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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
2 X# ^" H' }# W9 g: @: a1.5请参照本章例题,编写一个C程序,输出以下信息:) m# A$ A8 P$ ]
main(), r n- {" Q) q) {* \+ m0 R
{( n" r% \5 I8 G- |, h/ w' m/ D
printf(" ************ \n");3 C6 S. G8 G3 S' N- R. [' @
printf("\n");
+ O, ~4 g* d; P1 b7 x9 [! Cprintf(" Very Good! \n");
) M; k$ P% S) ~# ?+ t% P8 O+ R( iprintf("\n"); B- @$ A& @& Z* L. j/ d
printf(" ************\n");5 c. B* _; D, L7 G
}
* w4 c7 v% F9 h; G% B1.6编写一个程序,输入a b c三个值,输出其中最大者。
4 O8 Q2 d0 `, s" q; [0 e解:main()5 N/ x9 Q7 \9 e8 S u6 R+ ?9 V
{int a,b,c,max;( U" _ R! k, m. U* y, j+ {
printf("请输入三个数a,b,c:\n");
@4 Y& O$ V: f$ escanf("%d,%d,%d",&a,&b,&c);
( S$ A& D; F' i+ c" [9 _! }max=a;
& T, L9 |2 B: G- j6 g; I: Tif(maxmax=b;
; h, D- v4 Q8 V% W9 Wif(maxmax=c;6 v% Z; `: f! P4 m7 Y$ o
printf("最大数为:%d",max);
5 l$ c" y% U" i" Q) l}1 J. t) w+ o+ _" G1 x, L1 M2 `
第三章. z( E1 C5 }1 `/ m* x
3.3 请将下面各数用八进制数和十六进制数表示:
- X# O+ D$ j6 L2 v(1)10 (2)32 (3)75 (4)-617
8 W+ f. l3 ?8 _& ~' h$ {4 _8 Q(5)-111 (6)2483 (7)-28654 (8)21003
. b% ?5 Z3 |9 e6 N: J4 Q: u. J解:十 八 十六4 K( t) f3 k4 p/ |& Z6 {7 L0 I3 v
(10)=(12)=(a)1 L$ D1 ?/ Y z0 r' H3 S: F
(32)=(40)=200 D( X. z/ x. I0 e( @* T2 m
(75)=(113)=4b
' V" D$ j5 t$ |6 v/ ]' a9 c (-617)=(176627)=fd97
, A' f" m7 A A7 p; u1 j5 j -111=177621=ff91, D: C, C* x1 U# K9 w
2483=4663=963
5 H% R3 P% j- s: i8 y* L: G. i -28654=110022=9012# @2 g' c! y. r t4 q2 I5 t
21003=51013=520b1 B5 @; X+ O3 F6 w7 H. h
3.5字符常量与字符串常量有什么区别?0 a- `0 Y( P2 J0 }9 }
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
. G7 G+ n7 `) V8 }. w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'.7 b0 C9 Z5 g- [! w
3.6写出以下程序的运行结果:
+ i/ ^) [4 q+ X1 z) f; i! O#include6 S1 } U* _* |
void main(): ^; s0 R" R0 o, G
{4 M( i6 `3 y! o- Q6 T
char c1='a',c2='b',c3='c',c4='\101',c5='\116';/ q& w- Z6 L, [' V. s) C" `" B& N
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);4 {) z7 g; X7 m, M2 A
printf("\t\b%c %c\n",c4,c5);
4 z; \/ \) I4 [解:程序的运行结果为:& ^4 Q0 V, r& {# l4 z) v$ b4 C
aabb cc abc0 v x/ h+ S1 @1 q% z3 q1 k
A N. s( f1 L3 y6 J T& U' w
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,( s: O7 O% {; f; |$ T- d
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".! e- n o' X) c7 \5 X5 w2 W
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
j. Q+ L2 z; g1 W( p: T8 z输出.) d1 P7 r4 x, U K4 H! V# W
main()9 k. }) K. @0 p: v3 F0 H9 g
{char c1="C",c2="h",c3="i",c4='n',c5='a';7 |; x- H0 z0 a" u
c1+=4;
( S# n M, Y$ W; J# Ic2+=4;
, v( k p: v% E+ cc3+=4;
8 Q/ L* ^" y( xc4+=4;" Q3 z* |# ?7 r1 w9 ]# s# B% b
c5+=4;- H3 S9 d! d# i; ~4 s
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);
( ]; W7 J4 J* k( z- b- g}
' Q) w K5 D: w# s0 i3.8例3.6能否改成如下:
! E7 y7 z( J- ~0 E#include
, [. r6 [! b0 r% m6 w0 H( H& Vvoid main(); d2 k3 w9 S: @5 J
{0 Q8 @6 U! X# I5 q7 O
int c1,c2;(原为 char c1,c2)
- U: x7 E: f6 H J' w7 L$ L! {c1=97;
$ e3 w/ q' a3 T; B) rc2=98;3 Y. x5 V \( S6 t3 m$ v" p
printf("%c%c\n",c1,c2);
& g; g6 z4 y) T7 Dprintf("%d%d\n",c1,c2);3 `$ d9 T1 L9 `3 E/ V: e) F+ u
}/ m' Q" n2 L3 `/ C( X8 o- V I
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.% m5 {+ p5 \$ b, T# H" i& G
3.9求下面算术表达式的值.
. i9 ~6 n1 B! |6 E u(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
' u, U' E `6 o) t4 q/ t(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
9 j6 V6 F- S$ s9 m3 H) n3.10写出下面程序的运行结果:
7 U( g! ^, D7 S#include
( V. F" C+ Q2 [8 I. tvoid main()8 V1 F6 q0 P. M6 K" g
{5 w9 _4 q. G/ H3 g" Z5 a$ r4 O
int i,j,m,n;7 E" y. U' e3 g9 a2 r& e5 D1 J
i=8;
+ w A4 ?" {( ^+ N q* ej=10;
$ {7 x9 U) X# L& T! [m=++i;1 K4 |+ u @9 g; {) R& h
n=j++;" q$ |+ l- F( I( w8 }0 O
printf("%d,%d,%d,%d\n",i,j,m,n);
' V/ ~: c) ?9 k! W8 F' R}4 L8 o4 {' x& l' n
解:结果: 9,11,9,10
+ \2 C; R& K" w* M第4章
* u" q7 d6 i2 |6 n$ i* d4.4.a=3,b=4,c=5,x=1.2,y=2.4,z=-3.6,u=51274,n=128765,c1='a',c2='b'.想得9 Z# D- ]0 K) H- K
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:
* o* Y/ E4 q% ua= 3 b= 4 c= 5
& `% g; f. J# i( W$ px=1.200000,y=2.400000,z=-3.600000
9 X# e5 x$ z" ax+y= 3.60 y+z=-1.20 z+x=-2.40; v( D/ L& }# M6 {
u= 51274 n= 1287659 }" `" U- P! {
c1='a' or 97(ASCII)- a# M& Y$ j% B) \# V* a8 t
c2='B' or 98(ASCII)
) ~4 @1 Q5 Y* r. N# |& {解:
4 C4 f1 _- E# o5 ~6 omain()- Q) c6 C% o$ t& X& E" H
{
8 Q$ ^5 k* M; g1 @0 B( t! R9 bint a,b,c;: j, a) E/ S6 Z: T5 J0 r
long int u,n;: S( k0 r; E5 Z
float x,y,z;
p0 @8 N3 E# M& Q' f, Mchar c1,c2;
" @1 I+ H' X2 aa=3;b=4;c=5;
, D- i2 G+ z+ ?5 L/ i) Qx=1.2;y=2.4;z=-3.6;- X* Y5 B$ ?8 |- Z# d
u=51274;n=128765;
" O# }6 d8 p9 d4 i" ic1='a';c2='b';
$ [; A+ ` q% P2 U7 Sprintf("\n");
% k6 X: _7 I: Z' Vprintf("a=%2d b=%2d c=%2d\n",a,b,c);% C5 ]5 f% [9 ]) o% Q
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
" Q% @# y$ Z) Y3 lprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);7 A, L6 ]9 M$ h' B
printf("u=%6ld n=%9ld\n",u,n);$ z( I; g- @+ ^1 S
printf("c1='%c' or %d(ASCII)\n",c1,c2);
; b$ R7 K) A3 m/ f. |2 q$ Xprintf("c2='%c' or %d(ASCII)\n",c2,c2);
( }% ^1 a8 @* U- [}+ c- n, m0 `% b' O, @
4.5请写出下面程序的输出结果.
* @- D7 @& A, j) H结果:! L% t* c& z K8 @0 D
57+ M% {* o0 y3 u
5 7: }' c9 L t4 P2 B/ w6 }- U' }
67.856400,-789.1239629 t: T5 a! s g9 a# ~0 s- [' d
67.856400 ,-789.123962
' J1 \# P4 H" E 67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962
( g' D7 H ]" O1 E) d5 I! `6.785640e+001,-7.89e+002: n' r' o8 [8 {8 `5 [. x
A,65,101,41
! r1 A. y& k7 e# M1234567,4553207,d6871 ^' A% N4 p0 g: Z# ~
65535,17777,ffff,-1
/ I) E+ c {1 L/ E7 y+ B) J2 mCOMPUTER, COM5 s; V3 ~! H# E" S5 _
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',
; l8 }6 T7 u2 ]2 {问在键盘上如何输入?
3 \& W8 _, ]+ D9 `" K8 Umain()! X/ T/ C R( w& F, I
{
8 _; C3 Z) F* }' lint a,b;* Y. C7 E, w2 k, I& k
float x,y;
5 b6 r: K t7 o: {) F+ Ochar c1,c2;2 g t+ K/ r7 `. v0 l6 y
scanf("a=%d b=%d,&a,&b);
^2 B. {! z" b4 A- H7 |! gscanf(" x=%f y=%e",&x,&y);0 K( [" d* o) h8 @: S O# Q
scanf(" c1=%c c2=%c",&c1,&c2);
- L7 H1 Q( h9 d8 x) c' T7 Y% z}
* B+ t2 V6 \. C: Q解:可按如下方式在键盘上输入:
: x+ }9 S- \. `4 u9 Sa=3 b=7$ Z* N: {2 _3 l( x* F1 F( o4 S
x=8.5 y=71.82- r' R% n7 ]7 c8 N
c1=A c2=a
) v" E( \+ L, `8 q1 |9 k- z- H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"被第二3 V* W G/ o; i/ ]. j3 V# i
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
* h0 H3 z6 E+ K5 g. W: y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% E( e3 M/ d( C9 k0 j; _% h
初步概念即可,以后再进一步深入理解.
8 M1 {2 P7 g( P4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-
8 [- T6 A9 k8 S3.75,z=57.8,请问
. Y' d. H% D. k ^1 f0 X1 ?" P6 k在键盘上如何输入数据?
! h! D4 }( L* E; ]9 p2 @/ ~: Q Bscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);) K6 t1 B+ J% C% J- `0 c5 q% H
解:- H% D2 D0 E7 p2 m0 q, d
main()
' K0 _% T4 H6 A' D6 Z% Z/ v! t{
; Y3 |, E" s1 H6 e G+ C& Wint a,b;
+ t' q2 ?: ~/ s" z5 y& nfloat x,y,z;
. X: q1 t8 b0 v3 q+ C6 l3 y# G/ [! lchar c1,c2;. K* O" [& `$ S6 b' e: M
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
: S" g" D+ F, L/ p# y& r0 `1 h}- K+ U# t5 G' ~( g" ~$ W
运行时输入:
1 l2 l9 L' S6 q6 N x10 20Aa1.5 -3.75 +1.5,67.8
6 k/ [# J3 Y S4 L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
/ h5 v! x1 \6 _6 ]/ L! V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# c9 R2 L' N; c7 d0 c
会赋给任何变量。5 L5 r4 I; B. q6 { b
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,
# I; o0 U- w# M9 w$ ?4 U5 [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
. h/ {/ P1 i O$ G; J1 [程.
& |' F5 V5 c) Y" u% _ ?2 W/ E解:main()
! Y0 V! ]6 J' p+ a1 c& y% y{: E- L5 G# P+ }8 [
float pi,h,r,l,s,sq,vq,vz;7 K6 \% O; K4 j0 ~, f6 n; L
pi=3.1415926;
$ |+ O: O4 S7 S6 t. R9 r% d1 |8 Dprintf("请输入圆半径r圆柱高h:\n"); k3 a J! c' G
scanf("%f,%f",&r,&h);
! w2 x( b. I j# ^: ml=2*pi*r;5 _! J& B! f2 @1 G
s=r*r*pi;0 J ]" L: e' e1 a2 r* u% }& L. q
sq=4*pi*r*r;7 C% A* x6 f$ p% H
vq=4.0/3.0*pi*r*r*r;
1 b+ k9 `8 I4 c0 P/ Vvz=pi*r*r*h;
/ `) h% @2 s8 h& x- N" `! ^printf("圆周长为: =%6.2f\n",l);/ S; _& ?/ X2 I/ @
printf("圆面积为: =%6.2f\n",s);% @. S) ?. W$ e* a( c
printf("圆球表面积为: =%6.2f\n",sq);
2 n+ p4 d1 ~' E6 l9 oprintf("圆球体积为: =%6.2f\n",vz);
# z9 Q" K L; G$ y( T' h" c}
/ d* Z9 u T8 u6 G$ M4.9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,6 W' g: \# ^6 c. H
取两位小数.
A- ]; Z, X/ P解: main()
6 J3 h; E D% ?. n{
. Y4 o' `/ U {5 l' i1 b; Wfloat c,f;6 C) P* B4 [# p3 z
printf("请输入一个华氏温度:\n");
5 g5 e9 }" Q7 M, F8 D1 [scanf("%f",&f);6 M, J0 O. Z- L2 l+ }
c=(5.0/9.0)*(f-32);
7 w9 d4 @$ U1 l/ M5 ^printf("摄氏温度为:%5.2f\n",c);5 ~/ z$ `2 D! d5 p/ f$ I- M
}/ }0 t- v1 w' O* J
第五章 逻辑运算和判断选取结构
/ q* n7 E- [& A: N9 G5.4有三个整数a,b,c,由键盘输入,输出其中最大的数., d" S4 e! O O
main()
2 j7 N5 q' J# c6 ~; M5 z* @{# U* V! ^! [ _7 X
int a,b,c;
" ~- J7 E6 G {' oprintf("请输入三个数:");- V! _! _2 Y( p/ b2 G
scanf("%d,%d,%d",&a,&b,&c);
, L; G7 w' I( k' @if(a if(b printf("max=%d\n",c);
' l5 t2 f; p, D: P4 _0 ? else. X* p" s) K# u
printf("max=%d\n",b);- q6 ]- P' k- e7 Z- D: v4 U
else if(a printf("max=%d\n",c);5 K0 P' {! E! U `; W. d1 b* J
else) q' i E. i, c) j2 Y- H2 l* N
printf("max-%d\n",a);
* A( z( t6 C1 R7 _. U; R( y; o}! G% _$ a2 C# R5 E3 ?; Y( l
方法2:使用条件表达式.
; [% E; d: F6 p2 d; `% Y W$ Y! i: rmain()4 n4 V# d: f, o/ H0 T( D! p- A
{int a,b,c,termp,max;+ T9 q( `. c5 b3 i- W
printf(" 请输入 A,B,C: ");% _: g, [/ {! \6 g
scanf("%d,%d,%d",&a,&b,&c);
" I3 q6 T6 d1 {6 x; F printf("A=%d,B=%d,C=%d\n",a,b,c);; N' w/ x7 ^ z, {+ i5 S* V7 h& B
temp=(a>b)?a:b;
2 S2 O: V. z9 R0 z7 Q% J5 U" f1 _ max=(temp>c)? temp:c;$ U: u* j) m/ H: H# a
printf(" A,B,C中最大数是%d,",max);
% a( ^" c2 v6 h8 }7 U$ Y}- o, M8 q: @$ ~- e% t2 u2 {
5.5 main()
- Y; V1 w) w/ _7 b5 Y& L{int x,y;
/ l$ B; ^' w( `2 Y. Z9 yprintf("输入x:");8 }4 ^1 `( V; m
scanf("%d",&x);( Q' a3 y' ^6 w1 z$ r2 W# f
if(x<1)
( _/ j2 d: ^1 j, @" ] {y=x;
; W1 i# a; c6 m, M printf("X-%d,Y=X=%d \n",x,y);( G; Z5 q7 P1 y/ K1 z8 W. R \8 e0 U
}
# u% F4 V) b. m& l, Delse if(x<10)
) b# r* R& C: X$ d6 | {y=2*x-1;7 K, L' I C- k/ J7 {
printf(" X=%d, Y=2*X-1=%d\n",x,y);$ ~" V$ ?* G$ n5 `, a
}8 \) D* g( F" X
else0 x( I& v; L: M: q
{y=3*x-11;
l2 ?' |# X: {; m printf("X=5d, Y=3*x-11=%d \n",x,y);
5 }% R5 ?9 T; L: Q: }, d, o1 y* j }6 |9 Z0 v" J5 v# H
}- H+ C- K1 g3 U5 w* c0 r, N
(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! - E C4 K; `1 ?0 i" H
void main()1 v# T: z, x- ]! j. u$ T
{0 b3 h, I% H6 _! \! y) @
float s,i;
/ U9 [/ o* U) L' Ichar a;
$ b3 |3 q; G5 I4 _0 S2 Wscanf("%f",&s);
) p: Y4 I2 L8 ]# {" P1 z. zwhile(s>100||s<0)
5 I. R, O' ~, g& T% {{
' B: w; U2 t. [# U" W+ G6 Iprintf("输入错误!error!");3 G! d# G4 a; k3 H0 z
scanf("%f",&s);* }( A5 w# v0 c; ~ k4 t
}% N* K' l/ }3 Z& Z/ r5 o: W
i=s/10;! F8 r* O; E2 k$ ?8 \
switch((int)i)5 R1 b3 X2 l& a4 B2 d+ p
{
- x ~; M3 M1 F: T1 @) U4 dcase 10:- s% R e* Q; p0 p0 x
case 9: a='A';break;
$ Z9 x, @$ i6 `9 ?* p$ ncase 8: a='B';break;, Q8 q1 i, O/ a# n$ \
case 7: a='C';break;& p/ `% A* `6 z- T1 O
case 6: a='D';break;
5 U; g, M" A1 i4 ?- D! s$ Ncase 5:
* `2 }5 U8 l; [: _6 H- Qcase 4:
! w3 s7 E! J; m6 V: J& mcase 2:
. {( y4 `4 w& E5 fcase 1:4 h- n0 `; B! q' F
case 0: a='E';
* f. d' P- w0 v- y2 y4 x; K9 j}& U4 D& j' g( k1 S. d
printf("%c",a);
& w; t6 s0 }/ t" z% g5 w}
% A9 k+ X4 i- M$ h5.7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.
* K' Z+ q5 i( ~- w: d* S按逆序打印出各位数字.例如原数为321,应输出123.# q8 E2 b' m% ?. g
main()
9 V" B7 J; N" t+ b' B: x6 y& C( f {# n6 P- r$ H/ s# l
long int num;0 L, q' [( }+ t1 v9 m
int indiv,ten,hundred,housand,tenthousand,place;6 `; V9 c2 I p* H; ~
printf("请输入一个整数(0-99999):");$ c4 o, Y& B5 G' h d* D
scanf("%ld",&num);# p: q" `0 }9 F) S
if(num>9999)
: a8 l& V" p2 s2 y6 L5 J1 D5 @. T place=5; y+ ^/ |0 z$ m$ c" _: ^
else if(num>999)
' h% l% l5 u6 a, I" b! m place=4;3 Q% C& G. ]6 f
else if(num>99)! E# E, m! r" p
place=3;1 N" n7 p5 |' H. F
else if(num>9)
" F# v2 c0 U1 e& b! W, R9 V4 ^ place=2;
d8 A# L1 T2 F0 |3 A0 kelse place=1;" {: l$ w. u. O8 k' I8 I3 |
printf("place=%d\n",place);
% N& W3 ?9 L; U- ~- M5 Rprintf("每位数字为:");" J+ {, F* G/ U# c1 O) q8 L
ten_thousand=num/10000;
$ C2 |8 u+ n4 ]" T6 G: Ythousand=(num-tenthousand*10000)/1000;
. [0 m1 d y( {% d' J/ e. _hundred=(num-tenthousand*10000-thousand*1000)/100;
Y* V) U" ^6 E; H" m7 rten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;- G7 t" ]& s7 n1 {
indiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;
" V8 Z# L+ ]( N9 @ m6 T5 F" _switch(place)
" k8 g1 ~: S. J2 X1 o{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);2 n! P) S) k+ ~4 o% a3 A6 P
printf("\n反序数字为:");
5 R5 \6 T" @2 X0 T printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);
; G/ ?9 o: I( j5 K- ?4 P' C break;& }7 C# r: ]; ~" p7 y3 D
case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);& L! V( @- t' f" }+ v
printf("\n反序数字为:");
+ t" Y: K9 C' D: h printf("%d%d%d%d\n",indiv,ten,hundred,thousand);
+ s7 v4 r9 t6 r6 b4 E2 i2 t break;' Z" e$ f* M/ d6 M
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);
; t8 P- V+ K) E! P1 r- o7 [- d printf("\n反序数字为:");* L$ z$ Z- n1 ~7 y. S# d. c @
printf("%d%d%d\n",indiv,ten,hundred);
$ N1 S: w; A2 G8 ^case 2:printf("%d,%d\n",ten,indiv);
) _& A. O( v4 g) { b2 k' O printf("\n反序数字为:");' ~/ a. y3 Z, r6 O6 A
printf("%d%d\n",indiv,ten);
4 {$ Q; i+ k- c8 I; q8 `case 1:printf("%d\n",indiv);& b; _0 Z+ P* E1 a f
printf("\n反序数字为:");
8 v4 u, y& X, B5 r" b' Z7 S printf("%d\n",indiv);7 V/ d# S* n. [! m% X
}0 y4 E. l$ j* G0 O+ T# A
}
- z% P6 D- B! b& P1 ^' I5 Z# Y5.8
2 u9 A: l/ [* L# G+ O1.if语句* b' `. b0 t4 h5 O; i
main()
2 U, j7 m6 e) x{long i;
% z2 N+ n+ P6 ?- i) T. a5 h float bonus,bon1,bon2,bon4,bon6,bon10;
9 G+ ]- ?' ^: J( y* |7 t bon1=100000*0.1;) P! Y7 d6 \: n( k
bon2=bon1+100000*0.075;
; [/ E$ l$ f7 g/ ~ bon4=bon2+200000*0.05;' Q' i' ~% j3 L8 d: F" w# U" v
bon6=bon4+200000*0.03;
2 m* f4 f" @ D4 z0 n1 E bon10=bon6+400000*0.015;
( G. e; ?9 W! ^ scanf("%ld",&i);
* n% }& @# x- B3 X) Q& Z if(i<=1e5)bonus=i*0.1;
% t5 g) v9 s' A4 B else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
6 |! g- ]8 ~8 G' l4 ` else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
" a' i. V# s; |8 V5 p else if(i<=6e5)bonus=bon4+(i-400000)*0.03;. T/ z& J# J8 R* H3 B- i* W4 s, ?
