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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
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签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案+ `, k- \1 K4 z# G' P
1.5请参照本章例题,编写一个C程序,输出以下信息:
" R1 w0 S) \# g+ }3 y, Fmain()7 N8 y* r3 E6 k
{. j9 b4 @. |, P4 B% Q
printf(" ************ \n");3 ]0 S, v* F1 x: X8 w1 _* T* t
printf("\n");
" K* b3 j0 y V6 y8 Qprintf(" Very Good! \n");
# K1 p( C x5 _) G, \; L2 z: kprintf("\n");
' a- S. L, ~4 L1 L$ i ?printf(" ************\n");
/ W0 e' R; I% y# C$ P}+ b0 O3 q+ G; H- I; M: D! i
1.6编写一个程序,输入a b c三个值,输出其中最大者。* M9 I! x: n4 ?% {
解:main()3 n" u1 A: X) k: b1 _
{int a,b,c,max;
+ e# g0 `: Q* Y: Hprintf("请输入三个数a,b,c:\n");
9 i. U8 h9 D! c: t/ ?$ t- x% H: C1 bscanf("%d,%d,%d",&a,&b,&c);
: B+ [2 g( B4 z8 j7 Z; f) [max=a;& a* x0 j* T% F$ z
if(maxmax=b;( Q* E1 r% v) w0 ~" v
if(maxmax=c;
, u& T- J8 G9 l zprintf("最大数为:%d",max);' S2 y, b( v$ B. `
}
% O& ?' {7 P" x3 y- H: ^第三章
3 I/ Z) R5 h* |8 n) c3.3 请将下面各数用八进制数和十六进制数表示:
- K2 d2 r6 B+ K. D2 h(1)10 (2)32 (3)75 (4)-617
J7 h% r& I0 U& i0 m(5)-111 (6)2483 (7)-28654 (8)21003
* y% U6 W) _, a: c0 Y( H解:十 八 十六
! F6 |! x- f' @: P1 O0 ?3 Y# F (10)=(12)=(a)
2 f8 |4 S7 X; B0 k2 }9 Y. x (32)=(40)=20
9 c- S, _8 f! t7 n (75)=(113)=4b
3 o" _& |0 J" y. m) ^5 G9 ~ (-617)=(176627)=fd97
5 O/ d, {* \4 g$ T1 W5 k i3 T: z -111=177621=ff91
9 N0 n, `& [6 g* J( e 2483=4663=963: u( T. n9 Y# ^8 Z" w
-28654=110022=9012 o4 E# d' _& B H; T* p) Z: ^# f
21003=51013=520b% q; M% {. r' |. a+ Z' @
3.5字符常量与字符串常量有什么区别?, D; W( g7 Q; f1 \/ M
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
% ^. d4 L$ x6 [% ^( T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'.1 T: q# E: t0 {/ v+ k, M, O
3.6写出以下程序的运行结果:- n* {9 ]8 f1 g$ y& y, J
#include$ Y3 I/ m% L! B" K9 `7 [
void main()) `( U9 B5 E# p) c# P' v5 I
{4 B, \0 f4 b3 I& b
char c1='a',c2='b',c3='c',c4='\101',c5='\116';, h+ u' B( Y4 _5 t
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);5 O$ s) n$ c9 I
printf("\t\b%c %c\n",c4,c5);
7 L( T" c- |) \2 `7 {* x解:程序的运行结果为:
! x, U! _- K9 l, saabb cc abc! P- D2 r# |& W! T9 g- y3 I
A N1 R0 h; s( x6 j, K9 o
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,% e( g* t& ^, U# l
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".
$ o5 p5 g, J, V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# N6 |3 W7 k5 X- D9 L' c) `/ J
输出.$ b/ K( d# o# Y: X
main()
8 d2 L& t1 \: c9 f0 d{char c1="C",c2="h",c3="i",c4='n',c5='a';) x. o$ X! v: K# f2 _
c1+=4;
{7 z/ l3 m6 W" `# Dc2+=4;
) c, N z9 R3 p- |- A. l% N, e0 jc3+=4;4 R" x% I- k" U2 Q3 E
c4+=4;
& S; s* S' H2 z7 r$ Y/ \# Mc5+=4;& l/ f' [( s- a
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);5 W3 O) M8 i" A" g- q/ f
}$ U/ g4 h7 F, O9 V1 V- l9 m; `
3.8例3.6能否改成如下:
1 n7 k7 {# V- z3 j6 V#include4 _5 J( P) r; Q7 |6 y0 B
void main()6 a6 Q5 I$ Y5 Y! M9 r8 ^. M6 V
{
* \% M: l n; @5 y- xint c1,c2;(原为 char c1,c2)- e4 S J$ I& }
c1=97;1 R6 u+ H! w' ~' h6 k
c2=98;! o y7 B8 N9 k! X0 U! m
printf("%c%c\n",c1,c2);, ~2 P" E T2 ~4 W
printf("%d%d\n",c1,c2);" [9 u1 e7 I0 s% k) L* o
}
" a9 Q4 m4 Z Z( }, F8 |9 F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. y `: @' L; N" v+ j$ D2 a/ w
3.9求下面算术表达式的值.# ^( \, J- `! Y
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)( i6 F* |' Z2 q
(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)# R5 ]% W3 c/ Y9 d) O% f
3.10写出下面程序的运行结果:7 @& S' R! b I, ^' ~$ \( M: K5 n/ O
#include
4 \6 `- z! ? Mvoid main()
+ N1 B% S2 x; m: y{
4 v8 T Z$ t. @2 jint i,j,m,n;
9 @& d) [4 E# P' k5 M' L. `0 n: `6 ti=8;
" d% n3 k+ ?; n; u) z7 Kj=10;6 M7 s/ N! \& c4 |
m=++i;
5 R/ g9 F7 B7 L; B5 `n=j++;
4 J! k- {3 p: _: G3 Vprintf("%d,%d,%d,%d\n",i,j,m,n);) W$ c, h( M$ q! Y( s" B# n% M
}
6 p# a0 y! @$ K1 q2 l: g& Z解:结果: 9,11,9,10
4 n- ?2 S/ E' F4 ^: a5 v5 R( F第4章; U) g7 I/ a3 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
) G0 @3 L9 l1 P! M; V1 o6 A* q7 x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: a; s' N' A4 Y, X8 R
a= 3 b= 4 c= 5& a( Z8 A% j$ o* a
x=1.200000,y=2.400000,z=-3.600000
( Q1 O+ e+ {! P3 M4 Lx+y= 3.60 y+z=-1.20 z+x=-2.40
% N$ V; o% k2 x2 u1 k/ ju= 51274 n= 128765
8 G' d& H( @6 H! M3 k3 X) Pc1='a' or 97(ASCII) V: b* [0 P- L
c2='B' or 98(ASCII)/ Y9 f- S9 P- r* Z6 v$ b
解:
" O* c V X/ f% ?7 R( {- Gmain()
7 ]* Z' @+ G9 E{
/ t, ]; y) m" [2 ` Wint a,b,c;/ ~' E$ d7 X) c
long int u,n;! Q9 |* o3 Q1 }! v4 H+ z
float x,y,z;# b% B, u j. Z
char c1,c2;
1 h: E4 ~2 d- Fa=3;b=4;c=5;" R8 c; p' X( |
x=1.2;y=2.4;z=-3.6;8 ] W+ {' ^& y
u=51274;n=128765;* i# u. v5 H/ k) E
c1='a';c2='b';' f" j& w7 r( s
printf("\n");
- a) l2 Z t! z4 o+ I3 m6 B; Tprintf("a=%2d b=%2d c=%2d\n",a,b,c); } S/ G. D; o6 t6 u
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);+ R8 c* M* q6 ~
printf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);4 F' q# [! S- u9 Z
printf("u=%6ld n=%9ld\n",u,n);9 |% ?; i: [+ Z
printf("c1='%c' or %d(ASCII)\n",c1,c2);6 g% R9 v0 W; _: N; ?4 q
printf("c2='%c' or %d(ASCII)\n",c2,c2);
" W. u; ~! P9 o, w/ C}
$ d* D, c. s) G& m- B+ K4.5请写出下面程序的输出结果.
y5 t! S: Z+ y2 B# }6 f结果:
+ R M* u) X% g: J57
b, X1 g# W) e 5 7' u& \! i7 b6 U6 e- r: l
67.856400,-789.1239629 o- \$ V- @$ Z/ M
67.856400 ,-789.123962( w% [# F3 k# R! q" B! a
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962
. d" I/ T- h7 f1 u/ e3 Q" n* u- E6.785640e+001,-7.89e+002
/ e( y+ l7 v8 T4 N9 a3 W/ WA,65,101,41
/ l8 n+ i, N8 F; I" {" V X1234567,4553207,d687 T* b% {7 P( X* ]
65535,17777,ffff,-1- m3 b5 E! @! p F
COMPUTER, COM# I8 X- _' H1 v3 ]6 O; m) P0 Q
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',$ G4 W8 F+ b: a; @3 ^: M4 o9 s
问在键盘上如何输入?3 b* ~& i% h4 c4 _: y ?( d
main()- u% y# h- \ u5 _* N
{
0 k2 L: W' V) ]( X4 @( W, @int a,b;
5 ` c% j( Y( V0 T5 J4 @ Z! k+ Hfloat x,y;
2 i% D& K3 C1 G! z8 h4 ochar c1,c2;
. [$ k( f5 e/ cscanf("a=%d b=%d,&a,&b);
* a3 X/ d7 @, v* [2 Y' ^+ U* hscanf(" x=%f y=%e",&x,&y);
* I$ \+ L$ k! t8 H: P! rscanf(" c1=%c c2=%c",&c1,&c2);+ |6 W3 x9 q# r; x+ i9 l5 w
}
5 J( e0 M, q( K8 p- V- ^3 r" n1 u; _解:可按如下方式在键盘上输入:
* S9 i/ M# k* m, P* r- |; Ya=3 b=72 W! J' n8 T, `; \/ ~8 b
x=8.5 y=71.82
3 ^0 T1 q) q& H z$ hc1=A c2=a% b$ E, p; ~! A) b
说明:在边疆使用一个或多个scnaf函数时,第一个输入行末尾输入的"回车"被第二& \ @# u$ w2 ~- Y2 f v) _
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& r6 x* }5 C% G8 T# P: k6 U
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: U$ E" _7 Z% ?$ T5 Y, @
初步概念即可,以后再进一步深入理解., V8 @4 W8 j4 F4 w
4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-4 M$ j4 `! l* y$ Y
3.75,z=57.8,请问
; ~$ H1 i% G& P! B# J0 m$ N在键盘上如何输入数据?; ?, x, `" a" _
scanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);1 J. c& o. Y( r' }/ H) n" g
解:* O1 v2 h2 \. B# J1 j G
main()# G4 U* B/ n! |8 H
{$ \7 W; n: P: |; T$ ]5 i6 q
int a,b;
& p' i, J4 L+ K( w, o. T; I$ mfloat x,y,z;
5 |! q7 n) {. {9 Tchar c1,c2;% f$ M! G- T4 R: _! o
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
: v* u2 K: K; a& V9 ?" h}
3 M. ~* [' Z' h! o/ W, h- @- K5 M& ?运行时输入:
* J6 j) V/ P( ?10 20Aa1.5 -3.75 +1.5,67.81 h* R, r3 {$ }' [% ~
注解:按%5d格式的要求输入a与b时,要先键入三个空格,而后再打入10与20。%*f, h$ l9 s) L3 r
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
; J% `+ t- z2 b$ @7 t* }会赋给任何变量。2 L, s6 Z3 F8 G! A* t: H2 l
4.8设圆半径r=1.5,圆柱高h=3,求圆周长,圆面积,圆球表面积,圆球体积,圆柱体积,
& |# \: L# V+ G+ F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/ g; N6 f7 \3 S, e$ e
程.
* w! x0 J5 M$ @# J3 \0 a解:main()
Z2 `5 q! y8 S# F G/ o{- \. P% q5 T8 P/ H
float pi,h,r,l,s,sq,vq,vz;
( n" v1 A- s! p% upi=3.1415926;
& p- e9 i9 a" t1 @7 t# Uprintf("请输入圆半径r圆柱高h:\n");
* Q5 l: A' F9 a) Vscanf("%f,%f",&r,&h);+ Y! V$ i7 j a( v+ f) a
l=2*pi*r;
/ p0 [. O4 O9 X: `$ p' q; `s=r*r*pi;
8 G6 g' y7 U$ K% T" Nsq=4*pi*r*r;
, P" e4 C6 I" K: |8 Gvq=4.0/3.0*pi*r*r*r;
% y0 n* @ Z* |5 Nvz=pi*r*r*h;
# h' H3 Z \* Fprintf("圆周长为: =%6.2f\n",l);3 G1 h' {4 o4 l2 h# H2 w
printf("圆面积为: =%6.2f\n",s);: z; f! G9 h5 q- ^; Q
printf("圆球表面积为: =%6.2f\n",sq);
$ T) G( `5 R" z, o Vprintf("圆球体积为: =%6.2f\n",vz);
; @6 p2 y" H4 J" G3 Q}
/ a, ]: z8 S; Q% ^$ L& K9 \4.9输入一个华氏温度,要求输出摄氏温度,公式为C=5/9(F-32),输出要有文字说明,, i# p. i& h: l1 s
取两位小数.
6 O5 a; T0 Y. U- U解: main()8 w5 ~# s1 |9 C D" S6 S- k
{
1 m" T4 l" r, w( U+ B6 a+ H( e/ T9 Qfloat c,f;
. w0 R% ]( T7 h, N/ P" X- nprintf("请输入一个华氏温度:\n");" y# O9 W8 Y2 [# r8 t8 Y: n
scanf("%f",&f);
" {9 F V2 N5 _ s z1 p4 Tc=(5.0/9.0)*(f-32);- a! F$ L, c" m- j; k. _: I
printf("摄氏温度为:%5.2f\n",c);
|7 p/ K/ e) Q}( w* N+ ?1 R0 |
第五章 逻辑运算和判断选取结构
$ O$ o8 V% q# p; @( g) u' q6 m5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.3 r2 c' ?, U# g
main()
2 n3 v3 b% H: y' J% J{/ Z3 w5 A9 |" P! p5 f% n. M2 o
int a,b,c;3 V8 N0 o' O* D) J! @
printf("请输入三个数:");; N* \: `! F8 p' g3 [# ~+ _# z7 z
scanf("%d,%d,%d",&a,&b,&c);5 _, n D- `* x# c6 x0 S/ j% L% O. E
if(a if(b printf("max=%d\n",c);9 k* F: ]1 Y. P1 Q1 ?
else2 I" c3 U G# ^, f
printf("max=%d\n",b);
" h7 o6 B$ F1 Q& m3 |6 Delse if(a printf("max=%d\n",c);
7 s2 C/ q8 `, Y- U* S2 \6 k* W# I/ L: B else
6 h* [+ v, K" U+ e6 s T printf("max-%d\n",a);& @+ b' O0 p8 L$ i2 @7 ~+ N) {' ^
}0 h1 n: K+ e4 D0 `' p- K# o; u k
方法2:使用条件表达式.
6 w5 t& c, Z: \: smain()
+ y/ G% l+ u q; q; {, B2 a; T! \{int a,b,c,termp,max;
2 a0 e3 a V9 m/ ` printf(" 请输入 A,B,C: ");4 u3 r/ x) c$ @: p
scanf("%d,%d,%d",&a,&b,&c);
; o8 L5 B! M1 U printf("A=%d,B=%d,C=%d\n",a,b,c);
|1 U, `6 }& |0 [. r! y! a temp=(a>b)?a:b;
- Q w0 u% u/ C+ Z9 q1 Z2 I+ f5 F max=(temp>c)? temp:c;
: i+ _" M0 e/ f8 h- w8 Z5 r; D printf(" A,B,C中最大数是%d,",max);
) U* T# V! M [+ x$ u# ?5 u9 {# b}
+ E% B7 h& ?" N& _( l# ]5.5 main()
z% p! Q) X2 K6 d, o; z{int x,y;$ T) B8 f9 u0 @4 E2 ]- f
printf("输入x:");
+ j+ C% u! ]( Hscanf("%d",&x);
! c D6 C0 i- |. {# w: aif(x<1)
, `; _3 G: V- R$ b1 D2 ] {y=x;7 m9 P0 _4 R* t1 c! @8 `+ A: c H
printf("X-%d,Y=X=%d \n",x,y);
i5 O" J" Z8 m; ~9 F# X3 K }" h( g. ~* o% M$ P1 n
else if(x<10)
5 v% `) k! [0 y7 s Q8 S4 s {y=2*x-1;
_7 r# k! v# b9 @7 |0 [+ E printf(" X=%d, Y=2*X-1=%d\n",x,y);. v z) i" e6 s# n
}
! z* S. i* P, N. H& delse& P, r- F$ d+ k
{y=3*x-11;
' I$ J- j9 C" V printf("X=5d, Y=3*x-11=%d \n",x,y);6 l; P2 e, g2 @) m y% l
}
- d8 j7 T; q8 D) U& E5 r4 n, L}( W; a: c) L0 y- f% s6 p# t
(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! 0 O u4 i7 F2 j& s. M: E1 N
void main()8 N" J8 E9 T, U' U3 ]
{' t) N/ W" ~6 N$ U& T
float s,i;
" @. @) f' L9 T& t. Ychar a;
l3 s9 O$ e' g+ Mscanf("%f",&s);: S2 W3 e7 W$ L6 f( F" T8 ?; Q* {
while(s>100||s<0)
4 N+ h5 D& |4 E' {* o{* q t- \: v8 {
printf("输入错误!error!");1 N6 _3 h' B8 e; c8 \
scanf("%f",&s); N9 d5 j- O9 L; h1 l ^$ x1 j
}% A9 s5 f, `8 p u. g
i=s/10;
8 E$ M# p% w5 V, S& x) \: f% cswitch((int)i)9 l' ?! o: Z( s
{
. [$ e. B# Z4 s# H4 z/ u" @6 Qcase 10:
+ K! z; ]' ^/ R. C7 E( K1 hcase 9: a='A';break;
6 H9 x, Q& _; d# i) L4 Ccase 8: a='B';break;
/ q, t! }" c4 K) k# V* g0 acase 7: a='C';break;
/ M; j6 k: N# U# j7 `case 6: a='D';break;
9 }& O0 c/ X' W, V' p Xcase 5:" I% g, n* M; ]
case 4:
* ]4 U; W5 o( q) R/ \ k4 fcase 2:& g% Q, L; E3 \+ b0 z
case 1:( L) c+ Y" F+ _
case 0: a='E';% q% O! P! {" s3 y( i C' o9 w
}/ o( R W6 k5 e
printf("%c",a);
/ s, E7 S! S, T" @( z4 |}4 p7 v$ {6 G- w' R; D: |
5.7给一个不多于5位的正整数,要求:1.求它是几位数2.分别打印出每一位数字3.
