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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
. w( {, d5 u& }- c6 i8 A1.5请参照本章例题,编写一个C程序,输出以下信息:8 A1 J1 n+ L0 j" f1 N% V
main()
) y2 n, y, j% [) Y1 y& n{
% @8 j% s& B. J3 Aprintf(" ************ \n");0 ?# P! u: l* O. W, p6 _" }
printf("\n");; l! p3 y' A6 O. }; o9 x" X
printf(" Very Good! \n");
! z) g" h& M1 u4 L: @printf("\n");7 a5 I; @5 A, r; J
printf(" ************\n");
2 \, B0 E6 c5 T- k}) h1 H4 S0 S1 m" I% U- b0 P
1.6编写一个程序,输入a b c三个值,输出其中最大者。9 V4 E- _# r9 q
解:main()
( P- q0 G" x3 y0 x4 c{int a,b,c,max;+ g& O2 Z( i. b* i
printf("请输入三个数a,b,c:\n");% S: \, ^. D9 S9 D
scanf("%d,%d,%d",&a,&b,&c);5 ?- P, O: v! Q' C
max=a;
; Q9 K/ ~ S) uif(maxmax=b;
" J$ o$ ~( N. V1 U$ @( Tif(maxmax=c;
$ d E$ T! H- f4 j! Y' kprintf("最大数为:%d",max);
( a3 h; f* P. w: L! S}; [3 `( @: g9 C, _: e
第三章7 q& J9 l& G4 T
3.3 请将下面各数用八进制数和十六进制数表示:
4 i# a0 O4 x9 ^9 r* J(1)10 (2)32 (3)75 (4)-617
/ ^$ @, d$ z5 m+ s4 F(5)-111 (6)2483 (7)-28654 (8)21003
, w# x6 J! q# ~# V" f, d解:十 八 十六
% T( I, L4 {7 l( I9 K2 r (10)=(12)=(a)% |2 W( }# a/ H, j0 S& [3 L7 ~ j
(32)=(40)=20* [ Q, t% q$ O' p) ^" U/ H; Q
(75)=(113)=4b
' ^2 Z5 h/ e0 }8 i (-617)=(176627)=fd97
9 y0 _, F" b8 d# `( z% L4 g! h$ z! ? -111=177621=ff91
" S' ^9 `% w2 M! |' \ 2483=4663=9630 A1 w, M: }( \& q
-28654=110022=9012& r, n# w4 l2 B$ u. q) I
21003=51013=520b/ M1 `0 j1 f) y. B& a0 C6 {( C
3.5字符常量与字符串常量有什么区别?4 N) b/ Y" ]# \0 s- B; Q, n
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1 }$ }5 `4 R t5 d
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'.
6 e' h) D8 ^+ @, O* y1 t6 z3.6写出以下程序的运行结果:
& {, @& c0 U ?& m- L* |4 Y) A/ K#include
0 t$ q$ J+ ]3 J8 c) s6 xvoid main()/ T5 Z4 }8 n1 E8 B7 a
{, }* F- X; }2 P( n$ v
char c1='a',c2='b',c3='c',c4='\101',c5='\116';" Z) R9 a' d% z/ t
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);
: F' `2 v& K! Yprintf("\t\b%c %c\n",c4,c5);
/ C- n; {- g" q解:程序的运行结果为:
2 n4 ^6 P6 C( ?aabb cc abc1 ?6 B, w% ~" V6 l: L* F% L$ c' X
A N
9 q& T" }6 S* g" i3 h I3.7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,1 c4 `9 \7 o1 ?4 B5 _& ~; _' `
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"., x& ^+ w5 P3 N5 Y/ ?! l, i) A2 t) j
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
; V# o, K$ p* N5 K- o输出.3 x+ G! J% e n9 `
main()7 s" S8 R. W7 l3 B4 M. X
{char c1="C",c2="h",c3="i",c4='n',c5='a';* Q- N0 A% z+ m( z" f0 R
c1+=4;
( w6 `: U; G, `4 cc2+=4;2 }5 W& q5 e# W9 s) X
c3+=4;- L+ m5 Z6 k" b; Q( R
c4+=4;