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;. n4 ~" @$ j0 j+ k) N' n
else bonus=bon10+(i-1000000)*0.01;+ B% f2 N, R( z0 o- g8 ]
printf("bonus=%10.2f",bonus);$ P$ ~! ], [9 }" Z- O2 g! m
}( ]* a6 G, D M
用switch语句编程序
; Z8 i% @. b3 B: X* imain()
0 s( B4 J6 S: G0 E6 T* R{long i;8 M' [& B- m% m% } I9 i
float bonus,bon1,bon2,bon4,bon6,bon10;
: o: O3 u& y$ Z! L* J0 U int branch;
( k6 k+ @. K" @- ^. X) @; J bon1=100000*0.1;
- z A9 U, P+ m bon2=bon1+100000*0.075;2 p- E# p {* y5 M/ l9 ~2 h
bon4=bon2+200000*0.05;
) w1 ^: W$ g! k% G8 k bon6=bon4+200000*0.03;
" a3 C' p( r2 s5 c! t& B bon10=bon6+400000*0.015;8 F) H+ u* w. X8 x9 K `! I1 P+ l
scanf("%ld",&i); A6 o* D* ?/ V/ e/ ~' b# ]
branch=i/100000;
% k. f' U9 A* T4 d# M* _8 W$ @ if(branch>10)branch=10;5 ~ G6 M0 C9 I! Q7 A
switch(branch)4 Z1 f% P, f9 a+ \
{case 0:bonus=i*0.1;break;
5 y/ O) f- o) F6 R, j( @ case 1:bonus=bon1+(i-100000)*0.075;break;
1 z$ Q# U1 J0 r% j P+ b% L case 2:
8 Q7 w2 ^, D3 T7 d5 c# O! J case 3:bonus=bon2+(i-200000)*0.05;break;
! G# \. G" D* o5 Z; c& s case 4:% |' ~5 ^/ i: g( E \, n8 W @7 [
case 5:bonus=bon4+(i-400000)*0.03;break;
9 V& l" j6 Y& ~5 d& C! v# ? case 6:( V! n* }) Q* \. |
case 7
* P% z: B- H k* `7 j case 8:* S8 \+ d% a9 u0 k! m' g/ e
case 9:bonus=bon6+(i-600000)*0.015;break;
$ N$ h/ \% h- c6 y) R8 Y: ~* ` case 10:bonus=bon10+(i-1000000)*0.01;
' S: O! v$ D7 B- b }& u) a1 S( |3 Y; m7 r6 A* A
printf("bonus=%10.2f",bonus);
X: O7 b# g! v1 z" R, B}
6 o9 ~9 R4 e2 B3 g- C5.9 输入四个整数,按大小顺序输出.
- l$ x- C( S# t5 q! ^: Rmain()( p- [; k7 x, n8 n# j* L8 M4 K
{int t,a,b,c,d;9 H0 ^- @( N9 }5 N5 l( t3 D
printf("请输入四个数:"); N, M7 [3 X. T/ R
scanf("%d,%d,%d,%d",&a,&b,&c,&d);
V* I# @3 { T printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
8 g7 S: `1 }/ Y! a" Y% L* ^ if(a>b)
5 X4 J' F7 a a* s {t=a;a=b;b=t;}/ L* R0 O$ u5 c# T/ ?
if(a>c)
* J, T8 m* A/ @& @* c( w {t=a;a=c;c=t;}
9 W: F w7 ? |: V* ~' R7 u Z* A7 G if(a>d)
/ c1 i. K8 ^6 j- v {t=a;a=d;d=t;}& u6 u( T! J, D0 k" q& ?! T
if(b>c)/ T- n+ }2 ?0 v$ t
{t=b;b=c;c=t;}
5 _/ Y: e I# ]( X! ^8 g: `4 Z8 [ if(b>d)
' r% d9 Z$ g; P1 \7 } {t=b;b=d;d=t;}
8 f$ r4 r! G% [% a! I if(c>d). o& U, A1 k5 O; a
{t=c;c=d;d=t;}
) ?! o4 \& H4 J2 ?' f% y8 @' {printf("\n 排序结果如下: \n");! z) T7 }! y6 u9 n! b
printf(" %d %d %d %d \n",a,b,c,d);
3 N" L6 L3 f+ i5 U' y {+ P}
4 I4 u$ A6 e6 ]5.10塔
: M: o3 A- S# _) vmain()
& h/ Y3 b' f9 l{
7 s+ H; s. v9 e, k$ ]$ q/ `- Oint h=10;
P, I6 r, E4 D0 wfloat x,y,x0=2,y0=2,d1,d2,d3,d4;# v1 u9 }" ?$ ~: P* R
printf("请输入一个点(x,y):"); E2 v- M: J. F U# Z6 P+ E& P
scanf("%f,%f",&x,&y);) ?) U; s9 t" ~( t" T. K
d1=(x-x0)*(x-x0)+(y-y0)(y-y0);
1 P8 F( K: P% `& {" V4 |( L, v- x) ~d2=(x-x0)*(x-x0)+(y+y0)(y+y0);0 t- h5 J9 |# M# z
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);* x$ a" k. [: P6 i" u
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);4 t# ?+ o% `5 O" q3 y6 E& f9 K
if(d1>1 && d2>1 && d3>1 && d4>1)
! D$ c* T( ^4 Wh=0;7 ?& s9 l( l9 Y/ t
printf("该点高度为%d",h);4 d# e1 t$ t1 {' `
}
1 k$ X k9 s1 b- i! ?. _; B; r第六章 循环语句
7 c" T1 j4 t4 l& t5 f) H% K6.1输入两个正数,求最大公约数最小公倍数.' Y( o6 _9 R9 X
main()
6 x. j: b( Y8 y% Q7 @ `/ C{
" A) A' d6 @* r: W7 h! |/ t0 S' Aint a,b,num1,num2,temp;
c) A& S4 ]: @printf("请输入两个正整数:\n");
{2 S5 G; k7 i- tscanf("%d,%d",&num1,&num2);5 G- Z7 w6 r7 u/ D0 @5 i6 q- P6 M
if(num1{
1 O. s$ w5 O5 C j. M K: ?, ^temp=num1;) _7 S: f) ]; |) v' ^+ R; x3 b
num1=num2;
4 j- A( f5 S) O& lnum2=temp;! j5 N8 \* C8 A; ^4 u, s
}
2 o d' e; z0 L4 j# Q9 aa=num1,b=num2;& a; V! X( N: F F R: o1 ]8 q5 A
while(b!=0)
5 Y0 J U1 K( m: b2 ? { F: J6 ?8 _6 q
temp=a%b;) L K5 z+ H& X+ l
a=b;6 J( j+ k/ ] K! ^+ {) L e) q; |
b=temp;- {; k% ^) v/ k
}- S. [9 F$ b8 J& c& \; W) L1 Q
printf("它们的最大公约数为:%d\n",a);
. H. _/ [! W" `8 Vprintf("它们的最小公倍数为:%d\n",num1*num2/2);
& @# u6 ?' O7 x& p! z' w9 Q}- P2 f6 I2 n1 z; d4 H
6.2输入一行字符,分别统计出其中英文字母,空格,数字和其它字符的个数.
; j4 T ~* J- F解:0 S" U5 c J2 i: y
#include < >
/ N: H: }+ Q- q8 ]6 jmain()
8 n9 e) ?, w2 ?{% z/ ?3 B, }9 S
char c;
K" a/ C$ _, y$ ]% x8 \, Tint letters=0,space=0,degit=0,other=0;
0 {8 E0 E" @" V Sprintf("请输入一行字符:\n");0 \, p+ ]6 G! v& \6 G( D
scanf("%c",&c);2 m: a3 I% P% P7 l# _$ v
while((c=getchar())!='\n')
' D! P' @' W1 e0 d: N{
: J1 ^2 {9 _/ R: f" nif(c>='a'&&c<='z'||c>'A'&&c<='Z')
; f6 A: z) Q' `5 M/ @% Jletters++;
) l3 W: m. o& r W1 \else if(c==' '), e- O% L( s/ V( e: B \2 m
space++;# C# }7 P9 E4 V- n" y
else if(c>='0'&&c<='9')
; h0 C. R- A wdigit++;
$ B9 z6 ~+ b0 l# p$ j9 A8 @/ H$ ielse" C* G9 N1 N- _1 Y
other++;
: l0 g' Q" S5 t" s' n7 l# v' M; r, x}
# H* G4 H5 c) X; |/ C+ ~& I" oprintf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
8 x! s" x- j+ Qd\n",letters,space,
* a+ v% v$ c! p r2 bdigit,other);$ D5 o) l& g9 O7 g5 Z' g
}& I0 N& L& ]$ G. ]
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.
& B& n9 [+ Q! m3 x$ z" P, Q7 K解:
2 `3 c+ v: |, E( i# |! Y) ?main()' I+ d6 m3 H* z4 A( z: I& W; [" @
{" A0 v x2 P: N+ E$ n8 P! ? C+ }
int a,n,count=1,sn=0,tn=0;: f, a t0 d9 n9 U
printf("请输入a和n的值:\n");8 s1 V6 T) d3 @. W* K4 e
scanf("%d,%d",&a,&n);
) x' }# W( W& ?4 R1 Cprintf("a=%d n=%d \n",a,n);
% d8 U& h6 u; C. |while(count<=n)8 z5 G; q# w- o$ B7 B" L
{( I: v3 X1 {( n$ ?" p% h
tn=tn+a;' H x8 ~/ v4 @& G% B: X
sn=sn+tn;
5 L. O% d0 ?! `a=a*10;
0 C7 H! v4 ~: O% B) O++count;
2 R8 ^' J6 w# G" v" n}
& u- b5 k& Z5 m4 }0 l# e4 Nprintf("a+aa+aaa+…=%d\n",sn);
9 _, i% L3 @4 r4 B}$ a. c. a" j H
6.4 求1+2!+3!+4!+…+20!.& @( t) K# v( K* M
main()( P" c& e4 Q! v- o9 b0 ^; _7 Y
{5 u; @ u, Z& P
float n,s=0,t=1;
, Y8 a3 q) O3 G& c# k. |for(n=1;n<=20;n++)2 y# Z2 G' V: N `7 Q
{0 A, V! e. O4 c$ u4 L7 M
t=t*n;5 o: f5 E" w3 u. j
s=s+t;1 C& Z w; h6 o9 C. B d3 O
}
, g. ?; q6 |0 h% D9 J+ `printf("1!+2!+…+20!=%e\n",s);
# d. ]3 c4 }0 k. ?( m& N. `) y k}
. G# S( _) t* W" t. h3 {, }0 K6.5 main()
! R5 i+ [7 E/ A- s+ x+ D1 _{7 O1 O- n) ]1 |& m& w# t' t2 L
int N1=100,N2=50,N3=10;' n# R8 P2 h7 `. q0 h$ {. p
float k;
: R) E* P+ [, [5 j2 i) T9 Bfloat s1=0,s2=0,s3=0;
b) l. z& } ?; Lfor(k=1;k<=N1;k++)4 J; l b& N+ @% C H0 x( @. R
{
: ?4 Y1 p: G. D9 a4 Ss1=s1+k;4 ^4 z4 N6 U a$ V) F; Q' C
}& G* v8 |1 Y. w' R0 A, \
for(k=1;k<=N2;k++)1 A: U1 P6 G8 X# L6 m
{
% k3 O6 G9 k7 F/ G4 us2=s2+k*k;! [9 s9 a- ?8 k! t* L1 c0 i8 n
}
( D! Y5 Z7 T/ T8 Z/ G8 dfor(k=1;k<=N3;k++)4 C" M( a. v0 `2 M# j
{6 t2 W2 n1 T1 R5 t
s3=s3+1/k;
$ P e( t" z# r}
1 h) ^: Z- D. j+ ~printf("总和=%8.2f\n",s1+s2+s3);
8 m! k, u4 k$ `6 c8 o}1 r) ?3 Z, W+ i2 W! \0 X" |
6.6水仙开花
6 j& j7 r8 d# amain()
1 | z) j6 s5 h5 q s- f, W, @4 h1 M% H{( f" e ] d1 p- k3 [$ z
int i,j,k,n;
+ x3 }+ `* @3 {8 n9 x3 L- ^printf(" '水仙花'数是:");4 c' K$ N5 g! V% f% x! L4 O; n
for(n=100;n<1000;n++)
% f; |' W( f: F, |. V{
+ @2 p W; E& f1 H+ I3 k9 b7 yi=n/100;4 W& k+ u |% q/ g2 o4 A- E
j=n/10-i*10;; X# C c5 I( {& h# c$ u
k=n%10;7 i) Q2 k6 k: ~$ s) p
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)9 K2 y, |5 E) F( ]
{ z6 X5 F* w% M) }7 ~
printf("%d",n);
, I ^# H* q+ }' |- G) W% c}
+ i7 ^$ l5 H* X}4 G1 C; q1 c" u. `4 K- R! }& r. h- s
printf("\n");9 [1 m( _+ |( h9 |( Z6 p6 R
}* A8 X* o m' f6 P0 w- C! y9 W
6.7完数
~" f2 N( p& z B' G; Zmain()' k" p% @! E/ h4 K& E
#include M 1000
" f4 M* e2 u8 emain()8 s4 U X* i! l6 ]
{7 U5 Z7 U( [: H5 [, C/ c3 E
int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;' V9 s& _( i/ Y( g
int i,j,n,s;: G! \2 d5 B5 \' o8 r% Q1 b. J& m
for(j=2;j<=M;j++)
0 u4 M- z$ X& X- g7 U$ d+ q) c{" K% h! V9 R( |
n=0;$ i9 ^% ]$ [8 X4 G* y0 R0 a
s=j;
& k; S7 e% A2 d0 P+ y- F+ rfor(i=1;i {! _& j W/ k3 k/ _+ r5 p2 O
if((j%i)==0)7 h8 U& B3 C" W' z# N
{* P! d3 g7 u7 {) g% j
if((j%i)==0)
; r6 W9 ]7 i# _8 r' P {
$ C- I+ W, ^6 N9 p$ D9 I n++;2 V3 T; f) E8 _9 [1 i* }3 i+ D5 T. W
s=s-i;6 R# N9 X# Q9 L# N& O
switch(n)
9 {! _1 z: f. n5 @8 c: } a N {' F- n" G5 D1 L
case 1:7 x& F& \7 {8 z8 U! u* B
k0=i;
! f9 M5 D5 Y. L. x: U break;
6 S1 D3 V5 ?: o9 v case 2:
. e' R! o- |8 w' L4 o k1=i;
8 `+ ?+ q# B: k' C& ]) \ break;) R- T. Q+ ^* p9 E5 v
case 3:
8 B2 X9 e& Q; ~/ E1 L' R9 T! q k2=i;
+ S+ s, n+ F7 D. A( D break;
% G0 W2 l$ `' c( _: f9 j0 A' R# x case 4:8 [2 R. p! I# G* v n
k3=i;
8 O' s4 a8 x+ G; [1 J: a- R J& D break;6 B( T6 f8 Z4 C, D# d2 E
case 5:
]5 e" C$ j7 F5 ]% e8 _ k4=i;
% s e! g! J* F/ j3 m break;# H& y: m" {* d
case 6:
$ u; o. L+ o7 @: B; T; C H k5=i;& @& h& s: j& i! ^6 o) x" `
break;
1 j. g7 P- U( E case 7:
/ c+ d! `: J7 A1 n) ~ k6=i;# O' e' B, L& b2 B! l
break;
7 o7 p4 a7 u) j$ l. T2 d case 8:$ O2 @; o6 i% S7 c' Y
k7=i;
. `* B0 V# S1 W+ x break;8 `; d% `( F! O1 p# O1 _
case 9:. e7 s, S* D* S( l
k8=i;
% [' g+ L! Z$ g* F0 s( c$ l break; _7 a' P5 q7 m7 e' d
case 10:6 n" o& ?7 T1 m7 H" R1 J! g3 g. w
k9=i;. k# F Q O0 o% V* E% U9 B
break;* m$ E6 }- _ g( z8 \4 _
}
: l7 G K F% G& n4 J0 A, b% V: H }9 v* z* s: m. i$ W8 X- k* I: N
}
4 E$ `! ~' H3 ~* U* H$ xif(s==0)" D/ ~% |% u4 s$ D( d" Q3 T
{
; g0 x3 u+ y6 e2 J9 T7 H/ E4 V" s, |printf("%d是一个‘完数’,它的因子是",j);
$ R7 l& ?; }, O: j0 Q5 wif(n>1)
- Y" B" h- j4 E4 o printf("%d,%d",k0,k1);
" D8 {' m2 F# ?4 o0 Z$ V. cif(n>2)2 Q+ A2 x5 L$ h7 A x
printf(",%d",k2);8 e, T# q+ y0 n8 }" K3 L; f
if(n>3)( g' ~ u! W; M0 \) `1 ?' n# `
printf(",%d",k3);# }, S8 L8 C3 h" I) p$ x! W; \/ y) d
if(n>4)& ]) E; e2 S* K
printf(",%d",k4);1 y0 e) Q/ l; P9 x+ j
if(n>5)$ b# v& F# S9 I; ?+ q
printf(",%d",k5);: d; P0 I( e+ t- x2 Q9 U
if(n>6)- M! `# z1 [& c K5 C8 c K
printf(",%d",k6);
, ?: \2 J! E7 L' [: Pif(n>7)2 T6 C8 b2 J( y8 c+ E4 v# l( f
printf(",%d",k7); v% H" t2 V' v4 U3 ?( U- ?
if(n>8)/ u/ G: j5 a( y! W* h+ f
printf(",%d",k8);9 K, L' K0 N; G; W' t0 j
if(n>9)& ~$ a- O- n8 l5 R
printf(",%d",k9);
: H2 Z+ q7 o6 @& {! J( xprintf("\n");% \) V% N, U: w# V$ h7 y
}
1 [2 w) l# I c# ^( E% y- |& f' V}, Z- O0 {5 ~. L3 I" s
方法二:此题用数组方法更为简单.