0 L* t. w% Z+ C% G2 l+ S按逆序打印出各位数字.例如原数为321,应输出123./ [9 R6 U$ ?; @' x1 S& Y6 X
main()
* R% W+ x5 Z" r- p' L {
. _4 _ H. X) y5 }8 v long int num;
5 u' e0 ^$ n0 p: Q) l0 @* O" P int indiv,ten,hundred,housand,tenthousand,place; J8 A2 ]9 w7 @" X% z
printf("请输入一个整数(0-99999):"); d9 t. F: W$ B- y! d; I$ c- x! c
scanf("%ld",&num);
5 H" `6 l, i8 ^4 c' c if(num>9999) x: I% a7 c; k, |& b4 Z% v
place=5;- \2 d+ r3 n- P, k p
else if(num>999)
% L5 N, F' x' z place=4;2 H6 r+ _3 a7 B( ]8 i" y
else if(num>99)
0 p: y3 O3 C$ K place=3;
. A5 [" r5 `: P7 V$ S ?else if(num>9)
# ?% @& e6 {: g. o place=2;
& W" ]7 p7 D3 [2 Selse place=1;
4 q# p( E6 `0 N% ^. [5 z% zprintf("place=%d\n",place);
6 G; c. z2 i+ G" C# nprintf("每位数字为:"); N: N5 N9 _1 T8 h# s+ C
ten_thousand=num/10000;
6 G$ U7 n2 t- ?) |thousand=(num-tenthousand*10000)/1000;
7 ?' w+ e2 D3 {: X! Whundred=(num-tenthousand*10000-thousand*1000)/100;, R h# c+ B W5 x2 L
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;/ k1 Y) y4 T9 ~! V6 _. V
indiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;0 D* \1 }# F. F% H% Z w
switch(place)
6 v1 e& b1 e' L% i, W{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);3 @) o4 q2 l( t! F
printf("\n反序数字为:");, W0 F1 _. r" S) C/ z
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);
8 P$ P% l# u5 X) R/ F break;
1 @& i0 g- C& U' ^5 i3 Q/ P! Ecase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);+ o2 q+ e$ B# r
printf("\n反序数字为:");
( z1 W4 ^! u' u* C printf("%d%d%d%d\n",indiv,ten,hundred,thousand);( k2 j3 }# Q" R, }" ^3 B
break;3 i6 {; D6 G5 J) f: j1 D
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);6 p8 o. p- H: {, I
printf("\n反序数字为:");
C: l d6 Y P1 G! Q% l" {# F; a printf("%d%d%d\n",indiv,ten,hundred);* ?2 b) f0 t: t' \
case 2:printf("%d,%d\n",ten,indiv);
3 ^0 }: k U2 n) {# J7 U! V printf("\n反序数字为:");
, v0 X* M- Z$ m7 L9 O printf("%d%d\n",indiv,ten);
3 a4 n% w6 A& t8 W9 ?case 1:printf("%d\n",indiv);' p+ M( W n Q- ]) f- c' [
printf("\n反序数字为:");- B3 Q. N- Y7 Q) S% a( d7 X! W
printf("%d\n",indiv);0 g& j* l. k$ d$ c4 f
}2 C+ e4 p' z1 c# q5 O, M1 h$ g u
}0 R8 G5 c0 E, G K2 Z: y
5.8
" T) p2 I5 h6 [8 D: q/ u2 A+ O1.if语句
, l* ]) e: K3 a& `8 ?. [main()
6 w4 |* p) B+ B{long i;
: O- G" T, U( ]6 ~# ? float bonus,bon1,bon2,bon4,bon6,bon10;
/ T; A6 E9 n5 L% N' } w3 N' F1 J bon1=100000*0.1;
1 R' D" F6 s* a; d bon2=bon1+100000*0.075;
! M+ Q _. f0 n4 C. L; ~# ` bon4=bon2+200000*0.05;
$ S+ ^. h. [ r$ J' f0 s" G bon6=bon4+200000*0.03;
; g9 K0 w; O/ D) K# X& G bon10=bon6+400000*0.015;
6 C3 Q9 G. v0 R scanf("%ld",&i);
6 K; S$ W5 U6 P. c# J/ X/ ~ if(i<=1e5)bonus=i*0.1;
+ w V; c& Q! a3 z) m3 F else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
" W9 _ v- t" V( x( {. b else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
8 S. N( {3 l4 j q5 [ else if(i<=6e5)bonus=bon4+(i-400000)*0.03;4 N7 Z8 f& Q. Y! l5 E# ?7 J( ~
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
# c# B* U2 Z; M2 q8 P- P. N, Z3 \; c else bonus=bon10+(i-1000000)*0.01;
; {+ m- h- \8 G6 t& d' O$ n printf("bonus=%10.2f",bonus);
: {4 W6 C: ~2 {% g}: K3 G( \! J7 X. y, g- Y6 K4 Q& b
用switch语句编程序
6 {' j# C1 f% [$ W$ Hmain()2 N" [+ W0 k3 e5 H e- }
{long i;
8 i! m- M9 Y( ]( J3 L float bonus,bon1,bon2,bon4,bon6,bon10;" M! W6 b# Y4 u: \
int branch;: f9 w/ L5 C1 u4 I6 q% c
bon1=100000*0.1;
0 M4 ?$ ?! V1 c5 g; q7 u% [ bon2=bon1+100000*0.075;# y* t* j# w7 B
bon4=bon2+200000*0.05;
3 D8 I' ]& Y# R' j bon6=bon4+200000*0.03;- j$ b t8 M# i. a s) h
bon10=bon6+400000*0.015;
* k1 g. _, I- v r6 { C( K& ] scanf("%ld",&i);
# C; P' ?& k# k! H branch=i/100000;! {$ A: k% K. ~" |7 q. |
if(branch>10)branch=10;
3 R. i& r. }/ v4 y$ z/ f k7 @+ E5 { switch(branch)
3 g4 i% L" R) O/ r" O* {" p {case 0:bonus=i*0.1;break;5 P. u) [, `! u+ C A1 @
case 1:bonus=bon1+(i-100000)*0.075;break;5 H7 S7 L# K" n: g% w5 I
case 2:
( i2 b G; `( v. _$ ~ case 3:bonus=bon2+(i-200000)*0.05;break;' [' c H. Z( F. O8 G# h1 ]
case 4:' I* t, B2 Y- o$ O
case 5:bonus=bon4+(i-400000)*0.03;break;7 `: B2 i9 }; A8 U
case 6:
; ?& @! A) Z+ R case 7
+ X* B9 M4 O; o* P case 8:
! d1 w. p% y: `* N, B case 9:bonus=bon6+(i-600000)*0.015;break;; X! f* h8 D5 v; E9 \+ k1 M
case 10:bonus=bon10+(i-1000000)*0.01;! c8 ^7 b, E y& C5 }/ l0 ?
}
X6 \$ Y/ V( j8 z printf("bonus=%10.2f",bonus);
* O. @& ^0 a8 v' @8 x}
1 F9 Q+ H% [* j' s5 a! t0 P" S) u4 L' U5.9 输入四个整数,按大小顺序输出.
: ?& ~/ h, h) m0 A# Z% ]8 l' b4 Y$ Nmain(). p: v) _, ]2 w+ o2 m, Q' d
{int t,a,b,c,d;
) N0 ?+ t, ~1 S7 [" K: F0 _ printf("请输入四个数:");
1 S5 L L$ I3 A/ c( B scanf("%d,%d,%d,%d",&a,&b,&c,&d);
% |9 L6 r6 z6 d* ?, O printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);) x7 l2 X8 {! X$ ], [7 N
if(a>b)
! _- S1 K$ s& g; |' ~ {t=a;a=b;b=t;}
# c4 L' b# w' i% ^* b1 T8 o( w% o- T if(a>c)" u$ }$ p; a- W0 c$ ~; _
{t=a;a=c;c=t;}
0 q) s+ @8 @( M1 C# S2 i7 M: F0 U if(a>d)) \ |3 R# z- N U' }
{t=a;a=d;d=t;}
5 H6 G( m1 G+ o if(b>c)+ ~3 F8 K2 M: R- f& l1 o/ H4 o) M5 w
{t=b;b=c;c=t;}
7 Y! {1 u [5 G9 `5 p+ o- Y3 h if(b>d)
9 z; @) r3 B9 D# R. F8 J j5 [ {t=b;b=d;d=t;}! C: |" A9 q% l6 Q( f
if(c>d)
: A% w$ [6 v9 i) }8 n2 p+ } {t=c;c=d;d=t;}
! ^7 I8 T% t5 U4 A# I& n7 E! h5 Fprintf("\n 排序结果如下: \n");7 k- W' f1 ~+ _" q% T' R9 u F8 j# W' Y
printf(" %d %d %d %d \n",a,b,c,d);
+ o M, i; q. p5 p9 b5 _}: v7 }/ ?" d# M6 V4 h4 D& d
5.10塔. R$ ~8 H- n- W5 [% p/ P8 p! d
main()
! j; X" j9 `5 E8 G: C3 l" @{
% I$ S0 E# t) b, {0 H1 Bint h=10;7 u" g, [- [0 U- P8 g8 t
float x,y,x0=2,y0=2,d1,d2,d3,d4;
" j3 h+ c4 S' P; Cprintf("请输入一个点(x,y):"); z& }- @8 u/ O- f
scanf("%f,%f",&x,&y);, Z8 _) [9 S* p( N
d1=(x-x0)*(x-x0)+(y-y0)(y-y0);
7 D: q# o) F& N5 k! Xd2=(x-x0)*(x-x0)+(y+y0)(y+y0);9 F, j/ ?: M2 ? z% y9 p) V
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);* j! U+ G+ \! u
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);
5 p6 E6 T5 M7 k3 ^/ R* Z3 i1 hif(d1>1 && d2>1 && d3>1 && d4>1)
# V1 e# _6 r D# Qh=0;5 S/ M8 K" J4 G5 K" h
printf("该点高度为%d",h);
$ C8 {2 Y& ]* A% H/ q' _}( Z$ q. M2 v$ U+ M6 R
第六章 循环语句
; J& y) B7 d+ l# G4 s f+ r6.1输入两个正数,求最大公约数最小公倍数.3 t$ V& E* d" A! F) ~+ X$ {# Q7 U
main()
; J0 ^0 m& b5 N{
5 D' P4 L4 n* p8 G% @$ T. pint a,b,num1,num2,temp;
& S& |( i" _7 D% d( a) g' _printf("请输入两个正整数:\n");
4 y2 V2 k. g2 C& p5 p4 yscanf("%d,%d",&num1,&num2);0 d7 R7 N( ]+ g }$ m" e2 q4 r
if(num1{
, @8 o" Y" t2 E- h7 L. ftemp=num1;+ W+ g9 n' G, Q3 H! m
num1=num2;
6 W- |) X' f( _+ Z! m3 ynum2=temp;
1 r3 T# ~3 B: q6 K& S5 I}$ M, \& i! x: T% |
a=num1,b=num2;; W( l! g) D' X: B- V
while(b!=0)5 P4 r; u3 j- V8 f3 K: u& r
{
. W+ D5 S+ ^, d" t7 F5 A- k2 S temp=a%b;5 B9 k+ Y4 [( G( H9 ^
a=b;
6 E1 _4 c" Q* Q) T7 ~' Y; f b=temp;
0 h+ r* W$ W5 T* a* P; W5 x& s }# f4 t H, P% X5 D, ]% F7 @
printf("它们的最大公约数为:%d\n",a);
8 y7 s0 e; i9 L/ M! _" Mprintf("它们的最小公倍数为:%d\n",num1*num2/2);
5 n/ w+ g3 B* d- R! S- d4 }}
6 l; `) N# X9 T& j+ r& V% z6.2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.
! ?! T7 X; E+ N8 Z解:
. C) g9 w9 D5 s, u; w1 B#include < >' N( E: [7 F7 H
main()
9 G* Q# S2 E$ ? B{
4 i/ k3 ^1 m3 A6 \3 [char c;
1 K0 s; W |$ D5 K7 V1 ^8 ?+ Qint letters=0,space=0,degit=0,other=0;. Z' o6 i! X) n1 p0 H
printf("请输入一行字符:\n");
9 \% w8 Y4 x; f8 bscanf("%c",&c);
* R0 F X, Y% C3 N; @) N' cwhile((c=getchar())!='\n')2 z3 ?; o( H# \9 Z( m
{
! B7 i8 ^) d- y) W" x/ R4 s% lif(c>='a'&&c<='z'||c>'A'&&c<='Z')
% \6 T% A! \" j9 ?! f, cletters++;3 x/ T5 u: Y& _8 \, |6 I! p
else if(c==' ')7 J5 H: o. P0 O/ i& n
space++;
: o2 w- V. |+ e0 ~5 p' x+ z6 Nelse if(c>='0'&&c<='9')( C& b4 ]5 `; F! u. ]) Y2 B
digit++;
+ K; Z! v# [- a* ~" D' e# zelse
# M# z& E( i* i$ |9 ^0 @other++;9 M) U$ f' \$ }- S
}) ~0 b9 s: Y" t
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%# m8 g. g8 y1 y1 U
d\n",letters,space,
, L7 a# d$ T. g, @* H) j0 ~digit,other);" u6 h' w- B' Y+ f- n6 p
}0 C" X9 F0 R1 s* d& N3 }6 V F( H
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.
5 h7 m, B9 J; a1 E1 ]# ~/ |, O解:6 C( i( _& d; n" L
main()
: K) M8 v7 P6 g. x; i{) @! k8 G, D$ e8 p4 ]4 s/ ^3 u
int a,n,count=1,sn=0,tn=0;
* P* X, g7 ~7 g1 x$ Wprintf("请输入a和n的值:\n");( C$ N: Y5 L& W
scanf("%d,%d",&a,&n);
0 J/ {5 S& X2 e. Bprintf("a=%d n=%d \n",a,n);
. _" m- z" O# N& Swhile(count<=n)
: S7 b$ d' x; u' g- w{
; ~6 p( x. I, ]1 R9 Y- jtn=tn+a;+ {' K) }) m; `
sn=sn+tn;
6 V2 `2 z a0 I* n# b' H# ga=a*10;7 x% ^9 ]- {/ ~; I y. G
++count;7 V$ n. f) L- G. ~
}
0 X" N0 }0 t% ?4 @! X; o4 @printf("a+aa+aaa+…=%d\n",sn);
+ y8 V5 i( H7 o( @8 `9 Q, R7 S; d}
# O: b/ X0 o3 X+ C6.4 求1+2!+3!+4!+…+20!.; k. s9 g$ G1 X7 }- s9 {
main()7 L+ e; k3 E! ?0 D }
{* b) v3 A. Z2 S7 `7 z. ]* o
float n,s=0,t=1;
% A K- v% |; \% }. t$ P- _for(n=1;n<=20;n++)1 y5 S2 _. \9 [% V5 [
{ r/ B% i5 j; Q0 U0 a. U
t=t*n;2 R+ Q8 _3 ^1 E+ d
s=s+t;
& a4 C( Y6 X& q4 W4 A1 ]}
! |2 I& _# k4 b6 B& Aprintf("1!+2!+…+20!=%e\n",s);
1 X2 {8 u y9 L4 q( M/ @: c# N}
8 w% c( v" a: x1 h6.5 main()9 V) E& e3 `) F6 a/ L
{
' F8 A$ I$ ^% T+ k* L! k# ?int N1=100,N2=50,N3=10;
7 p7 `3 q5 | ?5 `) Ffloat k;
$ e2 m9 R7 a0 w6 }+ \: ffloat s1=0,s2=0,s3=0;
" G q! {( m8 m0 G1 U( Jfor(k=1;k<=N1;k++)
# p/ D8 W1 s" E% E/ c! H{7 n$ a# n4 m% D9 l6 j1 z
s1=s1+k;- e; Q- Q' J9 F. ~, b* J" b
}
% t" W* l. Y8 b9 x- x+ f/ pfor(k=1;k<=N2;k++)
4 p7 U1 q! }0 L6 \5 H `) G{8 C2 P0 i, ?* q4 q( N
s2=s2+k*k;* r w1 ^; W, @3 h: m; l# X
}
2 U) S7 L4 C, M( w+ r3 hfor(k=1;k<=N3;k++)
5 u* y% u" ?! m( f: ^' G, z{
7 \. |7 c: d5 G3 `s3=s3+1/k;
% f; a# m7 l, v; v5 c5 Y}8 }# h p( k7 X" ]% n& `9 g; b
printf("总和=%8.2f\n",s1+s2+s3); r9 B+ h" ^( u% W
}8 v' _" @3 {. e* W; \
6.6水仙开花5 X( n! Z9 S- u) a3 Y
main()* [2 r% c4 s- o1 l' j
{
4 ^- `$ _ v1 c, n+ M3 \; U Nint i,j,k,n;
1 I" Y- J3 O$ m/ ^' D' sprintf(" '水仙花'数是:");
) `; M" }- ?% ^3 T! c; z0 Afor(n=100;n<1000;n++)
3 H. r2 Z# \' D. t5 \; d* C7 \{
0 @$ A5 N) \, X' j$ o3 E$ Ji=n/100;/ e1 Y* V0 z7 F1 U
j=n/10-i*10;
& r6 K# C* M" w' @$ N/ rk=n%10;; b7 M4 d# a! a! c* Q
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)1 P" S7 |1 e0 j: \, d# q
{( @5 H% o/ G: q4 V
printf("%d",n);
/ Z3 Y8 \2 M- W- y! F) d}
8 |$ o# [5 q* R$ w# L, b) c% L q}' {# `" H8 W3 s& u/ G0 Z
printf("\n");
9 A& s6 c% h& o8 H% h}
0 Y; O3 S" G1 f$ j: B7 V- S6.7完数
; j" z* h; k) _: S: lmain()& } Z) c8 |! Q. P1 M9 z
#include M 1000. @: A! j3 B V# @/ p2 j( K* \! D0 r7 K
main()4 b# i2 {/ l) G1 Q: K# _! S
{
) m) [" q6 n9 L+ U% \int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;
+ f' K. V/ ^9 M7 p1 h5 }int i,j,n,s;7 T) O6 V' {8 @7 H
for(j=2;j<=M;j++)
V) H! Y8 A! b: l( K \! r6 @: t{
1 F6 m- s3 K# Y. B, e i" A @n=0;* b; F9 Z8 ~# m$ Q4 L% \
s=j;
0 Q+ N8 ~% \- l4 |0 v2 Mfor(i=1;i {
6 J7 y @9 Y4 R) gif((j%i)==0)$ }$ z7 y6 _: U' A9 y. M- b
{. @; ^! ~0 L0 ^9 X% g* V
if((j%i)==0)
+ ~) w# n3 G) x4 K {* s R/ j4 C( Z- i% ~* A9 \8 J
n++;
# {% w6 T, l. q1 y( q0 \) b s=s-i;
$ c+ O1 K. c6 a" P8 l9 J5 c switch(n)# C, @3 ?! p) \8 C# N+ p* J
{( v- U+ y6 e& E; c, t: h
case 1:
& j) K4 f2 W4 A1 m; y4 q k0=i;) Q2 @4 A6 b8 h$ O& d$ D9 @5 }1 ~1 S
break;
) F5 v7 o/ B6 `% M. G8 `5 K case 2:
9 J) f3 P. ]* e/ T* U3 L k1=i;, U5 K) C$ A* L$ z8 l$ v! m
break;3 ^ @* H8 H8 y6 B
case 3:
* J1 v( ?4 Y) I k2=i;
$ }& t( s! Y0 ?6 o. |" n2 J break;# k; R6 z! `6 F3 I, [- l' c
case 4:
# ^6 x# g7 L& m+ p+ P k3=i;, t) j. {0 _3 z% ?- C
break;
/ ^4 Z6 }- u! Q) H$ ^ case 5:
- i- C r: C6 ^- I: w/ j k4=i;
8 u7 G( R; s2 O- e break;/ V! Q, c4 y) x2 n. [8 U7 G9 J2 A2 \1 A
case 6:# `/ q; V; K3 J2 H: J; [
k5=i;, q/ M D* A$ P+ v: w4 p" |
break;" Z7 Y _( \4 [1 k4 s; ]& m
case 7:) ]. i k' V2 a5 x) S
k6=i;' v. q5 e4 W3 I4 E* [
break;2 d$ l0 Y! R( W0 ~5 u0 c, ~8 X
case 8:% y- ?( G) [3 I, P6 x, g ]
k7=i;
* X, ]! C& @: F I: R- W, W% J break;
8 x' @: D: d6 p/ w! U( t3 P: n3 Y' V case 9:
, c1 z- f! p0 b' X/ D k8=i;
# g% L3 r( E( a9 ]1 ` break;4 B: q- n$ n# r' ? U
case 10:
8 _" n7 \" ?7 @/ U: @; x k9=i;
9 R! W2 K/ b9 F- o4 B/ ^4 A break;2 v9 P/ R- P4 z5 t; C0 i5 J
}# Z& G8 j' v& d. r& j! y
}
+ I( S8 [0 p7 s n }0 [& E/ Z/ x8 F% x2 M- W1 O
if(s==0)
& a2 T9 X; `; I7 Z' X& \1 |4 p {& l' V1 B, c5 Z& ?; w
printf("%d是一个‘完数’,它的因子是",j);
* ?$ D; h9 g8 ~! K+ n0 yif(n>1)
$ h+ n% \6 g9 J8 A* z* a3 a printf("%d,%d",k0,k1);
$ K2 K7 c, B# o! o( Gif(n>2)" `+ u7 O" }# ^# L/ W
printf(",%d",k2);
1 m: E2 q! X5 u2 a1 w! d2 Zif(n>3)
+ Y8 A2 m; J5 t* {! b& Q0 k, l printf(",%d",k3);; V4 ]+ [" @) h- Y7 u) [. T
if(n>4)7 T1 C7 k, H& q! ~: ~& I- d4 V$ o
printf(",%d",k4);) {/ E* @, x) f0 }* y; C1 B
if(n>5)
/ j: h3 t& y# G/ H! C7 S7 p$ v/ g& ~. | printf(",%d",k5);: p2 h3 U o# k* f$ H+ Q' i
if(n>6)
) Q% L8 ~; `2 D/ G7 u2 o printf(",%d",k6);
' o$ p# e) \/ P$ A4 rif(n>7)' A$ f* i5 l$ E
printf(",%d",k7);8 p }6 A9 }" B+ d4 ?
if(n>8)
4 ]. w( o4 q! K8 r6 ] printf(",%d",k8);4 s# w# U! x9 u/ c, z1 j5 j8 e
if(n>9)+ g8 w: N7 K: g6 L0 v$ H
printf(",%d",k9);, u/ B7 k/ h) m' B k* J
printf("\n");
# C' i8 y* o% }( O" t }( N' b& b1 ]1 }* `/ {6 Q
}5 U& v z& F6 W
方法二:此题用数组方法更为简单.: k+ F [0 C/ \! }+ h4 m/ U
main()
/ u0 q h! T9 ]# ^& e5 l{
* d& l/ O* U" K$ _3 J# Z# jstatic int k[10];- P! _; b* r! F3 v
int i,j,n,s;
( _; _- R3 K. \4 z0 rfor(j=2;j<=1000;j++)3 r" H1 G# \ R7 S/ Z8 n9 D8 V4 X
{/ J6 q n4 W" H
n=-1;
9 \/ F: T1 ]0 L3 q" l5 ns=j;- ?9 q; j; s9 G2 V' ~# v4 {; `
for(i=1;i{
) g# t2 @ h9 H' ^if((j%i)==0)
/ J! j& `3 J6 {8 m{2 l+ b( ?; d4 G, w5 l' f; e6 M; P
n++;& U% K; l$ C6 a: S4 m# }& x3 m
s=s-i;
) _5 w. }0 Q. m2 j& lk[n]=i;; z0 e! W, P+ s$ U; V
}
4 G i" w; K/ f& X) Y% E% ~ }
/ q1 J y7 y( q% v1 _if(s==0)
1 w% x5 P0 m6 g3 @{( h% s, {: [7 V- z8 ?# a0 h
printf("%d是一个完数,它的因子是:",j);
* U9 G; J1 U+ A8 {1 Vfor(i=0;iprintf("%d,",k[i]);
1 i+ h) W; p' r* i( V1 rprintf("%d\n",k[n]);
: w, o; n0 V/ N( v2 _1 F; a! T( H- J}
. m5 ~, [0 C$ y3 O7 g- O2 ^9 D}
' R) O( O4 i! j6 S' x9 ^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.