! [! y' v& _% p) c+ tc5+=4;& n- S' p1 V( g7 k- `- p
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);
. w1 y+ E: ]( }: r/ d& E& W; p- k}
0 }- i, b: B* u! R3.8例3.6能否改成如下:+ P( W( P: ~. c- W; E" H
#include
- v& C' |* ~+ S) p5 kvoid main()7 v1 t8 f+ c$ \4 ^+ n4 F
{: M. W9 X$ s: e& v
int c1,c2;(原为 char c1,c2)
0 l. d. [( j& `6 e0 a" Lc1=97;
0 s. B# E: q% B+ C# i& z7 Y% Pc2=98;
1 U) M0 x% s/ x; w4 bprintf("%c%c\n",c1,c2);3 M+ ]0 ^! u* \+ n
printf("%d%d\n",c1,c2);
! k7 Y2 w+ q5 E# _ {}1 C7 O8 r0 k |
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.3 s" N5 |& @( R
3.9求下面算术表达式的值.! r1 D9 x9 q: A: Q& D% a$ }8 h
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
4 f7 R1 l; s5 ]) s/ R(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)' F+ J0 y- d$ R- Z# Z6 {0 T5 q
3.10写出下面程序的运行结果:
" G, S3 m, E% R1 O: E: \: `. A& z#include
/ i4 A3 K- }1 j4 E( kvoid main()( T7 ~# f# z2 h5 q- h
{9 t+ V! H( W4 P( @) I8 k- @
int i,j,m,n;
2 v. K9 K$ l- P3 O, Ci=8;1 s0 M6 \4 O* ~$ k, H* f
j=10;
& w. X+ |3 u- j% i7 I- nm=++i;8 q- @$ E, C7 r
n=j++;
) {1 L' v2 T$ W* nprintf("%d,%d,%d,%d\n",i,j,m,n); A* @/ Q9 s% m* x
}
0 c/ i/ I3 P9 f) ?) [解:结果: 9,11,9,10& O) E8 s7 P' }- @' u2 }/ F
第4章
5 z. N3 `, J3 x( h) `" m4.4.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! g5 C3 ]$ {( F- ^% d0 ?$ 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:! b9 A; h, c. M. X7 y, n
a= 3 b= 4 c= 5* B+ y+ O+ C4 R _0 r
x=1.200000,y=2.400000,z=-3.600000
- d% v9 U3 C* y4 m+ W4 Bx+y= 3.60 y+z=-1.20 z+x=-2.40
" U; k2 ]! U. R2 {u= 51274 n= 128765 w$ w4 Z# {. B8 Z2 c
c1='a' or 97(ASCII)
$ ?+ i9 n6 l0 z6 [* W; y, Bc2='B' or 98(ASCII)
0 J0 I) `6 N/ M解:
' m3 t4 u3 y* N0 J6 Dmain()
6 ~( ^4 H# g( y2 f1 c1 t& u, P{& R, U; d$ ]8 p
int a,b,c;
8 D) r( _2 }, r8 b8 K4 z5 flong int u,n;
$ d7 d2 H1 O* Y* J) N Vfloat x,y,z;$ b* m$ k5 X( @* P
char c1,c2;
7 J* {- {5 H8 E' D% J' ha=3;b=4;c=5;+ ~3 V; C& Z3 |2 s1 D6 h$ ^+ D
x=1.2;y=2.4;z=-3.6;
2 l& V, U5 {5 T. S* w, ^/ X( x, `. Iu=51274;n=128765;6 Q! |( ~9 a& o% l# J) J/ ~* `
c1='a';c2='b';
6 y: Q! F5 f, G$ Uprintf("\n");
( j0 }9 r i, D7 Cprintf("a=%2d b=%2d c=%2d\n",a,b,c);) T" N; p6 W/ a: L/ C6 e
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);8 L7 v! `' i6 z- ~
printf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);$ C: S0 e* `' m) I/ w8 n
printf("u=%6ld n=%9ld\n",u,n);
+ ]7 {# z5 m! m8 V a0 p5 D! tprintf("c1='%c' or %d(ASCII)\n",c1,c2);; g7 c, y' v2 J. v7 @ @; E; ?
printf("c2='%c' or %d(ASCII)\n",c2,c2);
$ A2 c! E: ^% l1 t0 f}* d9 g4 n% i! T1 V% n
4.5请写出下面程序的输出结果.
