {+ |+ L* f0 I: D. ]( W& o {flag=1;& @- e$ L* v! s& {
t[i++]=c;
) n1 `$ h6 h+ M+ O! C! n }% }3 ?" k/ ~$ b' m* u/ c
else if(flag)8 i- \, ?9 ~& e& L' D
{
$ `3 d- S1 I3 E$ Q( e: C# ~ s6 T t[i]='\0';
) d' w% I, H: X printf("\n 十进制数%d\n",htoi(t));1 V! d6 O' r$ g
printf("继续吗?"); x; s- l/ E- |2 N2 `+ q
c=getchar();
* }' y M3 g. s if(c=='N'||c=='n')
3 X* Y* U6 g r% K3 r' B0 p flag1=0;
4 Q9 d; R$ _7 e else; _ s! Q4 S. ~
{flag=0;8 P1 m' D1 ^# b9 r$ L* u" g
i=0;
% X% O- {+ V- `' `1 T- i0 _ printf("\n 输入十六进制数:");
' Y5 c+ W- m& X y2 X' f' a ` }% H- C% b: h' y# F& R ^0 T& V
}
8 z" ]2 ~5 f, V2 ~+ P" p}8 c4 D, t$ J4 P
}) a# _- v6 D: n5 i9 m
htoi(s): s- {! l: P8 G, o( {- C& d
char s[];
; A/ J# j, G5 S: }, N5 V{ int i,n;* k0 J ?% k6 P' p( }! j( ?- ]
n=0;
9 T A I4 l/ h7 V- c1 N for(i=0;s[i]!='\0';i++)% k1 W& _4 L% k( G
{if(s[i]>='0'&&s[i]<='9')
! p5 Y5 _& f( d2 u1 C5 e n=n*16+s[i]-'0';
# { W, f+ G. N if(s[i]>='a'&&s[i]<='f')
! Q( E) o/ e7 o% a n=n*16+s[i]-'a'+10;
4 b: e7 S$ O2 n- ]6 k if(s[i]>='A'&&s[i]<='F')
# b+ D. O4 D+ ~* q; V n=n*16+s[i]-'A'+10;$ Y( V- o6 ]! b/ q6 A
}
7 j/ D; X9 J+ a+ J return(n);
]' n& ]8 P4 P( P* Z, w! O. s) h9 h) Q}
5 ?, u6 C0 b9 l! \0 A9 R! K
8 D8 T! T. T( Z; z5 E0 G9 i
# h2 o5 G/ F. D$ S9 I4 Q& v, f4 i* _( o) @# \6 P, q" Y% L2 X `) ?
8.17
- `" T J' @8 z3 G+ @* [#include! P- `/ e. t6 Y4 \7 t
void counvert(n)
' r8 M8 |5 L A% h( y: S5 Lint n;; b7 ^( P' s4 H k& R; D* W
{ int i;! n& @0 L2 |% m2 b2 q4 M" b
if((i=n/10)!=0)
0 ]% {6 M9 C( R0 c convert(i);
9 b- Q- R; }. X putchar(n%10+'0');
- O8 {% g0 f5 H1 |- E! ]* t: O7 d}
( C" a& e2 d% wmain()
$ F) o! X/ @" x& F c7 ]9 C{ int number;1 C$ `" ^+ \! n6 W
printf("\n 输入整数:");
/ D& K. \ L( f1 @ scanf("%d",&number);
9 s& R/ E" R- g; u printf("\n 输出是: ");
9 E- T! a2 _. Z# { if(number<0)( \! C* A( B- `, J2 d0 ~
{ putchar('-');3 P! ^8 x# f' r/ [+ O
number=-number; P Z& }1 h8 ]
}* e F* H! ]: b% Z
convert(number);6 \* C, y! z% \0 R A4 g. i
}
* V( f0 s$ G& d$ l
+ A0 @" ]: ^) B' |- y. L. z- [7 B; g
$ @5 d/ k+ H& w, ?5 q
8.18
. k0 M8 l T0 z: A7 ^; Vmain()
+ b9 x" x, \8 _6 q4 d6 n: d* e4 f8 g{
* `4 ^" u0 ?& R! e1 y$ \ int year,month,day;
4 t" ^! Z+ A( H( K& z; z5 D' U int days;, n' L1 T* K' {6 ~3 H+ \
printf("\n 请输入日期(年,月,日)\n");
) B' \+ k# E) P5 T$ s; b+ w- ? scanf("%d,%d,%d",&year,&month,&day);
P/ h O* Y6 x4 N( ` printf("\n %d年%d月%d日",year,month,day);
% s( R( Y9 r& ^; Q0 @+ k0 { days=sum_day(month,day);
! X& K( \' I0 E$ j if(leap(year)&&month>=3)
" }) ~0 ?6 g9 b2 d1 V days=days+1;0 E5 @" y& M# ]7 ]7 a9 B
printf("是该年的%d天.\n",days);
) [+ J% o! K' y4 W% p3 `5 ] }
- H0 L3 z2 ~4 m+ I, b0 @ static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
: O' x0 T( w/ [& g4 T$ n int(sum_day(month,day)
$ H' b, m6 e1 p- D! r int month,day;
! m0 T& A, [: R' z {# g% T5 x- O6 t. f, R
int i;, t% _4 T9 r, E# \- W8 I& [
for(i=1;i day+=day_tab[i];
: H E, B/ Y5 ] return(day);
; E$ B7 o; Q# H+ a" i; _* L }) R8 d+ r" K4 Z+ f9 T! ~
int leap(year)3 J! N* }1 R8 B
int year;
6 h0 z: V3 H6 D7 O) l- { { E6 u4 V* A) K& Q( k
int leap;- Y2 Q3 ]* l8 L7 o [4 ~* B; r
leap=year%4==0&&year%100!=0||year%400==0;/ T F4 _+ o- ~1 K% L7 O
return(leap);
3 \7 n- X+ J/ ]: R! s* ~ }
$ K6 A8 i9 {9 E8 s- g第九章 编译预处理) J9 h" Z( `/ m9 G8 Z9 {. l5 u4 Y
9.1
0 |: b8 ^7 s" ]/ p, |#define SWAP(a,b) t=b;b=a;a=t4 j2 [7 ^. `4 q
main()0 t! W8 v7 G; O% u
{
; j0 a* T6 s& O' gint a,b,t;
- ]* S/ S# O& u- f' K/ w) M0 nprintf("请输入两个整数 a,b:");" \3 T4 J) x7 B9 Y/ c( j
scanf("%d,%d",&a,&b);. l/ m; ~6 b* f% P- {: |9 I8 y
SWAP(a,b); ^) A' j+ s) F% m0 e3 g
printf("交换结果为:a=%d,b=%d\n",a,b);
3 ^" @3 p: r, j9 N}
) H* c' E% f: ]$ b9 Y, A2 \
4 W; t* }4 E( o* n! j8 A$ Y
3 K# z, h5 t5 k9.2
6 g4 t! T; e- W: u$ z- W#define SURPLUS(a,b) ((a)%(b)) {1 T) S* L9 \' Q7 W2 _
main()
, o) J2 H! ]( {: I4 y9 G {
$ e9 F; m: ^0 Q# ] int a,b;$ L: V! u5 N! J
printf(" 请输入两个整数 a,b:");
% y" Q' i' D4 ^% t& x scanf("%d,%d",&a,&b);
I6 X- }/ |# p8 [% J3 f# Cprintf("a,b相除的余数为:%d\n",SURPLUS(a,b));
" C- _# @# r" F! `" z1 A4 c }
" L2 y- {8 _; T) `6 w/ x. n& `) K, _& e/ [8 o8 ~
8 v9 u' Z; N9 X/ u5 R8 [ y) o9.3+ C6 ~ C/ J q6 c+ w( k
#include
* q) F$ `" H6 y' b& Z" s#defin S(a,b,c) ((a+b+c)/2)
}( a+ t% O% k+ h#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-" k; v. \0 u0 I+ L4 K
c)))& i7 J3 J# z3 Q, E
main()
0 Y+ ^$ x0 O. |+ J( u+ M- [/ ]. P9 b {
, H1 l% K7 h+ K9 ]9 G% [ float a,b,c;% J, ]3 E) T9 P4 |
printf("请输入三角形的三条边:");( y# h( l% H8 C0 T( l" }5 |
scanf("%f,%f,%f",&a,&b,&c);
$ y' e* y. n! \ if(a+b>c && a+c>b && b+c>a)0 Y0 j' ^: T% t3 @; z3 W& r% o4 A5 K
printf("其面积为:%8.2f.\n",AREA(a,b,c));
" x7 f2 g5 I( I; V* e else+ R5 h* C) Q8 N h% F
printf("不能构成三角形!");