# f4 Z0 e9 D" w2 fmain()
) f/ B* Y8 m! ^+ Y/ K, N7 Z{$ c7 ~8 b2 J2 Z9 I' B
static int k[10];
2 u2 O8 z( M6 ]int i,j,n,s;0 }4 X0 s! ?# E% {% f- h, ?
for(j=2;j<=1000;j++)' L+ J6 ^. m) ^- C, L0 t+ }, ]5 O
{- v! j3 O" v9 M4 w
n=-1;
$ e3 r! \& U, A% @ r& D" ts=j;
/ A. D* M1 G( V' W. h& zfor(i=1;i{
0 ^" @. o/ f" Q1 @* v ]8 B/ x- pif((j%i)==0)4 R. ^2 Z( {9 s
{$ e' o& L5 l0 a/ x: ~
n++;# J* E, L# i' l: {/ S
s=s-i;5 F, ^" L7 t2 S* t$ S
k[n]=i;/ ?. C+ W/ F% U' y# `, ^3 [0 c, D- _" q
}) |! U3 M, ?6 g/ L/ U
}
8 L4 | t. M4 `3 c( Yif(s==0)
1 w2 m* v! x. ~7 z3 Y{
% J* x2 b' k, d6 X" K, W! }! |printf("%d是一个完数,它的因子是:",j);
6 Q0 d5 a) c, D; X# [for(i=0;iprintf("%d,",k[i]);
" R! @" k" k' ?. g$ dprintf("%d\n",k[n]);
% s6 R. l& R. {! E6 M# q& G7 _# E}9 V% I' r' h& @' g% p" Q
}
# X1 M S9 y; h, `+ i) ]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.1 }$ g! `2 n t6 ]& _
解: main()& j B4 {' e" S( Z$ X
{
% R' Q# J5 g2 P" F0 nint n,t,number=20;
! R4 O0 ~2 q5 Gfloat a=2,b=1,s=0;$ Y: {: M1 _) `& J2 `: T& E/ F
for(n=1;n<=number;n++)3 c5 o. S# }' F1 }0 y
{
6 q- U: L1 B/ E/ as=s+a/b;
0 B/ \, u( W+ Q2 P5 zt=a,a=a+b,b=t;# y. {( B7 P& V4 e& b
}; ?3 O4 {" q$ A( u) c
printf("总和=%9.6f\n",s);
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6.9球反弹问题
4 D4 s, T3 N* ] Ymain()
% ~: v' H8 X" M" ~* R{
0 W; {+ v- ^0 b. c3 P2 @float sn=100.0,hn=sn/2;# _" R* c& }6 }7 [
int n;
7 {0 E" B0 v6 A) r2 C1 u& j% }for(n=2;n<=10;n++)
6 {5 J; p" j" t1 N. F5 ]{
% S" o* G8 Z& N9 Usn=sn+2*hn;
4 n8 }9 @% p1 H1 E9 [' rhn=hn/2;
* A7 U# K& r% S0 U( W) l- J6 S$ Y}5 V. F$ T' ]! X- S
printf("第10次落地时共经过%f米 \n",sn);
! D. R1 ?- E0 b$ K" b6 \printf("第10次反弹%f米.\n",hn);
: s/ f1 _! j5 J$ K}4 J3 h, u+ I5 I$ n8 M! ^( y( K
6.10猴子吃桃0 m1 x$ O1 `4 ]9 J
main()
- t$ d/ ?/ F* b$ t{( i |3 _, F9 R" @! \3 ~
int day,x1,x2;
5 z% H6 [1 M; A7 O' ~1 N) Vday=9;& Y1 h- H8 n( k
x2=1;5 v& s6 N x( h- W
while(day>0)
+ y( q0 u5 w+ F d7 Z{
3 W a; e- v, B& V, nx1=(x2+1)*2;4 a; U) i% A1 E2 |, o* E9 A8 e# c
x2=x1;
7 e! ?; e1 h; I- fday--;5 P( s$ {. |/ ~9 b+ C$ ?. e
}' V" {+ Q# O( z9 l5 R
printf("桃子总数=%d\n",x1);, j# r5 _5 g R) \# o- e( {" W
}
& d5 N; W. P( S5 w+ x$ w) p
( o" u- W% Y8 U( t/ c6.12: h8 i4 w- ~' x5 l
#include"math.h"
/ H$ g2 y# ]. o; `8 e4 |main()& V5 Z! z; B6 B( i
{float x,x0,f,f1;8 D [) I8 _& T0 R; w
x=1.5;5 K) B* ^6 a1 S
do1 }. O, f6 T+ J' t5 h/ r6 G* k, W, A; I/ |
{x0=x;
' t7 S1 b+ i2 _$ l6 K9 Z f=((2*x0-4)*x0+3)*x0-6;
: H3 V% ]' f. I7 x f1=(6*x0-8)*x0+3;% m4 M/ p/ t5 N0 [+ H
x=x0-f/f1;
' t* F7 H" h/ n2 g7 ]( A7 q1 V+ Y' d }$ L' k" {( e4 [- f) p
while(fabs(x-x0)>=1e-5);
% J# h& W, |5 [: T4 G% w) t printf("x=%6.2f\n",x);1 T* d3 N6 e- j |" b @1 a
}
* {$ z9 U& x# [4 t5 ~& N, j/ ^7 S0 }. J' w
6.13
3 z! u" @4 f1 v3 `2 V$ m: L#include"math.h"/ u) d6 T9 Q9 ?
main()1 u' f. B5 q/ ^3 N
{float x0,x1,x2,fx0,fx1,fx2;
3 s; R3 j5 _/ o2 [. H do2 a5 V8 Y, x4 \ u0 [; D
{scanf("%f,%f",&x1,&x2);9 R* v- R, ~3 C( S' [
fx1=x1*((2*x1-4)*x1+3)-6;
" u+ G, Z2 G" y0 r fx2=x2*((2*x2-4)*x2+3)-6;2 r+ W! E: V; l- d# i
}
2 p; L0 [6 y8 | l7 ?% A0 d( @ while(fx1*fx2>0);
- x' L5 ~1 `8 f0 G+ [! Z0 p3 ]* R do0 l; U/ r2 C; ?
{x0=(x1+x2)/2;: ?7 W. r# L5 C2 b, n( |
fx0=x0*((2*x0-4)*x0+3)-6;, o4 B$ r# k3 I
if((fx0*fx1)<0)
* l9 ^9 c; Y( K( W$ k {x2=x0;" d) `3 v" t O' G- B. y/ w5 `
fx2=fx0;
! t: ^; ^3 e2 O6 w& ?/ k }
& d3 y }& t: L9 Z/ a0 y else
_ e3 O: t2 G, m {x1=x0; x5 t" z8 D. D- I3 v2 V
fx1=fx0;
# ^0 B5 [; s' { N }3 N; {) H2 h5 [1 A
}1 | L, B' [/ r8 w" d
while(fabs(fx0)>=1e-5);% z) J; X& ]6 f
printf("x0=%6.2f\n",x0);
/ c. \8 N# t% V8 n( I" ^! x}
* e, j0 y$ f1 `; T6.14打印图案
' S6 d. J3 k2 l4 Bmain()
6 u; W N3 }# N$ `+ e, |; U{int i,j,k; D( C7 m2 p2 Z- T
for(i=0;i<=3;i++)/ H q( Q. h- p2 q) L4 R
{for(j=0;j<=2-i;j++)
% V) o' e! u! k+ L5 \! N# I- D" x$ q printf(" ");
: K& a& l8 ^# N" B5 s/ |$ k for(k=0;k<=2*i;k++)
8 l9 D' z( H8 v( Y" a4 z. C" o( Y printf("*");- G! [# `9 Y: T- N+ d; r$ p! J6 a0 E
printf("\n");$ G& n2 ^, U* Q" w& H& o& ` l
}
. R, b; Y6 Z0 a, v1 x5 d for(i=0;i<=2;i++)3 j: c$ s( i5 K! X$ J4 g L& s2 d0 q
{for(j=0;j<=i;j++)3 _" @3 p) F/ n
printf(" ");
8 V& ]' H7 V/ c8 X7 k# r% b for(k=0;k<=4-2*i;k++)1 k5 w- ]0 Z8 n7 b4 p/ t
printf("*");0 o5 |& n! d% B+ z
printf("\n");
* K# \( h" {4 a* q1 U0 i, \ }
/ r; `- L( @/ d+ k}
' y. n% w- {. U9 b: N0 K. Y6.15乒乓比赛
5 F. d' ~+ d' Y" Wmain()
8 n( m, z4 D; t9 Q{& S" _- ]/ R9 G4 _1 S: G0 h
char i,j,k;8 {! i: {1 ]+ e) t0 W
for(i='x';i<='z';i++)2 H# Q, @) z, j# |1 s- S1 z5 x4 K. Z
for(j='x';j<='z';j++)
# a/ p. V2 i3 l2 h1 ] {$ `6 L; n* s5 M, t) X4 b U$ V
if(i!=j). q' w: R' D, Y) g8 Z
for(k='x';k<='z';k++)
/ `, x) d( d) J' L {" h6 x: Y* X" d" m9 V
if(i!=k&&j!=k)' O$ q5 ~5 x5 ]4 G# D
{if(i!='x' && k!='x' && k! ='z')$ h' W. i" }% V3 @+ t
printf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);
# w) j. }# H. w5 k9 k }
( V4 V& ] b3 C9 L }& L& \, X3 ^+ F" q* V
}
% k# o7 ?- A- a I. ?, `) |}) A. x8 R7 z7 L2 m$ j# I6 o
C语言设计谭浩强第三版的课后习题答案
/ d+ \3 j$ k& U$ I4 _7.1用筛选法求100之内的素数.
" X8 V8 `) @& |- f#include
# J- s+ `& P3 Q1 n" \* v#define N 101
- y) X3 d" l# |$ f/ rmain()
- k2 i0 _+ j/ Y. Y" Q{int i,j,line,a[N];, j# Y) y0 f, n1 s9 M
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)6 t3 K, y0 J! o4 a1 e7 k
if(a[j]%a[i]==0)8 t% Y+ _7 U8 q
a[j]=0;; Q) e: S" E0 _' E. k9 K* i
printf("\n");
& e/ A" u1 P% V0 o3 f' nfor(i=2,line=0;i{ if(a[i]!=0)
6 F, L# h$ i/ j1 b6 w {printf("%5d",a[i]);1 F# E2 D/ i, ]; M! `" T
line++;
/ s0 U7 K# [4 ]8 ~1 U( v/ G if(line==10)% [- e* Z+ v) Y, Z$ T
{printf("\n");
2 ^* n' Z+ K: F3 n. L- O line=0;}
2 @4 }% N7 F: {& Z5 s/ n } |' I' G7 D4 a( V
}
# Q& K% W, Z# r4 M4 X7.2用选择法对10个数排序.
2 `; h" C$ m; E4 `: A#define N 10
8 b3 ^2 S6 k `! k% v+ i9 l+ mmain()
& a1 L5 h8 x$ k* j{ int i,j,min,temp,a[N];5 @0 Q7 w7 D$ T
printf("请输入十个数:\n");
7 i7 p& [% w- q' x4 c# r5 bfor (i=0;i{ printf("a[%d]=",i);5 _; y# ]: T+ n7 |0 g1 p& }6 U
scanf("%d",&a[i]);
" O4 ^* X& B/ L; H$ b. s}4 U$ j' i1 T4 P7 ^$ K1 i
printf("\n");, J+ \( @4 ?+ m( i6 T) F2 Y
for(i=0;i printf("%5d",a[i]);$ C* D# j I2 W; d* b
printf("\n");
7 W" n4 B& `: Qfor (i=0;i{ min=i;- B3 I& M- i' E' E; ]
for(j=i+1;j if(a[min]>a[j]) min=j;
9 |/ ^1 i6 @' h5 @ temp=a[i];. ?2 {' w4 q" A) L* m7 }6 C
a[i]=a[min];8 o# U2 D. ?5 X1 S. I; L2 f }( t
a[min]=temp;% k2 l7 d4 G& Q0 j8 I
}
( k _) a! R, f( L. O+ Oprintf("\n排序结果如下:\n"); W' e. j z3 |& v' J
for(i=0;iprintf("%5d",a[i]);
5 e- [" z) I' D! Y}' z8 z* b- s& N$ u x
7.3对角线和:
& h; i, i( q) g& Gmain()! S- H! m! r; W- ?. w) h
{
* I& Y; N x6 @9 zfloat a[3][3],sum=0;
7 `& k: g6 w6 gint i,j;2 c! U$ j! b/ Y. o0 C3 T8 t% u
printf("请输入矩阵元素:\n");0 e& K5 S% k/ p. q0 i) C/ l
for(i=0;i<3;i++)
* y/ B0 T1 U3 _' p% C2 U* F9 `" p/ A for(j=0;j<3;j++)* l( H5 P% q. `. b) f) M
scanf("%f",&a[i][j]);6 {% z; }" Z) `: L- I" Y" C! t2 B; i. |
for(i=0;i<3;i++)5 J9 X3 L# J* E7 y3 W
sum=sum+a[i][i];
: V, x3 z) k W* ?. Z printf("对角元素之和=6.2f",sum);
, h e4 v0 j( R4 R) k* F( a& q}
' @. \7 C" z! r5 |# V) g+ ]7.4插入数据到数组
5 r, S0 Y' X7 U; }. d. Xmain(); I; U u z; G3 w7 Z# [1 [! J
{int a[11]={1,4,6,9,13,16,19,28,40,100};) t+ V7 B( i# ?+ b
int temp1,temp2,number,end,i,j; _( {( ]: N/ k- W/ G7 Q7 j0 }
printf("初始数组如下:");
: Q/ f y8 @8 m) x) u) r; h2 A& _ xfor (i=0;i<10;i++)1 m7 b; [ l$ u
printf("%5d",a[i]);8 W6 x$ g7 D9 F* Z
printf("\n");- b u8 K- d2 U7 k
printf("输入插入数据:");
/ P; @% J8 m4 H0 P& iscanf("%d",&number);5 V( R9 x2 q! g3 e& V2 x2 M
end=a[9];
& s3 H. [( w" `: c: C8 j& m2 a% O" S$ J* _if(number>end)* P# K& W" m; F" B# T' y
a[10]=number;! r8 r( F& d" d( l
else
% Z: ^# C: X* w: F* h {for(i=0;i<10;i++)
6 b0 D/ ^4 E* G- c- i { if(a[i]>number)4 B( R- d! {/ h8 r
{temp1=a[i];
# J/ _* O3 ~9 w" x% \ a[i]=number;
$ w$ A& N5 p x$ a1 R5 j2 R+ z for(j=i+1;j<11;j++)$ l; Z7 F X: I* X3 J8 ~! k v" k
{temp2=a[j];
: W% [6 ~8 ]8 A" U a[j]=temp1;
! {! \$ i5 [6 D% b1 s9 Y temp1=temp2;
- ~& R5 O6 N: L }/ [7 e9 S, y; t! b4 ^
break;: _, c9 W' b- Z; u! \/ J
}
7 j; |* T* k) `+ W }! g. o8 L$ k: }
}% @, H7 p$ `. Y2 ]5 ~8 \
for(i=0;j<11;i++)
1 O# x3 ? q9 _0 R" { printf("a%6d",a[i]);( C( S9 @ Q: [
}8 Q) P4 h1 p% B: t
7.5将一个数组逆序存放。, d2 S( z$ D/ }
#define N 5! I9 M3 ^' u6 p- p6 V5 t
main()
5 x; _. k9 m$ j: s6 ?1 x{ int a[N]={8,6,5,4,1},i,temp;
# C' _% k$ |: R0 ?printf("\n 初始数组:\n");6 g5 c$ z4 y/ Z5 g: W! Y0 A6 ]
for(i=0;iprintf("%4d",a[i]);
# c: X6 x0 u6 ` E0 Z$ D% Y$ Bfor(i=0;i{ temp=a[i];- Q+ Y6 G) Y5 `3 P: G$ b: W
a[i]=a[N-i-1];8 g7 H' V7 b1 n
a[N-i-1]=temp;$ _ ]- {# ]% l4 G4 }+ f& ]" B2 |
}
% m& @9 x1 t% L! e( _7 _& t+ z9 `: Zprintf("\n 交换后的数组:\n");; I1 Z. o* _/ V2 C8 b2 _! S
for(i=0;i printf("%4d",a[i]);
8 S) q4 M* k9 ?1 F8 g) j5 \}
+ f4 |! n# s% B$ x7.6杨辉三角
- ^2 C( i/ X( T# @#define N 11
; c2 |0 c% B9 Lmain()
- ^" M; O: \0 E/ {, v+ b{ int i,j,a[N][N];' U3 p* K8 x* V9 k% Y' d; E
for(i=1;i {a[i][i]=1;
+ O) V$ i( ~# S5 ^ a[i][1]=1;
& Y/ T' B* r* c1 Q }0 L4 t! R! _3 a L+ e
for(i=3;i for(j=2;j<=i-1;j++)
C& P6 d9 _6 X. v0 X a[i][j]=a[i01][j-1]+a[i-1][j];7 s) m+ Y( O9 u! C3 T- P
for(i=1;i { for(j=1;j<=i;j++)3 H" h/ g: g( r% U1 ?' H
printf("%6d",a[i][j];6 V$ X! m5 `. v `
printf("\n");. S4 X# d" R( t! E, ^1 [
}
, O7 |" g S4 O+ c5 Y printf("\n");
/ @# V; f# v2 j; Q6 R2 ^}
, ?& V j- Y; g, L7.8鞍点
$ P9 i; o. d& z/ o1 F#define N 10( L; h, c+ R. W
#define M 10
, p$ i U( E% `main(): S7 s1 N. G% M$ R& k3 X' }8 d
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;- u0 l( U- Q/ g1 i- P$ x) O4 m* k
printf("\n输入行数n:");
2 {/ v# f3 Z- S7 @ scanf("%d",&n);
3 w2 V0 A! J5 g( r printf("\n输入列数m:");$ j5 l$ W0 f7 r" f) B
scanf("%d",&m);( U" F# O" j) G
; n2 k( N6 w; a) |; O+ ?$ T
for(i=0;i { printf("第%d行?\n",i);
- ^- q9 q* G/ c0 H8 Y0 ]/ q4 }% x for(j=0;j scanf("%d",&a[i][j];# v4 Q$ M, w* p7 p3 P
}
W, A% ]: ~/ b P8 K for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
x. B% L S# E: `7 Y pritf("\n");# K3 O" v0 q( S2 H9 n
}
( l: q) B* v3 l flag2=0;
* W. Q. I$ ^' x" a* S: S for(i=0;i { max=a[i][0];% e) D" K, c _4 g$ _3 h3 R
for(j=0;j if(a[i][j]>max)# T7 w N' H; @$ M b
{ max=a[i][j];
* |4 B$ x9 C& w. G1 V# J maxj=j;* v5 M# p! S/ r" y( u* a7 |
}
- F2 }" U- s; I" F' Z* S6 D, W for (k=0,flag1=1;k if(max>a[k][max])# f; Q- s" |7 _
flag1=0;
$ Z7 E) T4 ^& h" T; k if(flag1)
# ~5 J1 ~# U# m9 t3 f { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);
* y6 v- C* t) l- c flag2=1;# j" y4 M% q, H" \3 F$ K! _2 x7 I
}; v M4 E' {% V; R
}9 A0 |% b- h( d5 k7 u% D
if(!flag2)6 |2 j# f# q: ^' ^
printf("\n 矩阵中无鞍点! \n");
4 {! M9 T1 F4 B7 R, c}
: W) P& {& _" |) K: W
r5 n( v: w+ i: u$ R" P7.9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.9 N7 M& z3 L2 y" c: X/ h
#include7 ?/ D0 \% F% b2 Y6 k6 S2 w
#define N 15* _2 r0 b& M$ [, _3 j
main()
4 p' ^$ u% k) O7 u/ s{ int i,j,number,top,bott,min,loca,a[N],flag;8 m9 N! i& w3 M2 y9 w2 Z. u
char c;& J8 u( N) Y" g4 J$ t# h
printf("输入15个数(a[i]>[i-1])\n);6 B: D3 Y9 ^$ x/ i5 D5 S
scanf("%d",&a[0]);: n, H9 w& X. W6 ^+ \
i=1;
+ N7 [+ T5 u$ e5 q while(i { scanf("%d",&a[i]);# P+ X; y8 K2 E! a' k
if(a[i]>=a[i-1])& ?* ^$ V2 z7 M) F$ I" z p
i++;
4 C" c- W% D1 U0 e: x, { esle! i$ e8 ]9 w1 w# p8 j- u* s
{printf("请重输入a[i]");
% F* F+ x- ?# c' d% C printf("必须大于%d\n",a[i-1]);
; ^- Q6 v+ U" [0 u4 [0 U }
( k* t4 w) w J$ p4 n8 u }" \, y. z- n* y
printf("\n");( T" W. j. k% o
for(i=0;i printf("%4d",a[i]);
# v7 D8 M6 a7 ?% U printf("\n");
- u$ A6 e3 s5 c% `4 T% E( M: O$ ]
7 |6 x5 H3 `1 I flag=1;
( _ j" Q. f+ f' ?7 P while(flag)- X1 {! x( Y9 V6 `
{" x; w5 w' u1 `* ?