+ |. e, V; i" \1 \- X( v0 k% t结果:$ O( I3 F5 a/ D* d2 c8 O
57
! m, h7 t+ R- i3 J1 G 5 7, @0 a+ l1 H2 P7 b! G2 @7 T
67.856400,-789.1239622 R2 f/ [, c' C
67.856400 ,-789.123962- g1 a$ M# R8 ~4 r- R4 S9 b, Z
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962
! g) Z* G! L* ?6.785640e+001,-7.89e+002
0 r9 w. p' \ z( d' Z- GA,65,101,41" R' z8 k3 }' x( \! T9 [" e* b, \% N- ]% u
1234567,4553207,d687$ w& O# o- R, ]$ a# K$ f
65535,17777,ffff,-10 I r5 E. M! r# M- e
COMPUTER, COM5 z# c' A) k! n0 M
4.6用下面的scanf函数输入数据,使a=3,b=7,x=8.5,y=71.82,c1='A',c2='a',
4 ]4 K# x# p- X9 Y+ k3 Y2 r+ x问在键盘上如何输入?
3 C3 d8 y) t2 ^0 J5 K. hmain()
: T7 A; _, C& V6 G( X6 l{7 W( b5 G: i$ H- I, _* H
int a,b;" Q" N- k' [- E: R$ V4 F) G6 e
float x,y;7 M' E; Q6 w/ y1 m1 \8 b
char c1,c2;2 J7 l# D' ^5 y, t5 [3 f
scanf("a=%d b=%d,&a,&b);
4 t/ I/ B5 l) @7 \0 Q% gscanf(" x=%f y=%e",&x,&y);
- f. T5 H, E6 R. H! H" S( Cscanf(" c1=%c c2=%c",&c1,&c2);
3 b8 L/ E% y# K}
- b' F+ }4 F, E# y解:可按如下方式在键盘上输入:, z6 C3 a* J h: {) D
a=3 b=75 W2 F! t; d6 V$ v& L s* A
x=8.5 y=71.82
1 X! u- m- c. u; p Uc1=A c2=a& ^$ J! t' j* @" y3 M% }$ M( k5 g8 T9 Z
说明:在边疆使用一个或多个scnaf函数时,第一个输入行末尾输入的"回车"被第二7 u) u$ w4 Z& ]* {. w8 p
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9 i$ C; m7 O( F9 b
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
0 s! t% L7 d% n7 x8 o5 ~初步概念即可,以后再进一步深入理解.- ], U+ c1 g2 u0 ]& A
4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-3 z3 ?% B# Z) G* l9 F" A& U7 v
3.75,z=57.8,请问. }$ K6 t, Y# d O9 M* l0 w
在键盘上如何输入数据?
% i! v( |+ J2 m1 Oscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);
5 w( k& F: x) W* m解:
d& K, }# e9 x9 hmain()
* E& I$ z4 n9 g7 V7 | S: ~{
) ~1 n3 M, i/ |; F6 R; {; Zint a,b; ~' H0 W% `# _( t' t7 W' l& N
float x,y,z;
' @+ _. x' l7 o2 Y/ [$ Uchar c1,c2;6 e2 y. u% g* T/ I: H
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
9 |6 g( x$ \% B5 k}! `+ {. Q1 c' P( i
运行时输入:
5 t% z' F/ W8 ~10 20Aa1.5 -3.75 +1.5,67.8# c7 ^1 x7 w; u/ a) w
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
* i" ?+ | h I0 f! j; o* f( M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
* d; r9 B7 L0 w0 R# F会赋给任何变量。7 n, h8 m3 y9 v% ~6 u; j
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,$ p- c* r( V+ V1 d4 o: s
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; ]4 P6 ~7 f( Y( n* [4 i
程." {# S2 r+ Y+ ?( j
解:main()
$ ~* l4 P( d0 Q4 I C{
! a3 _5 O" L9 D5 H+ V; nfloat pi,h,r,l,s,sq,vq,vz;
& K9 R. r0 H5 N8 o# d: ^6 @% S2 A: ypi=3.1415926;% U9 E& L$ X) T. D
printf("请输入圆半径r圆柱高h:\n");( n& F( d6 y+ P% Q! y4 F; V% p4 l
scanf("%f,%f",&r,&h);
% f' j4 c. _ d* D7 @. jl=2*pi*r;
& }, L5 U2 \8 J6 C! q- Ks=r*r*pi;0 i0 M3 @! C3 l) }( b% j# \
sq=4*pi*r*r;$ D8 v U7 u9 ~, M! @. [
vq=4.0/3.0*pi*r*r*r;. K) _; X5 j) D
vz=pi*r*r*h;5 }" x7 d# ]9 X
printf("圆周长为: =%6.2f\n",l);
& \+ P% ^3 }. y3 M2 n" A$ A, {printf("圆面积为: =%6.2f\n",s);: z4 m, N% q2 y: a3 u
printf("圆球表面积为: =%6.2f\n",sq);: b+ C) C4 {7 |4 u/ z6 Z
printf("圆球体积为: =%6.2f\n",vz);
/ i+ O$ @8 G4 ~5 F6 X* A% ]}" y& b, E W$ [8 }
4.9输入一个华氏温度,要求输出摄氏温度,公式为C=5/9(F-32),输出要有文字说明,- B1 i; I7 w, T' E& t8 A7 w% ?