7 V8 u( n, R, r) I: `$ J }3 i7 t" D* ~5 Y# B3 e
# }. X$ J. Z5 w" s
; g0 _ M" v% @5 p5 T. p
5 E4 V# u. ?1 B/ ~$ \+ }9.4% ?1 X0 M) h0 H; Y2 x" k- h6 U$ F
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)
6 w' O( s* \0 Z: z- Gmain()
5 W2 F) A- E# ]3 J. N {
0 J1 o& a C( i' S$ d" D int year;+ ]* |9 A v/ {* U' V! _$ t" a2 f
printf("\n请输入某一年:");
. I+ Z k l2 T2 {5 k H# O; G scanf("%d",&year);5 m9 K9 Z" @3 d" y# H
if(LEAP_YEAR(year))
% a5 {+ F" Y* ?8 {* U/ e printf("%d 是闰年.\n",year);9 U, V! C) n2 M1 k( s- x3 Q
else
+ ~4 D% `- O5 o printf("%d 不是闰年.\n",year);1 f4 ^1 q8 Z# p$ Y" \9 ^
}
5 E* c; r+ s; P! h
' q% |$ C3 [$ X) M( ` ?6 p
- @; s) {: r4 G% l4 T$ [" b& l
4 P$ a% B1 t) p* C! \( O9.5解:展开后: @% k+ A' n. n& |1 S
printf("&#118alue=%format\t",x);
; ^% X$ Z* d( ]( u# h2 {9 ?4 N$ K9 Sprintf("&#118alue=%format\t",x);putchar('\n');
6 l+ U/ O9 Q# @- Gprintf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');2 e6 A. u/ A) ~
输出结果:
) V" W1 |$ K0 J# N7 a&#118alue=5.000000ormat &#118alue=5.000000ormat
6 O M/ o5 w7 _- a&#118alue=3.000000ormat &#118alue=8.000000ormat
1 A, R ]" c% Z/ l" e2 i
; O4 J& O# e1 c; X0 M, }
8 ~# \ t' H4 F) i: }9.8# f- A& X% y1 M; H# y
main()
9 g& y6 z n% H3 Y& G! Q {9 _$ V+ D! }: X" F- m
int a,b,c;/ R1 P1 o. S2 |, _
printf("请输入三个整数:");% Z9 }' f; U" l) Z: o# ^* H
scanf("%d,%d,%d",&a,&b,&c);: b) B" R. n/ `2 F/ C! S
printf("三个之中最大值为:%d\n",max(a,b,c));5 h; z3 R/ N+ l# W' u- V9 Q
}
* g" X( f9 L! @0 C8 V D* _ max(x,y,z)2 I# Z) W/ }8 A3 I& O( u
int x,y,z;
1 o1 C o6 a4 m7 g) [. ] {7 f: u- @" c3 O8 V# {1 K0 F; |
int t;
9 S) S6 j$ F7 u8 z0 }* F6 L t=(x>y? x:y);
{2 c7 N/ w Q+ u, v& V q return(t>z? t:z);
1 N! e4 G9 \ { }
* E( p- y; Y8 b' q" V
, F, g" Z, c5 x ~
6 t+ e9 w( y; B# H' ~* z1 {/ Z
7 x( h( S2 \ {: |9.10
. q/ F9 e/ t* h/ w9 _( }0 w2 Y. g#include
' h1 O& O* r, S! M1 A#define MAX 80
. G8 O3 l- q( D% t' y6 ?#define CHANGE 1- p6 B9 N+ I, h* k* S" P H3 h
main()" s- _: p2 i1 p# V2 u H2 b
{
# I9 O6 C7 S0 D/ T+ |( Y3 o char str[MAX];. M) P7 _2 o! Y1 M
int i;; Y9 Q2 G t* E! t, L& l
printf("请输入文本行:\n");
. y/ E( d; s9 z3 c6 \ scanf("%s",str);. T* H9 `( V; Q, s' t
#if(CHANGE)
2 j7 R; H# v0 l. Q' ^ {+ D* o: Y# D0 D5 M' u
for (i=0;i {0 B8 f( d6 ]$ m+ `' v0 w
if(str[i]!='\0'
- S5 {5 U: [' V4 Y Z- L0 H2 }3 G if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
0 l! \* y, A. r( p+ w str[i]+=1;
' U! A/ h1 h0 u9 _( _* M else if(str[i]=='z' || str[i]=='Z')
" j! r4 H6 P9 i$ b& }( } str[i]-=25;: y0 y+ N9 g( ^. q
}
7 j+ L, g) G% Q c/ y}
+ U% V/ A) g n* z! J/ @#endif
7 F- s7 r7 v% N0 ]printf("输出电码为:\n%s",str);
5 ~- ^1 a7 c9 m( Z9 U* _0 c' n1 G}$ q, j7 i% B! C6 N
第十章 指针4 N/ r* p3 G2 i) f+ C% x
10.11 q3 y0 h4 H3 z5 U# P. b6 t- D
main()$ T; q& f0 a2 m) x3 ~; T
{int n1,n2,n3;
* w' ?! r; @( m; C% `, ?$ x% x int *p1,*p2,*p3;8 G% b: `: [& Q6 b! P4 t# B; l
scanf("%d,%d,%d",&n1,&n2,&n3);7 }" E# B" K$ V0 H' L! e
p1=&n1;" A8 p6 j( a, h% h
p2=&n2;" B+ _6 T/ Y( X& v9 U. @! A
p3=&n3;" @- b8 X, s8 S
if(n1>n2)swap(p1,p2);
: B+ x$ v" X q6 t+ f* `2 i/ Y if(n1>n3)swap(p1,p3);
' d' k% l5 Y# M9 z1 r5 O5 ` if(n2>n3)swap(p2,p3);
: k" \9 w, F) N9 V printf("%d,%d,%d\n",n1,n2,n3);" [2 Y8 D- s& p! K8 w3 v
}
5 L% v# S0 f& G9 U: K- {swap(p1,p2)
6 J5 H; a# {. W: Q0 S3 {int *p1,*p2;+ K7 M2 x0 R5 O* W* |; J
{int p;
! l. u' k( s: {2 ^' i6 s0 v p=*p1;*p1=*p2;*p2=p;. t: {# b& v; U" P7 ~
}
1 P5 u5 o6 ?6 m3 }( z10.2) G8 f$ l p# Z2 ?, {* t8 w1 ^
main()
7 P$ T: [7 n3 L$ E* N{char *str1[20],*str2[20],*str3[20];
: h; G, @. r; t3 Y- ?$ ]9 S0 \ char swap();7 a" _9 L; H" s" @4 r9 v* ^. ], Z
scanf("%s",str1);1 W) ]- G; ]' C
scanf("%s",str2);) ^ x% @+ y3 O8 |
scanf("%s",str3);
3 \2 A/ b1 K( W/ P) { if(strcmp(str1,str2)>0)swap(str1,str2);
2 _, Z) r; C/ m/ H if(strcmp(str1,str3)>0)swap(str1,str3);
! F/ v/ v; g$ v- R1 \: @ if(strcmp(str2,str3)>0)swap(str2,str3);
. @" n$ @3 l8 ^' d- {; b printf("%s\n%s\n%s\n",str1,str2,str3);
: c6 G$ ~' F/ d}
9 e& S& ]3 n0 @char swap(p1,p2)
3 ^' y* n( o9 m' _& qchar *p1,*p2;* @- s+ u1 ~+ N. F* X$ a3 H: I( ~
{char *p[20];: |8 l" o1 {# L* b
strcpy(p,p1);( O9 O; x5 l' Y3 q4 ~/ V
strcpy(p1,p2);% H0 U& n/ T/ Y+ K' X' `+ b
strcpy(p2,p);" F0 i0 ]$ i% M2 b
}
4 }7 w2 P. Y' _# a7 }10.3. [" d( ]) T8 [5 |
main()" n3 k( @5 n1 M1 V7 f" c/ f* _
{int number[10];
5 `! T% X, o4 R% ?% s( D input(number);1 \" P% `1 p) \! j4 G7 J+ W0 p
max_min_&#118alue(number);
8 D6 _# `& D4 j$ B: U( A! L output(number);& L( p4 H" m* n0 O" }3 }' A
}
' N, t( {! w5 }7 k/ n- ~: S5 D; Winput(number)3 q0 ?- p! q) I7 X% O- ]! {+ w' Y
int number[10];8 w. z/ ]9 L8 `, ]$ q4 ] `! w+ \& Q( _
{int i;
5 o7 B6 o: u8 |3 k& g6 i: ? for(i=0;i<10;i++)- @; }. J( Q4 d7 U. {% |9 H4 _& ?