printf("请输入查找数据:");4 v# {6 t6 T* t* S
scanf("%d",&number);
/ x! j& [3 e; E6 o" l loca=0;
% u9 M$ [0 Q& t8 o% n5 e top=0;
7 q. h. Q; _. [9 b' U; H* o# M bott=N-1;
+ u- R: W6 P& Q% B2 m! i if((numbera[N-1]))( j4 H1 B- G+ q. m: x+ k
loca=-1;
8 H( i9 F) h* m5 J0 T9 F while((loca==0)&&(top<=bott))% B5 N! L; f& o! j' e. z0 V
{ min=(bott+top)/2;
: g# z" ^* z# G+ q if(number==a[min])
. m# t+ C+ _* U- V/ m8 w { loca=min; |0 ]5 G+ r n, g" v! I' O0 A
printf("%d位于表中第%d个数\n",number,loca+1);7 C r! {5 N1 @4 V
}
6 ^# R. t1 k! m1 S/ c& l3 i else if(number bott=min-1;
% k9 N; t( g" d else$ }6 d/ H( A1 f2 y+ A4 i5 [
top=min+1;
: u j4 P" x, }9 x9 B& H }
- t$ b% L- E0 \7 i if(loca==0||loca==-1); f& h- ?7 Y3 {
printf("%d不在表中\n",number);
1 \2 P3 M) l I* m: C6 X printf("是否继续查找?Y/N!\n");
; D5 Q; J v! m6 R9 l: K c=getchar();
$ V9 F$ Q B' L/ I4 X4 b if(c=='N'||c=='n')
+ D* z9 A$ c: g flag=0;
# O1 P8 Q% y; j* g }
+ D. F# q# X# i, K}
9 Q( ? y8 N: ]2 W6 p" s6 v
- q" |7 K# ]3 O7 q" l& z7.10: C+ N7 O! R5 e
main()
Q, p2 T' g/ C% s. E6 F{ int i,j,uppn,lown,dign,span,othn;4 s3 Z' R5 Q; f& ]( E! Q k4 p
char text[3][80];
0 j) L! L, r- y, L- {* ^* _' ~ uppn=lown=dign=span=othn=0;
$ V( w, Q. w1 D& u2 N for(i=0;i<3;i++)8 e3 d, c. S- r. F
{ printf("\n请输入第%d行:\n",i);8 t1 o" A0 W4 M$ s
gets(text[i]);! o$ J0 G( A' @
for(j=0;j<80 && text[i][j]!='\0';j++)% z7 ~; F9 {, v) \
{if(text[i][j]>='A' && text[i][j]<='Z')
0 i; Z' s6 ^. j) |; ]5 O* V uppn+=1;
, z2 p+ U; F5 q else if(text[i][j]>='a' && text[i][j]<='z')
' ^6 L7 n* ^* S0 k5 Z, A lown+=1;
) I9 Z/ l, t- i' Y3 x) p6 S else if(text[i][j]>='1' && text[i][j]<='9')
, Y0 I/ v4 B: D8 {1 m( s dign+=1;: n* U. X: K. O5 \
else if(text[i][j]=' ')% d6 E2 }5 f7 }: ?1 s2 z
span+=1;, e% ?: U+ ]7 f9 |4 B
else
3 W4 b- b& ` A8 x5 j7 o0 p othn+=1;, I& G" w; ]( R# Q
}; {4 @0 g& ~) h" N& a3 J; Q8 z
}4 `0 X4 j9 U+ U0 p; H9 i4 ^
for(i=0;i<3;i++)& X0 g/ @9 N6 Z, I8 ^/ i4 w; _1 K n
printf("%s=n",text[i]);
9 W# X7 O |1 ]; c/ M3 }4 T printf("大写字母数:%d\n",uppn);
/ J5 J' z& H! U ^9 r* J. Z printf("小写字母数:%d\n",lown);
3 [3 _0 f. P2 w. I printf("数字个数:%d\n",dign);
4 }1 @$ E& {( ]; H8 \* S printf("空格个数:%d\n",span);& W' \! T- q6 a6 J1 B( i: s0 J5 Y
printf("其它字符:%d\n",othn);6 R- c0 C$ ~0 i% ~% i
}
. Z. `0 P8 \, [3 @3 ^5 m7 D1 R2 M
2 {) t- t( L1 h; _, ?; T
2 S$ N) ~( r' u: |. a5 R$ ]7.11
. v. _' X) W1 N* U9 Hmain()3 Q$ o% |# O- ~+ U7 S$ `
{static char a[5]={'*','*','*','*','*'};
1 B! U. A# b2 Y int i,j,k;
+ i. M$ b' C& ` char space=' ';! V7 ^* Z2 i, R, u$ m- K0 u
for(i=0;i<=5;i++)
+ I4 Q+ J1 ^, ~5 O- @0 ^$ J {printf("\n");: e" t' A! g4 b
for(j=1;j<=3*i;j++)
1 ?: a a" N" X( N7 h( o printf("%lc",space);
! p% P$ W6 r9 g0 {. b for(k=0;k<=5;k++)+ Z0 a& n3 m5 @5 x
printf("%3c",a[k];' A& @ o- l9 A; @
}4 X/ v6 Y% y* [0 `
}
7 p h. m9 w* h8 F. Z+ z$ r; w; G7.12
! D; [% `# W0 |) [2 C. z0 F#include
6 z$ O T. g7 ^main()- ~; R* @8 U: I/ `4 H/ k
{int i,n;
1 l5 w- F: o+ c) o" d. _/ r char ch[80],tran[80];
4 D6 [4 g: G! r2 s printf("请输入字符:");
: e( J' h) l+ [0 k1 Q8 j- G gets(ch);
( r& ^0 o0 d* K# K' lprintf("\n密码是%c",ch);
" _: p$ Z/ F. \) c! l7 Ji=0;
; l0 ~" j+ ^# |* y$ z3 ~while(ch[i]!='\0')
! ~4 g' W# \ C; |8 Q7 g0 @{if((ch[i]>='A')&&(ch[i]<='Z'))
8 S& J2 J( ]% } tran[i]=26+64-ch[i]+1+64;2 X. q @1 p& p$ T" S1 G; V g& T8 o
else if((ch[i]>='a')&&(ch[i]<='z'))5 G0 N7 B& ~5 C! t% W9 N0 T
tran[i]=26+96-ch[i]+1+96;
]! K7 q" e6 }else9 u" T! ]/ M7 G0 S4 D
tran[i]=ch[i];
( z8 M1 s+ G3 i2 {9 H i++;0 Z" S5 U! j/ N" N
}3 I( f" D7 e z& h' s( Z7 n! n
n=i;
* P* R( X& z% z+ qprintf("\n原文是:");4 c. M8 N( f: R2 ^2 ^' y( n
for(i=0;iputchar(tran[i]);
, K4 @0 e x1 c: v4 W4 G# c, Z}' U5 A8 b, G( I: v! k
7.131 P6 U. N. R$ \5 p
main()
! v$ C/ \# f* l, l {
; f5 j7 B' g) [; d" u char s1[80],s2[40];+ w7 m# ]. `3 `+ O& ^7 h) {& p
int i=0,j=0;
3 ]+ r* u9 A9 _' B8 V( v5 l5 V! R printf("\n请输入字符串1:");) o! f! U2 r! M) r$ D0 e% L" D
scanf("%s",s1);
6 v8 ?/ G2 c/ H" G o* j printf("\n请输入字符串2:");, E4 F1 C( n# u( H
scanf("%s",s2);
T) m: T. [+ Q; H7 ^* q T; q: q while(s1[i]!='\0')
; ?: `# u# Z4 U1 F; X$ t i++;
|; d# e" M! m) |% |while(s2[j]!='\0'), i+ d' N9 Y, [
s1[i++]=s2[j++];" A+ B/ N% f1 a. B; U! D& n" A2 e) S5 z
s1[i]='\0';' b3 a" p; j5 `: h
printf("\n连接后字符串为:%s",s1); |$ ^! `0 M1 V2 h' w) L
}
8 n: ^3 v+ o/ y3 f3 }) r4 d5 ?7 i* r( ^4 s2 x4 n4 R
& @7 X6 ^( w8 d* E1 m2 U N
7.14
0 f( p8 `* o. H& F0 p/ T" _#include' B9 t- w3 w& Y0 @2 m( m/ e
main()8 W5 m Z/ g' o6 Q
{int i,resu;
. | u/ R) H* w q char s1[100],s2[100];6 m) w( r$ ]$ U" G. {# x/ Q2 M/ J
printf("请输入字符串1:\n");" n3 K% ~3 W% X2 F7 P: }' g2 `9 E @- u$ `
gets(s1);: [/ w5 O9 j9 Q( i" n
printf("\n 请输入字符串2:\n");. a U9 Y1 v; Y! D) t+ a8 R. z
gets(s2);$ Z c- _; ~0 R3 s" M
i=0;- ~' k# D& u# q
while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;
6 e/ c; q5 z% f+ y. s if(s1[i]=='\0' && s2[i]=='\0')resu=0;7 F9 o7 z5 O+ a( @; w
else5 f6 s q6 ]1 p+ D+ M; i! t/ J: |
resu=s1[i]-s2[i];
" O: }" p5 p2 o( ]5 j1 k printf(" %s与%s比较结果是%d",s1,s2,resu);
2 X) [9 U4 k* ?& d2 w}
. Q8 q2 x% C0 G1 O) D: P7.15
; s6 k4 r& N: ~- s/ p#include
; ^/ a+ @- r- y6 W" w6 @6 q1 T1 Cmain()2 o- B9 q% @1 M; r8 n+ E6 p ]
{5 j8 V% X0 L* ?. T! Z, d( B
char from[80],to[80];
) E S- b2 O& o3 {: n! P int i;# ]5 s8 T- Z D4 r7 k
printf("请输入字符串");+ ^3 }- [0 e' s1 G. Q
scanf("%s",from);% N$ T6 U7 X, }% s, V+ M _3 Q' Y
for(i=0;i<=strlen(from);i++): D9 ^0 r0 e" a
to[i]=from[i];$ Y) |' ?+ |$ v
printf("复制字符串为:%s\n",to);
- B$ x& }9 W& c- R8 x2 v }% S, H& ~7 t4 d2 J
( y# B+ d/ R) z6 S
. B/ n6 J; y/ T0 i第八章 函数
6 u G' a. f# [& N, p8.1(最小公倍数=u*v/最大公约数.)
2 T. N; i$ f) e7 M/ d5 u4 Q3 dhcf(u,v)
0 N9 \, l3 \% K) h4 v6 c- i5 B0 nint u,v;
7 v) Y% u4 p" ~! }(int a,b,t,r;
n8 l* S4 i" m" \3 k if(u>v)- I5 f3 p/ W' \# ^: e
{t=u;u=v;v=t;}+ J5 p( c* T# Z' U
a=u;b=v;
) w1 U4 P+ {* { i/ _$ k# Y9 F while((r=b%a)!=0)6 f7 I. [' J( I; i g2 y
{b=a;a=r;}6 T- S8 `2 }1 X" u
return(a);0 }" l: Z: L5 J6 `2 @
}, d: H) F( G9 `: T$ I, p" j! H) X# A
lcd(u,v,h)
3 |) G$ d( r; ?1 }- y) c- M" r int u,v,h;
5 ^ L& ?! q: ?# r2 y% j+ u {int u,v,h,l;& q8 P4 @2 ?3 p5 I
scanf("%d,%d",&u,&v); q5 b W6 F( ]& d
h=hcf(u,v);
3 w% m4 Y1 Z9 a; T! F# U printf("H.C.F=%d\n",h);4 @+ W& E5 E8 H) p- Y
l=lcd(u,v,h);
7 W a# C2 c# P( n printf("L.C.d=%d\n",l);
0 g8 u! V( ]! L$ i }
; T& z$ I$ Z% t: U {return(u*v/h);}# J. k( p1 U2 c
main()
7 c# ^$ r* c5 ?& t0 f {int u,v,h,l;# }" Y8 I% }5 E$ L o0 P- ^
scanf("%d,%d",&u,&v);- N; ]7 {' F/ d4 A8 }: n
h=hcf(u,v);" ?1 y" b# e8 L) K, a
printf("H.C.F=%d\n",h);/ N; H2 n5 ?0 X! G& O, P
l=lcd(u,v,h);% ]1 t% T4 }+ G
printf("L.C.D=%d\n",l);3 U8 x5 B3 ~+ U0 ]
}
2 \+ I) x9 C7 ^2 n; J; L( X* n
3 v* u p2 i. k2 B9 m$ w+ j M$ v
$ g: D/ R+ p! ^+ f4 [1 C' s" u/ a0 H
8.2求方程根& I F" `4 Y% X' a
#include
6 u4 a! i( c3 g1 k9 Kfloat x1,x2,disc,p,q;
3 _/ P: h* l& W, n! kgreater_than_zero(a,b)
+ D, L; I) a' w) _float a,b;5 s0 |$ c8 T$ O7 I, ]# P+ R1 ]
{& P7 p# c5 E+ w) m3 u: Y; u& e; h. W
x1=(-b+sqrt(disc))/(2*a);
* X1 V* Y1 w9 e1 ~$ M% f! K$ }x2=(-b-sqrt(disc))/(2*a);0 |8 H. ~/ A1 W: Y/ q" f
}$ \8 v8 c# R; w
equal_to_zero(a,b)
8 j! r! X6 j% o, E X* a$ bfloat a,b;
/ h( \. p+ A/ N" f# ?{x1=x2=(-b)/(2*a);}
8 L$ ~- V( g- C% e" @smaller_than_zero(a,b)
1 ?% H! Z( [+ c! Z' P9 x3 O' Efloat a,b;) y! \6 C6 q8 l) Y& s4 T
{p=-b/(2*a);
' F- ]6 y8 c, t! c% i: @: a; X- Y rq=sqrt(disc)/(2*a);
3 J. @ A' f; ~% B}* n! a6 s( }; Y- P. k R( d2 Y3 Q" S. R
main()& N7 n( _! r7 c# ^- _
{' k1 r( G' b5 p+ `
float a,b,c;8 S8 o7 T, K% Q7 t( T, g
printf("\n输入方程的系数a,b,c:\n");3 K' }& ~! ]/ D) }" v
scanf("%f,%f,%f",&a,&b,&c);
! u( q$ P, P! Pprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);
/ H2 Z# B: m" T b0 i7 ~- Vdisc=b*b-4*a*c;
" e) _! l! p: m7 m. U6 ~8 Xprintf("方程的解是:\n");
9 G) }0 Q# s6 V% f1 }% M3 nif(disc>0)
& ^, S: k1 n$ E& D }. H{great_than_zero(a,b);4 O$ p! c; [0 E" v7 y; S# s
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);& W4 M! e" a- @- ]/ y0 n
}2 P9 U' T% Z. F/ _+ a) Q/ m# D1 R
else if(disc==0); H) }5 ^/ ], y5 Q0 q
{- C, Q+ L. V$ G' }
zero(a,b);! R: W4 n3 ]1 ?8 R9 d) w
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);% e. _/ e% }9 E' W
}
$ D- i/ G& f6 u7 N: uelse
! v6 B+ K2 P( P6 N {
5 u; ]) C. b5 x small_than_zero(a,b,c);
/ b/ ]9 U7 n d$ x4 V- I/ E6 C6 c printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);
( D+ x! k3 x# m4 E" S8 w) E }
5 e3 `& n$ b/ Y+ U" ]- Z' a" t$ k: d3 l}
% W! V6 c6 |$ V1 I9 F8.3素数$ }0 U0 l4 A& l. z$ @' `# h
#include"math.h"" Q- D1 O: ]4 H |0 H
main()
+ G0 |. D: o- X$ ^4 h{int number;% j; S- e5 }# q
scanf("%d",&number);6 Q, |9 ?* h( P2 `6 _7 Y9 m
if(prime(number))5 `$ |) N! H# U
printf("yes");+ t( J8 W$ r& l+ B
else
. k; T$ X3 E1 t8 R0 { printf("no");/ P9 N c8 Z: F; ? O$ K
}2 W0 @# k+ @+ u( q }5 \
int prime(number): Y( M, e, V$ Z, n
int number;2 L/ x" c g. D, C* L: a% {3 G
{int flag=1,n;
% A# c( } w/ [3 L. ]& T for(n=2;n if(number%n==0)
5 B" b8 d/ g: t. R flag=0;
5 x3 d: g: ]+ X' d8 f4 M) O# w return(flag);1 _; S6 v3 p9 H7 ?1 {( p! s7 W/ m& P
}) S% R6 W1 P2 A: K9 F. u2 a6 b3 Z, W
( c( k+ C1 u) q. n' A$ l7 l; m
# _( q. f* p* b4 o% a" G8.4
9 T& h3 m: u, m" H1 R$ R#define N 3/ e3 U ?- j I p7 O; p. r
int array[N][N];7 E6 M: D% W4 D0 V; I2 |; e
convert(array)
9 k9 ?! m8 n% f! t! i2 ]int array[3][3];7 _! H8 A: }7 A- p( q
{ int i,j,t;
- h; a- V9 g0 e% g/ w for(i=0;i for(j=i+1;j { t=array[i][j];
2 A5 W" g% J) E* \ array[i][j]=array[j][i];- p' Q0 ^; m( L8 `2 w7 \) C6 K
array[j][i]=t;% ]* G' l1 d- U( ^" o
}/ l- D+ K+ m& p- l9 J% D; A; B
}
c9 B% d' E% b# A9 o6 umain()) t' q. n, U+ x, A: l9 u8 J
{
1 W1 i8 p& D/ n2 P int i,j;
3 G* {- R. A7 [) g printf("输入数组元素:\n");4 b+ l. O6 \5 \) C8 }: I
for(i=0;i for(j=0;j scanf("%d",&array[i][j];
4 N$ l/ F' O' B) q! }/ y! Y- _ printf("\n数组是:\n");
! ?: L- j) u3 wfor(i=0;i { for(j=0;j printf("%5d",array[i][j]);
6 M- |3 v% s9 r/ A9 J" S printf("\n");
4 ~2 F. n* {. x, u }9 Z0 i& o- u$ s
convert(array);1 q9 u" M7 i$ |7 P' O. V
printf("转置数组是:\n");' f* f" ?' \1 ^8 v6 Q
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);- q9 @+ D& X% W, n
printf("\n");
* i; s+ `, Q7 S% M a: | }$ _8 V6 {3 A( S7 M9 Y) L+ \/ J
}
- O/ @' P& z7 y+ u9 J" L7 E G3 O. R! ?# ^
( e$ S0 B2 O; U3 h/ `: Q0 l5 o
; A0 ^5 Q* k' C/ \* o8.5. m; h+ E# A- n$ O
main(); M7 g( b! w6 s4 h0 X
{9 y& s3 w9 B5 B2 z3 w( s$ ]
char str[100];
, C* c8 q$ R) D7 Q2 E3 j printf("输入字符串:\n");
M" P- E( Y+ Z* E0 V& ` b scanf("%s",str);
: j0 V5 @/ S4 T# y9 s inverse(str);" N; N% W- f' v5 P, a$ N, O* e4 W
printf("转换后的字符串是: %s\n",str);! N/ g* g9 V" `- `) L" E+ {
}7 [2 a1 I" E; X( z5 ~" r
inverse(str)
: _4 `4 @! v4 D- V* ?char str[];/ h: _4 `6 ~$ _0 w* J
{& j: A( V% w2 }3 d3 H# a4 s
char t;& N1 p- H2 B3 g. D
int i,j;
7 ~1 ~% [8 H& w, D' J% Y( s' y for(i=0,j=strlen(str);i {( }% D( j' |3 x4 k7 W- d+ e4 l
t=str[i];
* D" ?" Q! W1 Y2 i, C str[i]=str[i-1];1 S* T9 r% X7 P& b1 _
str[i-1]=t;
3 {+ r( d6 N$ N; y# a8 W }
3 F) w7 y2 {$ n3 m, O! H}* S b* m9 W" y0 {
( ], C$ W# O' r3 A. k# F
9 ^ V' z% k/ @* c5 h' ]
3 C" w: ~/ N$ ^) o8.6
6 K* I( c+ Q5 @" uchar concatenate(string1,string2,string);
4 V9 G: K! |. p# I4 T+ C& wchar string1[],string2[],string[];0 h2 a8 c% m8 M. e1 K8 y
{
( u+ R- U9 B) n, ?* tint i,j;0 E! g" M" t7 C" a
for(i=0;string1[i]!='\0';i++)
T1 e8 \1 M3 A7 @9 p4 L# d7 A f2 G string[i]=string1[i];
: f% f' Y' P% b, ^7 ?for(j=0;string2[j]!='\0';j++)
! D a1 k1 R4 K: B6 N string[i+j]=string2[j];
) V% n' N6 Y! L9 t2 a- U. @3 M6 K string[i+j]='\0';" G$ k6 H* V1 q& a/ l
}
4 n7 d0 Y+ M7 S5 Dmain()
1 F) B! e1 g4 |+ X8 j{
& ]1 z( Z1 `5 i3 U7 H( C char s1[100],s2[100],s[100];# |& p4 {7 G. n( C& H9 T" V
printf("\n输入字符串1:\n");* S0 j' [. p4 K; P. [% G
scanf("%s",s1);
/ ^# m3 J4 w0 [% c9 y, G6 E printf("输入字符串2:\n");
9 h+ J8 a" F7 U: ?6 L. f scanf("%s",s2);; y, ^ o6 J+ T9 e# E- u
concatenate(s1,s2,s);6 M# e0 l, v& l1 U4 p
printf("连接后的字符串:%s\n",s);
% M) C1 L9 J5 _+ U* A}
[! A. H/ p% F3 {: Z1 ]+ E; c a1 ?$ Q: ]- i1 Z, Z* I! }
) @# a/ j8 J [ g8.8
" r! d" s1 u' }, t# a$ B% v$ Mmain()) I; |8 x/ c6 k* x
{ f+ u7 x+ r% W0 R! J$ u V
char str[80];
& [/ v3 X" x* p( G printf("请输入含有四个数字的字符串:\n");+ w* c3 J1 I0 k+ T' N" h
scanf("%s",str);
# W% k: a4 a( t3 K2 U4 \6 ] insert(str); m `! j& E1 `, g; ]+ F8 I6 _1 i
}! V$ X; }/ z! k" Q2 [( W8 x9 c
insert(str)
G/ C: _5 d1 y' F9 l$ \ char str[];" v4 R6 [/ E6 D1 u. K* m
{& ^# `. k" N W( P; C% K# ?