5 H, q) P( d' L5 F) p. g E解: main()& i* x% w6 E5 \
{
" k5 L& Z* m: {int n,t,number=20;: @! C: K' K. d* m7 a
float a=2,b=1,s=0;+ F7 n* ]3 F2 Y% p C/ U# ~
for(n=1;n<=number;n++)! C% S/ M7 A0 Q* G9 T- A9 d- u- P
{
# }, }- d, \) m( us=s+a/b;
# l( Q7 v& I, B# j+ S& i$ H/ \7 ]+ wt=a,a=a+b,b=t;
5 Q0 O8 c" k4 Y- J; I}
! N! b' ?( ?! V( |( B0 S |! hprintf("总和=%9.6f\n",s);
& j& W4 Q9 Y% A- ?3 o}0 N0 V* `( K0 t Z
6.9球反弹问题
0 d# I' I$ v% r5 \; k2 D4 nmain()
8 s9 N# U* i9 g/ s+ Z{
; J4 `$ k) p# W# i7 B( rfloat sn=100.0,hn=sn/2;
- O3 ^% T$ P. W, H! q7 J7 P" vint n;2 \0 r! E N& O! E- ] Z# K7 V, A
for(n=2;n<=10;n++) k. Z* I5 f8 H3 {
{
3 f$ I2 W; ^- Z7 j. usn=sn+2*hn;
* B+ f2 `+ ], v8 [% U7 A: fhn=hn/2;
$ X2 y' v& g7 s7 n$ T# O}( W8 [; G# P+ w! l
printf("第10次落地时共经过%f米 \n",sn);/ J* x. }6 T5 m4 d6 b
printf("第10次反弹%f米.\n",hn);
; j2 E- w2 F: i* M- t8 U. y* \}
2 w7 X3 W% ^. B6 W6 l0 R8 l6.10猴子吃桃
. `2 Q7 [: r6 v2 M fmain()
. }4 T5 `( ~* ` h2 F3 X{! r# C \" P4 ?
int day,x1,x2;. |2 H3 ^! x2 L( g! v
day=9;
3 Z# q1 n6 u$ ax2=1;
6 b8 C b0 o2 B8 xwhile(day>0)
, S( u# `4 B( n8 I, J* Z; H{
2 g' A4 w2 n4 ?' M5 V2 g( Yx1=(x2+1)*2;% I( [ G# T( v# r$ q5 K
x2=x1;
. R* l' i8 v5 C- uday--;
9 e3 ~# u0 E. ~/ o4 p& O- X1 c}
/ T; u0 i& s/ s) a9 G# _/ Dprintf("桃子总数=%d\n",x1);
' [7 e+ ?/ v. h' l}
, h0 M: g7 ^) n3 o+ {& T7 v
3 ~1 U1 j7 X, e' o6.12
) I3 k. O& U$ k#include"math.h"
2 T% e$ e8 ^) v0 @ E7 `main()
; E; n9 d8 w, c- }/ X0 w& z# z{float x,x0,f,f1;
" T/ H. @, N2 t" G x=1.5;
/ [; P: e8 e3 J7 G do/ o, o5 J9 H- }8 h1 q! Z% k. L
{x0=x;' A- [. y. p! T) M0 W
f=((2*x0-4)*x0+3)*x0-6;9 H6 e% u) @, h* Z. q) k. H
f1=(6*x0-8)*x0+3;9 D$ d/ F" d* t1 Q! R
x=x0-f/f1;
# R4 T4 G# U; ~ t3 R }
' c: J, Z4 ~' E3 C( i; G while(fabs(x-x0)>=1e-5);5 o/ y' r& u" K
printf("x=%6.2f\n",x);
2 ~* z7 L7 i h" K}
?8 b# Z3 C6 R9 y5 L- c* a, K, x* x
|+ Z! x6 c! m; ?% t6.134 k- [. R: S9 d" R$ }) J
#include"math.h"# e1 m0 I0 Y4 c1 _+ j( E3 I, e
main()
8 L' a7 _3 @, ?% f2 Z7 s5 [{float x0,x1,x2,fx0,fx1,fx2;
; ^3 c1 {) m6 E, L: d( g! W1 t do
0 S" Y: ^0 x# ?/ I4 T {scanf("%f,%f",&x1,&x2);) R7 a7 e, T/ y- l
fx1=x1*((2*x1-4)*x1+3)-6;5 X9 i0 [( v# G1 ?1 |
fx2=x2*((2*x2-4)*x2+3)-6;
1 m) O' J( e5 V }8 L6 Q8 d' W5 T7 T/ J9 @
while(fx1*fx2>0);
4 ?% V$ I- Q) H) b do
. u7 f$ A4 }2 r8 N1 W4 a {x0=(x1+x2)/2;8 {& T0 i4 S1 Q' R: n, b
fx0=x0*((2*x0-4)*x0+3)-6;- \, V7 F2 @" J/ t4 E% ^4 N9 C
if((fx0*fx1)<0)
" U1 U- u! V- B( W8 @ {x2=x0;
6 W3 |4 V* Y8 Q$ A( X7 l+ J fx2=fx0;
7 V/ }0 G7 R. l. }; b( F/ @ }
; H" X1 L& b1 n( E4 }. c2 B+ H* F else, w- u" M* _" Y: a! q z
{x1=x0;: U T' l/ v7 J% D$ _# _ n6 h/ r
fx1=fx0;; Q2 g2 z' R7 S1 F! H8 q+ t( q+ R5 i
}6 s s1 }' N1 L% q0 ~ E2 c
}' n8 I0 G; p( b' ^
while(fabs(fx0)>=1e-5);0 U& W. y9 X# X6 A4 T* I* T
printf("x0=%6.2f\n",x0);
# X( e# C/ y. n M B/ D}2 _3 a/ K6 `/ l# C" a+ y% g) L; u/ i
6.14打印图案# J; j$ ?8 e8 G2 ]
main()
2 I$ p: I) {) O6 O( {{int i,j,k;
& Q5 h0 u# q2 O6 Y for(i=0;i<=3;i++): Y$ @/ S3 M2 K! S
{for(j=0;j<=2-i;j++)
0 q" H' F! B, J+ M a7 v- [ printf(" ");$ t4 \) K4 n( u* L: ^
for(k=0;k<=2*i;k++)3 C0 Q" h4 v0 c& H5 a+ u
printf("*");
' q# ] w3 z+ I# p printf("\n");! h& `6 f9 v2 {1 ]( e) Z1 u
}" t9 F5 {& ~1 ?' m- L5 ?# A
for(i=0;i<=2;i++)
- N7 m U- y! S' U2 D0 I) J5 n/ l {for(j=0;j<=i;j++)8 \& i+ a' r h+ s0 w. G. u1 m. h/ x
printf(" ");
2 F$ E! K. m( Y2 P$ ^ for(k=0;k<=4-2*i;k++)8 |& _5 U" g* C" @* Y- ]
printf("*");
, ~9 W; P9 J. `; \" b: l) h9 } printf("\n");1 q) u; h2 W) P' j7 B
}
+ ~ g0 L5 d. h, w}- l' W9 e8 F. b( I
6.15乒乓比赛0 R j7 V& ?# |) y) B, h
main() | p& E8 Q" w) q4 z
{1 ?8 x4 g) W! O* j
char i,j,k;
0 q _" y" J! ~# b0 G0 L! b0 g$ lfor(i='x';i<='z';i++)3 g: r/ | U8 w4 {6 H! h, _, l; ?
for(j='x';j<='z';j++)
8 C; z$ [- _6 f4 }% P6 U" ^3 L! b9 r9 R {
4 I: T( `: h' j, ^if(i!=j). {3 ?5 W& s" B3 ~( j. m
for(k='x';k<='z';k++)
% s$ E2 W2 a, |# O* n) z" o {
( f1 A$ b2 j S3 _. [( @if(i!=k&&j!=k). h; M% g7 U% G
{if(i!='x' && k!='x' && k! ='z')% ^8 v6 H1 Z; M5 P
printf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);
) W) X" W" B8 u% E& h) k }+ f3 R) u2 Z$ z
}; E+ @# a M2 a
}2 u; \( b. x0 x5 ~( c9 \
}
! t9 I5 u& B$ E! oC语言设计谭浩强第三版的课后习题答案
7 m# z7 g# B5 L/ J7.1用筛选法求100之内的素数.& L- Z+ ?1 G8 h; I) ~) a
#include0 M! F c6 y, ]. _0 n% G
#define N 101* h4 e- g) ~) S Z" z) I5 c
main()" q H, \4 K, S, V: R% n
{int i,j,line,a[N];
: `% r& ?' }+ qfor(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
9 Z" J4 z3 e) Y, G if(a[j]%a[i]==0)- C1 m0 S( D" n
a[j]=0;
# G1 K- J( p- v( H) n1 Xprintf("\n");& f+ B3 t7 A+ q
for(i=2,line=0;i{ if(a[i]!=0)4 g3 y+ [! F6 `8 n) |
{printf("%5d",a[i]);7 L& L. \8 q- S! f
line++;+ A8 g( e, r- R: W: c
if(line==10)6 S& g- U7 Z+ S
{printf("\n");6 C2 B$ Z2 P2 T7 L' _/ K* d. f9 I+ v4 `
line=0;}/ o B* X# s9 O
}
4 q" ] B4 { G/ W8 {0 e# Z5 F}& q6 S. z* _, q0 s
7.2用选择法对10个数排序.
7 @( e/ m6 k. W8 u#define N 10
+ z/ C1 z; _. u0 g3 o! l6 vmain()
" c* e$ \% K) T* u{ int i,j,min,temp,a[N];
$ n$ t, r0 n! w( q) Q* Pprintf("请输入十个数:\n");
+ O1 a9 f6 z% \6 B! p/ Hfor (i=0;i{ printf("a[%d]=",i);5 ~! B, W$ q* I- V* Q
scanf("%d",&a[i]);% X* I' H' X4 G1 a8 x3 Z
}
$ k& Z! z( m6 M) G) E& Q: hprintf("\n");1 d. Q6 o. A) c6 K
for(i=0;i printf("%5d",a[i]);6 {" \: e0 U9 H1 K8 |- z; X# {( ~
printf("\n");9 I- q+ ^; Y4 x7 a3 f% ]- ]: m) A
for (i=0;i{ min=i;
( h2 Z6 V; [, D8 C% ~ for(j=i+1;j if(a[min]>a[j]) min=j;
4 [/ N9 t2 `9 }+ y& l temp=a[i];
) R2 b4 a. o2 D& `! ?5 G a[i]=a[min];
9 j- ^9 e3 z# I" o a[min]=temp;
& P, L5 v$ `& O}) E# l. R B$ ^& B. l3 r
printf("\n排序结果如下:\n");2 d. ~) b( N$ B1 G, Q
for(i=0;iprintf("%5d",a[i]);
* X& _. Q7 Z) `9 d, t! y' ]0 p}
& B6 i8 m1 P# H1 U' f( Z7.3对角线和: b/ A. s4 b$ U" @; }. D
main()
8 B% x3 m) Q. U{
9 R' ]0 b& O) T1 u; A- k2 tfloat a[3][3],sum=0;
7 r2 S1 i8 J, \$ l9 Z1 |int i,j;
n" T% ]# g+ X1 K& p0 `printf("请输入矩阵元素:\n");
$ z% h9 V1 [( i( D: P: ]+ h& hfor(i=0;i<3;i++)
0 K* j+ t# {2 d2 H# i2 }! ~ for(j=0;j<3;j++)
# I4 H$ _4 ~# y, X; J) d2 t scanf("%f",&a[i][j]);. {7 ^4 E" g+ b9 ^; R6 C9 w* j
for(i=0;i<3;i++)( Q8 W9 h! S: S K0 T3 H
sum=sum+a[i][i];0 q0 q- p. ` U+ o5 ]! s
printf("对角元素之和=6.2f",sum);
/ U/ R7 z+ I& y8 [& A! f}
1 ]' C W' P j# e7.4插入数据到数组( z- V9 F+ v/ A$ i1 h9 @
main(): v. `6 H n w: K$ O# }) G i
{int a[11]={1,4,6,9,13,16,19,28,40,100};
$ `# r5 }* T! e2 k6 wint temp1,temp2,number,end,i,j;7 L7 C- L9 ?& J4 N- S
printf("初始数组如下:");, M) z7 I1 J* M0 A0 @* I8 J
for (i=0;i<10;i++)" V9 c x( O1 Y& S) C- P
printf("%5d",a[i]);
, I+ _: s$ l9 y' P% Lprintf("\n");
6 Y* u' @- ]: ~printf("输入插入数据:");
) m! w K8 `" s! iscanf("%d",&number);# a7 o$ j4 x( U2 K1 b
end=a[9];
$ s8 \2 h5 ?2 U' yif(number>end); M3 x2 o9 n, Z/ i& F- r. ?& ]/ O
a[10]=number;* K- e$ P* }7 l. ?
else
6 K: f4 E8 M( T8 H( B: E( j {for(i=0;i<10;i++)" @5 D7 B1 @& n0 e5 e
{ if(a[i]>number)% s- |+ M9 @" q8 k3 p
{temp1=a[i];7 o. |# Z% I& ]' [! n# i
a[i]=number;3 z: L; ?) Y; `9 _6 Z( X$ \
for(j=i+1;j<11;j++)3 a' @' m0 S/ J" A
{temp2=a[j];
' Q F1 E* Y) O% y: ^; [ a[j]=temp1;; E7 ?; r2 K8 b; F& P3 o
temp1=temp2; n8 |. ^. H7 F" m0 ^+ Z: W
}$ P ? Y( H2 N5 Y$ G. l+ z
break;6 d3 K Y3 o J, X( k
}# [# q. p y0 I
}6 W/ C7 F& I& d1 I& s( e' p
}0 v# i) ~) S# Y# A; [
for(i=0;j<11;i++)% C2 T+ \0 A& m0 q
printf("a%6d",a[i]);/ d% m: X: O/ j- ^& z8 r
}+ N, s) k, Z. P# F
7.5将一个数组逆序存放。$ T+ I4 M) N6 w, h+ Q+ O4 ]
#define N 58 F. Z5 x/ g X- D9 f" \* t1 g
main()' U3 A: U$ V9 ^ e4 n. J
{ int a[N]={8,6,5,4,1},i,temp;
# {* q+ \+ k8 o( O9 L Fprintf("\n 初始数组:\n");, Z1 I7 r# O; }8 _9 M/ J0 z% q
for(i=0;iprintf("%4d",a[i]);
3 j; \3 Q, U# C1 j5 T! [for(i=0;i{ temp=a[i];% n+ ? B8 w. V/ U
a[i]=a[N-i-1];* X& M* ~# U3 [
a[N-i-1]=temp;1 \0 t9 J0 C8 e& ^# W8 U: _
}
8 ?3 w! d5 B" R* \7 v7 Wprintf("\n 交换后的数组:\n");
: g1 `7 @* C3 x1 Q0 `for(i=0;i printf("%4d",a[i]);
1 f' p; }3 f" j/ v# x}
. }( l5 K! F6 I `( l7.6杨辉三角* c) R q6 b, b
#define N 115 E0 ?9 I' [9 K+ c/ t. T/ k5 r
main()
- }/ c/ f. S- T7 A/ f5 T$ ~, S{ int i,j,a[N][N];% M* r% y( F# X
for(i=1;i {a[i][i]=1;
3 ^3 R5 I& F8 P: @1 P9 h: w4 h a[i][1]=1;% T4 l6 C7 w2 |5 X! |. {# m, c* j
}
4 Z+ l3 b& D1 F' d7 e" C for(i=3;i for(j=2;j<=i-1;j++)6 h A' T7 B; I9 ]2 \; S! ?3 t
a[i][j]=a[i01][j-1]+a[i-1][j];! U8 g2 a: H( v* C) M( g: S
for(i=1;i { for(j=1;j<=i;j++)
: ~( H: L8 z5 {9 f t+ X; r2 v; [ printf("%6d",a[i][j];) S0 B4 o0 g4 z1 ~! P* k3 `: O
printf("\n");2 I; U2 ^- P7 l( ?; K
}, d; l( Z1 }$ o
printf("\n");# a. `; @4 G$ @! j
}
3 e$ O. W7 i3 v- t7.8鞍点
A7 J. R% r( f6 X& L1 Z; E#define N 10& u# J6 @# w, E" Z) G' Q# N& i* N
#define M 10
* N0 e# a- D- h1 J3 ?$ ~main()
' j# Y9 y8 }4 G( J4 H{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;) P. E9 L0 q& g+ g, p, @, `# {" d5 O
printf("\n输入行数n:");( ?" x0 P( z; i. M' C
scanf("%d",&n);- B; N( z3 ?2 [) R, X Q8 p
printf("\n输入列数m:");
$ e; u* {9 S! t* B* ~0 z3 u scanf("%d",&m);! D$ q, b2 g2 W7 K9 R2 ?( j
3 v6 q; D( ^3 O' Y: I, N e. i
for(i=0;i { printf("第%d行?\n",i);
, U. s* d# M8 P& t' ^ for(j=0;j scanf("%d",&a[i][j];
4 {% k) ]4 S# P+ q }. K+ ]; m1 Y( h; x4 U0 ^* l
for(i=0;i { for(j=0;j printf("%5d",a[i][j]);$ z/ a f1 O+ A
pritf("\n");
3 {4 z# g" I! R }
3 o L; V. e8 a6 E, L flag2=0;- J+ ?4 i* r$ S
for(i=0;i { max=a[i][0];
& c) L8 v. `! o) y: S7 v for(j=0;j if(a[i][j]>max)+ Z/ w- {9 U. @
{ max=a[i][j];& e& `- N3 m2 S8 U
maxj=j;0 h: q: ]6 R- Z1 f/ I
}# b2 k' A, d! D" d! p' L+ b' ?) q- J
for (k=0,flag1=1;k if(max>a[k][max])
1 z. w7 M3 u3 P4 p0 j3 A flag1=0;
3 A# G0 X; g2 M& ~ if(flag1)' c# V& V4 s, v) ]& b6 R8 x
{ printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);
' M" h7 x$ ^7 c& I flag2=1;
1 x( s# _1 ^# O4 X% | }" x3 u( f# }! _2 `. L3 [
}
* n" g# s# R. ]if(!flag2)/ J* u# _# Q1 K) K' E: A
printf("\n 矩阵中无鞍点! \n");
4 M- ~6 `9 g2 f) s) _2 k p* n2 k}
# Z" _; R' d: ?. T& b/ H2 Z( Y9 `) [3 O- v& \
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.