取两位小数.# | D# I# n; H" E' ?; c$ E
解: main()
" g x7 t% X' |. |8 \) _! j{
- B6 |/ Y- n+ sfloat c,f;) w6 K/ v8 W- @% C3 E# l, c
printf("请输入一个华氏温度:\n");- ]; q I4 I5 M: |- ?2 ^3 B
scanf("%f",&f);
. @; r/ U, p ^1 n) y' ?c=(5.0/9.0)*(f-32);) s/ W0 c: C4 t3 ]
printf("摄氏温度为:%5.2f\n",c);
( z$ k& M V3 F9 w# X}) v4 G) y6 ~0 R* `7 h' E6 W
第五章 逻辑运算和判断选取结构3 q* \" O3 S$ q0 ^' A0 o9 i7 G
5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.$ v, T& M5 V. U" W+ t% E! j
main()
- V0 J6 j ~$ V, r; a5 U* c{
. M6 O0 I# a- u+ e. y/ r6 y* sint a,b,c;
0 y& n0 o5 X. \1 h/ Zprintf("请输入三个数:");6 D5 i. ] f2 s2 E# U# j
scanf("%d,%d,%d",&a,&b,&c);( Z5 o! a+ L$ q t+ F
if(a if(b printf("max=%d\n",c);
x! [" j ~ L else! A$ c" v1 J2 t: C8 `* [& ?
printf("max=%d\n",b);
) a3 T; N7 o" l7 |else if(a printf("max=%d\n",c);. D# W/ r- t p O; T" B! B
else
3 c ?6 w: ]: t: `/ c printf("max-%d\n",a);* G+ H f& ~$ p2 B o# u
}
, {! a6 u) g! h3 K! B; s1 d8 V- [方法2:使用条件表达式.5 S* u5 @# e3 V8 N" b; ?
main()
0 b( \" {8 f# j{int a,b,c,termp,max;
6 H4 b+ H# R; N2 {/ d$ y: }3 v! E _ printf(" 请输入 A,B,C: ");" f% Z0 c8 _3 A, U
scanf("%d,%d,%d",&a,&b,&c);% g; ]2 x+ @$ m; ]! {
printf("A=%d,B=%d,C=%d\n",a,b,c);! t9 X5 }7 \5 v* d0 X: I
temp=(a>b)?a:b;
4 C) z( m: u8 G* X9 ~ max=(temp>c)? temp:c;- H7 r, Z" w; h0 G$ @$ D" V% {
printf(" A,B,C中最大数是%d,",max);
8 h* L; { t2 \9 R}3 i/ }! D' ]" b: a; L+ V7 I
5.5 main()8 u1 j4 d2 O+ Y& `, g ~5 {( S
{int x,y; f7 N5 J: X# f8 v. Z8 A
printf("输入x:");; k5 T1 B6 R5 p( e
scanf("%d",&x);1 D) m s2 h; f4 m8 K( b& Z/ ^
if(x<1). l1 `6 U, G- E+ `' U
{y=x;
; c* ?$ m. s& @' S7 | printf("X-%d,Y=X=%d \n",x,y);
0 F2 y+ e" \' o: [' e& x6 |& u2 | }
) z0 M; {, H. y6 p& a( f5 B5 m+ {else if(x<10)
5 w% V( K1 l w3 O- q4 l {y=2*x-1;3 b" `6 |9 G: Q( s& j4 e9 Y, V
printf(" X=%d, Y=2*X-1=%d\n",x,y);3 Q" C) a( C4 N8 G! J
}; }" j( O1 _+ u: o' L- i% i
else
, t2 X" |* b$ n4 \7 A# E {y=3*x-11;
1 ^+ d4 d, W3 Y x printf("X=5d, Y=3*x-11=%d \n",x,y);
% c; w! I0 Z7 \" z4 | }2 x. N+ Z. r& M Z* }1 r
}
8 ]; f# j+ C- `. |* c e(习题5-6:)自己写的已经运行成功!不同的人有不同的算法,这些答案仅供参考! . Y6 `1 H; M- k5 f) [ ^
void main()
' c& M3 l% ?0 T! Q5 V7 M{6 I' T/ W% ?4 i2 }4 M
float s,i;
( O# Z" S, ?, c4 M7 d& \( n, Lchar a;3 l2 n. u, B u' @- w
scanf("%f",&s);
) ^9 g$ Z9 {: F2 r( `while(s>100||s<0)3 l( w9 D4 ~ H3 `
{
6 {$ ]6 G. G* I1 e+ S' J4 Tprintf("输入错误!error!");