int i;
8 H0 D5 A- t4 d- n4 ?4 D6 I. D( `! } for(i=strlen(str);i>0;i--)
! ~6 J4 [1 L: |$ v+ W { str[2*i]=str[i];2 J' o7 Y) Z( d8 R% X$ I
str[2*i-1]=' ';
( A$ `& N* L3 _: \ }
# C) R: k' r$ N3 G printf("\n 结果是:\n %s",str);
: b1 t5 l6 `, G }4 y k" Q0 Y$ V9 g3 Q
; Y# u. G. f# G' j3 d0 R- R
! b6 o" l- z+ f6 K1 c3 @) z
1 a( g. O4 l/ t( R& e2 X8.9) |- Q& d. u2 k! `( B
#include"math.h"
+ d7 X/ H& X' y& [$ {) lint alph,digit,space,others;( P5 W& t4 r& Y) y+ U2 M
main()2 Z- b4 J5 p5 L
{char text[80];
4 p7 r, d+ z+ A, y/ r% T; U gets(text);
! w+ J8 Q2 }2 n alph=0,digit=0,space=0,others=0;5 {4 V8 d* x$ l
count(text);4 T4 Z1 U2 w% \0 V1 m4 z
printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);1 S% h8 o9 s2 T% }
}! t4 X- ^& y0 q) C
count(str)- X& J j* m% f( b
char str[];
. c% `8 ?# a8 e- ^; Z1 a{int i;9 y( t& D" [# P5 Q5 ? R0 c$ D
for(i=0;str[i]!='\0';i++)7 h2 t2 h! M4 R: ^; p
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
; `* e8 c) b- z alph++;
, D: c6 @$ A' N ^1 c. I) e4 s else if(str[i]>='0'&&str[i]<='9')
8 w9 {/ R! h8 o. D& K digit++;
7 r, e- p' y5 T+ O. T0 d1 @ else if(strcmp(str[i],' ')==0)$ a7 \& m, g! {
space++;3 U! Y2 Z. y5 U+ C+ Z$ v
else
3 I% k# ?% f& X. s0 v7 I. A( G3 G others++;
' x7 ^0 S6 l, }/ C+ s" w}
% _! h) L1 L* x. b' T# K
" W6 x3 c- {0 G0 l
3 O4 B3 Q# b. E/ X) L! i5 j8.10
% t3 E: }2 V9 { g6 eint alphabetic(c);
# f2 K4 i( H; `! U8 cchar c;
; Z+ A. g E& k# [{; h4 {' b( E" b5 i ?5 B% \
if((c>='a' && c<='z'||(c>='A' && c<='Z'))
, x: n& b$ F8 u3 O# \ return(1);
! G _; l: l; k else
, L2 c$ z; H" N return(0);
* _/ J, R% j1 w. b}4 O- [8 U/ }$ p
- F6 V2 ^8 w) c) H2 ?int longest (string)
% x9 Z& g- ^% n4 S" D) Ichar string[];8 p @9 w1 F8 z1 a
{
7 L' J% T& a* F4 @5 H int len=0,i,length=0,flag=1,place,point;, a' S6 E/ n) D& D& q: O
for(i=0;i<=strlen(string);i++)
5 T3 P c9 \3 A+ v if(alphabctic(string[i]))
5 b& v# P. U' y- X1 }! R if(flag)
! b; v( s5 [; ^/ @ {
2 J. k2 A* Z) R) ^( [: `3 ]$ t7 K point=i;
: B c9 Y7 f4 W" x5 ] flag=0;
8 w( }9 z1 @) L9 Y3 | }
0 [7 C7 }6 k8 l) f else
; y% `6 S1 a: J len++;
" U$ C! ?9 N5 }: L/ B else
4 ~! [ O/ |2 L9 E { flag=1;* R' K+ C; X* A# N% E% q
if len>length)
. G. ~3 K+ G6 }1 t' } {length=len;& Y2 u3 t' x( W1 {2 w) v
place=point;
! K* l1 N! q% t u len=0;+ [% Q" m- d; X- A( a" W
}
- U% }5 f: K: R3 g. ^ }
6 Z, T/ G3 y9 d1 k9 B% D return(place);
5 A) w4 X: Q8 ?2 X7 H4 a/ P }
- A3 u% j( y2 ^ wmain()
# F' Z) {6 _6 H9 j{
; x' y1 B- z' w0 a5 }' \int i;
n0 k9 }7 D7 ^3 v' _5 o- c* e' h/ z* vchar line[100];% r9 d: p; H1 [' w
printf("输入一行文本\n");+ |* f: ]. [7 o' J: P' T
gets(line);! x' U& n( ?/ x @# f: q6 X h
printf("\n最长的单词是:");
]" z5 f" J4 P: e3 Hfor(i=longest(line);alphabctic(line[i]);i++)
$ k& L) g- C# f printf("%c",line[i];
" s; X9 E" p( s: k. cprintf("\n");
7 u$ r" C5 j/ A; y( x- Z: ?" @, a* R}
+ d9 U' Y0 M- W' t' F" V
% }% S0 r, s/ Y1 V/ }/ Q9 S2 ^0 s9 \: c& s1 s9 k- u" @3 X- g
2 c/ y- }' Z" J+ W; W6 r9 }2 m
8.11
& E- \; _6 X! n s/ d. ^% P) _#include
1 e' m9 Z( l0 P% V+ v. u! q+ F) U: @+ O! Z" n
#define N 10
) e" a: }1 i4 s6 A6 o0 {) qchar str[N];$ G2 n) m; h, Y" g% s+ f
main(); I& T. S" G1 Y! K1 p) a5 W7 E: A
{ q7 j6 J S0 V+ ]( q2 l
int i,flag;
2 ~9 X4 M, y! m: Z( O8 t5 @* mfor(flag=1;flag==1;)- A9 ^. ~6 _# _$ n) t1 H
{
9 D: G$ c# p$ `( M3 t* E6 | printf("\n输入字符串,长度为10:\n");
* A% I5 ^7 \% O# k* P scanf("%s",&str);
8 T& v9 u# ~0 o if(strlen(str)>N)+ W* j0 {, F. x6 o4 \$ o* ? O
printf("超过长度,请重输!");$ Z' z, K8 ~& m0 R- _, R2 L
else
5 `. o" W( a& U* ~; n% R/ s flag=0;# t7 y- y8 Z, J8 z* ~
}1 w2 v; \7 Q6 z: M' M; F
sort(str);4 \% [9 w' X9 _* _2 f
printf("\n 排序结果:");/ w. V, S) e4 f! l5 D4 L* o6 @
for(i=0;i printf("%c",str[i]);1 n0 C- l3 N. B0 N
}1 n! M+ [! o, Q8 K* Q$ a0 u4 l- H
sort(str)8 Y7 N; C! n7 K9 @
char str[N];" k0 ?" Z$ W& u* w L8 d
{7 l5 v! G" d. r. P# `
int i,j;
' @' O% i; D; @: q3 z& q& }6 V. ichar t;; i% {# v6 i2 y7 ` v5 L
for(j=1;j for(i=0;(i if(str[i]>str[i+1])& _: B" j5 p/ `5 g' K
{ t=str[i];
0 ? R" n* r2 t$ R str[i]=str[i+1];
' F, a" n( B" F7 c str[i+1]=t;
2 O$ ~6 y1 n8 ]6 r, V }- y1 T) \3 u' t ]& H# m0 p; P, z
}( a+ J$ P# b5 H7 Z V. q" q' Y
8.12( ~6 Z5 y, @% m; Z# F" h( U
#include
) b! ?' d8 }5 a6 }7 R#include
4 o: t4 C2 W2 C& [4 k- gfloat solut(a,b,c,d), c% I6 }! L. ?; M. Q- [
float a,b,c,d;: X$ l, {# _* l+ g( @
{float x=1,x0,f,f1;. D+ `3 z9 \% S. k/ K
do
; F6 L6 ^6 v0 z7 _ {x0=x;& ^' w$ B1 P. i) p8 Q8 P
f=((a*x0+b)*x0+c)*x0+d;
* u3 [; f: V7 t/ N5 d( ]; r0 I f1=(3*a*x0+2*b)*x0+c;7 i/ a8 f* b7 r% H
x=x0-f/f1;
+ u1 ?- N4 D6 Y2 Z" V: m8 W }
& ^. p' X( E1 X9 F+ R while(fabs(x-x0)>=1e-5);2 n# E' R' N* A6 R. Y; s
return(x);
: d. A5 G# p3 f( U5 l1 Q, \+ I}
( L# p0 s& T6 X) u. t6 ]9 g5 Emain()
6 S, D# \% d, ~{float a,b,c,d;
& v# Q, h+ m4 t scanf("%f,%f,%f,%f",&a,&b,&c,&d);
0 X! f0 B: c. J- u$ J printf("x=%10.7f\n",solut(a,b,c,d));
# T* @3 J; g7 u9 ?2 b}
0 `- @6 Q9 m2 j8.137 o* f% h: k+ q5 o3 c# x
#include9 {1 t5 Y* q% Z/ W7 h3 M2 s8 i! W
main()( \) o0 T' Y3 ~0 y
{int x,n;
& `6 V, y% c0 C$ Y' P6 @ float p();5 `0 J- @7 O- w6 g5 O
scanf("%d,%d",&n,&x);( l$ a' C, x5 c' s4 Y" n' G
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));; G; a6 L# ^% D& _# B* N
}
; W, R8 B* n' F; Z9 K) B7 O# y1 f9 jfloat p(tn,tx)6 ]& z l/ Z1 L, ]& y! q8 j& c
int tn,tx;& r$ h$ w# ], E6 m& H% i
{if(tn==0)9 d2 M. g6 U/ ]$ m- r7 m$ R
return(1);
8 G4 E) T& W! W1 U: G else if(tn==1)4 i G! a# F f1 w4 D, e
return(tx);" ~! Q3 \! N+ d9 \7 m
else
! z$ t6 m3 Y9 ~% w; c return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
, m8 _# R k& A) c# G& S! _+ {}* H3 _0 _# P8 X) S* b
8.14
3 s! ], Y' `3 |6 c#include "stdio.h"8 J: l; W ]) V; l( M
#define N 10
5 ^# T+ @+ H% r: v' P#define M 50 B9 t+ ?0 P' S% J) z8 ]
float score[N][M];
/ ~4 U2 R( ~/ J+ Qfloat a_stu[N],a_cor[M];
4 a: P6 W3 b- i# Omain(), p2 a9 @1 E! C
{int i,j,r,c;8 B6 h* f/ Y0 `
float h;; E# j% q, U5 U8 I" b& B: ?
float s_diff();* j, I: T' D3 w j/ u3 w" ]1 |4 _
float highest();
3 ~: p# M" y( L1 ]0 A: \& o& i1 O r=0;- v3 g8 E# R) I# V
c=1;
1 e# D1 Q9 V8 [7 m4 d input_stu();
7 Q. w6 I+ o$ @2 c' @3 U6 F5 F0 w& @ avr_stu();
) Z9 h% p% w7 R0 [( j avr_cor();
1 A& |0 |9 g: Q printf("\n number class 1 2 3 4 5 avr");2 h" t( H- U. C0 d$ q& Z
for(i=0;i {printf("\nNO%2d",i+1);" u& o% e! `+ w* g
for(j=0;j printf("%8.2f",score[i][j]);
2 I1 o4 E2 x: A) e printf("%8.2f",a_stu[i]);( w4 h0 E- p( W; J- [
}5 {! K+ v& a" F' \2 i8 ~5 z
printf("\nclassavr");
. D2 I9 L x. G8 ~8 `) _ for(j=0;j printf("%8.2f",a_cor[j]);
) o9 @: i- S7 \" f h=highest(&r,&c);
* L" s& V% H1 X! }7 M printf("\n\n%8.2f %d %d\n",h,r,c);0 a7 V' q$ L! v, n2 X8 ^
printf("\n %8.2f\n",s_diff());
. L. d7 ^3 j& C" _5 A6 X9 E' t}8 P k! Q* e; V7 f- b- Q& x M
input_stu(): Q0 M+ F" x5 z5 W; ^8 _; x0 E" W
{int i,j;: ?) s) }; P6 h
float x;' P$ p3 c Y* c% R
for(i=0;i {for(j=0;j {scanf("%f",&x);
% L* j o! |1 r' d* J1 }% Q7 T score[i][j]=x;
0 D) o4 J$ T) r) j }! u. @8 f$ \" E; W# E$ d: m% D* ]
}# j$ U+ H. U, v+ K
}4 S- X/ `4 B# P1 ~2 y
avr_stu()
5 C7 J: C; T9 V& `; W{int i,j;# k, [2 a Z8 W q8 b- p) p
float s;: z7 {5 S" l0 b+ _: L
for(i=0;i {for(j=0,s=0;j s+=score[i][j];$ X ]/ _* ^' P+ ~
a_stu[i]=s/5.0;9 E: r( o; M- R
}/ \: P5 `7 K7 D+ P9 e/ ^' ~2 B
}3 E) E6 O1 D1 w% @6 _: m' ]" v
avr_cor()6 b( b0 ^. v, y& j n: y
{int i,j;0 E- K6 z! q1 j9 X; \! _
float s; ~* N6 K. z* h5 C. O' i' E4 h4 I
for(j=0;j {for(i=0,s=0;i s+=score[i][j];
" R5 T* ]0 n6 G3 |" e a_cor[j]=s/(float)N;
3 S3 t& @3 \( q! [ T4 J! ~ }* @1 P1 m7 K [, R% B; n4 m7 f
}
7 V' p6 ?8 P1 O7 @* |float highest(r,c)( m, h0 {$ f+ J. o
int *r,*c;
, c, y! x& i9 P/ Y* s3 s; U{float high;8 B8 Q3 Q! N. a& i: L/ S
int i,j;
0 B6 r7 }) D. \- }8 x high=score[0][0];
- G; x5 A5 F3 k2 s2 a% b+ m% i1 { for(i=0;i for(j=0;j if(score[i][j]>high)
9 Z$ _/ n# l% ?; T, I3 L$ l {high=score[i][j];& _2 _6 y) [/ g2 a0 B# {6 u' `
*r=i+1;
2 F- o0 O {- V. ] *c=j+1;" [- Y2 [- o. q+ P# g
}- B2 ^8 Y8 K/ p0 K' U" ?7 A. ?
return(high);; {9 X/ O* ]& w
}- g0 I$ P- ]3 ]. w
float s_diff()
5 W1 `8 G) ^, o% x{int i,j;) s% l) i/ a9 I7 E7 S( w R% _* \
float sumx=0.0,sumxn=0.0;
$ r- B, X. [# h8 B, X: k for(i=0;i {sumx+=a_stu[i]*a_stu[i];
s7 I( \8 l: m sumxn+=a_stu[i];8 v' ]* L9 k R, i
}
( r; X0 |3 F6 ?! Q0 C0 R return(sumx/N-(sumxn/N)*(sumxn/N));6 S- M5 G6 Q. |4 O K7 I
}* E" d* a4 G" F% Y$ W
8.15
/ x8 A& U6 D2 M$ m" x6 D#include
6 W; J8 t( q1 b, k }5 W( V#define N 10
& f3 g2 m: B) {/ K5 o9 Dvoid input_e(num,name)
& |: o y S, bint num[];& ^- J. T% U2 L) G2 U( N
char name[N][8];& w4 n: s2 |$ k8 W" G/ s
{int i;4 s) o; f L* I* n# p6 m
for(i=0;i {scanf("%d",&num[i]);$ H: I- k$ p' M* r, M3 b1 D/ w8 _
gets(name[i]);+ e% ?! |. j7 W
}* ^2 _, {! B( ?; z. A
}
) r; ~) d% [% F/ v+ rvoid sort(num,name)
' L e, } `) D$ Rint num[];
# f) a# t N! v6 b Zchar name[N][8];
* U2 K" M/ @. m7 p+ l$ P{int i,j,min,temp1;
5 R _' ^% V. L# E char temp2[8];5 R0 z& |( o5 b4 P, h0 _
for(i=0;i {min=i;/ d `& k3 g* R2 I. N7 N/ ~5 |* j1 l
for(j=i;j if(num[min]>num[j])min=j;5 W( h/ ~5 U; y C4 y6 E
temp1=num[i];% F( n8 R( }* T0 u; f# C
num[i]=num[min];
8 v# z! h, ~0 p% d5 {& o num[min]=temp1;
7 U& ^4 e( o: v) g# _0 | strcpy(temp2,name[i]);3 m M! k" M Q; k9 k8 B2 b* e
strcpy(name[i],name[min]);3 u8 R; U! M2 P, o
strcpy(name[min],temp2);
- v: x+ f, L0 O. ]6 M6 ] }) k2 {6 C# n: b: h
for(i=0;i printf("\n%5d%10s",num[i],name[i]);
4 `8 K/ Y' T! X; z}
6 m, `% u% t6 Z1 u. cvoid search(n,num,name)
. z" V4 J/ _* k: m% Y aint n,num[];" s8 X1 C! D+ ~9 ^; J( b
char name[N][8];
8 C/ l9 t: c, L% H# Q, X{int top,bott,min,loca;1 w" J& b9 |& U' q" t, \% @
loca=0;1 t* h* T6 B$ Q8 X/ t0 b
top=0;
% k% M; ?8 \6 p+ W6 E" @( _1 ] bott=N-1;1 f3 F1 p4 ?3 G, ~! U4 U3 e
if((nnum[N-1]))
( ]# K* R1 Z( k b5 d& V loca=-1;
. ]; u w/ [+ }3 q! n while((loca==0)&&(top<=bott))
) f. w' q% y. o. H3 T. }% i T {min=(bott+top)/2;' k5 J# A7 [( m9 w$ ], z
if(n==num[min])- ~- E+ J$ ?1 d2 N
{loca=min;6 K7 {! [2 J9 _
printf("number=%d,name=%s\n",n,name[loca]);# V; B m: B" c2 w
}
+ o. T# }. \6 s/ P) [" l1 G& ^ else if(n bott=min-1;$ G4 X( \6 Q# K
else
" `) q |3 d4 E# P, g, h3 Q top=min+1;
& V' H1 k# w* O6 | }
9 e/ f6 O0 r6 h# T k# L. Y# ^+ m% T if(loca==0||loca==-1)
- G) X* A) \7 a; H# X7 l! g$ A) X printf("number=%d is not in table\n",n);3 G( E: t5 I. ~2 G0 U1 `
}$ Q3 V; v. c% b
main()
6 U# \! L3 a- p$ ~$ C{int num[N],number,flag,c,n;' T9 v( J3 p7 }, R0 y1 T
char name[N][8];
5 `0 J+ {0 J- l" W input_e(num,name);
9 P$ v$ `! `" c; J: z# n# T sort(num,name);1 V8 s- x! r9 p, F x+ t& @ R
for(flag=1;flag;). ? L( E) X1 e3 [) Z4 `- e( ~
{scanf("%d",&number);
7 A* M# Y7 X! `: Z" a9 _- M$ k! E+ y1 g search(number,num,name);
7 D" D8 k( ]9 }4 A9 D( _ printf("continue?Y/N!");1 w7 X$ [& w6 s' Y
c=getchar();" S$ P- D4 n& }) i: k* D
if(c=='N'||c=='n')( a: q3 m; O g0 D8 K# I
flag=0;
7 A$ i6 @4 d% b7 x. L, w) Z e }: Z7 s- Q7 E: G$ Z/ n( t8 [0 R
}
^* Y1 Z6 s; ]0 k+ D4 K
/ ?6 }- K" W5 E* ~8.16
+ V i; o4 ^9 H6 @ q- d( c#include7 a! h% q9 k( O
#define MAX 1000: @( Q8 i: f9 Y& C& Y
main() m3 N' p4 z. c3 r. A
{ int c,i,flag,flag1;
$ y# y) b" p; M3 \0 X( Y char t[MAX];9 d+ B, G2 _. B' e4 Z1 M* ~' u
i=0;
/ X+ o5 Q5 U1 r* B7 s1 K flag=0;
' b. j* }8 G4 f. k6 ? flag1=1;' |9 ^9 Q: l4 k( A
printf("\n输入十六进制数:");
8 }4 y: p2 p0 a3 q2 d3 P while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')" k4 T) F6 A0 S$ k& r
{flag=1;
) n+ L; g( d# Q t[i++]=c;
: i. G9 M) L# M }- L4 g* ~0 I, y0 e4 Q
else if(flag)
9 r' F5 j1 y& O" L4 Y& r& U {) g6 }: T+ d6 |! |
t[i]='\0';
: x% X r8 c5 J$ k& ]: l' r printf("\n 十进制数%d\n",htoi(t));3 Y" B4 E( ^5 i7 a% ~