scanf("%d",&number[i]);/ y& q5 @6 A5 o! K! p
}9 r! d; n) i1 S: l
max_min_&#118alue(number)' e( s. S+ w u
int number[10];% a* Q# `+ t2 w. R
{int *max,*min;& \3 X. t* t6 A: J1 l' S9 D
int *p,*end;' G3 ~. u; H. i7 v9 J
end=number+10; m, u0 ?( ]) A5 ?8 {( }
max=min=number;
+ V/ t+ r n! o# h2 ` for(p=number+1;p if(*p>*max)max=p;# F8 I4 v5 b6 g( v- {
else if(*p<*min)min=p;' q. A7 I0 y) x$ H" G# s; B" f
*p=number[0];, |9 F2 @' X" B) T1 y7 [
number[0]=*min;7 {2 k2 z( x+ _
*min=*p;# l; E+ Q, ^1 E- E+ |3 Q1 p9 l
*p=number[9];! l5 M0 `" `4 m( u! @1 |1 E
number[9]=*max;
$ V+ d" l- I3 W! }+ `9 K% H *max=*p;
& B4 B7 o+ Y1 E+ F. ~5 H9 r1 u return;2 j* ]( Z8 D4 ?0 C
}+ S/ K. _" p* F/ h9 J
output(number)
8 R8 O. L+ y8 Sint number[10];" A0 A9 k+ \( H1 f" B
{int *p;/ U1 B: F! @2 \
for(p=number;p printf("%d,",*p);; X9 m) y0 L1 S6 w% L) n
printf("%d\n",*p);
5 y4 g5 N0 }. L}
J7 H- E" ]2 C5 m n10.4
, Q o D( `5 \1 n! o: ymain()' W$ n$ |- w, u C$ k5 C1 y
{int number[20],n,m,i;
9 U9 a% v4 _! W: e, ~: a8 H scanf("%d",&n);! x2 P/ n% c3 o" [" _; Y/ y
scanf("%d",&m);
: H. Z9 E# L6 J& L/ S3 z for(i=0;i scanf("%d",&number[i]);1 q- x# m7 z& k( J, M9 z3 _) ?, y
move(number,n,m);
1 ]$ {3 B# b0 u" d3 Z/ c for(i=0;i printf("%8d",number[i]);' V/ V/ P' V: B4 V) f' G
}4 [$ J: _1 k1 _9 H; p& f) }! J
move(array,n,m)
. ?7 G( S2 O/ t9 Dint array[20],n,m;
2 M, d+ ~. R" w; J8 o{int *p,end;& u7 l8 k8 [% o1 `% x- U
end=*(array+n-1);- J& W* ?# U6 V9 b: p7 b
for(p=array+n-1;p>array;p--)
3 @# [: M! e3 A/ o( I' p *p=*(p-1);- g8 ^$ P5 b" R3 R4 R0 S
*array=end;
+ z8 ?6 T5 [7 V( @0 ?/ ]0 [0 l m--;# b% S% g E& P. d
if(m>0)move(array,n,m);
Z) o5 I" `0 x* X: n% Y3 m}9 \! y6 j: J$ n
10.5! G9 e4 ?! Z( }& @8 X w" ^
#define nmax 501 f9 Y' l( D1 E; N# u
main()7 ]$ Y, k g8 R) P" o
{int i,k,m,n,num[nmax],*p;* L( _4 i7 p' A8 w2 Q2 W8 a, d
scanf("%d",&n);
) U. G+ Y3 j& { p=num;
' [" o3 B! B9 N+ h for(i=0;i *(p+i)=i+1;& B4 ~" I' H% w$ j; P* Q
i=k=m=0;4 p; t3 P g/ A
while(m {if(*(p+i)!=0)k++;( p% b% a4 q) K2 \+ m, d2 e# ?
if(k==3)) D: ?8 M1 E( s: D5 s) U
{*(p+i)=0;6 O+ B W2 m [6 K' X y
k=0;* a% J; a. u% j1 }, K O) i; m2 y
m++;4 G% S+ m3 M- N7 j! G
}
- ~, O* [" d1 D% l i++;
* |2 w$ c: C6 `! { if(i==n)i=0;3 P4 n8 R$ o$ N
} R* e7 g+ U# m" {4 a4 N( L
while(*p==0)p++; L8 i0 N3 z* b5 X* J6 R
printf("%d",*p);# \0 b0 b9 O' X& ~0 }2 c/ \
}
1 ^0 H/ l+ r, O; Y$ c! j10.6* |4 Q( s. `2 T8 }+ l3 p" [. ]" F
main()
1 }+ L; E0 W. r' U" a/ O" z{int len;& h3 i1 b# w: X" D* K: m
char *str[20];
0 f" E$ \; y- L. F5 X/ \: { scanf("%s",str);2 a) H+ \: t: p- D+ u; U) h/ w: e
len=length(str);7 y8 F& |( x% D/ j
printf("\nlen=%d\n",len);
1 i, j t m0 z* z$ E}
2 W2 X4 r4 {, O" U9 Slength(p)" w7 R8 k3 Y$ q# }% p2 V
char *p;, a( y) T o- k
{int n=0;
+ y5 I5 b/ o6 L0 @ while(*p!='\0')
* s5 u' [. _- L& I. r4 q {n++;p++;}
+ J0 B, o: L1 S4 W5 g+ ~$ _ return(n);# E8 V& j4 \: ~) p( r2 ~
}
/ f" ^5 w. D# N10.77 {8 m5 i2 I( O+ p f1 p$ E
main()
. q$ c9 a+ D5 i{int m;
7 ]. ]9 N0 y5 }- J2 x char *str1[20],*str2[20];* S) s* y, }! i
scanf("%s",str1);
* K D$ [8 i' |, ?- W' P3 n scanf("%d",&m);9 g% L$ O% @3 |9 y5 B% Y+ T2 M* F: Z
if(strlen(str1) printf("error");7 B7 e8 @# Y" _1 {
else* Z+ L( _4 u6 B. |4 {9 h7 P0 T! v
{copystr(str1,str2,m);$ Q/ ~! U- @9 T, Y. |3 i6 x& b) g- b
printf("%s",str2);$ o1 G- B6 }3 y9 C+ K/ W2 P
}
' F. l2 M3 O {7 ^}( g+ F! x( a8 v$ C
copystr(p1,p2,m)
/ {, f* P/ [2 k3 @3 ochar *p1,*p2;# d# a7 s) X) m3 I4 U
int m;
5 c- [! K( F& G2 j$ F: z{int n=0; |2 t+ U# G0 \2 [2 ?5 b, M
while(n {n++;p1++;}
2 \/ C l4 [5 l: v$ ?# i while(*p1!='\0')# }5 y" b4 B* I& }3 h! E$ i6 m) f
{*p2=*p1;