* I G( F- f$ y @1 v& ~7 @* Bscanf("%f",&s);. `3 p& K4 o) k; j
}
2 d9 y! u9 T: h, G% S, ~# _i=s/10;' }9 n- K7 j% r( x- P
switch((int)i)+ U, x; ~7 i% b
{- o: H4 g# K& _) \) y1 f
case 10:2 L9 r1 B. m0 P5 h3 O( X8 P
case 9: a='A';break;7 a7 p: U, K R1 ?7 Y% R7 B
case 8: a='B';break;; r( B2 R5 |- ?2 z
case 7: a='C';break;( N' M, h! j3 n, J# A
case 6: a='D';break;( a/ M4 s* ]$ N7 t B5 L# {1 h
case 5:
$ [, q0 ^. f6 W" {case 4:5 I) b' s+ L, l" K' f3 c
case 2:& l( P) O9 F0 }8 A% Z+ d6 _
case 1:
! `$ y5 o( I% f# h, ocase 0: a='E';8 \$ T% b$ E3 z
}
/ E) O- X* h& T: f/ _printf("%c",a);) `! y* Z. @& ? p
}0 T! ?* g: {. C1 \' N
5.7给一个不多于5位的正整数,要求:1.求它是几位数2.分别打印出每一位数字3.% E( U& t9 B) p: Y* j
按逆序打印出各位数字.例如原数为321,应输出123.9 h) h) M# Z" O
main()4 ?7 W$ f- i2 Y2 S" K, @& X
{
0 J. n% D5 j$ `% | long int num;/ s/ H, F% x3 ^
int indiv,ten,hundred,housand,tenthousand,place;3 I8 t8 z4 F, [
printf("请输入一个整数(0-99999):");
$ F1 ?6 D% [/ {7 B/ {: t scanf("%ld",&num);
2 d6 I" u. b+ E E) R3 I if(num>9999)( P D i* r1 F+ F9 g. i
place=5;
1 m. d, \6 s' I: Delse if(num>999)
0 v, p: e$ ~* T place=4;, b0 W- O4 S, H2 a9 X0 |
else if(num>99)
b* o! l2 N# q( m) K3 i place=3;
' S: k' M2 h5 L& `: i3 Helse if(num>9): t& {% p# _6 B% P8 c% _
place=2;+ q$ b- n& @/ K( \4 `' u
else place=1;
! p4 I& W. q) a' ]* A6 fprintf("place=%d\n",place);
/ a* m6 B; T( qprintf("每位数字为:");
, N. h- K) S0 w. V" |" Oten_thousand=num/10000;# E2 z8 w+ V. y& N; r
thousand=(num-tenthousand*10000)/1000;
& s, V6 c0 f; T/ [( yhundred=(num-tenthousand*10000-thousand*1000)/100;
( ?" Y2 w1 B. `* m" F" ?3 o* H. {ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
. R2 J: \2 A- C9 K3 L- Iindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;7 f8 x( S! o# C c( Z1 ]" l4 ~- _
switch(place)
) Y% x0 @9 G& Z. t8 A: ^{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
! F7 S5 q2 Q1 w L& @9 X @ printf("\n反序数字为:");7 R& j2 }7 S4 R) E1 Z, W7 \
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);
# {) ?8 D# \+ p& ]5 y) A# `1 C5 a break;
4 I* D, W" G4 K& A2 ~3 `) L' h7 }0 hcase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
* j% j& F' L! W* E* [8 P- e: `' E printf("\n反序数字为:");, }+ l, m7 ?0 B: `0 _! C8 z
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);
9 s" B' {+ [1 s! u- Y8 o break;
6 o: h. v" Q" ~7 m, O3 K2 F2 mcase 3:printf("%d,%d,%d\n",hundred,ten,indiv);