printf("继续吗?");+ h* `% ?- Q1 Z" m
c=getchar();
/ O2 a- i, v: o# \! r$ D if(c=='N'||c=='n')
4 a3 _) u- m! O+ ?9 O flag1=0;
( |, i! F2 \& ^ p else+ @6 T9 d% D* j- I; b9 e Q% g( g
{flag=0;
! }2 p. b/ I: d% F; k! [, P+ j8 x i=0;
! G6 ?! O( E& k# f- h printf("\n 输入十六进制数:");: @/ q7 O3 A8 ^. ~. x( D# O
}9 I: Z6 M3 g# I
}
8 q, t \2 `+ k& m" f6 n}% a; |! k- g5 V) G( Q7 A* K
}+ ^, y( I3 S1 m4 s' J
htoi(s)
' `; q5 A) I6 {+ k2 K8 W, d5 zchar s[];
: U3 W1 F" i5 q. ^, M3 L$ G{ int i,n;
+ j" [, }$ ?. c5 I; V n=0;8 U: t( s2 I- d2 ~0 U" _$ v! P. v
for(i=0;s[i]!='\0';i++)
S- S; ]2 Q, s* o {if(s[i]>='0'&&s[i]<='9') `# \5 l5 `/ @# @# ]0 T7 s
n=n*16+s[i]-'0';
$ h; H) j% A H' p- R if(s[i]>='a'&&s[i]<='f')
3 f0 K* ]' U L% L v n=n*16+s[i]-'a'+10;
& Q. Q/ P6 G q4 m if(s[i]>='A'&&s[i]<='F')3 S# u9 {- p) s2 J
n=n*16+s[i]-'A'+10;
+ K: m! `8 q( A a, L }
2 f9 }1 O) n* W" g6 `1 m6 M0 N: n return(n);5 d% W; r$ t* L1 m% |
}
6 M. O' c0 g) x5 G# R' ^+ M
2 h0 ]# H. o; _! p8 C$ Z5 z( t0 n, T% z J, E
1 [( v0 M- L* l6 Z
8.17
g. L* b/ ]! i#include9 N0 S; `" t. l U+ r" ~
void counvert(n)& Y- \7 t+ R& \ |
int n;: }/ r$ p( I% f! C2 ^8 K3 D4 J
{ int i;- c8 |9 B( c. Y: @$ r
if((i=n/10)!=0)4 b: S6 r* O2 L
convert(i);
! }5 I( e! d; M4 G5 Y5 y7 r putchar(n%10+'0');
$ t2 S3 g1 H+ P6 \}
7 u" o( x3 F! w) D2 [main()
+ z, P. Z: |3 m# O/ O{ int number;
* w+ s1 h L( E printf("\n 输入整数:");; A1 Q1 ]+ h2 ~$ X' _0 I, O" k1 r
scanf("%d",&number);
+ r4 J. d8 `4 o) F+ j printf("\n 输出是: ");
5 v) S. N9 G3 K& ~9 } if(number<0)
, R7 S" k$ E9 M ]3 x: L { putchar('-');: O6 V/ E. F- a6 \8 L
number=-number;* ^; I1 r7 ~) h" {: F! F/ K
}
. r4 M; O6 f7 ]6 U$ Sconvert(number);
' Q# r/ k: I% [6 b/ {} a0 o7 l9 B5 c: f: s: L
6 u5 e# l# f2 o. a' U8 R y5 y! o5 A" ?% B( g
, G. m! O8 h5 V* \5 K) t }
8.183 I( I" }! {0 h* {+ @
main()* a9 G9 i' w9 g
{: ]# j6 E' J1 l9 d1 n6 |9 t
int year,month,day;3 F; i1 I# T; n
int days;/ U; v9 }$ z; o
printf("\n 请输入日期(年,月,日)\n");- k/ O, r5 D9 x$ [# b7 \
scanf("%d,%d,%d",&year,&month,&day);
x3 ]* ^* A! g2 f7 K/ ^ printf("\n %d年%d月%d日",year,month,day);% I9 V% [$ W* g+ ~2 {8 ~: T. a
days=sum_day(month,day);' E" w- N d$ G% t$ l. `) q, q/ o- f
if(leap(year)&&month>=3)% G8 q# I3 r6 @5 Q0 x+ Y
days=days+1;1 o% r/ z" T/ g* u8 \ W
printf("是该年的%d天.\n",days);' Z Z2 \8 v: N0 n* J3 w% J# C
}5 z% W6 C* s. F8 ~- v, r
static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
) I. v' }1 Q6 q int(sum_day(month,day): d: E) q" z- M
int month,day;1 Z/ } a7 \; F5 O
{
0 Y% T5 A# z1 o6 K; k( S7 { int i;
2 B' z( k, |9 [6 f+ I) Y( H for(i=1;i day+=day_tab[i];6 Z' \6 P! U: N) }" b
return(day);. c5 j, c) | _- l4 I
}% B9 T g" v) S3 _
int leap(year)/ k& ~% u" O& H$ }
int year;
4 \) {' P! ^- a& H2 I* K/ R {: h9 j4 e/ l9 C6 P# L: I6 @
int leap;
) K9 F/ n2 k0 L( q d" V leap=year%4==0&&year%100!=0||year%400==0;
. U4 g9 E' z% { a5 o return(leap);- G* f& H( `3 y
}( }7 X( Y. Z1 D( m" \" [/ l" L# h
第九章 编译预处理
) P0 l6 V" k' Y: g2 m7 t9.14 M4 M, @; R' [. o' v% T
#define SWAP(a,b) t=b;b=a;a=t. f+ l( t8 D. b' d) S/ v1 w
main()
" l$ t7 r G& I: ^6 b{& S G9 f q/ N( Q {9 n1 B
int a,b,t; @8 D) {/ H( ]: Z, ]
printf("请输入两个整数 a,b:");; ^- V1 p& v* @' {
scanf("%d,%d",&a,&b);
6 {0 H" U- v, I# c# QSWAP(a,b);
% D8 e* z6 {* H8 e; d0 C% Uprintf("交换结果为:a=%d,b=%d\n",a,b);% h2 m7 q) W7 v( m
} 5 K8 R8 x$ z: S- z1 w
, N) A% g3 k4 }
/ e0 b% ]5 C) l9.2# I" x/ R8 l' i6 I- j- ?* b+ V
#define SURPLUS(a,b) ((a)%(b))
* O% `" [! S2 P1 fmain()
% S, \6 T7 k% L& S {* H7 O) ^8 q1 C" G" n4 p
int a,b;, V. ?* y2 j4 H
printf(" 请输入两个整数 a,b:");
9 ?9 L6 u) K" n, ?: { scanf("%d,%d",&a,&b);
. ^& Q5 \; u+ {/ B9 C3 ~printf("a,b相除的余数为:%d\n",SURPLUS(a,b));
6 L1 H: c% q7 x9 r- P) x6 y: m }8 m) d9 q- C S1 X2 w* H
7 v- \7 ?! t* a% j7 @1 J
9 G" T/ G. R9 h* ]
9.3' k% G( g1 |5 {" D; a7 |2 B0 m
#include( u8 B6 m5 b" _0 U/ a( O% g: _" t
#defin S(a,b,c) ((a+b+c)/2)% O4 h3 Q3 s; |# s) 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)-
7 e9 p, L. B- a, K3 c; kc)))* b1 i: g3 Q( r5 }) `
main()
/ T" d1 M) f- ]) {" x! [* o {4 t3 x$ m. e: p0 t; x; U; t
float a,b,c;
4 ] k6 H# `& c" c6 M5 t printf("请输入三角形的三条边:");0 m" D1 U/ I; E" E% K/ x
scanf("%f,%f,%f",&a,&b,&c);
7 k4 S& Y% A2 U! F1 \$ h if(a+b>c && a+c>b && b+c>a)( c1 o+ b) z; O+ m: Y, I2 s- P5 ]
printf("其面积为:%8.2f.\n",AREA(a,b,c));$ E+ b! f: Z1 d5 ]* ~* O4 r
else9 E/ `, q( ^4 M! e
printf("不能构成三角形!");" }6 I& i8 O( {; ~
}" w5 P6 N2 A- r
; O- n+ Q; g$ W0 M5 p! r' E" g, f
' u( A* z. Q3 V. ~" }, M1 N
# e% S1 Z3 P; T$ J( T& O9.46 b* g7 {3 w# Y+ X
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0): W9 d, B+ o- F9 R
main()
4 n& r6 m3 x6 M; J5 A3 _ {
, M; j9 H; @! ^, T3 Q int year;
) C3 }& I+ `% ^+ { printf("\n请输入某一年:");
5 |8 d: x3 N& _ ? scanf("%d",&year);0 Y1 V% n% @% ?; U: ?( t
if(LEAP_YEAR(year))/ r2 s; O/ s7 h# t7 U9 s6 D
printf("%d 是闰年.\n",year);
# m4 ~9 A, b9 Z1 q) } else5 q. W8 l6 k$ J# H9 N1 @
printf("%d 不是闰年.\n",year);
8 J$ ?( b w* ?5 [. N( a/ y }; _+ E% f6 ^7 w& V/ S
: z4 @+ n3 t: f1 q. L/ U3 a& M$ H
2 X/ P( h! V" b1 C% L
4 p; k o+ ^, W# A9.5解:展开后:
3 q8 {: q, j- Z& }# @1 Cprintf("&#118alue=%format\t",x);
+ W' s) X4 M3 {" m, {9 J, \1 Lprintf("&#118alue=%format\t",x);putchar('\n');6 d& K+ @( Q* K& i- A
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');; j0 H L7 `1 a% }' Q
输出结果:
/ s0 s) A' o1 ~9 u9 n+ s) O&#118alue=5.000000ormat &#118alue=5.000000ormat+ `6 G; ?6 G+ @. M. Q
&#118alue=3.000000ormat &#118alue=8.000000ormat
4 [+ p7 ` y1 q; V8 y$ ~0 Y" h- \* ~
1 u0 S0 Z0 w, {2 b: R: Y4 n# I9 s
9.8
" G' n1 J- q a3 nmain()
0 `/ y- [5 }! z {) q( _3 J, c. f2 H7 F( h% f
int a,b,c;
: I5 ~/ Q/ l5 n2 U* U printf("请输入三个整数:");& S7 j$ M; A) @2 \1 T' s4 F( S& y
scanf("%d,%d,%d",&a,&b,&c);/ I! A6 @0 S2 ^6 a5 ?7 _
printf("三个之中最大值为:%d\n",max(a,b,c));9 l9 L3 C O; H- ?: P
}
, e. l* l0 D1 A" @# F. @1 _9 X max(x,y,z)5 t) r. h+ [& }# R* i
int x,y,z;, w$ i! b) _- w; k5 L
{: K* \! P) W) z* Q' _6 M/ b& D% v4 |
int t;
" u5 `& h- N0 S. P) f t=(x>y? x:y);' l3 g% H" j0 Z7 O. Q
return(t>z? t:z);
( L8 @( Q. B: c9 c) E) p5 c }% \. M( L |+ s5 y$ h# E
/ C, V2 D7 `+ v6 h) G7 O3 d* e- e
5 r% C V5 u* U& y) j( ]" ~2 F. D8 `
& U& j9 X+ s6 R4 z% {6 b
9.10
; |* u0 k7 R7 q8 g* o% T#include
5 f$ p( K! V1 D. } J( r#define MAX 80
% z0 V( h5 K- S8 A' u#define CHANGE 1
8 v& R% y% O3 T# U# c/ C+ ^main()
3 h! M$ ~5 Z: T l2 e# N- f{' V: _1 T# \; V) A+ ?
char str[MAX];! X k" t5 ^! J( [$ C2 ~9 h
int i;
3 i, c4 c- ]% R printf("请输入文本行:\n");- @% k" c! ~. b8 ?2 g+ ?* n, {8 g
scanf("%s",str);( `- m5 R/ D1 _5 q' V4 }
#if(CHANGE)
$ E6 {2 ?+ R% ]# q& | {
+ d' E+ Y" t9 w, F. M for (i=0;i {- N" @% L+ b: O+ u% r U& h
if(str[i]!='\0'
5 O" e9 A! W: e/ W* T& P if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')/ H( {6 m9 W2 s& `: I" C
str[i]+=1;
" o, d7 S# g3 e4 P' R else if(str[i]=='z' || str[i]=='Z')) n0 M% N# c3 j) B% q4 T# |
str[i]-=25;* G* {2 P/ M$ Q2 O+ G9 Y+ S" B
}
Y8 D9 S- N# y5 }- e& w# H) Q* Z}# i: p& U% K; @) f! u
#endif
% m9 S# k9 G& W1 Cprintf("输出电码为:\n%s",str);
+ I6 g6 g2 J& d}
8 w$ J) V' S# v9 o第十章 指针. \1 A( F5 E9 t$ i. ]
10.1* G6 ^# u+ c/ z. U! m/ B4 X
main()
) r" ]0 T0 }7 x f1 I{int n1,n2,n3;
" h0 b, p+ N) ^8 B1 _ int *p1,*p2,*p3;$ c4 c9 L+ f3 Z X
scanf("%d,%d,%d",&n1,&n2,&n3);
$ t5 |3 s8 N, S! E! L) ] p1=&n1;5 c+ Y6 D( k& ^+ x' J8 ?
p2=&n2;
' _) N' s6 Q3 ^( u p3=&n3;
. \4 G: Z; z# V1 J9 f5 P if(n1>n2)swap(p1,p2);6 G5 @& `# s; c9 S" ^
if(n1>n3)swap(p1,p3);5 J3 ~- W0 B: Q" {# s( f
if(n2>n3)swap(p2,p3);
: o+ Z, _ ?) ?7 K8 x printf("%d,%d,%d\n",n1,n2,n3);7 ?; g! z4 M! r# R
}, `: R6 A) l& H7 E
swap(p1,p2)
& `! x& C g3 F, ]/ q8 h% U% Gint *p1,*p2;
) l& i, \, ~$ T% j9 {) z% n1 B- O1 }{int p;+ O; W& u; G( I$ Z; b3 k$ Q% A; G
p=*p1;*p1=*p2;*p2=p;1 ]- y9 a. N3 P! q
}9 Q( c8 G# n! g7 W5 d' h2 x8 M
10.2
) M) k+ P# G" z3 l5 N& ~main()
/ O" U5 g% Q8 w% G. J{char *str1[20],*str2[20],*str3[20];2 Z ?: f, g) h+ F1 g2 n+ V2 F0 m
char swap();
2 l! G. z# E! H# g1 M4 L( ~ scanf("%s",str1);8 _! b2 k) D+ B) F/ b
scanf("%s",str2);
9 v5 I1 W* l" j$ s! p- q* U9 V# A scanf("%s",str3);
, T( ]3 J* W% g# H: t: Q if(strcmp(str1,str2)>0)swap(str1,str2);
' \# [' \4 |6 G4 s* q if(strcmp(str1,str3)>0)swap(str1,str3);
0 q, R& n% w6 {6 o8 y: O5 V) M3 B; ^' S if(strcmp(str2,str3)>0)swap(str2,str3);8 F3 E7 X) L) t7 h. H
printf("%s\n%s\n%s\n",str1,str2,str3);
2 Y' P+ v- O& F1 k}* @6 g& \ w, W. K/ @5 [, |
char swap(p1,p2)' \3 g0 l0 u$ h2 l
char *p1,*p2;
) n J6 F* c S: b y{char *p[20];
6 v7 y; S) ]) z& b3 b strcpy(p,p1);) T' S; n( n1 b& {% o" D
strcpy(p1,p2);
' a( _' {2 W( N* o strcpy(p2,p);
3 @+ d$ m6 X$ p) X2 b}
: ?7 H; p" h' M9 G5 X* k9 X10.3
+ A- k( Y" c1 l% l- Y) c8 Q# _main()
% [. l$ U& w$ Z" Z7 e& O- J5 d- K3 n{int number[10];
/ {% {, W1 k& j. ^" c* a' S& x4 @# n input(number);
3 g4 s) c S# h- W max_min_&#118alue(number);& p0 n' n- s9 b
output(number);
. c3 V( q9 \$ z) D& ~ }
1 i" j: H' V" `8 A L" G( ainput(number)
9 w3 t- J1 p# h3 T0 c9 xint number[10];
! \. \1 T7 o' N6 K7 M; q{int i;! M `$ k$ H f' T: O
for(i=0;i<10;i++)/ x w% t; @* @9 [& g z+ P
scanf("%d",&number[i]);
. D8 B. v5 [# ?4 r}% k3 @' T$ }, s3 b) C/ j; Q8 U
max_min_&#118alue(number)
% }* e L1 W+ O" R" g, x% }' Iint number[10];5 w% o, [7 s$ `2 T* p( h; E0 p
{int *max,*min;" A z! f* I% I( A1 P* J
int *p,*end;5 u5 p/ u* s7 b" e
end=number+10;
% C" g& K, p% Z+ L$ l9 H p* p max=min=number;
) r' z( M% I9 Y/ t- l# M for(p=number+1;p if(*p>*max)max=p;
1 s! f" T0 p% k5 m2 i4 ^ else if(*p<*min)min=p;' u, T8 Q/ Q4 J' y
*p=number[0];
- S; B# V" x5 @# r8 ] {7 p number[0]=*min;
& `- ~+ k$ W% c8 ] g; n9 Y8 K: _ *min=*p;
, B& Z- ]3 W# V+ q6 A' u *p=number[9];
5 p* V) v: t( c6 I- F" _ number[9]=*max;
- a: r6 G* Y% E *max=*p;7 o1 E- E) s& l& P
return;+ R$ ]" W8 Y5 {' m2 D: H
}8 @1 ?) J. I% g" Y% y
output(number)
- b$ y2 K$ N5 f' @4 _+ R' ?7 ^5 a& hint number[10];+ F# S v. T u1 K, X, u
{int *p;% E. D9 F( R5 G( i' n- w
for(p=number;p printf("%d,",*p);
: Q3 C" b7 g; r/ N _+ b! ^ printf("%d\n",*p);0 N# {" z. `; L+ ?9 d0 A& N
}
' Q% P+ k+ z% I G10.4- U% ^" |& E7 L4 O+ X% k
main()
/ [2 k7 z+ ~$ c2 i{int number[20],n,m,i;- y/ T- K) N0 ]* u8 r5 }
scanf("%d",&n);8 Y C( z) A! y9 k
scanf("%d",&m);3 x) M+ `8 d0 k" i
for(i=0;i scanf("%d",&number[i]);
) I/ U" y3 k- @+ F$ D; V! x& ? a' p move(number,n,m);$ _# k* T, F3 ]. W# }
for(i=0;i printf("%8d",number[i]);
1 g! ~) v1 @- t" }3 i}
$ [" P7 ^. T \$ N7 o6 `7 K8 _move(array,n,m): D5 _4 {; u/ \0 L- q# {5 n
int array[20],n,m;6 ?8 I& D, Z' p c3 O
{int *p,end;5 g" o1 V% i2 h, i( f8 w3 z- g
end=*(array+n-1);
* m" s- K; I, k3 h% u for(p=array+n-1;p>array;p--)
4 p3 P( I1 i! W *p=*(p-1);
3 [1 b4 P" E* R3 P/ B/ F5 z *array=end;
; [; b; [" ?# D, b) m7 Q! {, a m--;
( ~& Q6 ^/ Z+ ?5 `3 R1 u% e% K0 Z" A if(m>0)move(array,n,m);
- @1 `/ _9 X2 N! Y2 F9 _! [! Q}
4 V- `. y7 E& K8 [* Q8 L10.5
( H [2 ~1 K8 `( X, O- q#define nmax 508 Y; {. u2 _6 |" v5 H8 f: Q
main()
' `/ X/ S4 A" p+ V- g/ [& [* g" Y{int i,k,m,n,num[nmax],*p;
! V: }: e1 g! ?" s scanf("%d",&n);
7 z( F8 J9 c3 q( S0 c p=num;
4 p. |4 C" t. R for(i=0;i *(p+i)=i+1;
7 t6 ~1 y6 w& y- Z; \: E' i# s& { i=k=m=0;
6 e2 h' x0 H# o) q$ o while(m {if(*(p+i)!=0)k++;; M: d! x! c3 p e5 l& {: z
if(k==3)
, u1 T8 `8 N8 a3 _ {*(p+i)=0;8 t( a) T. q2 i. U9 A# ~
k=0;
' `; d- h3 a4 w0 \* s& Q m++;, e1 z5 F0 v' e( q: u, C' s
}/ l" Y% c8 t6 v9 f5 |4 d
i++;
! y+ D6 u9 N5 C. Z7 ~. _& G. {& U% | if(i==n)i=0;
- }* H W( P% I- i+ H+ r }0 d7 G$ j% S" @! u1 c. v. q
while(*p==0)p++;9 G' e+ j3 b( J5 p7 q" b
printf("%d",*p);
$ b. G. O$ [" G0 e m1 ^! [}