5 o6 F9 m, ^! U: @#include' Z- ^+ g3 {+ ^9 S3 ]: r
#define N 15
4 ~. _8 c# }5 y4 z5 P9 w9 lmain()
' b- K* t4 ~2 [7 A{ int i,j,number,top,bott,min,loca,a[N],flag;
; v2 V: N% d. n2 C% [& t char c;; M- Q7 n7 U2 R E: t7 v/ c
printf("输入15个数(a[i]>[i-1])\n);4 T( H, v/ E0 O6 N
scanf("%d",&a[0]);, i! j/ _4 n0 {$ T$ |0 v+ t
i=1;, ^1 I: o. |+ p) e* Q' @
while(i { scanf("%d",&a[i]);
9 p3 j; r! \3 r if(a[i]>=a[i-1])
* o, Z/ [$ `( w' @ i++;9 ]- r$ ^: h( b5 ?0 {5 D- \7 x
esle
% O; y" I ~) z# r2 J- x {printf("请重输入a[i]");
" |, u- d% O, h5 Y% I printf("必须大于%d\n",a[i-1]);! v2 Z& C5 [+ M0 o; W( [& @2 X
}0 F2 _' z. K, [* m/ x1 D% \
}& s: z! K8 x- Z
printf("\n");
7 f* b+ f% T C+ Q' D' u for(i=0;i printf("%4d",a[i]);
) ]* _0 ]* X' l3 M printf("\n");
7 ~" y" p. z/ B6 m" w! @ D: S! t
& n; T' F' F! y5 D flag=1;
: v8 i6 X: ^$ c) q% t/ I/ r, F* J while(flag)$ O* O! a! ~: T2 }) ^5 ?, Q! S
{$ T: G! e4 Q% n, u7 }7 p( c
printf("请输入查找数据:");
$ _7 W$ U8 M H9 r3 M: D* |2 j3 C scanf("%d",&number);7 ?# n( y4 i5 ~# }1 T* \
loca=0;
& ?+ } M n3 L7 g top=0;
5 P7 J* T. N- R6 G1 H5 D bott=N-1;
. C! ~& v. N' m P if((numbera[N-1]))) U( a& e. [6 C/ `
loca=-1;
: K3 j4 ~$ G6 {8 B$ | V while((loca==0)&&(top<=bott))+ @& G) U: e) v0 z. Y4 C
{ min=(bott+top)/2;& O! U% H1 o% P8 u
if(number==a[min])
0 R; q" H1 k! `) f; i+ j { loca=min;6 [5 V( {" b* \% _
printf("%d位于表中第%d个数\n",number,loca+1);# s6 K- e z' b4 Y4 j' Y: G
}8 M, s0 ]; c- G3 ^) u
else if(number bott=min-1;
' t l& v Y+ D/ C7 t else7 {, Z* ~ l9 H2 H/ v9 N
top=min+1;8 u& ~, J: o* A, \) s+ x% F" c
}% k3 ?8 z" f4 N% b4 Y$ M' G5 k
if(loca==0||loca==-1)% \4 ] B' P+ @" T4 h4 B1 G
printf("%d不在表中\n",number);/ ^. C& Q- y" G; u( ` I# L0 E
printf("是否继续查找?Y/N!\n");
* a# ~6 J9 u/ j- G5 a2 m c=getchar();7 E" N, L% ~1 @" P' E
if(c=='N'||c=='n')
( p* u5 e: L, o2 R# Q/ D. L$ _7 { flag=0;$ k! d2 Z- e5 p/ s/ d7 }
}
* A' R2 { P' g; R; ^}3 V' L8 c! C6 `" L( B/ n% O8 [
$ u# k% {- a" U: d D n! |
7.10
2 M9 W- G# h# q$ Q: Z. ?" |; D6 emain()) C6 S! i. Q+ `/ N/ F8 S
{ int i,j,uppn,lown,dign,span,othn;
% [4 [7 N+ I: M: F; t char text[3][80];
4 D" P+ `' S4 A% m8 m uppn=lown=dign=span=othn=0;- s6 d0 W. d1 _+ J" }; [
for(i=0;i<3;i++)
2 N0 N2 B U! F: H, Q+ v4 w { printf("\n请输入第%d行:\n",i);
9 O0 {* B6 y5 u; j: M2 ]; c gets(text[i]);2 d) g) J. k. _. ]
for(j=0;j<80 && text[i][j]!='\0';j++)2 V" g* N& u( [. J1 W
{if(text[i][j]>='A' && text[i][j]<='Z')
S( ?) N3 v& v( V uppn+=1;
' `; E8 @1 \: _' H: o else if(text[i][j]>='a' && text[i][j]<='z')
; A6 x& J) W+ m lown+=1;
* B% V. ]* ?" l: }# ? else if(text[i][j]>='1' && text[i][j]<='9')
y" O9 o3 o/ Y* b5 t3 A, E dign+=1;
% u1 ]; N& x$ |: g- W! ` else if(text[i][j]=' ')6 e7 W1 Z2 O& \) M
span+=1;
* D& D3 {6 p& U9 n else
8 L8 f* ~ M$ p, B2 I$ r othn+=1;
0 ^" X5 K( z. [ }4 X9 w* p! T% X" F
}
0 ~8 H" k" a0 N0 T. q- e- { for(i=0;i<3;i++)% ~3 Q" x- I9 m' Z0 o- C
printf("%s=n",text[i]);
1 w. Q$ R9 i! I; f# ` printf("大写字母数:%d\n",uppn);
3 O( n3 P) j3 G; n) n# c2 K) y printf("小写字母数:%d\n",lown);
! @3 }/ v0 Y/ [9 e' ` printf("数字个数:%d\n",dign);+ V- M6 {" | b; N( w. s
printf("空格个数:%d\n",span);8 l/ ~! \3 Z( {! a; x: u- x
printf("其它字符:%d\n",othn);- p" `0 E) h; N A" W- v
}
* t0 z0 q! |! k4 @7 U# {
8 g+ A5 Q W& B5 g" ]! y4 G* L* H' G/ H I% I1 n
7.118 O, J( d: H0 S# A/ h& h6 ~
main()9 T; B; Q: N' ^6 N$ i% S6 J) E
{static char a[5]={'*','*','*','*','*'};, J Q' W o8 V- c6 g+ d: s
int i,j,k;0 j" g& G# ]$ S6 L; R% E0 S# s. f! z
char space=' ';
- @5 {5 e% E4 m- c" |) \/ H& J& [ for(i=0;i<=5;i++)
6 j7 a5 R' v+ v" Z& N3 h {printf("\n");2 T9 R* W5 Z- D) t
for(j=1;j<=3*i;j++)
7 G M; Q* k2 \ l0 }& ~3 a; V printf("%lc",space);
$ O# u# D4 O# R6 K for(k=0;k<=5;k++)% u# X( y% ^8 r( R! ~% r& _
printf("%3c",a[k];. g; L- j |) ]( `% X& @! H6 O" [2 J
}
7 u8 F$ m; [8 f6 V Y}
# A# y0 z8 ^- [. i6 B: e% p7.129 Y; l: T! ] b0 ~9 H. b. ~
#include6 K9 \8 a+ b' Y$ p
main()2 \; F+ F5 g* n! x+ q: [1 l! P
{int i,n;8 ^/ }; Z! ]/ o s5 }
char ch[80],tran[80];
9 I) I% ^$ L' o3 J0 Z9 Y printf("请输入字符:");
5 g' U) z5 N8 v) |& F' W gets(ch);
* `0 b/ a# A5 i( \- w5 dprintf("\n密码是%c",ch);/ T3 t8 s/ _( `
i=0;) X6 z6 J- V+ h
while(ch[i]!='\0')
. W& N2 E6 ^4 D n/ N& N{if((ch[i]>='A')&&(ch[i]<='Z'))# x* ^, U7 c2 M
tran[i]=26+64-ch[i]+1+64;3 n: j( J$ Y. `1 v6 K3 o
else if((ch[i]>='a')&&(ch[i]<='z'))+ n. w' v; x. J
tran[i]=26+96-ch[i]+1+96;
: Y, \, S3 E6 o8 |) belse
7 C" R! b1 e6 d* \' _ tran[i]=ch[i];: h3 Y3 G) s: n5 c
i++;4 f3 z& \% E, _9 y* O( t6 X
}
% u6 m$ R4 x8 Hn=i;4 Y3 n1 C1 [9 K% s0 C9 M
printf("\n原文是:");
8 |# R, m9 Z4 k/ I# {- d" Xfor(i=0;iputchar(tran[i]);- g: F0 F! `/ P$ [ W# a% P0 L. u
}
8 u" A! A$ D2 F! A1 C4 X) w7.13' A, V/ Y* u1 D* m2 F( d
main()" {; B( l6 G* f! e; g: I* H% c
{0 _$ L5 d T) u' g3 P+ h8 f
char s1[80],s2[40];( X0 }' a6 j* v# m1 D
int i=0,j=0;
N# h( M) o' g: C$ n% c printf("\n请输入字符串1:");
/ `* h* X% w- S" `( |8 C" h5 o; @ scanf("%s",s1);/ L6 b2 ?* a' P$ d2 R* ~
printf("\n请输入字符串2:");
. }- c. Q9 [, G- |/ D+ x* p scanf("%s",s2);
0 \) C9 F+ @1 O; o. p while(s1[i]!='\0')
/ L! D% @4 O9 q+ E3 }% j i++;
6 n4 \5 d! A- m8 i4 N0 ?while(s2[j]!='\0')
i' ^! c& t V4 A P1 t$ t s1[i++]=s2[j++];
' B) v5 @$ L3 vs1[i]='\0';
7 I# x) B& y* G' S' l6 c t4 Lprintf("\n连接后字符串为:%s",s1);
$ ~. j* {( O- w5 ^$ a }
' H) C7 a. k: q+ ~7 z4 l4 Z# m' b, e; a# O- F4 q0 |8 w( |
: y; U8 d1 D1 m6 r' [. W7.14
{- z9 Z6 T; P' U#include( j% O U# r+ [- Z& y! K6 L) w- j
main()
* O5 I q# m7 G0 J: y3 e{int i,resu;# ^$ a/ i; b- S
char s1[100],s2[100];$ t, n5 j8 x! H: ~1 ^7 p# n- o
printf("请输入字符串1:\n");
6 t" e5 o/ g' a4 Q' N$ X gets(s1);
3 W9 Q9 _+ p1 B4 E( r" ? printf("\n 请输入字符串2:\n");
% J$ J, e7 e* P2 G. v) ^# M; l gets(s2);
/ P$ ?' o! l( h# ?" i i=0;
- P* F; z* x; V4 d2 J) Y while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;
, }0 G* y3 q$ K# U if(s1[i]=='\0' && s2[i]=='\0')resu=0;
5 O; v) p6 e6 n else \) P" ^1 {$ p* O2 s; [. V
resu=s1[i]-s2[i];
8 i) s2 b, X8 F J' q* A, T printf(" %s与%s比较结果是%d",s1,s2,resu);& j" \ C. C2 l9 d/ Y& m2 S# K
} E2 S7 J, v M8 y; C
7.152 b7 n2 f/ m5 I. @! T
#include* c- F6 C0 \8 ? O- |* \
main()
: J# Z0 l u) e) X/ v3 } {# Y" t6 u: ?/ ?, L ^6 L) `+ p1 f
char from[80],to[80];1 x! @5 J, G; W8 |' x4 F
int i; O, e4 ^" b/ F! Z5 k, J. y' K
printf("请输入字符串");
, H) |* P' v/ a" g( p2 o scanf("%s",from);" R- Q* T0 s$ {2 U( B
for(i=0;i<=strlen(from);i++)
. R4 m" r. P' D" {4 y# k7 T to[i]=from[i];
' u4 N: P$ P% P Z printf("复制字符串为:%s\n",to);
' a# a# J3 o1 w4 z1 _ }7 T! ]1 Q8 Z* N4 p' C1 I- k$ R2 w ?
- R3 f& B/ j( h; J5 l% c; I
3 }- {* L7 l$ w1 I5 ]# k第八章 函数+ i) y1 C, P3 E
8.1(最小公倍数=u*v/最大公约数.)! j0 S8 _! Z& }- D9 @& U
hcf(u,v)
; B( |1 R4 S2 s8 F j8 w8 jint u,v;9 ]% u, b- H6 H
(int a,b,t,r;
$ ? J7 M: o x% ^, f! Y if(u>v)! K) M: K" ~4 q+ E2 V1 ~
{t=u;u=v;v=t;}& v$ \+ E0 F: Z' U" e. `
a=u;b=v;
6 G8 B: r. t) \6 t% I while((r=b%a)!=0)9 N! J! M! @# m' [" g. B
{b=a;a=r;}/ o" q+ ]3 b h! T: M8 Z5 {
return(a);
' G% L( `- B9 ]; c7 ` }
; i2 E# T- E1 z# n lcd(u,v,h)
* H8 @1 @3 r7 o4 [ int u,v,h;
: |; A" r; G0 r$ g {int u,v,h,l;/ u+ ~& o h1 `
scanf("%d,%d",&u,&v);
# B/ O% {9 {% z h=hcf(u,v);9 }- z) o7 C* G& }+ G
printf("H.C.F=%d\n",h);" |6 }6 b5 G0 Q
l=lcd(u,v,h);
/ N( o) I. v% A# I printf("L.C.d=%d\n",l);
- }9 I( `4 X$ _1 Z+ B7 y9 u4 c }
2 X8 a, o; Z% M {return(u*v/h);}
# z9 z. K/ w! i* L8 K& u main()' Q+ G. Z! G; e( N6 v
{int u,v,h,l;
4 M% k( s8 I) z( V, z: T; x* F scanf("%d,%d",&u,&v);
: _, P# @, B) [( Z- S X- Z h=hcf(u,v);2 q- |. z. I) n# h* N
printf("H.C.F=%d\n",h);
4 N5 Z9 L4 @7 r l=lcd(u,v,h);
& m* G4 {" r( r1 {( t9 ~4 A printf("L.C.D=%d\n",l);& R8 }+ f- ~ h
}
$ R7 v4 h# q: Z) H7 x6 y8 |2 k: N8 F$ j) O
8 R# }. j: m, i# v
* o% G4 F8 Z* b& N; h8.2求方程根
/ U5 j" Y% k/ y#include
2 n+ }6 v5 z1 j8 Y) ffloat x1,x2,disc,p,q;$ Q- o9 `% p }+ |
greater_than_zero(a,b)
& K; E" c: [# ?float a,b;
. ?8 c5 T3 y7 z4 Y! P3 g; X{8 q5 V {$ Q* q
x1=(-b+sqrt(disc))/(2*a);
' [) E* \: W3 F5 Z+ u$ b1 C9 g! }% Ax2=(-b-sqrt(disc))/(2*a);( `6 B e$ ]0 P8 }( X9 h+ Y
}; b9 H2 K, `0 P2 ]: ?( h- i$ V
equal_to_zero(a,b)
- q& I# n# [; g. c. N% ?6 Bfloat a,b;4 g6 P- J9 g7 B- b9 E' V
{x1=x2=(-b)/(2*a);}2 B0 V- K, k" x* a" D8 z
smaller_than_zero(a,b)
7 N- X) T+ k3 V; B0 mfloat a,b;
; k$ Y, \% z O0 }3 f{p=-b/(2*a);$ A( J$ f# J& a) |1 Z
q=sqrt(disc)/(2*a);! V) }: W8 i+ d8 G+ i
}3 p3 F7 ~0 z/ f) T7 G
main()
6 o. v* p% v. ]* e) E8 G# f/ {4 p{
0 z4 A1 c: l, G4 f9 r1 V& T8 q* Wfloat a,b,c;# H4 I% ]9 {3 J+ s
printf("\n输入方程的系数a,b,c:\n");
5 H$ x) D5 Z5 X9 ]3 p2 dscanf("%f,%f,%f",&a,&b,&c);
1 Y% F! u$ N/ F4 G0 d# Mprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);8 H/ O0 k6 n8 F$ F
disc=b*b-4*a*c;+ |. r, n; `4 _3 q9 C1 L- o
printf("方程的解是:\n");
( A- b9 w1 g2 S5 U# m; z( @if(disc>0)! T$ h+ J. S% d ]( }& J
{great_than_zero(a,b);
6 p8 Z ]1 j9 K# V# }printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
3 B5 J3 Z" M2 B}( Y- w+ U4 b& J5 V2 C
else if(disc==0)
# t: }* R3 D. q! p/ q {* N$ j8 @: o5 U$ o2 Y/ u" B" P
zero(a,b);
) A7 l4 |% C( J$ P Rprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
% D/ O: a' C/ O1 ^: x }
: g9 L9 D2 i. U! c' d" G% _* Ielse
2 r: `. \5 ~3 f1 D( y$ z {2 |# h+ v* Z* ^% m
small_than_zero(a,b,c);
; k ]" t- U7 W! T3 U printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);1 U& l: }! k! n; |8 h3 i7 b, F
}
1 p+ U8 [# d" ^}: |/ V4 k* c# y" z- k
8.3素数
, w& r. A' ?$ t#include"math.h"
8 [6 |/ D: {! Pmain()
7 F) Q) `- @) I( B1 M+ W( R{int number;9 v! K H. \: p' i
scanf("%d",&number);
. N5 ^: O& W9 n; M( X if(prime(number))1 V; H6 R2 q4 ^! U' C6 O
printf("yes");
c& u1 z1 Z/ }. t else
. y* C8 M- ]7 R printf("no");4 Q- E: O6 S# m; d% d; _7 _
}, H2 V4 O; x# _- T t3 E$ d
int prime(number)
# R/ g3 [7 o7 _3 R% vint number;
6 v C8 z" A1 E8 I{int flag=1,n;) d9 j5 Z3 p. {/ u0 F
for(n=2;n if(number%n==0), M. \! l% U6 ~6 b
flag=0;. b: ]5 |" T# Q- Y
return(flag);
8 v) u: Z. I& Z}
8 ?/ H3 x( l) e+ r/ x4 H. g
! m' k; j4 S0 ~( i* D
; T2 h0 j+ B5 W# r/ j8 w; m# W3 G( {- o& z, \" W
8.4+ y$ r4 m t. R+ E3 k" E6 {
#define N 3
; z. V% T+ p/ o9 Hint array[N][N];3 I# { t! \- Z; V0 R
convert(array)
1 j7 x: t n) a2 E- B6 [int array[3][3];6 b6 k; x! F) L& {
{ int i,j,t;
. J- [9 W8 N3 F) T/ M for(i=0;i for(j=i+1;j { t=array[i][j];
: Z2 T1 r& o3 ?' U array[i][j]=array[j][i];
% r3 f) ?4 V4 l# d/ ] array[j][i]=t;! U8 e" p3 m3 M* M2 Q c+ C
}) V! W3 |4 u7 k6 N
}
; a8 S3 d0 H& b; kmain()1 n0 l6 L5 R4 z( t: B/ E3 E
{7 n! W* _% |8 @0 Y+ ?+ h6 c" {9 O5 l
int i,j;
/ {% v* l+ K8 |, `+ A/ z printf("输入数组元素:\n");( L$ r; q M+ L3 X/ P4 I6 p; L
for(i=0;i for(j=0;j scanf("%d",&array[i][j];
/ R$ ~2 F1 q6 V) _ printf("\n数组是:\n");! o; a' L8 J5 D+ K( L0 x0 l4 I4 O
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
F( q! R& @5 H: F printf("\n");
1 I- j' Y2 f" v3 q9 U4 A! h. {+ d }
7 K# d+ K; y/ t; e# q- ^ convert(array);. o7 Q# G' q5 N, B& U
printf("转置数组是:\n");& n: Y. s" y! y% p; G# u! ]6 I
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
8 Y& P e5 i) }. L* l) b y printf("\n");
& r: P) A- H& T2 H7 F2 F8 }& e e }
3 d: W8 G" _: |4 u- e! Z}! A* n1 k0 k* d" e; E1 _- H
! O. m/ @! w0 J/ H( p5 r0 @' s+ }" G/ b L1 D
0 |" i' a. `$ q9 }2 L8.5' s' M' f( `( R; e+ x% F
main()) s8 o+ [; P! q
{
/ i" y7 L! \# f9 m, Achar str[100]; `9 d6 @9 v w# ?5 i
printf("输入字符串:\n");
+ \( ]1 O0 a) R( M6 }1 Z scanf("%s",str);
# K1 s) J( S0 _1 e+ s9 M inverse(str);5 o! t3 f, z& [/ K2 g
printf("转换后的字符串是: %s\n",str);1 H9 S# s8 v6 S, H1 p0 C
}
: j. U) o- r7 pinverse(str)% ^4 g) \! W. L" Q1 O
char str[];2 `0 b2 h& Y, g( v% M
{# I/ u5 z4 w, E: }# t1 m1 Q
char t;1 `& s$ e9 q4 [ Y% i
int i,j;
* A5 R- p' }7 H6 b) j for(i=0,j=strlen(str);i {* e' s7 Q/ H$ D @. O2 V$ Z, u
t=str[i];
2 i* z, q+ r6 R* B! @7 p6 |& {& M str[i]=str[i-1];
/ c: [/ b) A$ E4 c str[i-1]=t;
* _2 \& g! z2 O$ p% A! D }/ q0 R4 {0 V/ p! S
}
$ s8 a m+ ?1 o8 ^$ M
3 W+ J) C7 o' m8 f J& ]
; Q* j4 o! @9 W- ^
, {3 @1 d' i$ Y$ F: \; `: h* S y8.6
0 \! S# g1 k) _char concatenate(string1,string2,string);
/ D' m' d2 F* k# G2 |/ X" Pchar string1[],string2[],string[];
' s/ |3 _4 A) a{
5 m3 b+ I+ W1 Vint i,j;" H1 n1 H* w) a& Y- Z& g
for(i=0;string1[i]!='\0';i++)2 ?; r4 R# |- p% h
string[i]=string1[i];) a+ x3 W- z$ Q' O% v, a- t
for(j=0;string2[j]!='\0';j++)
, s+ C3 M" G* W- ^- g' H* ?7 y: b string[i+j]=string2[j];& r. H `: G, a" G! ^+ m( x
string[i+j]='\0';! ]' e3 n6 @) M
}
3 X9 G: g3 o; w$ l+ X( K! C4 Dmain()6 v2 E% e0 R' _
{
2 y8 C: y. d: J! s char s1[100],s2[100],s[100];9 D; E+ U) a8 D' p
printf("\n输入字符串1:\n");! \* f1 U' T; l* \ G4 T; l+ U
scanf("%s",s1);
5 f) M& u' m) t* u printf("输入字符串2:\n");, W. |% r: N( N* Q$ T
scanf("%s",s2);
! M& R5 y& S3 k$ e6 ]8 t0 J concatenate(s1,s2,s);% }$ l2 M: ^# \6 M6 J4 K/ f
printf("连接后的字符串:%s\n",s);
* G" e, [( ~' z! P& H; z}
! A) x& y& o' F9 r9 [; y5 d& v! n7 C
; t5 @& J6 p, @- |$ R7 R3 }2 ~: M2 A9 F- T
8.8- @& v3 `" i" E4 d: X' _
main()" v, u% |0 f- d h
{
6 h4 g, V$ h. l9 | char str[80];- O9 R' O" C ^) H" Z( U% j$ |
printf("请输入含有四个数字的字符串:\n");
4 s4 v& X8 P4 N |. _) O/ a( j' w/ Z scanf("%s",str);, c9 X1 o4 l' ~" V- L( a
insert(str);' j/ f' e, f" W4 R! k
}
# Z# a: K1 s Z rinsert(str)
- X& c4 i3 u- Z2 O4 _2 U char str[];
* x4 ]0 W, v5 z+ V5 r# P+ g{
* V- l" l5 @. z* e; L' Q int i;# `% s: s0 d- F8 X9 c$ h8 @7 e I* Z" Q
for(i=strlen(str);i>0;i--)
# b {0 S4 x) @$ S# [4 ^ { str[2*i]=str[i];" \% t# Z% S4 P
str[2*i-1]=' ';
2 `1 _! c! I4 z4 s2 A( M, L }
( h/ s0 a ?. |2 T$ C o printf("\n 结果是:\n %s",str); p- W; v; e$ I( a- [1 e; i
}# E; u& |: ^/ r5 | \; y
$ r z8 \& y. l
; H6 z1 _! f* w# D. g9 O2 a, h2 C9 @6 u0 |2 Y3 K
8.9
% Q! a' o( p+ y; b7 Y- U#include"math.h"7 \5 H( \6 f) K" q9 j4 y
int alph,digit,space,others; L) L/ `& {; B8 L
main()+ x3 Q# |5 t9 S1 }! ] Z
{char text[80];
8 u5 g$ S4 l( Z! ` v0 O- D gets(text);( n$ y7 u. _5 f) r" o) T3 t
alph=0,digit=0,space=0,others=0;4 p* s& r$ ?9 o ?' Y
count(text);
- C/ u8 A3 @' N2 {* ?- Y printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);
6 ], M* [8 a+ }+ I- V) u* c4 W}* @4 W: E* r2 O- a4 a
count(str)
# c7 S9 T* e H& Z- Ochar str[];
; R) m. {3 E! P{int i;0 Y, m/ E4 M5 |, p* t; _
for(i=0;str[i]!='\0';i++)
% s W2 x% a1 ?1 V; T) u if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z')); y8 P/ E" x' ^; T. Y( t
alph++;) {# ?) u) q- Z& R$ {1 R: }" o
else if(str[i]>='0'&&str[i]<='9')' a4 `2 E& v) O
digit++;
% n4 y, V6 M1 b9 @. [! P% b$ R% Z else if(strcmp(str[i],' ')==0)" j N. e: P7 _( c B. _. g6 U
space++;
/ J$ \2 o" k4 o* t3 d; Y else$ G( {4 X# f z
others++;3 `1 u3 W5 W/ e. G
}8 B6 G* ?0 w# o2 b! }6 e' X3 J
3 X+ Z+ `% \ n) E; }9 P5 {) g8 {. q* n" V* \( n3 Y, f
8.10
' m$ Z' o9 U3 Q0 Uint alphabetic(c);# ^$ Y; x+ v/ D! s, V+ y; t
char c;
9 u2 M6 l. S& Z! J2 N{& c* b2 ^. [. k1 w3 S) l% T4 u+ e9 }
if((c>='a' && c<='z'||(c>='A' && c<='Z')). g* T4 V5 k9 ~( V4 ?