3 O G4 [5 z) O# X, N8 g p1++;! {6 Z9 j( t% E8 t' Z; ?
p2++;2 I% D7 B: x+ C5 s$ X5 b {
}( V% F( \- e# y4 ]
*p2='\0';+ D$ X4 v+ W7 a* I, G- ]% K
}# h9 G: J/ `( g. S; j, H0 l
10.8; H$ Z1 Y+ H5 z" D; u, |+ c5 t
#include"stdio.h"
( {1 [$ I& r2 F# j0 h% p$ S4 amain(); u; `( b5 H& b8 W0 H
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
: ?0 A: C5 U# s# x% a char *p,s[20];0 |' E& [0 C( t4 ]9 i% ]
for(i=0;i<20;i++)s[i]=0;; [( Z( J) n$ r9 m& c- w/ G1 V
i=0;. r$ h& Y1 b( z3 a4 \0 j& L
while((s[i]=getchar())!='\n')i++;
5 W6 f) g, F+ O$ Z+ Y p=s;- y/ [5 w" u* E1 I* G9 s/ a
while(*p!='\n')
x' o5 d4 T; b1 q& N* ^3 Q9 { {if(*p>='a'&&*p<='z')4 X7 o; I$ G/ G' o& Q' {
++sle;* U, k8 l% W) ` v+ H
else if(*p>='A'&&*p<='Z')
* i& I1 V+ \4 ^" y9 j ++cle;
1 B0 ]) s- _9 X else if(*p==' ')
0 O9 B3 _) P0 c9 ?. i; X ++wsp;" `! ?5 a S& b5 f) h2 M
else if(*p>='0'&&*p<='9')
$ d) O( z! ]. ^# T ++di;! r" L4 p' |8 n' A, Z/ O$ ?1 E: p
else6 [. S0 i8 m+ U; }* \9 P- ]
++ot;/ S) A& } }1 p6 l! ~) L( j
p++;/ h7 Z% p7 B" B- g4 E% y1 x; |
}. y: I4 a; w% ?3 u l/ i8 U
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
: L7 C; h6 j2 L}1 e4 H" G# P6 H" O1 N
10.90 J9 o) F& K9 x+ x) ~$ Y, t0 L4 t
main()( \9 h; l3 D+ Z) {, d
{int a[3][3],*p,i;' b, L: m# ?* Q' }
for(i=0;i<3;i++)
% J7 t; n- [( p% F- s scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);
1 P$ x0 N! X- ?1 E p=a;
; D8 q" J9 e; R0 k; T move(p);9 ^, b- Q" G4 i( {; m$ ?
for(i=0;i<3;i++)6 ]( B! M* a, G
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);
! b% ^$ _8 ]5 M* ^* {" }8 V( o}; i6 s: x2 j: u3 Y
move(pointer)* a' s# q; Z9 r6 i, Y3 _# N, p, z
int *pointer;
; T, g( h" ` M3 I% J) D& H- ~; f8 x{int i,j,t;, \: z5 x8 X7 k- T
for(i=0;i<2;i++)6 r9 b' S+ w# d
for(j=i+1;j<3;j++)* `2 m- _# K( l; z$ z1 z& q
{t=*(pointer+3*i+j);
D, |( g1 P1 V* p7 w( c8 o *(pointer+3*i+j)=*(pointer+3*j+i);
" L9 x; A# c# Y ~4 @ *(pointer+3*j+i)=t;) X; O K1 @# Z' _! Y% i
}
U, W% S9 e; c5 H8 m6 F+ [}
! n/ Z( M3 o( ~& f8 G b" g10.10- l5 p; E9 u# ?
main()
8 d# f, }# N$ Y; T4 R" {9 ~" ?{int a[5][5],*p,i,j;4 o* h) P. P2 K
for(i=0;i<5;i++), e& K+ @, ]' o0 V; i( _/ [
for(j=0;j<5;j++)
/ u+ K9 ^0 g. j. o } scanf("%d",&a[i][j]);
; c& B$ ^2 i7 J- T% g* u p=a;
6 ?2 R: \8 f6 n8 e; [ n change(p);
0 l3 W5 P8 w2 T" r5 v for(i=0;i<5;i++)
- q3 Z1 y& s! _3 K: W) x {printf("\n");: r" \8 e$ | k" b8 x: {7 w7 E$ N8 v
for(j=0;j<5;j++)
: b6 M l. ]; ]% `: q" q printf("%8d",a[i][j]);
* m8 ?: R Y' ~" Z* y+ V+ J }8 ]3 M' r- Q) O) C+ Z" I& u0 i
}% G' J4 m+ v# h& W1 K5 N
change(p); N) c, m6 y q+ N, [5 Z
int *p;
7 a+ s! \; h- E M{int i,j,change;% h7 ] u0 p4 \/ J1 r4 g( z) n5 X
int *pmax,*pmin;8 `% Q. q6 Y) ~& A0 L Z
pmax=p;2 j l8 v2 ~3 _. a M
pmin=p;( ^( j, z" O9 G; h, l J6 k2 A" H
for(i=0;i<5;i++)
& d* v \3 d) R for(j=0;j<5;j++)8 x7 w6 |; Y9 H8 x, D* f: W
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;2 W- r; O- }/ a
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;2 h( R: E* o" p
}
+ @/ H; h% P/ Y3 n) k* |+ ~9 z change=*(p+12);& T# q4 l( L) F! W+ A6 a
*(p+12)=*pmax;: v# q% z3 o) e3 e; [
*pmax=change;
( v2 M# t$ |# w- q change=*p;
+ _5 P& j& C9 m, x6 x4 Y, D# b$ k *p=*pmin;2 }) r: e4 }# O2 v5 s
*pmin=change;( B+ }8 r. h, F
pmin=p+1;
* U1 }5 O8 @! i9 D for(i=0;i<5;i++)8 x( m U7 _/ C* w
for(j=0;j<5;j++), K( S4 P( k6 `4 F
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;+ @5 C/ P$ y. u) x6 V8 E6 m
change=*(p+4); [$ h& \4 b, y/ [2 {, C
*(p+4)=*pmin;+ o" Y+ n! P) w, c H! y% v+ x
*pmin=change;5 X! O6 E- C6 |6 f! @- W
pmin=p+1;
c+ B8 E9 m$ b/ T for(i=0;i<5;i++)
" @1 b- l4 [ I for(j=0;j<5;j++)) [: Y$ X; g% y$ Q+ D* K: x+ ~5 u
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))) [* e; W4 O, m
pmin=p+5*i+j;
0 K. x; `& ?! x, E6 P change=*(p+20);- V$ S8 J- V2 [& I1 g q* l1 q, }
*(p+20)=*pmin;
4 |+ h8 c. T3 m! E7 g, J2 | *pmin=change;. S w, E! a3 n9 i% _$ a, c
pmin=p+1;
& N1 s i% ]* H' n for(i=0;i<5;i++)
, g# y; j# f2 |3 I6 y, M2 f+ b5 Z for(j=0;j<5;j++)
( E! Z- K" W$ ?- }. l if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
" j# B5 t" S5 W. ]! A" J: a! X &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;! C4 |+ {" M2 }( _2 N) h
change=*(p+24); H- T6 S; o6 I$ j
*(p+24)=*pmin;4 ^: L, E. }$ j; Y
*pmin=change;7 L5 D* h- B9 \( E. n9 c [8 z
}
& z0 s. H7 G9 W1 `10.11
6 R% g/ n( L4 g7 Pmain()1 a$ p! z* h" k/ m5 ~
{int i;
# {% \8 O2 y& F char *p,str[10][10];
6 n3 q0 s6 G4 ~$ V e for(i=0;i<10;i++)8 @! s5 d5 V5 P$ y& N& {
scanf("%s",str[i]);
5 F! j& P& i" O! R0 K' b p=str;
2 W8 u: n8 k) M) n% f- b2 w sort(p);3 a( y" P. {: |! m, A/ O- B/ U
for(i=0;i<10;i++)
5 b2 P; |$ \: v, l0 ]$ [# }2 B printf("%s\n",str[i]);+ Q4 Z$ I! |3 e% Q% ~) Y* D( q
}; A2 i, p4 Y8 {. A
sort(p)
; H+ d- f5 a0 U schar *p;( }5 Z* ]( @+ V6 \+ {: T
{int i,j;8 a" ~5 O. n' D/ w1 Q- D5 A
char s[10],*smax,*smin;
) v/ q5 [- R. F M0 r for(i=0;i<10;i++)% @& T0 Y! ^) l: W2 K, j6 ]+ T: l
{smax=p+10*i;
p! V4 t& ?/ c9 c for(j=i+1;j<10;j++)
2 T3 d* }4 K I" x5 Q% l" e {smin=p+10*j;( {6 @8 n2 x" K, p8 h ?( X' g- o3 U
if(strcmp(smax,smin)>0)
" O2 d i* n! `+ ]6 R# y {strcpy(s,smin);+ m2 f. |4 t5 @ K3 v5 w
strcpy(smin,smax);
3 F* c" p1 ^" U7 x3 ~6 e% g strcpy(smax,s);3 L; @! l# b3 o
}) f9 N- S8 ^) s0 T/ b5 J$ h
}) [2 X3 E: g9 N5 }
}
' y/ m" p5 q+ B1 o6 V* C4 _}
6 C4 p" ~7 \4 @5 C" \# l10.12; @' a }) |% g7 C) I' V
#define MAX 20
* U3 S+ O3 r x6 T, {$ kmain()
6 E5 f0 o; }- l5 c7 @{int i;
0 A+ O% V6 |1 x L# X char *pstr[10],str[10][MAX];8 t2 G/ i5 h O
for(i=0;i<10;i++)
% R: @+ e x0 O+ H pstr[i]=str[i];$ l7 F& e/ `7 F4 z- v3 Y
for(i=0;i<10;i++)
( k0 ?/ K" ^/ \6 d+ _ scanf("%s",pstr[i]); z* u) E" j" L% h: h0 r' O; b
sort(pstr);' b! ?3 S: [$ I/ r! Q
for(i=0;i<10;i++)
/ _6 I5 @) z; J3 h& l; P printf("%s\n",pstr[i]);: R& u+ x' k; d8 |8 ?