- Q* I8 O( ]) R10.6
) v- m: G0 h& emain(): ]: u% N0 {0 d; A: n
{int len;
6 W* b. \" _3 s- [% f' H char *str[20];1 I& A6 T" L7 A f
scanf("%s",str);6 ^5 h. ^4 m) {
len=length(str);
3 T; J3 J! {( X1 l) V printf("\nlen=%d\n",len);
# X- V2 H$ X; M- O% R; z. D}, l# ?- F0 i4 h$ E1 B7 G# f* x
length(p) b. q8 O+ u; _' U# f6 p' B' a
char *p;' Y) J* Y9 y* B
{int n=0;
: {3 x" H; ^7 T! N& j/ e2 @7 |4 p8 x while(*p!='\0')
8 g" K! f4 Z. H' C2 w) `' } {n++;p++;}
4 Y% Y! ~& M' B2 X8 B+ o% [4 T' [ return(n);
9 d1 o1 P! H' N% m, ?9 e}
' _0 Q. t t7 j! Y/ v3 i10.7
6 [* }+ J+ R+ s, nmain()5 B! ~# v2 s8 w
{int m;, ?4 O/ d# l# r
char *str1[20],*str2[20];
1 n7 @( H+ W3 I& O( p. Z" y scanf("%s",str1);
' M2 @$ C0 X6 N9 t scanf("%d",&m);
' q. ~( L$ p' Z8 q5 d6 |; \& n5 g if(strlen(str1) printf("error");. Y: \4 |$ _7 @ @ `$ W' w
else
" s% [( Z- x; R$ I8 \7 e- } {copystr(str1,str2,m);
# u9 a- C- R6 c& h' [! n) I+ X printf("%s",str2);/ n$ m; E5 u* ?, H+ i. q9 O+ M5 g
}
$ H, m5 x" _8 O: t. [3 c m}
' }3 N1 Y1 B0 f6 ~% [5 E6 fcopystr(p1,p2,m)0 {& _" \' g% ]5 ^$ ^ n" `
char *p1,*p2;
6 W% Y& R l# p( Y+ Wint m;8 E% V- K8 |! k% j2 b' D
{int n=0;, c6 o; `* }9 J& J# x, v, Z( d$ B
while(n {n++;p1++;}
% W. d- I: w7 x) B3 }9 J% i+ ^ while(*p1!='\0')
3 A- Q# E* M( x* g$ P {*p2=*p1;: n/ J, L2 l9 O0 k' J7 F; m
p1++;
& }& P7 U" S+ V" G* Q p2++;
3 z8 l q$ c3 t' l, [* h- P }
3 }0 l0 n, ~, h8 C* q1 A1 n" s *p2='\0';" ^6 I* v, r( g1 R; B( |/ |
}
( k, C) t. ?: t3 f10.8
* f3 G$ n& W! j' V0 N3 U! n#include"stdio.h"; \$ I! x$ F% M- p j: d: \
main()
* f' I* W: L5 A [" Q8 B{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
. N; S8 U- g6 k | char *p,s[20];$ ?- l. L: \/ X) R! `6 F
for(i=0;i<20;i++)s[i]=0;% }- R" A! D0 f1 E t' `
i=0;: ?* X& N% t: M9 ?$ k
while((s[i]=getchar())!='\n')i++;
) s o% z. D3 n* c7 ] p=s;# \5 a3 y4 S5 i& r: N; z
while(*p!='\n')
" N3 k9 L" [: L( D, { {if(*p>='a'&&*p<='z')
4 p' ^! p4 n4 b, h4 J! B) P ++sle;6 E( t9 Z, ]- l- J; b/ ?& ?8 \- o H' q/ m
else if(*p>='A'&&*p<='Z')* k- y$ N" F9 }# L% _
++cle;
7 Y0 V, @5 U, f9 I3 G8 J9 F* n else if(*p==' ')" x& v& x# A4 s. @1 u
++wsp;1 l4 N& `$ Q' o4 }. h
else if(*p>='0'&&*p<='9')% d' @) N, ?$ f+ p( F) Y k
++di;
. y7 Z. j$ ~: Y else; C4 |6 \& g) L! B- L) h
++ot;9 j+ T, D$ W; V* [
p++;9 {+ g6 @ l9 I' _- g# c) m, t* t
}3 ]+ u. O( }. J
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);8 p, F1 {* v! z4 I9 U) t, k3 d* D7 [
}" U3 D3 s7 P; x- s
10.9
0 V m; b" @+ D3 v, E. @main()
1 b& s! `9 W, p/ e{int a[3][3],*p,i;
% r* p$ }7 [" g7 f" C0 M for(i=0;i<3;i++)
. l8 m% ?3 V) R scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);3 u' r u# V* G6 \9 O2 Y- {2 ?6 \
p=a;( W7 e, |8 k9 _- \
move(p);2 |$ N( J- a/ t# K7 w
for(i=0;i<3;i++)0 r% `( Z3 n8 ^% k" m+ [
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);" @1 u( d5 W: d% T2 ], e& H
}( U/ B8 d3 P. f% Z [' J( `
move(pointer) M' ?2 i6 k1 w3 ~
int *pointer;
4 O6 ~# i, ?* x5 o! Q{int i,j,t;8 G! ]+ W# U3 `. G) n, V
for(i=0;i<2;i++)/ y/ d- v! [6 T+ P" J
for(j=i+1;j<3;j++)
; N9 m4 m9 ^% Q- k5 _5 K- j {t=*(pointer+3*i+j);3 D$ b# h# A7 Q
*(pointer+3*i+j)=*(pointer+3*j+i);
! _! X9 k$ W$ L3 ?0 U. I2 }3 s *(pointer+3*j+i)=t;" |4 o# k- r& P, g5 Q) Y
}! B' U1 i) _5 ~' M9 F3 }' O
}
" s; e% @. a( g2 c" m10.10$ `3 `" @) N& a
main()
& `6 Y$ `: E2 u5 e& p0 _{int a[5][5],*p,i,j;
: a" \" y3 y6 ] O0 {# `/ o for(i=0;i<5;i++)
( L7 m5 s2 L% @4 A" j. Z1 } for(j=0;j<5;j++)6 c1 q: ? r" r: |
scanf("%d",&a[i][j]);
+ q5 v1 E! C' Q. h! l p=a;5 j1 R# B( A; r3 B" Z4 l. A
change(p);* {# z+ W/ U' R/ |9 p! v# y
for(i=0;i<5;i++)$ r$ n- x2 m9 N- v
{printf("\n");" d, j$ a' |. U. I" f
for(j=0;j<5;j++)
; {3 o( N3 W* f4 E/ b. {: y printf("%8d",a[i][j]);2 ]* O" I. v( W4 W
}& L, K! H4 e9 h- r; I& L/ s
}
* Z S2 q$ x; |8 Echange(p)
0 E; N' I$ u# F4 T! @+ ]+ @0 g! ? Sint *p;+ b' ?" K" h2 t- V- b
{int i,j,change;
# I8 T( G! ^& y5 W$ h int *pmax,*pmin;
8 [' Z& l6 c. j% e9 a6 D* ]- ~1 k pmax=p;
, X! e4 d d g2 F1 D- | pmin=p;
# Q4 X, ]% M! O% g for(i=0;i<5;i++)
' k* i# ~5 ^* i8 r7 ^! m& k for(j=0;j<5;j++)" b. V1 G' e d9 Q2 ?: i( _( m
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;, ~- R% c, k" J9 l
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;
0 x6 A( z9 r6 r9 x% ~ }
E- W5 X) v0 N. z change=*(p+12);
' A% D Q2 } X" T; S *(p+12)=*pmax;& V* G$ t1 O c. D) \- \
*pmax=change;
3 j1 c A% H T( t6 p8 l change=*p;
7 X+ B( f+ S4 e7 a' N *p=*pmin;* |, C7 @& S( Z! d J& c
*pmin=change;
3 X/ p$ T( Z, W' v2 f" w pmin=p+1;
7 m# L4 j; O; l7 @' l for(i=0;i<5;i++)
8 e! [9 w. Y1 v4 p8 A5 i for(j=0;j<5;j++)
6 D# V% K7 c0 L- R- ` if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;5 x$ H* m' F$ ^, G* m' x: y) ]
change=*(p+4);. I1 Q: ^. [0 U8 Z5 ^% |/ c) M
*(p+4)=*pmin;
' D& k) w/ B8 k/ b0 Q5 s; w *pmin=change;+ v Q$ i& s' j) x7 Q, U2 |, u
pmin=p+1;1 \% M5 H1 v& V( Y
for(i=0;i<5;i++)
# u: N& N+ W7 }% c3 O" o for(j=0;j<5;j++)- ^. c- A0 [- s% r& F& g, Q$ M3 Y. l$ o
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
7 B. {- B! q, C8 J0 n9 f0 v% v pmin=p+5*i+j;
/ p& g, H# I* n2 }* w change=*(p+20);
/ q- G2 V; y3 k* Y7 r2 Y$ f *(p+20)=*pmin;% T I! s* O" C9 x# C
*pmin=change;6 D$ w. v& K: M* p/ S8 {+ q
pmin=p+1;$ f' k ]# w7 N7 d7 M0 Q2 P
for(i=0;i<5;i++)* l# }1 b6 m Q- s+ G, U
for(j=0;j<5;j++)7 I+ L# }: c! O
if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
8 v. S, M7 J& o) b3 Y1 M &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;* l B1 L4 j, C' D9 H5 Q* i
change=*(p+24);
1 l3 U, e/ q T/ C, C *(p+24)=*pmin;
9 `: L0 w9 m0 J8 v/ g *pmin=change;
" x- E7 R8 w: N7 J9 I}
/ `( b: O7 w7 I* O H1 U10.11
) k2 w, n" j9 o3 Jmain()
* q0 _+ @/ q; o' T0 M{int i;1 a; b2 f# g* J# J
char *p,str[10][10];
5 Q5 g$ i9 k. { for(i=0;i<10;i++)$ b6 _/ Z1 H" L3 e9 @3 i, D% M
scanf("%s",str[i]);
1 h- a1 R& G1 O) E p=str;7 m0 c1 U& X- l4 [ V
sort(p);1 L3 C& G& ?: r) _( Q) q
for(i=0;i<10;i++)2 j% S( r2 q; v# g
printf("%s\n",str[i]);3 r' ^$ k/ m0 x
}
% y! {; t3 |6 [0 { isort(p)- w8 j/ m; K3 _; \6 e* \
char *p;7 {$ M* k4 [5 T: r+ Z$ t
{int i,j;9 b4 l- ?4 q$ \% P. H2 {
char s[10],*smax,*smin;
. J s/ s" A7 @9 U for(i=0;i<10;i++)7 ~- ]$ v* F) o* O. [7 p( }
{smax=p+10*i;
# F) K9 s8 I+ e( c# E( _& R( Z for(j=i+1;j<10;j++)0 [ K" j3 o. j9 F4 B& d
{smin=p+10*j;
2 A5 z3 @2 T( a if(strcmp(smax,smin)>0)+ j9 D" [- Q. @
{strcpy(s,smin);( {9 }7 l4 \1 v) x/ z
strcpy(smin,smax);$ Z' S& U% t' Q0 h# f5 |* N5 E9 u# @
strcpy(smax,s);
. ]( w$ F, e: W- P4 H* |* | }
* f7 Z7 \+ R4 a- c; B% s1 M }7 h% f2 L# D. s+ K- U
}. Q; t& R& g/ F; z3 i
}
6 k' b- T+ T" Q2 e& I10.12
6 u# ^8 H( R' V1 u0 W4 Z#define MAX 20
! }+ \5 y9 k: S) ^ D, |main()
& K4 ~* h! ]9 p/ W! E X- Y{int i;
1 g- A) ]9 p6 B N* l- k& c char *pstr[10],str[10][MAX];% p9 |6 A: ~2 J' G4 e
for(i=0;i<10;i++)0 H& N& c1 w" ?' f. O! J( \
pstr[i]=str[i];
; V0 H" @) e+ b; Q* A for(i=0;i<10;i++)
6 h8 n' e2 R9 r! ]2 z' h scanf("%s",pstr[i]);
4 ~& @' X: v6 d- {1 g. u, x L sort(pstr);* Q) j( e* ^* w3 C% J+ w- x4 B
for(i=0;i<10;i++)8 ]: ?* y- e3 h0 D0 `7 A' i
printf("%s\n",pstr[i]);
/ d7 L" V2 S2 S0 I# I' |}
1 L0 \; g: x$ @, P. j/ m* rsort(pstr)2 D: p; c1 r4 a1 @
char *pstr[10];# L- D! m$ g0 m) y* A3 n( \
{int i,j;
' ], E( q3 M; Q, a char *p;
% K3 }- f. @" m; H! O$ G for(i=0;i<10;i++), G8 @: q3 \( e5 M' ~+ K1 h
{for(j=i+1;j<10;j++)8 w9 e7 _+ Z5 D7 }/ \; W
{if(strcmp(*(pstr+i),*(pstr+j))>0)
0 Z/ x! k. c: \ {p=*(pstr+i);2 s" i$ K0 a" Q6 v0 G m4 Y
*(pstr+i)=*(pstr+j);/ [% E% |3 p' x9 _ l
*(pstr+j)=p;
7 |0 d6 p% o. S" S; G }7 v( C8 f/ D2 n4 L( D; b
}
' |+ A6 j2 h# b; ^- a5 H; o4 u }
9 G; o( U$ ]: n y' S8 Q! L% ^}
( m" K! I- c6 P8 B3 H% I10.13
' E& S3 _$ }/ a0 b) P) b4 E#include"math.h"5 x3 f9 b0 U9 `$ Z5 x" ?$ Q4 e
main()8 [/ i$ G( f5 L- [( k& N
{int n=20;
. d5 T' O3 ?" `0 m0 v1 i K- x- |4 l float a,b,a1,b1,a2,b2,c,(*p)(),jiff();0 n# ~0 S' @) O) J0 c
scanf("%f,%f",&a,&b);
s# }+ V& d" f! ?9 ^6 s5 x- X. m scanf("%f,%f",&a1,&b1);2 `* |0 _4 w4 f! E
scanf("%f,%f",&a2,&b2);
+ N, S$ H; R6 P0 {6 n+ f p=sin;- J0 ?; ^9 i' [: s+ o% t. }
c=jiff(a,b,n,p);3 e u! x1 ] F' k1 d
printf("sin=%f\n",c);
. `& o T/ _ ]% h6 Y3 ?" H p=cos;7 ~% D9 Z, S, D" w) T
c=jiff(a1,b1,n,p);6 o" Y9 T `" p% ?1 z# B9 ^- H/ T
printf("cos=%f\n",c);9 G |* A3 c+ ?' r( q. W/ a7 D5 @
p=exp;2 r/ j' v0 v: ?6 H- O a
c=jiff(a2,b2,n,p);6 u" V) I+ ^+ g' J6 K0 H
printf("exp=%f\n",c); w/ Z3 w; P& ^8 t! p7 v
}
5 R6 i) z- B( C7 R) k( rfloat jiff(a,b,n,p)
% h R8 N6 L: tfloat a,b,(*p)(); z7 f8 a/ E9 V
int n;
8 G: G: T5 Q* ~* ]- ?- g{int i;1 S: Q4 |9 s, P' A0 X; ?8 U4 [
float x,f,h,area;3 q, w' [+ o; o
h=(b-a)/n;
: \% o% v- h s x=a;3 J9 k) W& {1 y, B
area=0;# @1 b& t) E- r e* w" G! ]! _6 ~
for(i=1;i<=n;i++)9 d; \$ w( B* s# N) a/ ~) J
{x=x+h;
2 ?5 f5 u3 @+ X; N1 a2 l) N( Z) o area=area+(*p)(x)*h;
7 I, T) D; Q F! S* {% _9 H6 g! {, J }* E3 p! N/ b; [8 F- v" ?8 l
return(area);/ h5 Z( S9 c( H
}2 N9 X. T( `& j8 k1 i8 c. T
10.14
+ s( l2 g: V; F2 \- A" Cmain()
: t+ [( s5 d% a# X' A3 t{int i,n,num[20];8 |1 ~& |7 L: A
char *p;# E( @2 m+ d6 Y4 T6 D# L
scanf("%d",&n);# V9 c, Y" e2 G8 u; U# ^) x8 d+ Z
for(i=0;i scanf("%d",&num[i]);
/ k9 u: U0 }/ N9 s: B; q! m p=num;
7 h3 s6 j* N5 ~; ]5 U& n. B sort(p,n);
; ]5 c0 G1 R8 n ^1 _" ] for(i=0;i printf("%8d",num[i]);: h4 ^: V2 }6 `7 T& Q3 a
}
8 x+ _& I6 F, U( G* c# bsort(p,m)
0 P, n* ^# I$ O0 K! I. P1 qchar *p;& z% W8 R: e% R! ]9 U
int m;! R/ B; ^/ C, ?3 b3 B: i1 Z! H
{int i;
* v' S( T# R$ L+ w char change,*p1,*p2;! G: W3 S+ ^/ A
for(i=0;i {p1=p+i;
/ p! W. [: t; A2 J2 L8 h' o0 K. @ p2=p+(m-1-i);
, `- K% j, \2 _6 {. J+ v9 v, J f change=*p1;
9 L+ B% k/ ^/ ]% ]2 P *p1=*p2;
, ^: Y0 ?; k; M" H. Q) n. {; n *p2=change;% B) a* u; @8 b0 O8 `$ V: A# D6 d
}
- W5 e' U3 V- S9 Q O, v}
6 v5 c/ K" i1 g( X% u10.15
m6 }5 D% E" h cmain()
' w. X" w5 O6 B( @/ _( s{int i,j,*pnum,num[4];
' y+ v5 l+ s- ?0 } float score[4][5],aver[4],*psco,*pave;$ v0 e1 D& A# \9 I$ a' `3 I4 Y
char course[5][10],*pcou;/ t" @! T2 v/ P& `
pcou=course[0];
2 ^) B) l* T3 `0 V% D# q X9 s for(i=0;i<5;i++)
/ z, [% b+ l9 ~; N* \ S { scanf("%s",pcou+10*i);4 `1 Q J/ Q9 k% B& y. y
printf("number"); `' R# W4 R/ Y0 ~. \8 b
for(i=0;i<5;i++)
% B1 q" y5 z8 w5 `, b) D printf(",%s",pcou+10*i);0 _1 }+ C$ p7 x& a
printf("\n");- j& [% i8 a1 N# L7 i
psco=score;0 ?0 J: c3 \1 T; T: v. V& {
pnum=num;" H1 ~1 G: `: k2 U+ G( }: {
for(i=0;i<4;i++)
! q5 h& h+ z5 l7 F {scanf("%d",pnum+i);
# F0 G/ C" g3 }1 r for(j=0;j<5;j++)$ m8 g' L4 l4 ]) `, T
scanf(",%f",psco+5*i+j);
! u- a5 E4 l4 } }
6 e s! P) Y2 G pave=aver;- ]2 |% N4 r: H4 n; g/ O
printf("\n");
3 b- L2 g- Z. a" v& o3 r+ L- w avsco(psco,pave);
$ u; B6 U# ^, h. M avcour1(pcou,psco);/ ~) X1 E( F! B z$ X
printf("\n");* }$ A# ~$ h4 F0 u+ Z
fali2(pcou,pnum,psco,pave);0 o) v. Z+ f- v1 r0 r! \
printf("\n");3 Y ]6 v1 y8 h5 f* b: k
good(pcou,pnum,psco,pave);% o3 k ~( }. Y1 @' H& Q2 w2 ~/ |
}
2 E6 N4 Y6 M% w& Kavsco(psco,pave)
" N& R' e: J6 B: o) dfloat *psco,*pave;. P) B; {3 m1 ?% _3 ?* [ W* p$ \
{int i,j;
# j& l# U/ \. E5 l; v float sum,average;, g \4 f! V% p; d# J9 [
for(i=0;i<4;i++)
# X9 b$ m4 V8 M9 q8 W! y6 O+ w {sum=0;
1 k( d5 v2 \( }, j# j for(j=0;j<5;j+)
6 Y5 w4 J0 l8 z+ [4 ]3 t sum+=(*(psco+5*i+j));/ h @3 u! u) c
average=sum/5;5 d0 r1 }4 d* S
*(pave+i)=average;1 B2 f: c2 G6 |0 u
}
7 N& H" J, Z5 {0 a}2 W& Q, E9 d( v5 k
avcour1(pcou,psco)3 U) Q/ t4 \5 ?) l( T9 k
char *pcou;
$ o, T! ?4 ] f$ |: G p/ F' I; efloat *psco;
) T4 q/ K! o: M. Z{int i;/ Q! Q; @7 r4 Y0 t9 s& z' W5 L) @
float sum,average1;5 V6 T4 x8 W. e0 f8 Q7 g5 h$ d
sum=0;& j- m0 k0 v' {4 ?