return(1);: @8 v/ n1 Q7 S0 F
else, g% t3 d, Q: s1 \3 M: t
return(0);
2 H5 h \4 k2 x- c}/ j3 |5 Z# A4 C& t8 p3 i
/ b5 D6 U2 h! h! @int longest (string)
8 S% G8 A& Y: k M5 V* X/ Echar string[];! m- ~7 c! l+ o7 \$ O
{2 f2 F* |9 Y& x; p
int len=0,i,length=0,flag=1,place,point;
& \+ n& |! R! d- Z+ s5 g6 G: S: ` for(i=0;i<=strlen(string);i++)- z( s1 Y4 w5 |# N
if(alphabctic(string[i]))
& \! r9 U3 T; @% [5 l5 t H' j if(flag)
& i4 \7 P! |8 j {+ i5 F5 g3 s/ ^' h8 _8 @* F
point=i;
8 u6 z( z: f) T4 y0 { flag=0;
8 w9 y0 X3 j5 w }' S/ q3 Z% r# ^9 n9 K4 L4 f
else
" D9 z* i# ]) @- X# p, a len++;8 s/ s& R# z3 N3 t
else
5 [- A! I7 F/ u; r8 w& ~ { flag=1;# k4 I4 E5 |( v) p; Y/ v2 N
if len>length)
$ s W) S) k2 y0 W* q {length=len;
8 R) B4 x) C' a! U' u* ^5 y place=point;
- s- v! Y& |7 Z: \- C len=0;& e( N4 U4 X( _$ B
}# }* B, A- z2 y$ |
}
2 r! z* E5 P" {9 {4 a* ^ return(place);
( Y2 s" Z) I5 Y3 ~ }
, s: }$ U- N) _# g+ h7 f$ _main()) s2 n' i1 z6 i6 q9 X* j
{
- Q/ p8 x. o4 pint i;
6 W( ?) E' W9 f8 A! Wchar line[100];
# I& o5 W4 m5 nprintf("输入一行文本\n");
5 ?6 Z# R4 p3 `gets(line);/ M, D6 A9 `0 `9 {
printf("\n最长的单词是:");
% `+ C7 U) h- Q }for(i=longest(line);alphabctic(line[i]);i++)! \6 V( `8 S1 B% w' t6 R
printf("%c",line[i];5 E8 P& `8 U* f2 {. G- Z& G! ]
printf("\n");2 B$ u0 A: q1 c& N
}
v9 U ^8 i, p( w* I- z/ ?
- } R- U8 d: Q: t, x
. G6 F/ ]" `: x( Y6 P: d$ E+ E3 e6 q
8.11
4 X* `* G N0 N1 g#include
' P, }! U8 t2 R6 N& |; H) x, R2 P3 h1 k2 H4 m4 i( D* V
#define N 10
2 i' i& }7 a) D8 [char str[N];
/ [: o6 k% l/ n. Lmain(); d. {' f/ {0 P
{1 Y" W" x! x* @: Z% d; c
int i,flag;
: n+ o' C# Q$ X! u5 S( u6 J3 W& Nfor(flag=1;flag==1;)
5 n& _: Y. _" S2 S2 u{
( s3 p- I& t [3 v$ n5 b printf("\n输入字符串,长度为10:\n");
; R, f8 B$ E" \% c4 n% q b scanf("%s",&str);
6 w3 P" \6 y7 o) t if(strlen(str)>N) a8 g; S' s- O1 b- X2 @
printf("超过长度,请重输!");
: B, t# s$ V& N+ T1 b9 X3 K9 X else
( X, K, o/ n- s8 K3 a5 r flag=0;- C4 r3 H, g l
}
" ]5 u2 t2 [# ] K# }$ s5 v$ nsort(str);4 @, D/ s8 {# b' }- v
printf("\n 排序结果:");
+ ? V/ x& d# F& f. |for(i=0;i printf("%c",str[i]);
% t% Y7 _7 q& c: q& I}
2 U* [5 T4 b# q( vsort(str)
& w- P8 P# {; B2 Zchar str[N];" j+ K( l9 S; R% C8 b" ~7 a
{
/ o2 s- O/ s% Yint i,j;% g* l* ~' X2 h; B8 K. v
char t;. C% \% A$ B# y: Z. P
for(j=1;j for(i=0;(i if(str[i]>str[i+1])
0 y; y4 h+ v8 M2 b" p* h& r, z% K { t=str[i];8 P' Q; l! `( _# n
str[i]=str[i+1];5 k+ R: `. J& I2 p% ?
str[i+1]=t;
: E2 M2 I' D) @1 w( f }
% c( H; l3 Z8 y}. B( }. ]+ ]; L1 V
8.12
0 o! v8 c; Q4 M; b9 ~; D#include/ Q' ~7 i7 R4 |. k; h) j
#include: l" G$ A) {* [. W4 z
float solut(a,b,c,d)
- @0 d% s4 Z/ J& d8 Z ufloat a,b,c,d;1 F; m+ P' B- r: A. n8 I( }! R, Z' q
{float x=1,x0,f,f1;9 ^0 ^7 L3 d0 K4 P/ N
do7 ^; g! _: ]5 Y5 \
{x0=x;) B% w0 s1 u; q( p5 q7 }7 H* @
f=((a*x0+b)*x0+c)*x0+d;
3 F! `4 s' T- C; ^; E; F0 l/ U f1=(3*a*x0+2*b)*x0+c;) u4 z% X" e! h
x=x0-f/f1;
8 a `; Y- a+ x" X7 x }7 N7 m8 S# N+ J9 s9 O& x8 V
while(fabs(x-x0)>=1e-5);, b2 w$ b1 C% W
return(x);
5 \1 p# t; y6 Q# D; F& |/ J" r* [}& A5 |4 k8 l& O7 @* F' H
main()
0 l- d8 k0 M G/ f) W1 _{float a,b,c,d;
[/ L8 s0 z/ @, t3 Z& Y; K* K V6 s scanf("%f,%f,%f,%f",&a,&b,&c,&d);
4 D1 H( P! k2 Z1 ~& |* q, _$ [ printf("x=%10.7f\n",solut(a,b,c,d));
( W, |" o' l" B}8 z* U d. V; ]$ e! F: d4 h
8.13
, N# I+ ~" x* R! h. i#include
4 E4 r. P1 r) w. Z4 q% gmain()
8 p: f+ u# C, b{int x,n;
: F- P2 J- F0 C' f2 O" s: J! g8 A# ` float p();
2 O% K2 _6 J' y; P8 _& V8 ^ scanf("%d,%d",&n,&x);
! D8 v7 ^6 N' J; k8 }: e* J$ }+ D printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
; A- t( s; N; ]; N' J! [}1 k; g3 ^0 y7 a& Y: K* w
float p(tn,tx)* o# x7 |: c4 n4 n8 I, X \+ V X
int tn,tx;
& p* Y6 z, {& C Q9 J) n& i{if(tn==0) W# N$ d( R, E8 J* O# I2 o1 u: k5 q
return(1);
0 y1 ]: `9 i! l8 R. y+ ~ else if(tn==1): R4 P- ]# N9 V! o6 o2 Y
return(tx);% Y1 {" `7 o$ ?7 s4 Y! S7 U
else
# ?# c0 y9 J) W% i return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
0 M7 z4 a" M. X0 ]7 d! D4 K" O}
$ s% D$ }+ U6 |. L$ `8.14
0 X9 K$ C: |' Y) O* s& z#include "stdio.h"3 V( ~5 R* v2 {, [+ _& U& o1 t* J
#define N 10
3 Y5 q+ G) e0 U L. f E# R#define M 5
' Z$ A3 }7 `$ L% }' ?+ ]6 wfloat score[N][M];' q) C$ k2 E7 ?5 G" G
float a_stu[N],a_cor[M];
& Y5 r1 t( s. `" O4 Q/ d& i4 ymain()
, |* {3 t6 k' Z. C" c7 P, _! W{int i,j,r,c;
J' l3 S& i! k/ s7 A% U5 e5 C9 W- B float h;
# G8 K5 G4 H+ t) j/ d5 k float s_diff();1 r' m8 c, N) H. n8 z
float highest(); e+ M- B& p9 ]/ @
r=0;2 k8 h# V$ V! W R& z2 z# U9 \# M
c=1;2 G# J Y* @8 J9 J" C8 A& o
input_stu();9 t. }+ Z- }" x2 v
avr_stu();
; A9 u l7 _1 ^* _; W7 P) h avr_cor();" x# d& g( e! p: |3 p+ d
printf("\n number class 1 2 3 4 5 avr");) D. v* T/ W# L. f
for(i=0;i {printf("\nNO%2d",i+1);* R" v3 D( P' _% ?9 p
for(j=0;j printf("%8.2f",score[i][j]);
& t: O+ ^# }6 N1 ?' D- X printf("%8.2f",a_stu[i]);
+ I1 ]5 S( J( q1 d3 b, Z }
: [: S$ Q7 X' U% V b! w printf("\nclassavr");! G5 K$ C/ C1 Q4 c! U
for(j=0;j printf("%8.2f",a_cor[j]);. G. x4 G$ g# W7 N' a, v
h=highest(&r,&c);# n$ S5 s5 Y& P0 e5 ^6 Z
printf("\n\n%8.2f %d %d\n",h,r,c);
2 D* L; l2 l3 g- K2 c printf("\n %8.2f\n",s_diff());
& c# W2 ^. v6 }8 e. k# V}: t) T' q' P4 T ]# p2 R# L7 w+ M4 ^
input_stu()6 U9 c1 X: c- S3 w3 X
{int i,j;7 ?6 u. b, H5 ?9 W' }2 r
float x;# g3 N. M3 o% q- n6 E$ L# m V) K
for(i=0;i {for(j=0;j {scanf("%f",&x);
A h8 t; o6 n0 e$ w9 z score[i][j]=x;
) }; X6 u( E3 V1 p3 ?, M }
/ ~$ e" O0 ~4 O- d) j }
) G. Q. j+ L3 |5 M$ t! J) W}
# b3 l8 X5 {' c! O) z3 r; eavr_stu()% ?3 s" G* Y: g0 G3 @
{int i,j;1 l8 L: i5 w- N4 p# h2 I" L
float s;
9 v; }6 O* i! N" I) u4 Z for(i=0;i {for(j=0,s=0;j s+=score[i][j];
; g) A6 K0 ]7 |* x B a_stu[i]=s/5.0;
% w+ r7 m4 U W8 H6 G }
$ g. f0 S, ^7 S( ~4 D) l Z, Q; K}! D# s5 E( f/ n1 w
avr_cor()) ?) ~9 n: S; B/ Y6 j. I7 M
{int i,j;
$ C& y7 @. w( Y, I0 S float s;
4 k a$ u% @! ~/ r% j; j: a8 E for(j=0;j {for(i=0,s=0;i s+=score[i][j];5 X7 v1 `. R! x
a_cor[j]=s/(float)N;0 _8 {, y* P4 {( o
}, A/ @) ` `6 C6 H- \8 r# O- I z2 c, t
}2 F- y+ v* L" S+ \+ U( `& t7 V" j
float highest(r,c)
& e& D/ t! {5 xint *r,*c;
/ P( { L. z! u2 b5 P{float high;5 w4 z- b( M' F) a1 ?0 p* j3 s8 F0 v
int i,j;
, m: W) b+ o( D$ a4 ?4 Y) a high=score[0][0];
, Z2 i/ j+ v/ S! c! ?; }7 l' D9 i" k for(i=0;i for(j=0;j if(score[i][j]>high)5 k d7 h% f; G4 ~; W
{high=score[i][j];
" M3 r9 \; T& f& u; Z) l; ^( d1 Q *r=i+1;' U: ^) i- y8 {
*c=j+1;, o: o- o) E# Z/ s
}4 a3 L1 v; t3 C% W$ ?; _ a+ N* a
return(high);
* l* `: V \. P. X}# f! r p/ B5 N/ a0 T
float s_diff()9 u, e2 r0 E$ h0 `8 D: Z- q0 B
{int i,j;# v" c) h& N8 k- q3 c1 F1 e
float sumx=0.0,sumxn=0.0;
0 t" d, h# S3 X" p9 [) X6 o for(i=0;i {sumx+=a_stu[i]*a_stu[i];
* x) O' O6 R/ q9 _; I9 W( K, F, i sumxn+=a_stu[i];
& s4 G! @8 u! T% u% g* g3 E }
! X! s# ~ i5 }, y return(sumx/N-(sumxn/N)*(sumxn/N));
5 }# g% Q+ d' G+ r, i9 s}) ]( t. z+ o* s/ K8 G; {% V
8.151 [8 j. t/ l c
#include
' ~ y' Z: w* L( @3 J#define N 10
- n. s3 W3 t) u5 h- F2 r: ^( Dvoid input_e(num,name)5 E) U( ]) k4 | M
int num[];
6 b1 E/ X. L d! Hchar name[N][8];
. J7 g3 _# A9 ?7 X( `, G{int i;2 {7 S. o) ?5 i2 K- v% w2 {
for(i=0;i {scanf("%d",&num[i]);
- T8 X; b( Q8 {/ V% T1 l2 l gets(name[i]);
& K1 ]/ A2 K5 x4 k/ L2 a$ `" g" P }0 x N; C+ a, H; R( f- A! n% q/ o
}
# q# m8 s# m, B% pvoid sort(num,name)
( z; s. S) a" [% Oint num[];
' M' p3 \. @5 d* A& fchar name[N][8];
+ d7 B0 s; j2 U# R, Y* j0 Z# ]+ @- l4 K{int i,j,min,temp1;
' r- ?9 A# f6 S$ k; i; Q* G char temp2[8];( }9 j6 w6 l8 G3 S# t
for(i=0;i {min=i; }# b" \3 E9 a; v
for(j=i;j if(num[min]>num[j])min=j;1 m9 v3 J: m4 i' t" b/ Y6 _1 c
temp1=num[i];& k5 c8 D# \% U4 m* R0 h
num[i]=num[min];
- g5 A; c- I( ^ c, T/ T* g num[min]=temp1; k% V) j4 e1 E9 D2 A
strcpy(temp2,name[i]);- l, p1 a/ @- S# B
strcpy(name[i],name[min]);5 S) `6 j6 N3 z& q; V# w
strcpy(name[min],temp2);
& K) Y; y' N4 z: \- o! R) s }
- c: P" U7 p: |& q0 z9 p6 f for(i=0;i printf("\n%5d%10s",num[i],name[i]);& w( \# l) x, K7 L6 v. H8 J; g
}/ S1 a5 j0 z1 P3 o6 b
void search(n,num,name)
4 z. u6 \5 g1 Oint n,num[];
" d1 j% x+ L) `, k: J' [( _char name[N][8];% E3 Y9 N: x8 I4 a9 `
{int top,bott,min,loca;
7 x+ q! {" h$ B5 Q4 Z loca=0;! ?2 `' S5 `/ ? X7 Z3 g+ S
top=0;/ h4 q9 w, ]' `$ H( O
bott=N-1;
. a+ v: W2 o1 x/ s( E, y if((nnum[N-1])), x# `6 Y5 L$ s, b B
loca=-1;7 ~ u2 V$ p& r, q5 u4 d. e
while((loca==0)&&(top<=bott))4 b/ V" ?1 _" e a& G
{min=(bott+top)/2;
/ ?* O# s0 m+ F! @- p7 Y if(n==num[min])( B0 l# o7 N4 @6 x& ]
{loca=min;$ @( c. l3 |& \0 ?$ h
printf("number=%d,name=%s\n",n,name[loca]);
- K8 p( T N8 N+ w# M" j }
; T9 C) J4 r3 ~8 V else if(n bott=min-1;
5 u0 C. `7 w- H" Z+ ? else" j* u9 i, N& I8 g7 y) b: f# S- v
top=min+1;+ c" L# R0 {' A3 p% c
}' J6 @( r$ k, d1 {2 p E0 h/ ]& w
if(loca==0||loca==-1)* {3 O$ I& W) {; X
printf("number=%d is not in table\n",n);: c: u+ `1 g5 Z$ i
}+ S- b" _6 j9 m; Y8 h9 D- V
main() l7 ?0 ]1 z! z! W* X) i% {
{int num[N],number,flag,c,n;8 E. g( n/ {% z+ N2 _1 h- b
char name[N][8];+ f' F8 W- p$ u, T, v% _
input_e(num,name);
% w! S! X3 d7 ]( V sort(num,name);
) X, y; t) w) p( b; \ for(flag=1;flag;)
8 f/ ^3 h6 i0 o2 l0 f6 X {scanf("%d",&number);4 p! n. E' |4 ?- M9 t: O
search(number,num,name);
& n: R, Y0 H; r printf("continue?Y/N!");
# W A# \; h) r. Q" P c=getchar();/ E% Z8 S* O" N) ]. E/ S( R
if(c=='N'||c=='n')) Z0 O/ i6 w3 U' W
flag=0;
: I" J& E% y: }! t* j5 W* e2 H }9 o5 G) n8 ^/ x8 F, n
}* `( [% O* x# c# @
" V( w) ^9 {# `3 u
8.16
& I1 Z8 B' ?, R) v& O1 n#include3 q% d' e! x1 D1 J9 a# g, _- Z
#define MAX 10003 T7 U$ [1 n; O; m! b2 P1 Z0 Y
main()
$ T) a: q* q- D1 z& v# r{ int c,i,flag,flag1;
3 j9 b% Q, U% Z- R) l char t[MAX];% n8 A: n g. R3 ?6 c" z8 G+ P
i=0;: L0 V: I5 j6 J+ w4 c/ ?+ s9 y' X
flag=0;
& q5 h1 Z% d9 [( l2 A; K) E flag1=1;
0 P" O: t, M3 I$ b& c printf("\n输入十六进制数:");; I E& M; l! w/ P+ m: H7 o7 E
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')% [: c3 I6 K! K
{flag=1;
& B' ]2 S2 {1 k2 n3 ^9 F$ [0 I t[i++]=c;( Y: E* l, @- ]2 w
}
4 M* F1 v5 `" e else if(flag)/ r7 K: V% l2 p" a
{+ W) h1 |3 w+ V; @6 Z6 o ?