}
# F5 M8 Q$ j* p1 Isort(pstr)
* y; H! ]2 F, J% Y% O0 u- n$ ^5 G4 hchar *pstr[10];
/ X6 T: ]& M; ]+ k3 F{int i,j;$ F( x5 c5 X& ^3 N; ?/ {" ^
char *p;2 c! `6 o( W1 p
for(i=0;i<10;i++)
/ r0 ^+ Z5 N! e/ r% Q {for(j=i+1;j<10;j++)5 m& p( Q Q F$ y+ B l- O
{if(strcmp(*(pstr+i),*(pstr+j))>0) N: E" R/ ^6 v0 G
{p=*(pstr+i);
0 e, R& S# `+ \! { *(pstr+i)=*(pstr+j);
& q* E2 I0 i2 d *(pstr+j)=p;
, l/ `7 O4 B, I1 q }
+ Y' m' L% ?( }9 r: q$ ]; L0 \ }6 R, o+ [4 d7 ^: C
}
% v" t( A. t" ~9 l}
. b" ?' X+ {" e& N10.13 _# k) z6 T$ E! o" X' ]) q
#include"math.h"
% s- M' I8 X. }% Hmain()
" P* s9 x J( t7 y; s6 b8 D) G{int n=20;
. q! f" P3 t! {! `1 A- [% b# g float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
% A; M* \. X2 o M scanf("%f,%f",&a,&b);+ r% p& x) O. \ Y0 d7 n
scanf("%f,%f",&a1,&b1);0 g3 w& e+ ?; c
scanf("%f,%f",&a2,&b2);
7 u0 d: m' B! X/ N8 A p=sin;
) v# C, l x9 b& s) ~. V c=jiff(a,b,n,p);
$ S/ _, J6 I. t3 z4 s8 k! R printf("sin=%f\n",c);
6 O# A- b" ?5 x/ e p=cos;
1 a% J; g6 x! j& z7 r# D c=jiff(a1,b1,n,p);
J0 X8 Y. w6 U3 _* E# C9 ] printf("cos=%f\n",c);
& A" s8 Y0 E5 A p=exp;
/ E# m& r, G. Y: V/ h# ^" | c=jiff(a2,b2,n,p);
$ o( `' h* j" v' u printf("exp=%f\n",c);& m; }* V" I8 M w. B
}& `# @- ~8 N/ ?- s1 D
float jiff(a,b,n,p)
- l% u* P) w+ v, @, Q$ ?, lfloat a,b,(*p)();
2 X- |+ D* x* N" l) tint n;
- \$ v7 E* g& x" M. m; p/ U$ a+ {4 }{int i;* O2 n& S$ x! P3 H7 {6 o2 ?7 p
float x,f,h,area;
5 }/ H/ W( |/ [/ C# m8 P1 Y h=(b-a)/n;
: E- J3 |% P) |; t x=a;
* _ h' Q8 o0 U" Y area=0;) P& i+ I5 _% _) l& v* k$ Z% x* ^4 G
for(i=1;i<=n;i++)
" B; E+ w# p! A( K. W+ V; b {x=x+h;# K; D$ @) P7 i) w5 Z9 s
area=area+(*p)(x)*h;& \# m' V( O7 F8 y
}
# S N1 Q9 P1 V( X; e5 K2 G return(area);
, t, |4 r% Y0 n4 U* J}1 G2 J+ ]3 q0 E8 N7 L8 q
10.148 J/ p* I! i( \" j# W1 i
main()
9 h1 Q3 _- n. P. r{int i,n,num[20];$ _# u* U- E8 C2 k) i3 H8 P
char *p;
" u& m6 f9 I/ }+ K. a scanf("%d",&n);
5 E9 O; ] d- ~9 Y6 y for(i=0;i scanf("%d",&num[i]);
# |4 a% q$ _& w2 I+ A p=num;
. a5 K- X$ Q& B- W' E2 Y! E sort(p,n);
, N8 b) I& u; K: U& W for(i=0;i printf("%8d",num[i]);& \: @8 z2 X+ |4 S! U) _; C) f
}
4 q5 D) g! ?& E1 n) ?, Jsort(p,m)6 x0 i; ^+ a& X. F" f
char *p;5 _3 K3 Y, Z9 A
int m;
+ y' R- Z0 M5 ~{int i;) o& S' i+ M- c" D# Q, ]
char change,*p1,*p2;
! }- ?; H6 y6 B8 P, }; c I for(i=0;i {p1=p+i;
$ r6 y5 ?$ y# F( |- q0 b p2=p+(m-1-i);4 v8 _5 x \6 |9 D I
change=*p1;* J, B7 B& T L8 e$ w
*p1=*p2;- O9 e5 Q' N& J) R; U
*p2=change;3 h/ r9 X8 E. c) E5 _7 c1 w
}
; `4 F& @( |# V2 G" |' M}
$ N: @9 I, `% J4 Q+ Q1 M10.15. t% V s6 O# B% R4 T) B+ o
main()+ N9 W& R5 V$ f6 O
{int i,j,*pnum,num[4];0 X+ [* I4 i; ?6 |6 U# b/ o
float score[4][5],aver[4],*psco,*pave;6 m1 k) m5 P V7 P/ | {
char course[5][10],*pcou;
/ T% `) D$ {* p* [+ h9 z pcou=course[0];
0 i2 K% b* T0 i6 P- {! d for(i=0;i<5;i++) \1 W7 S; Q9 |9 S9 u# f1 R
scanf("%s",pcou+10*i);$ u9 T- z: h+ F5 q( I+ Z4 B. U
printf("number");
) t9 U+ P% q2 ? for(i=0;i<5;i++)
9 x2 B5 t( J5 D; r+ j printf(",%s",pcou+10*i);3 V; j- ?, R3 H2 z+ m0 I
printf("\n");
0 n# K2 s, J, ^6 C psco=score;! x9 v: }9 L8 V: G
pnum=num;
1 @: U1 X. g- P" _: O2 H8 ] for(i=0;i<4;i++)& b2 S. i. U# }4 w$ @
{scanf("%d",pnum+i);
5 F8 [+ C3 Z$ ^6 U5 R for(j=0;j<5;j++)
|5 Z4 ^. @! |1 A& O) c/ _ scanf(",%f",psco+5*i+j);
: a- V3 f4 z- q6 n0 y }! b) g* n) g: E1 k) n* E' E, n7 M
pave=aver; \* p( W( |" {3 x& n: N
printf("\n");
6 W; W6 N1 h V8 X0 j* ]' f avsco(psco,pave);
4 S$ Y; \8 u# Q5 ~. h avcour1(pcou,psco);
8 r& [: Q% S; R; H, {; S! ]8 N printf("\n");
F! V% }& Q. u* c: e fali2(pcou,pnum,psco,pave);
* @0 T, C7 s/ h% l0 ]% L printf("\n");
: @9 Z1 i. {' r8 V+ R6 y* h, y good(pcou,pnum,psco,pave);
; d% I' B8 p) ?# P) b. W3 x+ U}: s8 }% \! N2 Z# z* i9 [
avsco(psco,pave)# C5 J: R- ~' k, A( }) |# f
float *psco,*pave;5 t8 U/ v9 _/ n' D
{int i,j;2 ]' b! r# M2 U% d9 c
float sum,average;
/ d4 d3 g0 [6 a& w for(i=0;i<4;i++)* f$ i9 C6 B+ c# X1 F. K
{sum=0;
$ |7 b1 }. O6 J# O% `. Y for(j=0;j<5;j+)9 G1 p4 [2 C0 \6 e f
sum+=(*(psco+5*i+j));
( N! N. G$ w7 {+ ~+ N u! a& V average=sum/5;
/ \3 Y( Y. q# f *(pave+i)=average;' D* P% [$ Y1 b+ l7 ?