for(i=0;i<4;i++)* W8 q' ~9 Y4 K6 ]- T
sum+=(*(psco+5*i))
6 w3 \- a2 a+ h3 j' J average1=sum/4;! N/ `5 u) h" }0 ~8 }; m/ s8 C
printf("%s %5.2f\n",pcou,average1);* O9 i; V8 Z1 G
}
& y1 I4 E8 y) P0 R1 S. v% Wfali2(pcou,pnum,psco,pave)
" M2 j2 B$ e( Y, c$ i+ D kchar *pcou;
N0 r3 x5 F) @: M' {) t$ n- Aint *pnum;
2 L$ ^- c q0 Efloat *psco,*pave;5 \. m& m" A" v5 f' n( M/ l. T
{int i,j,k,label;; H% |- o- l$ m: W" j+ O1 _: M
printf("\nnumber\n");
8 d& B3 e. y+ `3 o2 f4 ` for(i=0;i<5;i++)+ P2 |2 s" x: d0 L4 _
printf("%-8s",pcou+10*i);
9 z5 i5 X2 \6 s1 h5 h1 {& f1 B6 J printf("\naverage\n");! {+ u% B% q% R b( g
for(i=0;i<4;i++), B& t' z9 p$ Y, O% o
{label=0;3 I' Q; ~/ {2 @1 y- o
for(j=0;j<5;j++)
0 X' b2 t; f% J2 c, y if(*(psco+5*i+j)<60.0)label++;8 D' |, k: R! D, V
if(label>=2)* a* u. [" U, G/ \& h* f
{printf("%-8d",*(pnum+i));
: p0 j- l, O! C# h9 j7 s% z: ~, H/ s for(k=0;k<5;k++)
, ?! `- x* I" x" a8 X/ G1 F printf("%-8.2f",*(psco+5*i+k));
6 u! J3 Z0 | L7 L% H printf("%-8.2f",*(pave+i));& S( V/ E) `8 z% y5 @) S
}
- ^. Z: a0 O0 G$ |6 E/ B6 x$ d k }
) t9 @0 }$ S( M* z; @9 f. |2 W}# @6 ^: p' ]: b' G
good(pcou,pnum,psco,pave)" t: o# g. V! j
char *pcou;7 [4 d) Z8 h, h1 s4 S+ Y9 s
int *pnum;
% P: Y5 r3 o* ^% I# wfloat *psco,*pave;- C9 P5 L6 |7 a0 y6 A
{int i,j,k,label;5 m/ e. r1 B$ o8 k: S' Q
printf("number");
3 O3 N9 E ~+ i4 J for(i=0;i<5;i++)1 D+ t+ R1 `7 ~& p( H5 ~! u! H: E6 V
printf("%-8s",pcou+10*i);
; V; i' a+ y$ {: h- s: A; G' C printf("average");, E- y3 D2 H& W' h" z2 Q2 m9 V
for(i=0;i<4;i++)
# y/ P2 ?2 P1 d" g' v" c9 Y+ ] {label=0;
& ]8 p- u) g( v- V8 v8 \! O# x for(j=0;j<5;j++)
) ~* j2 f6 m5 m0 }" O z% ?/ ~ if(*(psco+5*i+j)>=85.0)label++;
9 R( d) m2 Y |. q. | if((label>=5)||(*(pave+i)>=90))
, ~+ E, v3 A& W& Q$ I {printf("%-8d",*(pnum+i));
5 M+ n9 Z8 O0 w4 l: ^" G' { for(k=0;k<5;k++)3 h: M/ ^6 f8 |* S
printf("%-8.2f",*(psco+5*i+k));
8 l$ M: O2 q2 `% n6 v- W$ C printf("%-8.2f",*(pave+i));
9 {$ n9 ^0 y7 Q. v( B5 T }: O" d J# {! N6 ~+ f' y
}
; D& J1 N" ~ r/ ~) }}
/ b A" k1 ?& a; G) e3 j( ?10.160 |" a w+ c: h8 y+ u7 K% M
#include"stdio.h"+ M+ ~7 z5 w( E/ [, X1 i/ g6 a* S8 Y
main()% H8 D& s0 u! d- U8 \5 K
{char str[50],*pstr;
+ C% H ~$ y. L) z2 R+ _ int i,j,k,m,e10,digit,ndigit,a[10],*pa;
1 b! G4 z# M! d gets(str);$ A' v0 ^: |4 ~7 b& w# G/ ?
pstr=str;' r( z$ B' U' o0 G) T
pa=a;( O! t9 o# r' l: @$ Y* z$ o/ m
ndigit=0;( N) e. _1 Z9 x; N: b9 X8 D4 T M
i=j=0;
c) J2 h1 x) Y# e while(*(pstr+i)!='\0')* B# v6 L/ {& q* t# l( |* A
{if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))( V8 j' P1 F# K0 r" v6 G* T
j++;' W2 ], }2 W7 o0 F/ n, j; |
else. k& A/ }& K, n- ]
{if(j>0)! S4 n( S5 \. R
{digit=*(pstr+i-1)-48;+ L! x* d& ~$ j- H& t2 R7 h4 ~$ Q0 _
k=1;
4 H# Y/ E5 u. d while(k {e10=1;1 z" J1 S6 o! v5 m* s
for(m=1;m<=k;m++)
, O0 F2 W" m# T4 C) v e10=e10*10;' {. X/ H9 ^4 w0 Y9 }7 b$ B' u+ y
digit+=(*(pstr+i-1-k)-48)*e10;
0 t9 Z% B/ j8 f' a& j$ y7 ~3 D, o k++;
: R& H$ p; K5 y. H- k! y: S }
, l; K" e* Y: h. v5 e' F *pa=digit;
! O3 |2 F. { T2 i5 Y ndigit++;
0 @1 t5 E' {. p" B5 e; G1 k' s pa++;( L9 F. G4 B3 Q
j=0;
. c3 x8 [ W( H' T2 S }: q( Q5 t* n6 o3 ]. Q
}6 [2 X, d8 h) Y+ V- k C
i++;$ T, P9 ]: Q. F+ C) ?' I
}
+ I+ e0 I0 o( K: X" E if(j>0)9 G$ D) S$ B. d( J" C4 t+ N, b
{digit=*(pstr+i-1)-48;7 F t6 m$ U* A" u( u
k=1;4 `; Y4 K2 X& E: m1 O
while(k {e10=1;
% g. n5 K! } T! S; t* K for(m=1;m<=k;m++); E) M9 s z- T5 m$ f6 E/ x. X
e10=e10*10;' u z1 {' e8 d5 r- M
digit+=(*(pstr+i-1-k)-48)*e10;
. o- @4 w0 o$ b* ?& B k++;
8 `. ^4 ~7 k2 V( s" Q) Z1 E, \ }- i# U5 {; Y3 Z3 A/ R! e) T9 l: I
*pa=digit;
- ]. ]( ?1 ~1 s ndigit++;
; }8 J: {6 Z! I5 J j=0;
Z2 _* Z3 c: l9 g% p5 v' P/ D } % |6 Z5 |9 J' G3 C" E. `' P
printf("ndigit=%d\n",ndigit);
" s3 C7 z; Z \' M, c j=0;! Y' B7 E* A0 t" L/ P# z; Z
pa=a;& O7 T& ]7 N6 R
for(j=0;j printf("%d",*(pa+j));- f9 ~4 N( m9 D5 ^+ }
}
/ O3 g! G: P) ]9 Y5 r- w- y10.17
) ?. R8 L0 h; W- { g$ zmain()" J; k1 A. I) v2 ]+ R# V' l
{int m;2 T( U6 s) }( @* E
char str1[20],str2[20],*p1,*p2;
5 U, H4 j% `3 n- d. v G2 E scanf("%s",str1);& }6 {) N1 m7 c- g# Y
scanf("%s",str2);+ H; i6 Q6 U: |7 R
p1=str1;
9 K6 W j7 ^5 c2 F- J- H8 g p2=str2;) a3 Y8 W9 F3 v. L: }
m=strcmp(p1,p2);
+ L1 U7 {3 j9 b8 X printf("%d\n",m);1 N) k8 l2 S" S& Z
}
: t+ M# n) B" w, C) vstrcmp(p1,p2)
' ?, l4 r2 b$ d6 h7 A2 N) Q+ fchar *p1,*p2;* H7 F3 }4 C+ E2 o r
{int i=0;2 F" l6 q+ Y$ E0 G
while(*(p1+i)==*(p2+i))5 o" ?% Y" j# W
if(*(p+i++)=='\0')return(0);& C+ K) S& b( p( z, r% G
return(*(p1+i)-*(p2+i));
% ~- |( [+ b+ b( g' r! u}" F7 M7 Z- _. u9 c2 i9 r
10.18
% l+ p" _( T, Z; }' `" ^) Cmain()
* h, W I& }& G, l2 z3 n: {7 S6 K9 [{static char *mname[13]={"illeagl","January","February","March",
) ~4 Q. `( g7 o7 c. l3 J) ] "April","May","June","July","August","September","October"," `5 E$ I$ S# J6 L0 Q
"November","December"};4 L8 p0 `" h1 `2 Q. g
int n;
3 E1 B5 K9 \- @8 F scanf("%d",&n);
. Y% F7 I S! n, m; A if((n>=1)&&(n<=12))
" ^/ N) v* t, T. _8 _ printf("%s\n",*(mname+n));0 j1 |! y0 m5 L! f; `% o' B. X
else
0 p1 b" @ u x( ]" J printf("error");6 U4 B& ~& s9 h& n
}
# s( R1 p, E, v8 z* T10.20/ h2 v5 Q- h. W5 @1 @
main()) x* |: l; T9 }9 [) a+ W
{int i;
. E- ?$ J) D" R char **p,*pstr[5],str[5][10]; q7 g* B+ I0 X$ g
for(i=0;i<5;i++)
! H* y4 R/ e$ R! F- q pstr[i]=str[i];4 i* m/ U; `5 l6 ]/ _6 n8 o2 u
for(i=0;i<5;i++)6 O1 G- ~2 k' J0 K/ z$ D; d" }/ i
scanf("%s",pstr[i]);) y4 {' T. _* K3 m
p=pstr;
8 W% D9 O9 [, D% {# j- b9 e sort(p);
8 r4 A9 d+ F7 F; I4 [. x6 n, M7 d for(i=0;i<5;i++)7 v( d" x3 d' |0 U$ [4 r
printf("%s\n",pstr[i]);
& h# a3 x' n" {2 z6 B: l/ G( p8 J}
. N0 Y" } K# l$ Dsort(p)" V& a" B# U5 ^$ F; i) {4 ?5 h
char **P;7 x- e5 g4 K/ l6 I4 }, k# c3 n
{int i,j;
% j& Z% h/ s- W ?3 F4 u6 A char *pchange;
/ e' z0 G% M8 [0 g9 L0 z/ [7 r: G for(i=0;i<5;i++)
$ W7 I6 }2 n3 z4 s {for(j=i+1;j<5;j++)( }' d- Z; |9 d- \
{if(strcmp(*(p+i),*(p+j))>0)
& s# Q5 [, A4 K {pchange=*(p+i);
. a0 D5 s1 Z. s5 k' h2 [) z1 ?# y$ h p *(p+i)=*(p+j);8 u$ K0 r, S6 N4 x7 o6 |2 N
*(p+j)=pchange;9 i3 [5 Y0 L; Z/ y F- A) G
}8 s4 D$ ]1 W+ a$ c
}
' |, R8 X$ t! S, B6 ?. I+ B- Z. q }
( l; H; F6 W3 Z( j0 |4 R* k# S}% h/ q- X" {0 l
10.21! Q: O) Y" O5 h
main()% b: N- r% P) p" o; |
{int i,n,digit[20],**p,*pstr[20];% O- [3 o% n, z3 ] Z3 L+ \: l0 W+ _
scanf("%d",&n); E9 V7 \1 j, e& n3 b' M C
for(i=0;i pstr[i]=&digit[i];
# c; L5 F% w( {+ w( B3 W8 M for(i=0;i scanf("%d",pstr[i]);
' m, P9 ]7 u# B+ V: i% G5 O0 p p=pstr;
1 s( u5 {1 V# q sort(p,n);
" b) _# _5 E: r* q for(i=0;i printf("%d ",*pstr[i]);
: K7 {' G1 w- i* \}
1 Q/ }, C3 H$ ~sort(p,n)
/ a5 A5 g+ u1 _" s: pint **p,n;
; [# L) r; Q# [! o+ y+ l/ [{int i,j,*pchange;% x$ q% k6 {4 D, x) R
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))) W% T) X4 o$ s6 e& E( Q" [) g3 p
{pchange=*(p+i);
3 f2 _- }6 ^% ?$ Z *(p+i)=*(p+j);
" ^; K% C5 Y7 K* U* c c *(p+j)=pchange;
8 b2 U4 ^: y3 U0 m6 j4 Y7 L: ?, ^ }9 t/ L; ]( s) N3 X. N
}! H; k- x/ ?! F4 B' q& \7 d
}
+ r |7 w4 |3 V6 U2 U* b! k}6 f8 g' H' k5 L
第十一章 结构体与共用体; u0 d |1 L u* y, f" l# k
11.1
. i3 s" `& ?" d/ p" a& ]8 c; E' Bstruct' k$ {7 |; i7 R: R4 U
{int year;, D4 J/ f w- P
int month;
+ |; e0 J# i. A/ C* s int day;
! R1 j+ K9 U# @' g }date;
+ k1 J: V! r! tmain(), I6 ?8 o0 y, n8 j1 N/ Y: H# l
{int days;
8 c: u+ h. ~$ y9 S' s3 D0 R scanf("%d,%d,%d",&date.year,&date.month,&date.day);) s. ~0 J" ~; P9 l% ^
switch(date.month)
" c( X/ j7 x- s4 h, |& y/ G3 D% _ {case 1:days=date.day;break;+ P* Q+ l+ K9 Q6 B& N% R- L
case 2:days=date.day+31;break;/ o- x9 L% `% A5 b2 d
case 3:days=date.day+59;break;
+ [- L5 C8 s$ ^6 G9 q6 e9 \! f case 4:days=date.day+90;break;% E" K8 n6 `1 z. i! t( r( H3 ]
case 5:days=date.day+120;break;
1 t. |& f: q* f2 c/ H1 Z case 6:days=date.day+151;break;
7 z5 H5 |* J- a# W case 7:days=date.day+181;break;
, w, U5 j) g2 e* T4 o case 8:days=date.day+212;break;$ G( }. S7 T. L
case 9:days=date.day+243;break;
0 l; C7 |9 O3 K* K( r& Z& i+ J; s' w case 10:days=date.day+273;break;
8 h1 |& m) w6 ~' d9 C case 11:days=date.day+304;break;
9 V9 Y/ W6 \3 f) ^1 { case 12:days=date.day+334;break;
0 Z( q& y$ h0 {! \* k }
$ O# H$ P7 M$ ~* K if((date.year%4==0&&date.year%100!=0||date.year%400==0)
# O4 R# m/ T" H* x* X &&date.month>=3)
. p9 f( k8 z9 J z days+=1;2 o( _; x! L7 z
printf("days=%d\n",days);# r9 Y/ T6 G. n% X: _; B8 B
}
/ z. N& r/ N1 ^2 s# E11.2
( o7 s5 O2 A" c, T8 ?struct dt
" l% X2 G' A0 T X% G {int year;& O, e% s7 O! y( G+ k
int month;- i$ K5 `) P s
int day;
7 B. t8 g% s. Y! S6 H+ f9 c8 f6 T }date;% {% d& ^2 q5 F5 I! v
main()* j' v/ R" P1 Y. @0 C& Y3 w0 v/ G, o
{
# A* q& ]9 O/ U$ z" i scanf("%d,%d,%d",&date.year,&date.month,&date.day);
( I+ C+ `. T) v0 K printf("\n%d\n",days(date.year,date.month,date.day));
! |$ J: J$ b# f}
7 g) |6 n' [7 z* {0 o adays(year,month,day)
5 V3 ~/ @) \0 m, {3 A4 D* Dint year,month,day;! Y( e- _: }8 G4 O% x
{int daysum=0,i;% C' M6 ?; m" ^
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}( @) r6 b9 j$ u2 F
for(i=1;i daysum+=daytab[i];. x; r; i/ v$ Z& D. h
daysum+=day;" T1 u0 b$ F) y+ s3 X- F
if((year%4==0&&year%100!=0||year%400==0)&&month>=3)6 t1 f5 h( O, `0 _& i
daysum+=1;& x3 i9 }, d* M! }9 }
return(daysum);6 U6 _$ `8 ~) }- r; ]& F( |# p
}6 q3 G4 ]0 Q' E0 Q2 d' x
11.3
! c* l% G d8 ~# u3 Z" a7 x' i11.4% }6 R @+ R0 W0 W
#define N 5
' j; w7 n8 b: ?, L% }& C0 Zstruct student
* q. v0 R; j, F* G6 [ {char num[6];" @/ b: |8 t' F2 E( I
char name[8];8 i& r! J) D- Q4 D3 [% W, z C! D3 {
int score[4];& U" ~$ q( {- B7 ~
}stu[N];% c g# L" n4 M
main()+ I( l' c1 b" Q6 L5 _( z
{9 f3 D3 ~4 M6 x
input(stu);
3 \# T6 K1 L6 d0 u print(stu);
. D/ `, o1 |: j}
$ o& _. ^( [# r% E3 `+ q% o8 S3 ~input(stu)
% ~! ^$ l/ q" ^ }9 h4 o, o5 ystruct student stu[];
: {" O+ S- |2 D, v b- X% y3 ^$ W{int i,j;2 i& Y( N: y9 s$ i( s5 t
for(i=0;i {printf("number");/ s# t6 P1 {2 J3 R% D/ l5 o
scanf("%s",stu[i].num);
C, G& r8 s, l. r% t" E4 R ? printf("name");+ d$ @0 K; P; ^& r; h* I
scanf("%s",stu[i].name);1 {' {; Z* j/ A& k# }
for(j=0;j<3;j++)
+ z& X& d% ~. H' \' l/ R( S {printf("\nscore\n");
' _* ^) S2 {" G# Z scanf("%d",&stu[i].score[j]);
- \, j9 B- v* G9 G }
. G0 q3 H. N: X$ }1 U6 g) J printf("\n");2 T/ d3 j. n3 N0 c7 r
}# [# Z3 X' E( y: ?8 C1 ~8 q
}
: n0 _( g" Q$ E! M+ lprint(stu)# I4 y3 X# Z8 A) Q4 y ]; Z. h& G
struct student stu[];
& n: q+ X% }, ?{int i,j;8 M0 z2 a6 r1 }5 m) E9 b) H7 X
printf("\nnumber name score1 score2 score3 \n");
& o; v/ d0 s; \% R0 R( e2 O7 G. [ for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
1 V8 [! E: w4 M A2 V for(j=0;j<3;j++)0 |# W: P) n2 O: b4 }
printf("%7d",stu[i].score[j]);6 Y" A0 `6 @$ t& q
printf("\n");
6 t: O; C/ ~6 M7 l( v$ W$ M }' H& e1 u5 o: J) V- ^2 p' F
}2 a; R- t& ?4 t) G: Y* L3 T
11.58 N" Y- f9 J9 k. @/ k, @8 `
struct student
$ @/ s1 m+ E9 v' A8 f$ }$ Y8 }/ n0 u5 l2 ] {char num[6];
+ M9 P7 ]/ x- x9 ? char name[8];
9 ~# a: ^: p; `+ \4 I' H int score[4];
7 `8 P y. g5 A0 J! y7 K float avr;
/ H9 _6 T: Z% d% D4 E( m }stu[5];
5 s5 M- l9 U- D- X8 |" T/ e: b. Qmain()
3 v1 n3 K% v4 K- t' n$ ~{int i,j,max,maxi,sum;+ p; Z; S& d! C. T: u8 y, n% l1 f
float average;
! U! H! w4 _; K for(i=0;i<5;i++)
7 L8 a) J d2 @* v7 l0 ]# \0 ` {printf("number");5 N$ g( i8 M7 S+ h1 Z. @
scanf("%s",stu[i].num);5 }7 Q& M: b5 ]/ [ e
printf("name");9 D# B: b& K3 z1 t; b- T+ n
scanf("%s",stu[i].name);- D+ z5 {2 g8 i9 c* V4 R6 N
for(j=0;j<3;j++)
4 F, e# J4 l. o. a' W- R {printf("\nscore\n");5 A- R3 Z7 B- T% Q# _$ R
scanf("%d",&stu[i].score[j]);
, }4 z% _: O. Z- Z }. q) n- E6 U; [! G* c) { h
}
7 R) Z* v& ]0 K average=0;
! D1 L& ]( i2 ?/ { max=0;) ?+ @# Z5 y2 A
maxi=0;
. n' V1 x" {+ ]+ C for(i=0;i<5;i++)% ]& V# Q" e* t$ f2 P
{sum=0;
) `. r: v- L5 z' J" ^, M% S% q for(j=0;j<3;j++)7 F* L- p8 A2 f3 j+ g2 z
sum+=stu[i].score[j];9 b& E: C5 g# O3 o) ]
stu[i].avr=sum/3.0;; z2 m( w$ {. X4 E; Q. ~& P
average+=stu[i].avr;: c. z3 Y4 U T: h% _% O) i* f' N
if(sum>max)% o; N" G& K" m5 {. h
{max=sum;
7 _, z8 o0 f0 _) } maxi=i;
* @) m" ]' J7 k9 E# P }+ A" p- b* g5 C# D% A [4 I
}
5 `3 q9 l. ]6 L. c average/=5;
" r/ Q9 t6 \9 U, O; S; m/ t/ w printf("number name score1 score2 score3 average\n");
7 p- k0 m1 K9 T } r/ s1 r for(i=0;i<5;i++)1 l$ v8 q* i( l- E8 C2 b: \
{printf("%8s%10s",stu[i].num,stu[i].name);
: R2 A& w r/ j- c for(j=0;j<3;j++)1 L/ w4 C1 T3 i, ^* m
printf("%7d",stu[i].score[j]);2 W( p# V7 k( ]8 T8 k, R3 Y8 f
printf("%6.2f\n",stu[i].avr);
' O" x3 [: Y- Z1 v: [: u- s }
3 u: D2 j. d" L1 h+ W. \& ]8 h printf("average=%5.2f\n",average);
8 \$ f4 i( n: y; A/ r printf("The best student is %s,sum=%d\n",stu[maxi].name,max);! `' `7 ~; E8 I$ f# C. E' a P# I
}- c3 g a- s: |, } b
* h& _ O: s4 v
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zan
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