t[i]='\0';+ [ N) K0 i P
printf("\n 十进制数%d\n",htoi(t));& i3 u* F S/ h# w
printf("继续吗?");' b% k& j4 A$ U8 y1 V
c=getchar();' T1 e" e: W( g0 C3 s
if(c=='N'||c=='n')1 Z2 [; {/ O9 j/ Y( U" p
flag1=0;# F6 H6 l8 v1 q
else8 x8 ^5 D7 ?" |7 y9 s6 x L
{flag=0;% M8 |3 k& q5 v! I2 G0 d
i=0;
- S+ ]8 G d' V2 g( p printf("\n 输入十六进制数:");; w. m% q9 x( D q6 A5 |! @
}3 B' P2 S1 P6 g: z
}
( K7 K3 M# T5 \# K% \* }5 @}
4 N& k, ~/ O% s) G: |# g6 m}
5 O9 G- P$ J- d2 c" ]htoi(s); J7 m a+ y1 V9 {
char s[];% O$ a0 ?, _" _* X
{ int i,n;; ]# c" z: U# Y1 z3 I6 q F! E' s
n=0;
6 z' A5 i$ A* N7 _ for(i=0;s[i]!='\0';i++)
0 v- {3 r, L. G$ h {if(s[i]>='0'&&s[i]<='9')) b+ L, `7 O7 N
n=n*16+s[i]-'0';. ~+ a% b! ?0 M2 l3 Z3 M
if(s[i]>='a'&&s[i]<='f')
' i. x% Y0 x! Y n=n*16+s[i]-'a'+10;
. R! c) `/ Y6 {( p if(s[i]>='A'&&s[i]<='F')# W. ^* E9 `! b4 s
n=n*16+s[i]-'A'+10;
$ }$ u* t# O0 F D z0 j5 A }7 a# G; B2 H4 D/ y3 w. v
return(n);
a/ b2 `! T6 }+ C$ D4 f: P8 \. C}
4 Z! H+ g7 S+ t) ?" E! g# t8 ~" v5 ^$ U$ C; h
" Z3 w# y/ \$ P0 H% E% Z K
. Y( d# \' R, x* X
8.17
9 | T1 ]" J+ A#include5 a b1 w9 ]( ^+ I1 I8 x" b
void counvert(n)- {* A) ^+ ~$ M2 O' s. `
int n; U; U8 _4 r0 }, V8 Y- m9 o& l
{ int i;
4 _7 X7 x, }* ` if((i=n/10)!=0)! F+ B/ P2 z9 U8 |
convert(i);/ X8 i& ?0 Q' |; S* m: a
putchar(n%10+'0');# v: s& U: g$ b3 R& c% F
}
; a: D% X9 \. m- U9 S0 Amain()' M0 n+ ^* R: m8 m
{ int number;
o8 U+ Z; b6 x# n# @ printf("\n 输入整数:");
) {% Q+ F& @- |8 u scanf("%d",&number);* ?5 Z- ?" P% k. l/ b7 y% w
printf("\n 输出是: ");
7 T: d+ j6 G" j* k if(number<0)
2 g" M8 |+ {3 M { putchar('-');
: F G5 \4 G0 F/ \ k( j number=-number;) @6 j8 n7 v1 G1 Y
}. m7 p ]" U9 N4 e6 U. M
convert(number);
: G+ G- q: x- v1 y- B! z}) ]- v; e. [! d& J. U( ~! [
) ^3 e: O0 i# Y: w( S- P$ F
! g& t( y0 R( Q4 g+ E7 a7 M) ?" _* `" S, `
8.187 |( d2 z- g# A2 g9 M
main()$ F3 H- P" X1 v2 y
{$ N! _5 e" }0 G' w
int year,month,day;
) s# W% ~' C' x8 J int days;, z4 p* ^& L! n& o; t. o1 i
printf("\n 请输入日期(年,月,日)\n");
) p, w( A' g: d* U2 S% `2 M scanf("%d,%d,%d",&year,&month,&day);% E# p8 H% l4 h
printf("\n %d年%d月%d日",year,month,day);% P. [1 M/ l/ i. B! o9 G
days=sum_day(month,day);7 U9 r4 b& l9 s* V% ~
if(leap(year)&&month>=3)- a0 _6 [' @0 j+ Z
days=days+1;$ p! h; u+ e$ b- @/ B
printf("是该年的%d天.\n",days);
; e; X9 H- d7 V$ v- r }
4 N% Q% Q$ Z( b4 [8 C static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
# A% i/ n6 r, t) G8 P5 S int(sum_day(month,day)- F R/ l0 s& v. b. I8 X* e
int month,day;
" O% h1 @7 l g4 H7 H2 O {
: J8 [. B4 d# `7 @5 S int i;* A0 K) w j. o+ q1 i- f: w' {
for(i=1;i day+=day_tab[i];
; F9 p- W, J0 y- Y% D" K6 j return(day);- s. K9 M6 c" g" e, }7 l
}
Z( m) X8 o8 R9 \; j' g int leap(year)4 [" W1 j6 }* V# x+ _
int year;
I3 U* Y; {* K! W1 r) v; \/ J {, [* i" D/ A N4 d
int leap;; z6 F4 W" T: g3 K
leap=year%4==0&&year%100!=0||year%400==0;
& N/ j3 t) s; P0 v# y return(leap);
I/ C. m8 c0 L3 O }0 X) C% a$ K9 R3 o* Q" Y* D3 e
第九章 编译预处理
/ T# B8 r; z. H% ~. I: G9.1
3 x: ?+ x$ G6 l& x; R#define SWAP(a,b) t=b;b=a;a=t' w- M; g; i K8 Z6 d
main()
$ j/ q3 O$ b+ @# X- }{7 n5 W! c% v) C6 D8 _& X5 F, J1 {
int a,b,t;
) f; `# x: @* g4 N% F Dprintf("请输入两个整数 a,b:");
t/ Y+ |) F, z/ H( |5 o4 Cscanf("%d,%d",&a,&b);
6 r& B( I1 J! X+ D) k% {! [SWAP(a,b);
; j# B0 M) g6 H* w" z+ A' cprintf("交换结果为:a=%d,b=%d\n",a,b);
8 s& Z/ E! |/ d0 X$ B, L) u, C7 ]} . \6 [9 ?" _* W! r
# i0 `$ i: j4 j, p+ b7 V
* C- J* z o! x4 q2 d+ i9.2* R/ ]0 a' G# H7 y0 S( X
#define SURPLUS(a,b) ((a)%(b))
N" y" x8 m" n/ _0 j# M6 tmain()" e" F' F) }. @0 s
{1 w2 I. q; e% _/ @- O. O
int a,b;
4 p7 S! a0 w# W, @; ^- u6 q$ r* e' ? printf(" 请输入两个整数 a,b:");
( d$ `0 _- {0 B5 ? scanf("%d,%d",&a,&b);) f4 _; `4 b6 S: W" p
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));, X/ e1 i9 K8 X0 N4 i& f( h' W
}
+ {$ S& l4 A! L
6 k) g* |9 u! k1 G! h" C5 J' e1 N4 C
9.3/ A6 z5 j+ v0 ]# e$ k
#include5 f& y7 m# f6 L5 h
#defin S(a,b,c) ((a+b+c)/2)
J( \7 I' Q: L#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-. `& c$ f' _" ~) `3 d7 y) ]5 ~
c)))' L5 q/ P: p; G3 I. j B
main() a* y( x* K, M! m( S3 i0 n
{; _: {( L+ v' s. E m d, _7 G' V& c
float a,b,c;
3 N1 \& V' I9 q4 ^: ` printf("请输入三角形的三条边:");) D- b$ S/ b" h
scanf("%f,%f,%f",&a,&b,&c);
0 M6 N7 c) O5 Z. U/ x7 Y( O7 i if(a+b>c && a+c>b && b+c>a)
- n& \; }+ X4 x4 V8 \3 [ printf("其面积为:%8.2f.\n",AREA(a,b,c));7 i9 T% G$ E: e2 `
else4 Y& k! X; D$ \ c* q1 u; \8 j# O
printf("不能构成三角形!");8 g( m' j5 @6 H9 x+ f
}
" z* \8 `7 Q3 n
: O2 G# H# e, U! U- l
! v$ L( M* k0 q9 g. j
5 F" E4 c6 H& S* N* O: G% t+ C+ [9.4& l: `/ n; k, v5 D5 G) Q
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0) z0 g( d& {3 _9 h+ t
main(). k& }* k* y* O/ f1 [; V
{/ Z+ K' { s' o6 `1 B
int year;
1 g" A" g) H! f* a4 z, g printf("\n请输入某一年:");
`4 |) ]. P& f+ p scanf("%d",&year);, R7 b+ D7 |% p* J; [* d2 A
if(LEAP_YEAR(year))4 z; ]- X8 e- \0 V
printf("%d 是闰年.\n",year);/ v z4 e9 j! O" ^! k, i3 a; X
else
; _# \6 ]5 {3 o7 U5 {' S printf("%d 不是闰年.\n",year);4 x3 v+ y7 g1 @( J2 u6 R( B& K
}
R0 B7 U+ I8 m. z( M" [& |) d, j i
! D/ j ~$ S6 K3 ~7 K: U0 |% M/ A
" e, |: Z( j7 n8 x- X! e2 f9.5解:展开后:
) t3 y! `5 @ A* K, V i' Y( Cprintf("&#118alue=%format\t",x);; S, y$ h+ c* d" c1 k( a
printf("&#118alue=%format\t",x);putchar('\n');
6 H; \* X1 }# sprintf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');' j* k! V! u/ [, H( B- r! g
输出结果:
& G& K% b0 a3 ?+ |/ v% O&#118alue=5.000000ormat &#118alue=5.000000ormat# t) w2 v+ Z! o; z ?
&#118alue=3.000000ormat &#118alue=8.000000ormat; ?. Q9 f+ _ h/ |
# [5 _ {7 L: G0 l
$ G7 c. V2 m1 k! R& c4 W0 K1 V9 h8 E; G
9.8
& A5 Q g2 q6 ]7 q7 ]3 hmain()* ?7 }2 y; H% D
{
6 Y! t& d2 B$ R1 p; I# [: h7 j. W int a,b,c;& P: t! Z* R7 V8 @9 G. `% X
printf("请输入三个整数:");
2 @, ~) x; I) u4 [ scanf("%d,%d,%d",&a,&b,&c);4 Q! O, Q+ m- l: L' }, \7 Z, i6 f3 {& d
printf("三个之中最大值为:%d\n",max(a,b,c));
$ [* X7 j1 R6 ]! P# I }
( j5 a& q5 Q8 ?0 A# E max(x,y,z)7 A/ e$ n. c; T0 G& ^- C
int x,y,z;+ ?/ O3 ?1 ^/ ~
{
2 A) x# R& P5 r: _; k. u int t;+ q7 k9 X* X& t4 R5 K/ j' z
t=(x>y? x:y);
) p# |$ V3 l& w1 x7 m8 c. r3 a return(t>z? t:z);
0 U; ?4 L: U$ m4 r }' U% A4 a+ ~" Z2 @0 H% W$ o ]" Y# X
: ~: K0 [# |$ X% Y
* s" F& N5 n, Z, o6 m( G' Y
- r4 j- W$ V6 `& W1 ] K" z2 p9.10, g' `9 d l) N) ~- J+ ?, E* \
#include3 O3 j" _7 h6 ?+ y. J3 s
#define MAX 80# {2 R) ^3 K( V: n5 t4 ~
#define CHANGE 13 T( `; m- p- o* F
main(); a+ F* d% q, E! A7 W
{ b2 l+ [# C2 F7 @" G
char str[MAX];1 c. @3 \+ X# o
int i;" [; w, I* B% W) n v2 r
printf("请输入文本行:\n");
9 F9 t% Z$ B( \9 o. H0 X scanf("%s",str);! U6 P; ?6 D$ s' V8 }* A& a
#if(CHANGE)
& p# Q$ l C; X" [1 c: k0 Q4 l/ V {
7 N1 {6 N0 a/ F/ r for (i=0;i {1 W3 C5 f+ [5 M* f% A
if(str[i]!='\0'$ z6 l2 o; P# |3 b2 K3 ^
if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')3 X* t& ^. z" `: @! x+ k8 C8 b
str[i]+=1;) E# F& P9 n9 l& X' O
else if(str[i]=='z' || str[i]=='Z')2 k+ c+ `# K+ _* \) D% h
str[i]-=25;4 }9 Q! B* i" V \- s+ Q
}
; ~: G9 r F+ w& T1 k7 `( X}
8 P6 z( w, `3 c+ [#endif
' U' n* Y7 g) M- P+ J) ^( iprintf("输出电码为:\n%s",str);% \/ C. T) K1 X5 `9 ~
}
9 i4 X. i A" B( m& ~第十章 指针
6 R4 M! s3 e5 g10.1
, b$ o4 ]* w3 `, n! x# @' omain(). ~9 l! P; ]* [- s( ?, x0 S
{int n1,n2,n3;
0 ?% E+ P t; Y; |( a e3 S int *p1,*p2,*p3;
: ?4 E+ H& K, B9 Y2 ?$ P& ~ scanf("%d,%d,%d",&n1,&n2,&n3);
, |1 k/ Q. c$ L& S% x) ? p1=&n1;1 r- t$ Q% q1 k& f- Z* B
p2=&n2;# r2 g6 S2 ?6 U. P2 g4 m' i( L
p3=&n3;
( }$ l+ P$ W1 J; t if(n1>n2)swap(p1,p2);
2 Y4 h7 x1 p* ? if(n1>n3)swap(p1,p3);
! ]' t* V. O1 X" c, m+ s- Z if(n2>n3)swap(p2,p3);2 |) z$ ?+ Q8 l5 h
printf("%d,%d,%d\n",n1,n2,n3);
x/ p* {( y) C3 D}
9 g/ z6 a; ` V( E& y! iswap(p1,p2); f6 R. ^& x, @
int *p1,*p2;" R2 C: M* U- d O4 u
{int p; f! F+ J! j' F. U9 J
p=*p1;*p1=*p2;*p2=p;4 Q" J# F2 ^3 ^3 d
}" |) u% |4 v) ~- I: [
10.2/ Y9 v m. y% _5 ^1 N1 [4 v
main(): u( `! m( o. E, j/ w
{char *str1[20],*str2[20],*str3[20];
: r- z' N% k* z x char swap();4 \( X ~9 A( h( y8 }
scanf("%s",str1);$ ~' [$ w& I g2 V
scanf("%s",str2);
* E/ |- S% H* E scanf("%s",str3);* \: M% n" l5 {
if(strcmp(str1,str2)>0)swap(str1,str2);
9 _- g0 `3 N/ M8 g { if(strcmp(str1,str3)>0)swap(str1,str3);: S6 ^& y( ?! p% L/ T8 R$ l
if(strcmp(str2,str3)>0)swap(str2,str3);4 n2 P/ g$ }3 n9 ~/ m, X
printf("%s\n%s\n%s\n",str1,str2,str3);
( }7 l" a% D$ l M: r. R5 H}& C0 R0 m5 z2 m, r
char swap(p1,p2)9 ]" ^# R6 f% c
char *p1,*p2;0 o5 V ^8 P; @' T& k
{char *p[20];
$ Q. w' k+ }2 o; N strcpy(p,p1);
. Z2 w$ D. s" e9 d strcpy(p1,p2);
9 J4 d# l* u5 q( O- z. ~2 G strcpy(p2,p);
) ~6 K. u& g# G4 }7 o, }; J}6 {8 B/ a; P Y1 Y6 v/ G3 y
10.3: P. l( x6 H; k, P1 @5 _9 R" ~
main()9 Y& {' l4 ]* _ N( N) D2 b, [
{int number[10];
; ]8 L( F+ \& m s input(number);6 S9 R$ R. D4 u- Q# E- j8 ^" f, _
max_min_&#118alue(number);/ G' I8 Y* Y4 G5 q* R# O) T5 U% G
output(number);& j8 K4 C9 P( m9 ^6 U
}. N- t/ H$ t1 D, M& H0 F
input(number)3 V4 @, G$ [! o# {# V/ W/ }
int number[10];
2 w' R2 U/ w7 D) X- X{int i;
' ]5 m3 [' E+ [ for(i=0;i<10;i++)1 N, x h% |4 ]$ u6 A
scanf("%d",&number[i]); V$ N1 V) O7 ~; H3 S( E. Y
}# M. R0 j* K' Z& F A
max_min_&#118alue(number): Q% H, D: ~ [, b5 [* B- N; ?