}( g$ T; b+ c. y Z' O. ` \
}
/ X8 w* E9 W0 z9 d; Vavcour1(pcou,psco)
# M! _7 @$ |3 R8 O* achar *pcou;, ~, \! Q5 _* a) v' x5 d
float *psco;1 H8 b7 e5 t- {7 _
{int i;6 Z! E6 _- j) [' ^' [1 X
float sum,average1;
/ s: X" c7 u9 F+ h sum=0;
) D; x# `4 ]% W* h% U" V for(i=0;i<4;i++)
& ^. ?" k. S K1 t% S4 K sum+=(*(psco+5*i)), U- I; c) K/ i/ A
average1=sum/4;5 s+ y# W4 x0 \- ]$ N- c
printf("%s %5.2f\n",pcou,average1);
1 j/ E/ e* _; F+ Y}
: k3 Q T, a" n0 G' v/ s6 lfali2(pcou,pnum,psco,pave)
- S' N+ A- b6 ^9 V, M# Uchar *pcou;
* B1 B% ?+ n3 b' Vint *pnum;8 L& ]$ d( `. h) n
float *psco,*pave;
( M) m X1 f: l# O! i: _$ J7 i{int i,j,k,label;3 [; ~8 b: e7 L# b1 W! C7 O
printf("\nnumber\n");0 @( c$ \' w! v- ^
for(i=0;i<5;i++)
* Q Y% b9 F' P printf("%-8s",pcou+10*i);0 `/ x k# q/ x( Z# Y5 \% t/ ]" c
printf("\naverage\n");$ [5 ?1 a C8 W( W
for(i=0;i<4;i++) }& x6 W& x; }% i5 q
{label=0;
+ i8 ]( E e5 L for(j=0;j<5;j++)1 Z$ X. H, ?4 ?( Q
if(*(psco+5*i+j)<60.0)label++;3 x9 @( g7 S8 Q- S7 P
if(label>=2)) p( Z* e4 o8 U
{printf("%-8d",*(pnum+i));
' D& ` {9 C+ M- | P) {9 s for(k=0;k<5;k++)( M; m, s L; i( _# B4 Q4 P0 m
printf("%-8.2f",*(psco+5*i+k));
3 [7 p; Y* K9 i, f2 s printf("%-8.2f",*(pave+i));3 X! B6 P+ @8 [! J2 `3 ]
}' @, H' A! A2 z; g: I
}
$ B4 V% ~3 R c6 y! u}' |% E* P2 Z9 Y$ |/ B. R+ Y0 d9 H
good(pcou,pnum,psco,pave)
; U" e4 ~8 D* R: m3 Q, _% ^char *pcou;
2 E6 b+ W3 Q; o$ sint *pnum;
% |( d. e0 u5 D c; v1 S# g2 sfloat *psco,*pave;! Z' X, ?' d+ D/ M& F/ I) K8 ~6 O& E- B
{int i,j,k,label;% v& D- k. A! h6 ?) O- [
printf("number");
6 K+ ~$ ]2 G5 E3 k# y for(i=0;i<5;i++)
4 e0 u: l. W; n% c3 H: I- ^ printf("%-8s",pcou+10*i);3 n& G+ \9 ?% U' ]- h- [
printf("average");7 c2 Y6 c% u7 k* e) Y; p/ `
for(i=0;i<4;i++)& y8 S& j/ d: W4 E# T7 i/ u
{label=0;6 W. X$ O; l" N b8 t0 v1 [
for(j=0;j<5;j++)$ \0 k: [ V: T, T0 a
if(*(psco+5*i+j)>=85.0)label++;: S' P' m: f( a+ {" @0 E& X' w
if((label>=5)||(*(pave+i)>=90))
+ @! U+ a7 P" C0 f0 y% i- W F/ ~" { {printf("%-8d",*(pnum+i));
: \) C1 V q0 A) e1 J for(k=0;k<5;k++)
; @( ~/ K V1 A$ } printf("%-8.2f",*(psco+5*i+k));
% r2 Q# [- u: Q9 K; b printf("%-8.2f",*(pave+i));/ t1 J; Z8 C* u9 Q+ v: }2 ?5 Z
}
) [5 l. U- S6 ?: \, t) j }
- y6 s$ f0 |2 V. V}8 G# j1 J2 I# v3 K: O6 t r
10.165 s9 _- C; a4 t% ^' h) N, d
#include"stdio.h"5 `" W7 j( b. @& X; v; Z
main()
: R' B! d1 x) [/ K{char str[50],*pstr;
% e3 B1 O, W, P% }# x int i,j,k,m,e10,digit,ndigit,a[10],*pa;+ `: o8 j3 C; Y( s
gets(str);( ~+ ?: i& p w0 Q
pstr=str;+ r8 A, ?6 ~$ c) o
pa=a;
9 m( i5 p% T3 R8 D* _) | ndigit=0;$ a; x2 {7 H+ B
i=j=0;
5 Q- d4 p- _( s6 p" R( A* m while(*(pstr+i)!='\0'), y0 a; f ~: e- n( d' V2 _
{if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))
2 D) } h' @1 W7 a! | j++;
) F& D8 K- e4 ]' D* B else' z( E0 c0 j( i8 W4 g
{if(j>0)1 Q7 r$ O+ u1 T
{digit=*(pstr+i-1)-48;; |! ], }7 A# t5 Q! u! h# [; Q8 F
k=1;5 ~3 g' U5 q. m7 S- N$ S; H
while(k {e10=1;
2 R2 N6 Z6 V6 d+ r3 v for(m=1;m<=k;m++)
9 S" G3 v9 j H" I7 b! l& u e10=e10*10;/ ` _4 P1 w) ^* ?; E; a# W
digit+=(*(pstr+i-1-k)-48)*e10;' b- ?5 f4 ~' t+ Q& o
k++;
+ W6 v' P) S% _, D! ], ? }
/ a Q2 y7 A2 W) v- ~$ G *pa=digit;: r% t# a8 X/ a
ndigit++;
8 m3 Z9 Z; Q+ R5 m# x! I6 x pa++;
5 ]- K! n E( U9 r j=0;$ R. C" S* y. S5 Z( n* u
}% R* M5 d- Z. z
} l: t+ S; r; D' H
i++;8 H+ P% Q2 a5 K: A# l% ^0 e9 j
}, P! R+ e. i C# ?. S
if(j>0)
* R' O( M( v7 g: J- F! |5 n {digit=*(pstr+i-1)-48;* j* Y, P4 t; c9 |: y
k=1;0 g: H( O& E8 @8 W8 R a. N9 t
while(k {e10=1;
. _8 A: S( `3 g' A for(m=1;m<=k;m++)
( F* E. o! h/ M0 L' Q9 i" q e10=e10*10;
9 m1 Z5 l8 I2 P7 u& f8 Y5 ` digit+=(*(pstr+i-1-k)-48)*e10;/ w6 Q( ^: L# ^* _) F
k++;! G$ I1 u! K0 `& S1 v/ X, _
}1 B' ]' z. o" F* q, O. Z
*pa=digit;
8 ?$ N) ^7 d6 V3 N ndigit++;
% z+ t$ h2 [2 j" ` j=0;
, z, r! K/ f0 H* { }
% g3 {: B6 H, S( N printf("ndigit=%d\n",ndigit);/ P( B N4 r4 t
j=0;0 T' e: v" \( H; y4 z( L
pa=a;) T4 l6 d/ N! S& I$ ]. l
for(j=0;j printf("%d",*(pa+j));
! S* Y7 X$ G( ?) Z% G}) I' ^6 J9 D( v2 U2 U- k& S
10.17
) |2 m) W; `5 G' }main()- |5 x ?; u" C* a: y) ^" W
{int m;
5 V. d- V$ X! A' T) g: `; P. {" d char str1[20],str2[20],*p1,*p2;
9 J" Q' F7 u) L, _" _( y' X' e3 g scanf("%s",str1);
! g7 W5 H* H8 h* J+ V$ \: t scanf("%s",str2);: D, C: y# _! B; {" {) U& x
p1=str1;