int number[10];
2 x9 Q8 M% q4 P) h7 ]7 m; a) f{int *max,*min;
8 E) @ @+ ?8 }& _5 y2 i8 z" h int *p,*end;
) K- _3 |& u4 q" C& ~ end=number+10;/ \" P" f! d& D( K
max=min=number;* K& m$ {; C9 O; l8 N5 G
for(p=number+1;p if(*p>*max)max=p;
) A" |( F" m: d2 W9 r' v$ X! K- j else if(*p<*min)min=p;3 r: R- O B) y9 R- l, a+ e
*p=number[0];
2 D4 ]8 j# W9 i. [ number[0]=*min;
5 q& n% d7 \, T' t *min=*p;
9 F+ v4 e b8 z k5 p) m# R *p=number[9];
0 J2 W, ]) a; @* s7 j number[9]=*max;
1 O- r# e! H, b p4 C *max=*p;" D5 D* o* G" ^/ S' F& A
return;
$ |0 z: }1 b, N- s. W, \}& y6 W7 S: ?5 Q$ \& [7 I0 c5 f. i
output(number)
$ g/ s3 |' x) { I1 f- r+ L* X+ @int number[10];0 l; R- J/ |0 y
{int *p;
4 S% g O) a! q5 c5 ~( Y for(p=number;p printf("%d,",*p);
+ Q" d# B) g8 y5 }6 w printf("%d\n",*p);
5 l# A/ U" q# J- E}
5 V2 G% T/ y& n1 p, a10.4( k, x( V+ U1 P$ n* c/ \' l- Y
main(). ?4 q) d- S6 E, D. u. t6 O
{int number[20],n,m,i;
# \+ E1 ^- f9 H- k) l" I( Q O scanf("%d",&n);# W7 f+ [( n: [8 _; j
scanf("%d",&m);
- n' |+ g- u' N" p! v0 ^3 {/ U4 e for(i=0;i scanf("%d",&number[i]);* V; k8 d0 E+ q* s
move(number,n,m);
% G2 D- L7 C( N7 N9 X0 u for(i=0;i printf("%8d",number[i]);
0 r& @( F3 R% h: S& _% G" X6 M5 K}
- R0 e3 n% _: |# @move(array,n,m)
$ F/ A t ]% k( w+ O7 o8 W) \* Cint array[20],n,m;
1 ^9 M) J' n( R4 ?$ u{int *p,end;$ s6 i" M9 [+ Q) V+ O. U
end=*(array+n-1);
_8 a, ^! a7 W4 ?' ] for(p=array+n-1;p>array;p--)
0 b0 q% u! e( ~" d; C *p=*(p-1);6 B; d! G5 i9 w+ U7 q8 l6 \
*array=end;/ G/ [" T' \- Y5 c' c4 X. d& [
m--;4 @5 Y( Q& P9 M. E3 e) f
if(m>0)move(array,n,m);8 P/ v) s8 W" h: u4 k' m& O
}: k& \$ f, B) U/ [/ i- `
10.5" H3 N( X1 K% @+ ^, S, G
#define nmax 50/ A- }, C" J) l( K+ c* u0 I
main()
9 ?: d) e7 f3 R Y( O8 Q3 x2 s{int i,k,m,n,num[nmax],*p;
$ I$ q9 s9 H- @" {% f4 L scanf("%d",&n);" v- h `/ ], B" s' n" I
p=num;
7 K, m6 i- c2 U for(i=0;i *(p+i)=i+1;( q8 ?" I" e/ o/ V
i=k=m=0;% b8 h Z. J- l
while(m {if(*(p+i)!=0)k++;$ C! @' y/ j1 c) R4 E' z+ e
if(k==3)
6 T6 t* t6 a1 J3 h' ~4 I& ^. l {*(p+i)=0;
# z J1 S& M1 Q( e* Q k=0;
9 P! L" I7 N+ Q' R2 f7 N3 } m++;
, T( a2 J! o3 y6 _4 y }
- r: a/ [: ~# G4 t3 h2 a i++;! r/ N% C) y5 g( A3 Z B% J
if(i==n)i=0;
% ]$ |& m( X+ `. Y, P3 T! X# p }
7 g* g0 _; Q+ \- M! } while(*p==0)p++;0 Z% M# v- u) F5 J7 b$ |7 E
printf("%d",*p);
B. B, f- S; e" y}
. N" u4 {: J: i/ h- U" R10.6
' H: ] w2 v4 D. Q; umain()9 \% J1 v: ~2 c8 \
{int len;
; G% \* F+ H! ~8 E5 C. K4 v9 K char *str[20];0 u7 k& @, J7 l z$ f
scanf("%s",str);# V2 n" U- U4 {3 e
len=length(str);
$ k! f. X) W# e+ b3 s7 a- Y0 z9 l$ { printf("\nlen=%d\n",len);
5 a! \; b6 q* p1 i}/ S6 O3 E* i" d2 I2 C9 K0 d
length(p)
}: V9 K ~+ a3 n. H8 g; ? H( schar *p;
; V, E0 w$ H. h' w' g; J8 K0 B{int n=0;
) J- `3 g9 T6 h" Y* z. x1 t while(*p!='\0')' o4 A" s- y3 m
{n++;p++;}/ e/ b$ R9 V8 p7 H
return(n);" {5 n0 t: {4 D' ], L2 k
}
7 r) d, s: |. J* B! D/ Q9 J10.7- L2 l5 |% e, R
main()# ~* a" k' P" t( u% \
{int m;& \) S: W, j) H( N- d; u
char *str1[20],*str2[20];: s- O, r! {* L8 l" g$ y" [
scanf("%s",str1);
/ k; i: D, T2 c7 z# @$ j scanf("%d",&m);# i4 v9 s9 a# i4 z2 i) K
if(strlen(str1) printf("error");9 s* B* f% }9 t% k
else' ]9 ?% I1 d5 A+ L6 v! R
{copystr(str1,str2,m);
: l9 `5 P! p) D- }+ W! ^3 T4 m$ ~" z printf("%s",str2);
2 V! I0 i' x# Q M3 i- @( H; s }
4 r+ o. d; i. v Z5 [ {% f7 C, Z}
% l) }, h- {; c1 b) ~% Xcopystr(p1,p2,m)8 }8 a8 F. j$ M/ p8 c; M
char *p1,*p2;8 T; c" p% A6 f
int m;4 G$ y2 C2 c, c1 Z+ @" j6 n
{int n=0;
8 o5 v- l) P! Z9 W7 [, M: @ while(n {n++;p1++;}
3 j6 }# c0 F7 d4 P! S" e while(*p1!='\0')
7 U$ T( R, s5 P# C. `+ x. S0 W {*p2=*p1;, I; y2 _9 M! [3 v
p1++;
3 e- t- C0 T( v5 w! g p2++;
) C9 ~- [( P, F. b7 x# ?# D }- r% W0 m0 X, |, ~' j; s/ ~
*p2='\0';- F: z: z# D% s$ B# n" ~
}
: e' X s2 A, Q. ?9 X, @10.80 g3 O* E, a+ t* p- K
#include"stdio.h"
, K1 a4 w2 K6 ?& t' ]; hmain()
) w/ u: e' T" S( S{int cle=0,sle=0,di=0,wsp=0,ot=0,i;4 @* H; I i* N) ?. \$ O9 r
char *p,s[20];8 t( C* D! q/ P9 p9 J
for(i=0;i<20;i++)s[i]=0;7 g" M" W5 @$ Y
i=0;4 s! Z7 w6 L$ R7 t+ t4 |" u
while((s[i]=getchar())!='\n')i++;
6 W$ m6 y% }% p% _7 P; e5 c/ ^3 v p=s;* K, W( ` W$ B) e$ s6 m
while(*p!='\n')# ~' v+ O7 }" F
{if(*p>='a'&&*p<='z')- X6 `, S/ o0 J% v6 ]6 Y8 Q
++sle;: _& W, \5 V$ x0 Y6 B' G; e' F
else if(*p>='A'&&*p<='Z')+ L* ?% J' t" m1 l0 ]2 b! l/ f
++cle;" R2 [; O8 H& l% b- U2 m3 v
else if(*p==' ')
! F% K6 [7 A5 n* r; y ++wsp;
7 S* H( l& J4 `0 _5 X* z2 q else if(*p>='0'&&*p<='9')
: a! ^; j! s$ g" P/ g. D9 @- h3 t2 g ++di;+ S: f2 C$ S, T8 Z' o' g; g* Q
else/ d$ p; G9 }. R" v
++ot;
4 n! N0 K" m- L: b# f3 \, k p++;
% o; v2 a: h# s/ H z9 S1 H }; O) }# W1 v1 G2 l$ D1 c( _$ |
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
6 \* U5 t( \! v6 Z+ ]}
: X* E% J9 {" u* z5 y10.9
0 B4 u) Z! ?0 ~- ]6 n- W! wmain()( V1 w9 e+ z8 K# q+ C$ l" u( l
{int a[3][3],*p,i;/ r& I8 j% N4 y$ h. u
for(i=0;i<3;i++)
) @" `! Y0 I0 E& H) ]& { q scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);' c( B) |& L2 K$ N2 _4 i+ m
p=a;) R/ F9 }) p e5 ?! k4 y
move(p);$ k$ X: \5 u' s! V) {
for(i=0;i<3;i++)
( y) ?9 Q& K: z1 S printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);
" q- d- i/ u7 R/ ^: i+ |}
0 s2 R4 q9 A* i3 E8 U L: Hmove(pointer)# a" n0 }$ I$ [/ S6 J
int *pointer;
% s/ c9 V* D! [: f* ~) z7 T{int i,j,t;! ]' I" Z- ^! H' o6 }" d
for(i=0;i<2;i++)$ x7 m- I$ ]; Z6 I7 u
for(j=i+1;j<3;j++)
( a# a% D+ R; b {t=*(pointer+3*i+j);4 c b3 w! T$ ]; S5 o }* x
*(pointer+3*i+j)=*(pointer+3*j+i);, c9 k% ^0 ~" @0 |( ~
*(pointer+3*j+i)=t;' C4 L4 |% V4 q( s7 C: v) L
}
9 L) W' E r$ H7 f: P}; u) u# d9 i9 Z/ v* M+ L. P q& A
10.10
- `, X+ v, o! R3 nmain()
* N6 I7 J) Q0 j( K{int a[5][5],*p,i,j;
0 e+ \ T" p- U6 `: B# I for(i=0;i<5;i++)
3 r, \0 Y+ E! x5 c/ L for(j=0;j<5;j++)
4 T: B3 r6 _& V! |$ G3 e scanf("%d",&a[i][j]);, z. ? H$ o( D! Z2 _! d3 m
p=a;5 J7 z3 h/ I% k+ E+ i& v1 Q! `9 }
change(p);1 Y* I5 X w( L5 Z1 \
for(i=0;i<5;i++)
) f* T) S% q- @! X {printf("\n");
% w1 ?. g$ Y" s0 b& V/ y; S for(j=0;j<5;j++)4 g0 i% S, O0 ]+ g# q0 Q
printf("%8d",a[i][j]);
; k1 Q0 U* u0 A" ?/ J# F }
) X6 O" k( Y# Y. G1 Z}% s4 e5 ]# q3 _' J' w
change(p), \' @$ `+ ^: F6 Z! E- L. b. G
int *p;
. y/ _4 L) e, C' ~& L# L+ n* @2 [9 \0 @{int i,j,change;4 o2 E& q) B: E8 }
int *pmax,*pmin;# x" g* c8 T6 J0 M
pmax=p;/ P( Q5 ^# ^6 C( a! H5 |3 M
pmin=p;( I2 m! u5 F) o0 }; h1 d0 C
for(i=0;i<5;i++)
# h n" ?& J# Y, q% k) R for(j=0;j<5;j++)
$ P H+ ~( r' K8 C* K {if(*pmax<*(p+5*i+j))pmax=p+5*i+j;
- _ [0 Q: m' p6 A5 ~ if(*pmin>*(p+5*i+j))pmin=p+5*i+j;8 x( R) | o3 `, ~
}2 D; V o- [% _
change=*(p+12);
+ \) C& x2 x4 b *(p+12)=*pmax;
; f4 M! K7 i# H" g/ a *pmax=change;
- ?( U6 j( u; r change=*p;/ T5 S0 d/ J- w( t9 q5 F
*p=*pmin;
- m& S; A7 T4 S! L; m *pmin=change;
$ u- Y Q9 K8 \8 |1 i) ] pmin=p+1;
. l' {% N+ d# }9 U for(i=0;i<5;i++)
/ q- K8 x# H# c6 L) i8 d+ d8 T for(j=0;j<5;j++)
$ x. D0 O5 H2 \2 D. c: H2 \ if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
% C/ D) }: X( K: Z- G/ {( K. q change=*(p+4);
% s) M7 ~7 b: _" s *(p+4)=*pmin;
+ n8 Y' K3 S# Y *pmin=change;5 p& ~/ l6 ^" y0 p; O' N
pmin=p+1;0 v* u; k% {1 T6 h# U
for(i=0;i<5;i++)' t" W6 k1 T' {+ J1 n
for(j=0;j<5;j++), p* \! b" L2 d4 B4 K
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))2 M: X- n. |# o4 N* f, n- `# A) R
pmin=p+5*i+j;- [1 @5 h6 I. V S& [( l' P
change=*(p+20);0 E, @. [6 ?0 Z/ _
*(p+20)=*pmin;
, M1 r s6 w1 Q *pmin=change;
! V; b2 m; f/ \. B3 k pmin=p+1;
8 S/ m' d( i' Y( K9 }: f5 s* U* F for(i=0;i<5;i++)
* h- R; N7 b1 C" d+ _& F6 D- N for(j=0;j<5;j++)
' w& B: S: |# c7 _0 v3 D if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
) h% F y% Y4 i &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;6 L! f/ Y) }1 Z; n
change=*(p+24);6 y6 g" H0 X, J5 h' J4 e# J$ `
*(p+24)=*pmin;
* q( K) ^6 k9 ~3 ~. M6 _ h+ R8 e *pmin=change;
# u/ u5 b% J$ N# H& o}
M" M; ]" _; w4 [10.110 ]. o6 r# Y- Q n
main()
" t" g5 ^# L8 j% c{int i;
6 }4 A/ X6 Z' f char *p,str[10][10];/ q; E) {" `$ |% ^
for(i=0;i<10;i++); {* ^# }: Y2 x2 V7 f
scanf("%s",str[i]);
; J$ D* _! p) o' a; G" }% [( P0 X0 Q ~ p=str;" h2 T) t n5 g$ o; w! G5 _; y+ D
sort(p);; g! z* d/ K6 G: [( r# }$ A
for(i=0;i<10;i++)' F& O8 S% s' E: _9 O
printf("%s\n",str[i]);2 r) h) x( B) A& @- X) x R2 Y
}
: d& Q7 ]! d2 D0 y: Esort(p)
* d( r# A; V, g) ~8 Y# Dchar *p;/ `/ r7 i6 f# F8 N2 m
{int i,j;
0 P- \2 l% P( q char s[10],*smax,*smin;
; O n9 V7 K4 ^! C) ^! S for(i=0;i<10;i++) l5 v% x4 |- f6 T
{smax=p+10*i;, {% T' F9 h6 O. r- _
for(j=i+1;j<10;j++)
. y) H( M5 x! c6 G9 _ {smin=p+10*j;
" m6 m* [6 O- o' o6 I; @ [1 V if(strcmp(smax,smin)>0)
- _$ A' e& a# ^! J) ~ {strcpy(s,smin); f. B# l7 h. b
strcpy(smin,smax);
/ c; d% K2 Q" ^8 h* a strcpy(smax,s);
0 N3 j) G+ q! B1 |: U" H }
! Q2 C+ \2 m3 g( |4 d% L }1 Y/ i A* @. r7 ^4 b
}/ h+ d% i' Q4 n) E
}
+ `9 |/ ^& n. @! _3 @: N2 G10.125 `# m1 [4 z! {+ x- C0 K
#define MAX 20
3 R8 @% h+ {' _5 l* bmain()8 U( H7 R8 |0 M0 d
{int i;
& r9 c% |1 r, f& y char *pstr[10],str[10][MAX];& x' Y/ Y/ p& g) h, n5 y- N) ^
for(i=0;i<10;i++)
. n* H4 x) ^3 ^+ Q! F* l pstr[i]=str[i];
+ a7 V) j! O5 X" f2 ?! g# _ for(i=0;i<10;i++)! h3 n2 M0 O5 |" X
scanf("%s",pstr[i]);5 `7 H) V* u, Z% F, b: X0 F
sort(pstr); W. }$ G& E' y# d, I
for(i=0;i<10;i++)
* f' f$ m- O4 a! C" D2 n5 ` printf("%s\n",pstr[i]);
6 F/ |7 p2 H1 G: g2 W} n' q K8 S2 W# ^
sort(pstr)1 ]3 c! A* Y5 ~$ `" l. R A, c$ A
char *pstr[10];& t3 h0 h, Z" d" }" a0 j0 v% u
{int i,j;5 D# ~9 i; m; h: z& M" L
char *p;
( A2 x: g$ s. l. e- t& y, i4 O for(i=0;i<10;i++)
! j }) V) O9 { {for(j=i+1;j<10;j++), |" v1 {7 A/ y! b1 C$ {& S* I/ J
{if(strcmp(*(pstr+i),*(pstr+j))>0)0 m. N6 G F& S% f2 t
{p=*(pstr+i);. @; n) a/ |4 ^+ I( S E
*(pstr+i)=*(pstr+j);, D$ n+ }* i1 Q4 W8 z
*(pstr+j)=p;
; l0 e x3 _3 N! X1 Q1 n }
- k1 u4 S# ?! A, ^ }' y3 y, O* k/ p
}
3 o' l& y- k( ^4 @5 w}
9 u% @* ^( [# \6 Y% _10.13
8 s1 r3 i3 G; {#include"math.h"9 U& o3 T1 e" B; l7 j. v. D
main()
* W9 H8 g9 j9 q# Z( d0 y{int n=20;2 L9 _5 H3 ^4 z+ o5 J- {& K+ \
float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
7 d0 Q+ o" [% U$ x4 a scanf("%f,%f",&a,&b);& |" j) M9 A0 ]) e
scanf("%f,%f",&a1,&b1);
[0 s/ d' d; e- t+ M& F; B scanf("%f,%f",&a2,&b2);9 t# u- r( _( P/ q+ [/ I: v
p=sin;
\3 T; r# w( J2 v c=jiff(a,b,n,p);' X6 p9 \! y; P- b% B. @7 X6 M3 [
printf("sin=%f\n",c);
4 ^: s7 ]' i# g# w p=cos;
$ w$ m% A7 q5 e' s c=jiff(a1,b1,n,p);; M6 U6 a B/ B; `& w0 y' z# X
printf("cos=%f\n",c);
0 k# y I2 u) Z0 ~ p=exp;/ x" R0 T" J ? W1 o( M8 o
c=jiff(a2,b2,n,p);
2 I! U8 I$ j5 }, u- ~# j& W printf("exp=%f\n",c);
b/ j1 R$ s# f}
+ b& u/ U, x Dfloat jiff(a,b,n,p)
+ x9 Y6 m+ L( A; W) T6 B: y$ Gfloat a,b,(*p)();
- ^. C; `, F% e6 pint n;
) S% C, X8 i! j c0 ~{int i;
: V2 G. ^( A u2 ]+ E; M float x,f,h,area;
& S* F1 K% h+ ^3 v h=(b-a)/n;
0 U, ^* T7 G6 J! c x=a;& ]# s4 b; g' r. h$ e# l7 G1 B
area=0;
/ f1 a1 Q2 W5 q' v* q for(i=1;i<=n;i++)
4 ]$ \: v4 u& l+ ]; k {x=x+h;
; ~7 c( c: ^& L% z, A+ F& P area=area+(*p)(x)*h;
2 U$ o L" u% t% l" ?1 a! u }
C3 z3 F6 f$ p' f return(area);: y* t% T. N S0 V# }5 p. o
}
, @' h5 n1 g3 e% s3 ~7 z10.14
2 \" [3 t7 Z. d7 N) S, cmain()0 l% c: m F3 \5 ~/ ~; I! [# B8 @
{int i,n,num[20];+ s" t; M2 v' Q7 \! z
char *p;/ c. b& e7 Y8 t7 S# A9 U! d- J+ }
scanf("%d",&n);
+ V. q6 A& E4 m# D for(i=0;i scanf("%d",&num[i]);
_. [9 u: L1 U) W p=num;
3 |4 V M: H* m( ^9 [; y( ?8 ~ sort(p,n);
9 p8 Y% N3 s, N1 f' n' F2 q! k for(i=0;i printf("%8d",num[i]);7 S" {7 l$ t' W$ P+ x* R: a# X
}
, `+ _, R; A- n3 J* J/ msort(p,m)
4 [. I9 R9 t" K2 Y% Z3 Rchar *p;
; x& R- V% Y, @8 S* d$ Sint m;
' j- H5 b& |6 w. S" |+ {{int i;' ^1 h# r ]" k; x; M% o
char change,*p1,*p2;8 Y1 v! m2 e) L; H* Z% i5 R
for(i=0;i {p1=p+i;
# Z8 N' g. P5 `9 K p2=p+(m-1-i);
# S9 o M$ Y! P9 `+ s( c change=*p1;
2 U+ F" b, z* I3 U1 b *p1=*p2;
8 n$ C2 V. I- j3 X *p2=change;7 A- {! k E" L- ^
}
) |, |/ k8 p$ ~- a' \}( ]3 m& t! w @4 w
10.15
% p4 P8 _. I0 c7 N% amain(), V% ]) u" a) g! r* x3 u& M! d
{int i,j,*pnum,num[4];/ A8 L6 j5 @! ]9 }4 T( q
float score[4][5],aver[4],*psco,*pave;
?6 z2 A/ b! t9 ~% k char course[5][10],*pcou;/ ]0 Z+ s& i/ t J3 U
pcou=course[0];0 W7 J3 _/ c ]+ @
for(i=0;i<5;i++)$ [6 {, i9 J1 G
scanf("%s",pcou+10*i);" n( \, `& R+ Q6 w" \( ~2 ]3 T: {
printf("number");
) m& {& e' y! n; ?& X for(i=0;i<5;i++)
9 `/ ]/ |0 }7 X) R" R printf(",%s",pcou+10*i);/ X# e; Y! {* Q- w+ O% a
printf("\n");
J4 Q) S: h- t2 E3 _5 y psco=score;2 x, ^, Z6 f" ?/ J
pnum=num;2 C# V$ q; {) p& ?* x6 C
for(i=0;i<4;i++)- G( z7 M' e y4 C
{scanf("%d",pnum+i);; v* J: K* J/ j! c, X
for(j=0;j<5;j++)8 a2 Q# [3 g ?* N) h3 h# x
scanf(",%f",psco+5*i+j);! [0 k' K1 |7 @! a/ l2 U
}
) l/ s( a7 G s% v! H( b& j pave=aver;( z: [9 R7 D8 y
printf("\n");' C% P( `2 J; l6 S t& o
avsco(psco,pave);/ O: O1 L1 Y8 y5 H- t% U; k+ A$ a8 v
avcour1(pcou,psco);
- S" r3 V k5 ?. h printf("\n");7 e4 }% h; W" u6 _
fali2(pcou,pnum,psco,pave);6 a4 B, b/ G" T: E* y; d0 H/ y6 Y
printf("\n");
2 ^. L6 X, M, u good(pcou,pnum,psco,pave);2 c( N5 _5 f+ H
}* D3 X* E' {) k# t# n& ]9 e, A3 L% Q
avsco(psco,pave)
7 S- o+ e6 D, z! g; P: ~float *psco,*pave;
% M2 N4 W. e" Y) D8 T& M{int i,j;' N3 F. |4 [$ Q) [ z7 A
float sum,average;2 B9 o: q4 ~+ @0 d
for(i=0;i<4;i++)
j* @4 C( [9 V0 L+ H" g4 A {sum=0;
5 o1 Z6 ^- s7 P0 p" }( ] for(j=0;j<5;j+)0 w1 c1 j' {$ g0 ~3 v
sum+=(*(psco+5*i+j));
: r: |- m# a' T' f average=sum/5;
, a% \- E* n' w3 j9 @' P *(pave+i)=average;7 z( ^9 }& s% ?) w K
}- T9 F/ ]0 E' {
}* r; t% \7 z, h+ F* N$ F* r
avcour1(pcou,psco)* g0 d- V) i* V; ?1 s
char *pcou;
3 V5 I: `7 u* n) p, [: Nfloat *psco;+ K s" ?/ _) e' w
{int i;
* }& _- R% O5 d2 W% J! w float sum,average1;
! }! ^, u9 k3 `: d2 u sum=0;
. U z. M: \# L4 \) s* B for(i=0;i<4;i++)
) D- I: g$ C- u: f0 i ^( t3 E) x9 d sum+=(*(psco+5*i))# ~9 u) F( W. b9 E: T
average1=sum/4;
/ y& `' y2 m# U. m1 \1 Y0 o printf("%s %5.2f\n",pcou,average1);
( L* L1 Z1 ^+ m/ |* d}- E* O- C' E3 _; m! m; ?9 P5 D+ t! w
fali2(pcou,pnum,psco,pave)
# g# r- B8 c9 Q! M- G: qchar *pcou;- P" L: X. P: E. Z9 F2 p: L
int *pnum;4 x8 E9 k1 U# ?5 w1 e5 \* s
float *psco,*pave;1 H) R) q; @: g3 u3 s, y) Z
{int i,j,k,label;
3 h- U; r N9 } printf("\nnumber\n");
4 e0 M6 b+ ?" ^. I for(i=0;i<5;i++)$ K8 x; L5 g; o3 u/ S L
printf("%-8s",pcou+10*i);- Z$ U# l) [7 Y
printf("\naverage\n");2 @5 d0 u; H9 t. O) Z7 X8 @4 x: M
for(i=0;i<4;i++): f/ m/ J$ F& a% h
{label=0;
, o; k- v3 P8 T+ g: ~, M% b h% d for(j=0;j<5;j++)
" k( Y% w; s+ P; D4 R4 w if(*(psco+5*i+j)<60.0)label++;: L1 Q3 L5 V; _3 m, m- N5 {
if(label>=2)) U( b' M, {5 Q. [8 B
{printf("%-8d",*(pnum+i));% D6 o" D' z" q7 _/ r7 X
for(k=0;k<5;k++). Y! x5 \8 c+ e- l/ C. S. X
printf("%-8.2f",*(psco+5*i+k));' u4 g$ b- j$ r/ l) P# \) T
printf("%-8.2f",*(pave+i));6 E1 ?# r/ }6 P( @
}
; J. P: O3 T4 ^8 d4 x+ y( d: F }% J' q% A! ?2 ^ K( u" W5 o
}8 \7 E/ U0 @3 p @4 B
good(pcou,pnum,psco,pave)
, j+ }2 J' n0 U# ?' schar *pcou;
: S ]6 m3 h" ?) W- ]; xint *pnum;# A& w/ x# I7 N& E& H7 y6 U
float *psco,*pave;
, p' }4 d7 T+ ^, u/ x{int i,j,k,label;6 a- b4 h0 ^; h( j$ |$ Y
printf("number");# ~1 W O, ~% ~7 l! [6 p
for(i=0;i<5;i++), H6 x4 H8 k9 q& Z
printf("%-8s",pcou+10*i);
# C6 q2 L% _2 a# N$ d/ @8 [7 b printf("average");
" V, ~5 @2 ? F$ G; k for(i=0;i<4;i++)2 x; ~ z7 t! u
{label=0;
7 m9 I1 \+ z. W1 }- Q for(j=0;j<5;j++)
% x; b9 H. |" C) N F! r if(*(psco+5*i+j)>=85.0)label++;! R* ?* a2 B1 w- k4 F
if((label>=5)||(*(pave+i)>=90)). p1 X& D- s& J& L8 X# x3 ]2 h
{printf("%-8d",*(pnum+i));
/ |/ a$ R/ m0 ~- a- P( A for(k=0;k<5;k++)
1 X( \1 i* m- T$ u& X printf("%-8.2f",*(psco+5*i+k));1 a, S7 v' G) G) D
printf("%-8.2f",*(pave+i));
9 h$ ~) f; [. [# h2 H) c: [# [3 ^ }
+ A% Q! A" O/ V( v ` }
1 @' J# K" T+ y}
" X( c: K/ s- P0 @10.16( @2 o. f8 t( L6 ?
#include"stdio.h"& m6 u5 _; b, i( U" Z! [1 B
main()
y" ~" _) q O3 L8 X* w0 p{char str[50],*pstr;( `! m6 M% N) ^
int i,j,k,m,e10,digit,ndigit,a[10],*pa; j6 c5 S) A+ {. w( w
gets(str);, k. ^# P/ i1 Q; u1 J
pstr=str;7 ~1 g( a( t& b" q1 K/ ~' o4 ^# i
pa=a;" a- ^ v: G) U( R( f; a5 b, O- l7 \+ F& y
ndigit=0;/ ?% o4 @0 H/ i
i=j=0;
4 ]7 x$ k5 o/ ]& ^) m# }* j while(*(pstr+i)!='\0')
: t6 D" l" _& M {if((*(pstr+i)>='0')&&(*(pstr+i)<='9')): c4 Z# W( ?' m, P% k o
j++;
- d: K) _" Z" E) L: H0 Z/ S else# |- [& ?, h P1 ~
{if(j>0)6 }8 h+ z) N5 T# J6 [; Z
{digit=*(pstr+i-1)-48;( d( h# p; T, H5 D
k=1;( i9 _! N% Q' ]" j! l* s
while(k {e10=1;* y! V% a, a3 M3 }' s5 P
for(m=1;m<=k;m++)3 ?8 D7 e& h5 {1 M1 c; j
e10=e10*10;
" E* V+ S" a2 U/ s! ] digit+=(*(pstr+i-1-k)-48)*e10;; t! O, \8 @1 @- R8 Q4 B
k++;5 q+ b* Q9 M/ o" G
}5 V8 P& A3 ?& V- `% ]( \
*pa=digit;, j+ @6 E, R0 d- u; ]7 y& w3 Y
ndigit++;- j! C' k+ O( ~3 L9 r$ l
pa++;7 i F' |. H+ P1 I5 v! x4 b9 M0 b
j=0;
! c; x- g! i3 F& ~! b, E }
+ t1 {3 p& c8 ]. P }
+ D0 {2 X7 a/ z* h3 P2 L i++;# {# l) [: N5 E6 r# \0 J+ H9 w" ~; c, @0 i
}
; _+ C8 Y, d. H& s; Z( F8 j if(j>0)
Z, i# m' s2 v' f {digit=*(pstr+i-1)-48;# ]5 |; X" Z* _3 R8 ^
k=1;4 c* W# H6 r: G7 s- ^# u
while(k {e10=1;
( M8 u( [, j9 w for(m=1;m<=k;m++)0 V2 @' S7 N& l- K
e10=e10*10;7 l6 B( {4 M% L/ H3 r
digit+=(*(pstr+i-1-k)-48)*e10;
0 Y& x G8 _- W( {# z k++;( H1 z* d: w2 y* o
}
+ `( ?2 v8 i. [! l *pa=digit;7 U. I- C! E! Z6 t
ndigit++;
, t# e! [' v6 {3 q! m+ ^ j=0;
, l! O5 F% Q* ^7 e R* c }
3 |/ W d- O. ~ p9 j$ \1 S2 c8 J printf("ndigit=%d\n",ndigit);
- { J5 L; o8 W* ^ j=0;
3 D3 w Q) V( E0 S1 T1 a+ ?/ m% q" o pa=a;
* ~; a* H0 \ W5 F* B for(j=0;j printf("%d",*(pa+j));2 w; ^; M4 `9 a
}
; k4 [" v: U1 H10.17
N* D; `8 a3 E9 _% H" omain()
' E- h7 x1 L: }1 A9 a2 H) T1 v8 @{int m;
: @, k' c" w6 T2 `9 t: [5 R char str1[20],str2[20],*p1,*p2;
% T3 `+ O7 p6 m) [0 r/ L scanf("%s",str1);( Z" k3 D% s4 P, n
scanf("%s",str2);2 x, @% Q6 F$ R* {- q+ ~
p1=str1;
+ ?; i( V+ b+ U* q7 m& y: t# @ p2=str2;
& ^ f& O: o5 x- S m=strcmp(p1,p2);; U& K# v7 b( c
printf("%d\n",m);
1 M. D" W _) I}
' U `9 D( H5 l0 \strcmp(p1,p2)/ D. w1 J% b' T- [. w4 m( z
char *p1,*p2;# w/ j6 W Y/ ]; h1 Y6 r+ I
{int i=0;
2 a6 X, \/ ^$ O( j$ a. p) m while(*(p1+i)==*(p2+i))3 L) g' p2 |3 x+ ?
if(*(p+i++)=='\0')return(0);
9 a8 [+ k. f% R2 M2 _ return(*(p1+i)-*(p2+i));
7 g- G* E2 n+ [/ Y5 \# n6 H2 m}" Y% K# `8 v& H7 T
10.18
. K- u: ~' L' t" rmain()
/ L4 \: }3 X& N{static char *mname[13]={"illeagl","January","February","March",
: m8 q8 ]' J) f "April","May","June","July","August","September","October",
/ U6 \& i! `, G' S" q "November","December"};# l% l9 {) t& \
int n;
) ^" F, Z0 R$ b1 _ scanf("%d",&n);
- K9 Q5 T Y* H( Q* a if((n>=1)&&(n<=12))
, }- I6 b2 d! | @/ ]8 Q, C printf("%s\n",*(mname+n));8 p" o6 q7 f/ i* B7 a
else7 N2 S# h5 {+ L9 Y
printf("error");( g* w: O- o: ^0 @% x
}
- ^8 N: R5 G& l7 I K' U2 a10.20
3 F0 b3 A2 X" ^7 G, U( C# ]% lmain()
5 e8 M0 L9 N, j0 I; x# U$ l0 a{int i;
; a2 h+ m- q# G9 F$ K7 q char **p,*pstr[5],str[5][10];
0 o, c" I' r% J% | for(i=0;i<5;i++)- \. E) q# \) W, P) b
pstr[i]=str[i];$ O: e2 t) I p! N" X) |/ b4 G) y
for(i=0;i<5;i++): e% p6 F8 l- k+ z/ F& |/ v2 o
scanf("%s",pstr[i]);
( s4 u8 H& T! i/ Y* U p=pstr;
, u3 ]+ i" q' |: i2 ^. X3 P sort(p);1 j) F% a& o. ?# }; _5 b
for(i=0;i<5;i++)) z/ N' w$ C' [
printf("%s\n",pstr[i]);
0 K, D6 t7 L& H1 ]/ } m+ g}
8 L. T& k) R8 n) _0 I: Qsort(p)( ?& e* l1 p* S6 L3 i
char **P;
2 W/ e% q+ p/ D( ?, N" p0 q$ ~6 {4 p{int i,j;
" h! @# D- T, V1 f* E char *pchange;) ?" D9 {) @5 d' Y
for(i=0;i<5;i++)+ d9 ?. O! \4 T* k4 J
{for(j=i+1;j<5;j++)
& b+ x6 X+ H8 M1 _9 ^& A {if(strcmp(*(p+i),*(p+j))>0)
' s. n" J$ z* O0 k6 | {pchange=*(p+i);
8 [# r$ v) `$ u6 L) C *(p+i)=*(p+j);
( \8 Q# \/ p1 o* ^" E$ k9 C *(p+j)=pchange;0 i0 z& i$ d7 n; a
}2 O% I' q% U. j3 S( R
}8 w5 P& r1 M8 S
}% R3 Q( {3 |6 ~; S3 g
}3 e, a0 m' H \: }4 {, l+ e
10.21
1 P% E$ l$ U% h+ w! ?# Xmain()* s" E% b) w. [9 O ]
{int i,n,digit[20],**p,*pstr[20];
4 k1 A5 U* e1 B& m scanf("%d",&n);
' b$ W- H" J# p0 ^* ] for(i=0;i pstr[i]=&digit[i];# B# h. |+ e4 l, y" B) f; L
for(i=0;i scanf("%d",pstr[i]);
1 d2 W' x V: t# t1 l' D$ Z p=pstr;
* y0 t. p. I" q; ` sort(p,n);9 M4 i" K( |$ N: n2 K- C
for(i=0;i printf("%d ",*pstr[i]);
b$ G- i) g' j! W}% Y9 g" G" U. X' K; C8 `7 O" \
sort(p,n)# }0 e# o# M' v R4 L; s$ ^0 h( X
int **p,n;
: ?3 ]2 z( J( h; G! [2 U6 W& l{int i,j,*pchange;
2 F8 I# U( \0 O; s7 J for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
- Y7 g/ p# x$ m9 j8 D, j ? {pchange=*(p+i);
5 w, k& W5 a3 }# O/ s4 S0 E, \$ \ *(p+i)=*(p+j);
, |. I! w9 ?" V) D) G0 I# N *(p+j)=pchange;2 [% \: q( }8 w& E) y# c
}+ N) {; I$ z' e# i% X
}: u4 n* @; W: _' t) j' R3 s( s
}
1 B# h) R5 x; C1 i}6 ~* \2 A: R! Q* X& q8 }) A2 }
第十一章 结构体与共用体
4 M0 \ E) W7 X3 n; W11.1$ J9 c) R: K# Y
struct& a$ P6 H9 A5 U5 @1 |
{int year;) y( j7 ]0 i7 c: B4 b
int month;
3 ]" _# X8 m" Y9 o' R int day;8 L. I+ T3 b; c/ d6 V
}date;
0 C% }( ^3 Q. f3 \8 jmain(): k1 V; x& V# U. K5 o5 {
{int days;
- M9 m) m: G* [: r8 U1 o8 c scanf("%d,%d,%d",&date.year,&date.month,&date.day);
$ q- e# Z1 [4 [- q2 f switch(date.month)+ @5 h( g3 T4 L- e. \9 B) j; z
{case 1:days=date.day;break;8 o6 y# }$ l! m, @: ^! |
case 2:days=date.day+31;break;
: b! U- Z4 H( M5 ]3 r7 y) Q case 3:days=date.day+59;break;7 H b; V, t4 j& b( i- W7 D1 G
case 4:days=date.day+90;break;
& L( v" {+ I! ~, x case 5:days=date.day+120;break;- ?5 J' c' j1 Y+ F
case 6:days=date.day+151;break;. J) U- @$ I2 P- t9 o3 l) z+ J, R2 H
case 7:days=date.day+181;break;" F" N9 ?( M% ~8 F
case 8:days=date.day+212;break;+ N8 N n$ d' H7 P2 j! P9 T
case 9:days=date.day+243;break;8 J5 c7 J2 q5 G6 H8 v
case 10:days=date.day+273;break;
8 ?+ P( C4 ~+ K% y; H case 11:days=date.day+304;break;
6 p5 i2 G$ l1 c7 D& Q! c& p case 12:days=date.day+334;break;4 G% z% N& d$ q! o% P
}: l- N) b0 M T& H2 Z: v
if((date.year%4==0&&date.year%100!=0||date.year%400==0)1 ]1 V6 \- o9 p0 Z. K) B* A* b
&&date.month>=3)0 _1 o5 d, o7 {9 b+ b3 W ~1 b
days+=1;
# Y3 t# o9 S8 F& ~& Z- S& o3 P; n printf("days=%d\n",days);5 F! V: I3 a) a* f V) K8 [
}
7 _/ l0 x( V3 r4 z11.2 V) A1 X1 n0 [1 f4 d" S' F3 e! r
struct dt6 D w! u [% c" r8 p
{int year;
$ |7 E8 v) K' }) W/ @ int month;- @% x% [3 a" o' w& f3 J- _
int day;
/ x. d; t( \2 y }date;4 D0 h) w; F$ {( k: V! q+ T. `
main()
: T3 e- T+ z, G4 ]{
" r5 F" t0 A6 L$ |7 s scanf("%d,%d,%d",&date.year,&date.month,&date.day);
% [* x7 T0 d7 B4 E9 r; h printf("\n%d\n",days(date.year,date.month,date.day));! a" _! a' v5 D( Z" s& A' f
}
3 l4 a: U* K% z8 l6 ~# n, gdays(year,month,day)
' T1 d# v1 L/ f( Q" u: B" Mint year,month,day;0 ^- g: e$ n+ g$ u/ f- }( Q
{int daysum=0,i;: W, k7 ~$ b! M+ `# A5 E7 A
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}( g+ p! X) n: z& P- i7 C' @
for(i=1;i daysum+=daytab[i];
; q+ q: D, r7 `3 q, `# P daysum+=day;' h1 I- \6 t' }; m. X c: [
if((year%4==0&&year%100!=0||year%400==0)&&month>=3)5 o5 H2 \2 V \7 g4 B
daysum+=1;
2 D8 n! }' E* X& [1 a. t3 Q return(daysum);
) h3 y4 _6 q5 S. T8 t4 o}- r$ `! r: E& s. j
11.3
( R5 V: @- g) A; s11.4) d6 ^2 b8 _# R) j! c
#define N 5
! ~: ?& `! P7 o. \, d6 p5 |/ xstruct student) w$ N. I* |5 \( G/ P8 X: x: w* ^
{char num[6]; G* I8 v) w% @# P; V
char name[8];
% i. B( a! L% {, r int score[4];
: y6 P, V b. h! \6 V }stu[N];& ^ V) v# ~, A; i+ r7 S7 {0 N
main()
$ m6 M1 W! S8 w5 N{6 k, C0 ?3 x: _
input(stu);
" H" X6 L4 z0 L. b( G& S6 b print(stu);
8 w8 Z8 m4 G: Y+ Z5 `} w8 x$ P0 a% Z0 i' w9 ^0 u
input(stu)
! q/ g& J; k2 I9 S: y2 k2 Vstruct student stu[];
; C# i1 F3 e. w; z x& H2 ?! V, l{int i,j;7 B, O0 ?, m4 z' \
for(i=0;i {printf("number");5 {9 M0 z3 C6 d* x8 b2 h
scanf("%s",stu[i].num);
8 {+ `, K# `; P4 `4 }- g4 _ printf("name");
9 S. ?* c- A4 s; y/ T, z$ @8 ^ scanf("%s",stu[i].name);; _- y5 V$ }3 F5 I5 h% u
for(j=0;j<3;j++)3 j3 V/ T+ o0 J% r
{printf("\nscore\n");
1 h: m$ v+ O, A$ v( a8 w: f scanf("%d",&stu[i].score[j]);) Y5 D+ w/ @7 c0 r/ T J
}1 a _' ^) P! K+ ^
printf("\n");' Q5 Q; {" r4 r; B ?6 G# s2 F
}
5 {8 a8 f1 c4 ]$ ?: I; e" K3 o3 J) d}3 T% p' B6 y# k
print(stu)
% J: k+ l9 ~* u$ Mstruct student stu[];
' j6 i# F. t9 `{int i,j;
5 u7 r$ I9 T) W0 p! w3 C' b. R9 x printf("\nnumber name score1 score2 score3 \n");9 f# K5 O. u. E8 n2 [2 c
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
: L. V1 I2 r- d; C2 Y: z for(j=0;j<3;j++)
" M8 _- Z, g( | o. y printf("%7d",stu[i].score[j]);4 z9 S# }5 D2 t+ u- [$ B, f
printf("\n");
7 \$ A; d. P8 n* t( s5 H* w }+ ^9 U3 C( Z: d1 L7 t6 ]3 B
}' X$ y3 F5 [' M
11.5) o( p+ u" @. I0 i
struct student+ Z! ?% Q& {3 _) [- q
{char num[6];
, M7 E: A6 O! g8 l. J char name[8];
3 V* p6 W0 H* a/ @* y; z, O8 I int score[4];
: H7 `6 I1 \+ }7 u+ o+ V+ b float avr;: j. ~5 o6 @! k4 Z z5 `
}stu[5];6 h4 T+ w1 p# \4 \/ h. P0 ]4 P
main()' c; J, }6 t2 \ w! ]
{int i,j,max,maxi,sum;
; P7 s& l% w$ C! G3 a! x9 O- z float average;5 Q* ?* h6 f ?- p/ ]
for(i=0;i<5;i++)9 @9 T# f3 A4 J4 I- Q6 s9 `
{printf("number");8 I, w; @3 a6 E1 `
scanf("%s",stu[i].num);
; M$ C+ ?6 ~; G0 L" R printf("name");, |' ?1 c& R. r1 E) W% {( E
scanf("%s",stu[i].name);0 Q' M- c: ^* g- [* T- x: R# X
for(j=0;j<3;j++)
" ?4 [/ d3 G: q: S7 R( b) B+ r {printf("\nscore\n");6 M0 ~$ D4 t$ U: ]+ }) B
scanf("%d",&stu[i].score[j]);
1 x: D) c7 ^$ M( P }- c/ C) ] [7 o3 q( e
}) Z6 o; k' E6 w' G* l, w
average=0;1 n! E3 c8 O" M% o
max=0;$ s$ `5 m% W& p
maxi=0;4 x. E! O5 H/ f) i% p9 Z4 u b5 J
for(i=0;i<5;i++)
: d) y8 t6 H! e4 c( l5 H7 @4 n {sum=0;
2 [( ^- C! r4 J7 M4 L for(j=0;j<3;j++)6 z: K8 @. Q# M z9 e( Z
sum+=stu[i].score[j];
2 E; K6 o& X- O1 k6 j. ? stu[i].avr=sum/3.0;
: m& m7 x5 u" [ average+=stu[i].avr;
9 ^& K. ]4 C; J9 d if(sum>max)- O4 H, [3 L j; S1 w3 z9 Z8 ^
{max=sum;7 _! u _0 O# a0 t' e9 N6 j
maxi=i;2 _( ]# n, l# l! m8 O
}
0 h* O; y7 t0 f+ P4 ~ }
2 S# [6 y* B4 w) H average/=5;
! R" Y# {9 x6 k3 X3 ^ printf("number name score1 score2 score3 average\n");1 e: Q8 b$ Y! n4 F" w. K: L7 L
for(i=0;i<5;i++)
/ V( a- r/ u3 o# f {printf("%8s%10s",stu[i].num,stu[i].name);
$ [* Z1 @$ a* U8 f for(j=0;j<3;j++)
$ s: v& D/ A ?% ~8 y- v printf("%7d",stu[i].score[j]);
7 U K& I: T' s& O G. v/ C5 C" T printf("%6.2f\n",stu[i].avr);
: ~% d, i8 Y: F: W* O) u( B9 Z }% P, L. l& c5 f
printf("average=%5.2f\n",average);7 m9 a# i7 H% l; y( `
printf("The best student is %s,sum=%d\n",stu[maxi].name,max);* }$ G9 t6 ]& W. g( m4 z
}, x: \7 F: Y$ H( P, g6 `
, C0 \% v/ q. i6 p
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