' e; }3 v2 e/ m p2=str2;2 W @, h2 ^+ Y% h5 |: i7 s
m=strcmp(p1,p2);
# @3 g6 ]9 e" |6 J# E+ q( s5 |( Q printf("%d\n",m);
3 { r: u0 l+ L, Q# |5 b, m9 t) k}: R4 p% b5 M( \- t' w
strcmp(p1,p2)/ n4 A1 m, ?3 c
char *p1,*p2;9 E, i" ^; c& T3 s6 V+ i# d
{int i=0;8 a9 W( V1 h2 S5 j5 O& O: m
while(*(p1+i)==*(p2+i)) P# J$ o+ ~- J
if(*(p+i++)=='\0')return(0);& f1 Q4 ]1 o3 p' C$ b4 e/ J, x
return(*(p1+i)-*(p2+i));7 g" ?! u& ?2 x3 E- g
}: f: c# @# K1 g+ ^# H
10.18! F" s8 A o5 J( ~
main(), h0 R. `* R/ E4 C2 L7 E
{static char *mname[13]={"illeagl","January","February","March",
& a0 D5 ?) c8 C, ~. [ "April","May","June","July","August","September","October",! Y% ^. h9 s7 d9 ]% K1 j9 `: T
"November","December"};
' m* Q7 H+ z! s7 r/ O int n;/ Q& _9 V9 E; O, }0 i3 {/ ?" S, s
scanf("%d",&n);
2 e3 Q6 E, y: d$ j+ s* Q4 L if((n>=1)&&(n<=12))
8 s3 B+ P$ F4 J% e$ @ printf("%s\n",*(mname+n));" `$ u; c& Z) ]4 W- j, v" O8 ?
else
# |8 W K- ]" k printf("error");
- j- Z2 @ n) n$ T2 x; R6 ^6 p}& W& N8 \9 `. c
10.20. F% {! m% T4 v, @+ Y) ~
main()" T1 g" d" Y/ `+ Q0 l5 ]! Y
{int i;
. l1 o0 f& G8 Q7 C ]8 g char **p,*pstr[5],str[5][10];) i l! {8 v' x2 \+ e
for(i=0;i<5;i++)
1 t/ j. m0 q& d; c/ w( E pstr[i]=str[i];
! c8 |3 X: S7 m* H; @# N for(i=0;i<5;i++)
+ `* j7 T' I: w) }) V. a& k' v! d scanf("%s",pstr[i]);
+ @/ S6 j2 K8 b. _ H# X8 b3 m p=pstr;
: o( l- ~& X* m( Q3 f sort(p);
8 p/ J) r4 w n/ {+ a! k8 ~ for(i=0;i<5;i++)2 x% e' q5 @! c/ d
printf("%s\n",pstr[i]);
0 d, T# B8 E' Z) B4 u. ]5 i. ?}
- j9 F2 g7 R ^& K! X* \1 y0 Vsort(p)
3 J8 `% D# o" W( {; r9 Schar **P;
% U, d! K" A$ X{int i,j;
. t+ J! I/ v. G5 s! t4 h char *pchange;+ V0 u* _8 D1 I5 n9 I/ v
for(i=0;i<5;i++)+ ?1 M4 ^0 W( G: C y; o) @
{for(j=i+1;j<5;j++)
. C t9 F! A8 Q }. e' A {if(strcmp(*(p+i),*(p+j))>0)
: T. s3 D; g% c% I. r {pchange=*(p+i);
. e4 e8 W) N2 H k# q# b3 D+ c *(p+i)=*(p+j);
* h* E' u7 o9 F8 F! m/ [ *(p+j)=pchange;* ]# u5 b3 {5 v! U+ u$ V
}
7 N2 A, h* w) t: d( _ }
7 C# h6 I0 w0 U/ A* N5 j' g( P# v }
! O3 l Z6 Q4 I5 |+ b}
+ X4 D5 \8 |/ w6 b10.219 \$ u: h9 M) L8 i- p0 w# `& s
main() z x; g* U* @2 G
{int i,n,digit[20],**p,*pstr[20];' P; W2 o" u2 V% X: K% u
scanf("%d",&n);; n$ T* e4 t6 b( A I
for(i=0;i pstr[i]=&digit[i];1 _* K2 k& _ r8 g; ~" _, Y6 E; w0 Y6 j
for(i=0;i scanf("%d",pstr[i]);- N6 e- D) G; B3 B; h
p=pstr;
2 D, x l" R1 K- m6 K sort(p,n);
" z/ E3 F- w2 D) R& ? for(i=0;i printf("%d ",*pstr[i]);
" I3 F Q6 n- @8 p3 P}
) [. L$ f3 U7 {# l- }sort(p,n)
' Z$ B, A# \/ i+ V5 H3 gint **p,n;. l5 f; Y% n9 r" E* |" ?5 S. r
{int i,j,*pchange;
. J( N4 T A5 Y8 m& T5 w for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
* [7 o. l$ o( z- D: _& y _! l: o {pchange=*(p+i);
+ f& m9 k, q6 w *(p+i)=*(p+j);* F5 r7 s' Q% @( [
*(p+j)=pchange;$ { I' z' u* x+ C# X
}- O0 [9 G( }# h# N" M
}
2 e- i7 C: X& \6 q1 @- Q }% K9 i8 C0 l- ~: i
} Q! j: i- _ m5 w* X7 b
第十一章 结构体与共用体2 \) U5 o9 `; s) O% o0 o$ W* z0 N
11.1
3 _) @7 a9 E- A. Gstruct
( Y, `! Y9 ^/ h3 k0 _, b {int year;5 h* c1 G6 C$ m& O( v" `
int month;: r2 d: b$ M' B8 }& O9 @ m
int day;4 Q& H4 o5 q" |7 ~1 q, s3 X" o
}date;* N+ m& a5 e$ _' R0 b
main(); `/ T& N2 f/ p7 x Z, n5 m
{int days;
7 H0 S9 h- K& x( [ scanf("%d,%d,%d",&date.year,&date.month,&date.day);: `. y. f2 {$ Y6 y4 T- t
switch(date.month)0 d* x" ^; `" P# p7 f
{case 1:days=date.day;break;
0 _, J9 w/ O' ^( s/ B+ W case 2:days=date.day+31;break;: o# g; }. f- j" V- K2 W+ L( l
case 3:days=date.day+59;break;
& ]# p; s8 C# y case 4:days=date.day+90;break;
# s6 n4 _3 y0 l; ?' Y& o! y case 5:days=date.day+120;break;
# ?" v$ p" D7 { y& Q case 6:days=date.day+151;break;& b* [4 p; p+ R7 Y2 q, i i
case 7:days=date.day+181;break;
" ^1 A7 q# c- P. [4 V& y5 R7 q8 ]& | z case 8:days=date.day+212;break;3 \5 g! p+ m- O# W0 C1 X
case 9:days=date.day+243;break;
- O6 g2 f0 \) F1 K case 10:days=date.day+273;break;
0 \3 k0 {4 `" ^" @/ ` case 11:days=date.day+304;break;; k, ^5 A8 _: f( d
case 12:days=date.day+334;break;$ ~$ x8 |9 F6 V# q! l5 Z
}
8 d- N) U/ {3 O2 _6 G& X4 S if((date.year%4==0&&date.year%100!=0||date.year%400==0)
8 o6 {8 O: C( U &&date.month>=3)
, l M o+ H" T/ H( C days+=1;+ f# Z' T( ]" ^
printf("days=%d\n",days);
% K* X$ {5 n) g1 @}
0 Y4 F5 D6 o3 E, i& C1 Y8 s) b" _11.22 U* F* c3 J8 a
struct dt
8 E+ ~5 r) x) C- d# E {int year;. \3 T, \5 s- k+ A- _3 G
int month;( o' @, s: d( ?, E# h# F- ~
int day;
. f0 v5 K9 r5 U# r& A+ T }date;2 x) R' Z5 s2 c$ f* d8 E% L/ B
main()
+ C/ t8 E" V3 t3 H{
, A0 N" i1 S; J2 e) H scanf("%d,%d,%d",&date.year,&date.month,&date.day);
& x7 N" c* P, h; A: T0 f& c8 g printf("\n%d\n",days(date.year,date.month,date.day));
+ \# w0 a8 \; z9 @9 i% K" J3 g0 N$ {+ u}0 E* r) E4 c7 ?, `, N" |' f
days(year,month,day)- R; ]# U& q: z h
int year,month,day;5 ?0 i8 O% O9 T3 @; \& f
{int daysum=0,i;; \5 u3 x7 G) b7 d* K
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}4 V F- U/ p d" {; v
for(i=1;i daysum+=daytab[i];3 [7 i( Z3 [( w! z# I: R, ?, u
daysum+=day;
5 x# d* H, B9 k% z/ f* K if((year%4==0&&year%100!=0||year%400==0)&&month>=3)
% w- U8 o4 p a4 @$ Z+ O: N P- r, V daysum+=1;
- N6 L O5 r( t$ O8 l3 y return(daysum);
! I% }# m& l" G4 w9 J3 C}( G8 r I5 S7 J9 ~* k1 e) D# J
11.3
5 Y6 ^) Y0 I% a- D5 D0 A$ a11.4
; m1 p* Q! c# O$ e7 I5 {0 J#define N 5
+ J) F# `. \+ }0 r% | Ostruct student
. S* u: e4 L7 |$ K {char num[6];6 o" z( v# l, U
char name[8];8 v0 N- ?9 D: p0 @
int score[4];
7 A6 L& e$ f" z }stu[N];
1 m- ]" c" }# U5 ?$ Zmain()( |# z( y8 c) H8 @' J; o l
{
2 ~) H# t+ n) [2 \ input(stu);9 F% q$ j' V- ]4 b, z% F5 }3 [5 |% b
print(stu);( G4 F. d4 Z6 o* s' j
}
& ]0 H6 m3 G" L# a/ @input(stu)" p& V4 P e: x4 V4 v# H
struct student stu[];* @$ \1 F( q* N+ p4 B8 A
{int i,j;
8 d) q% w$ ^- {2 e# ~( J for(i=0;i {printf("number");- O* L. f Q( L0 b" b1 s5 h" Z# A
scanf("%s",stu[i].num);
8 X" ?+ d9 s6 C' U8 |' R/ i) I printf("name");
) Q) s; G6 e3 K1 J8 K- F6 o scanf("%s",stu[i].name); s, B! j3 K$ L! D) T( E: `
for(j=0;j<3;j++)# p, L: G4 O+ Q4 g
{printf("\nscore\n");
5 \; }; O* U3 m9 L( ~ scanf("%d",&stu[i].score[j]);
( \# i0 p# t: X: ^, Q4 ?; o }% k! D) c, m: w# V
printf("\n");
! u9 p( L/ V# n& Y7 w2 F* ~3 q }
0 w5 g' V4 R! V: K t# E" j}) F. o* s2 O2 A* B
print(stu)
+ M' Z; J* e7 V8 e9 }# @4 G L( ustruct student stu[]; l- Z2 @/ ~. Y. t
{int i,j;# K. ]- J _3 u& E9 V2 M2 ~" W
printf("\nnumber name score1 score2 score3 \n");# Y# }/ T( _( F/ D
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
: U, l( e. _0 I5 h: M for(j=0;j<3;j++)0 k6 m0 k/ H+ h5 G* X# K4 U0 L: ~
printf("%7d",stu[i].score[j]);& y* P5 c) O& V5 L$ e) M2 a
printf("\n");. N9 z0 y- }0 G
}
' I N/ }: h$ H5 K" t D# f/ m}
) h" Q, n4 f, { x7 W9 {11.5: [/ Q7 R% K9 H& c( K* m6 a2 p$ M
struct student: S2 S! ~5 c) `: Y# k
{char num[6];6 q1 F- u" M$ r! N
char name[8];
' j( x' Y' [2 q: {" f2 C int score[4];
$ h4 G; X$ I; a' J6 o: k# w float avr;$ u0 }) n$ @2 Q6 j( F8 c V/ T
}stu[5];. {" i6 b+ N7 h5 K! p
main()& n" B4 e0 U+ N6 J) B4 b* y e. j: h
{int i,j,max,maxi,sum;% U1 q' s+ e2 U5 L+ p
float average;& M: @4 f8 ~& ~
for(i=0;i<5;i++)
8 I5 ~( V- ~( E" U8 t {printf("number");1 w+ m7 ~( G8 R/ l0 e' _
scanf("%s",stu[i].num);
A* o! N; k$ G$ t3 N printf("name");' }+ o1 W' u4 O; q& e
scanf("%s",stu[i].name);7 ^7 R' G8 Q( T2 p+ Y( q9 y
for(j=0;j<3;j++)
/ F4 w& C; P% E, M9 u4 C6 F {printf("\nscore\n");8 z' v5 ]4 g' x5 i0 e
scanf("%d",&stu[i].score[j]);6 w: ^5 X& l" D# i2 Q
}
. }$ R" f& a6 d$ @0 o% W }8 {9 d- T" k; W, `, W- _
average=0;
) {' R4 v: l% s/ L0 V1 y3 U max=0;
: e3 ^# `6 r$ v6 W& z, k$ _ maxi=0;7 a- F* y& h1 D7 r" d! f
for(i=0;i<5;i++)" m) c5 E( j1 \4 Q
{sum=0;
; e* I4 @( I7 l) h for(j=0;j<3;j++)0 \1 q r0 V: I9 ^3 N
sum+=stu[i].score[j];& G/ r! Y0 D% C; s( c+ |( }
stu[i].avr=sum/3.0;
- y* V; `7 J( E3 {2 \7 M% s average+=stu[i].avr;
' P+ G( S& h2 |( x if(sum>max)
# ]0 R2 T: u/ v7 O/ p, L4 _ {max=sum;
- N' X0 {9 H4 o L- g( c0 K8 K7 |4 u maxi=i;( O9 i" a1 c* o {/ }
}
{& E1 `$ S+ N) ^2 n, O% u, J }, x' s+ `3 s* |+ ?1 v# |: A5 O
average/=5;; U0 w( l0 N5 q8 a( @ C2 v
printf("number name score1 score2 score3 average\n");2 X# O; Z+ d6 d; p& b/ S1 H4 [
for(i=0;i<5;i++)
& m5 }0 R4 F1 x0 i8 A0 d% B! M {printf("%8s%10s",stu[i].num,stu[i].name);
+ l# ~+ M) T a" g. I0 Z% \ for(j=0;j<3;j++)) n4 Q" L/ L- ^2 P
printf("%7d",stu[i].score[j]);$ L/ c9 T T) I! F8 }
printf("%6.2f\n",stu[i].avr);" f& B) `2 `7 A7 ?! l, c! y; D% G
}
; [- Z- @9 U# G, r printf("average=%5.2f\n",average);
! C( P( N% S; B, a ~3 r( _ printf("The best student is %s,sum=%d\n",stu[maxi].name,max);
& k- g. t2 e" a( ~. |2 ^}/ s' }& a) g5 v8 q6 O5 O
0 t6 m5 `/ q* W# Z( s |
zan
|