; r: O! ^- W7 n: F8 I1 F( H* D printf("\n反序数字为:");
5 p7 A& U. i, P2 o7 S j. D r" ^ printf("%d%d%d\n",indiv,ten,hundred);
& g' u5 {/ x" K( G! @3 i- Gcase 2:printf("%d,%d\n",ten,indiv);6 T! [) e! e- Z# d7 B3 o
printf("\n反序数字为:"); T- o: [' v- N% q6 Y( F
printf("%d%d\n",indiv,ten);
" V& z- F2 k& K/ ?7 icase 1:printf("%d\n",indiv);0 S4 s, w$ p8 z" x( H8 U; J
printf("\n反序数字为:");
, y8 Y' `: ~+ Z# L) V printf("%d\n",indiv);
; S. e, [$ Q( d/ `; ?( k. ^ }$ K2 M6 B* c O W" w ~3 a. ^! c
}
! x0 s* L2 C$ H- X$ @5 a9 I5.83 o5 y9 O% z% q" f
1.if语句
0 D# u5 J; l, M$ E4 C+ C. Bmain()
8 ]' C$ g; h, ?. a, X{long i;
: `1 r7 f4 H y! ^$ N% F float bonus,bon1,bon2,bon4,bon6,bon10;
H, I! E& e0 y5 z. ~: K4 s bon1=100000*0.1;/ E4 e* Q* z( R( Q% V8 t2 p9 q
bon2=bon1+100000*0.075;* D: G; p7 ~7 D! k9 S+ L
bon4=bon2+200000*0.05;
4 e+ ~6 Q/ q$ i* t& U' t bon6=bon4+200000*0.03;: M; [+ G' c! P/ ~+ x2 K
bon10=bon6+400000*0.015;8 y- u% y+ Y& A
scanf("%ld",&i);( ] T: D( r N' I' h
if(i<=1e5)bonus=i*0.1;
/ J* J8 m2 E0 z' Z6 Q Z& e4 Y else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
9 j3 t5 w, A1 l% A7 l) V0 O else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
2 M; |* X8 a$ n# D else if(i<=6e5)bonus=bon4+(i-400000)*0.03;
n! L% c2 n- J/ ? else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
6 p, {) j3 \0 v8 K else bonus=bon10+(i-1000000)*0.01; ? |" z" Q2 J; N+ Q# P
printf("bonus=%10.2f",bonus);; h2 |- v9 I7 F" [* c9 I' ]
}
1 T3 _& ?* Z: Z! f用switch语句编程序
2 d6 B7 U% G' ^5 @/ p8 X& Nmain()! s0 Q3 R7 `- F# g2 a X
{long i;" b# ?+ ^. j* C ?. H- b
float bonus,bon1,bon2,bon4,bon6,bon10;
9 p2 m; k' A f0 [) I" c1 [" M9 O' U int branch;
" x: L) n4 c1 S1 ?: t, @ bon1=100000*0.1;: l. C6 F3 X- l; Y6 K$ |
bon2=bon1+100000*0.075;% W# o; Z7 n! a0 t+ K8 p/ X
bon4=bon2+200000*0.05;
G/ g; v* C, z- q bon6=bon4+200000*0.03;5 X( g8 B7 U7 Y
bon10=bon6+400000*0.015;* r- i. B6 B! `5 L) J" i& J% m! U" d" S
scanf("%ld",&i);
. E$ x% C2 D3 P' \: Y branch=i/100000;4 G; r7 Z* Y8 ?& m; f
if(branch>10)branch=10;
5 u- x- r R V switch(branch)
' @1 s* A& H2 K {case 0:bonus=i*0.1;break; Y7 q% v" u% c5 C, f% k
case 1:bonus=bon1+(i-100000)*0.075;break;9 N! B5 M) Z* V' L+ i
case 2:
* w& l4 r; u5 K+ | k; E case 3:bonus=bon2+(i-200000)*0.05;break;4 B# r( T' ]) k5 ~8 c; e
case 4:$ s4 c$ b8 w, a5 _
case 5:bonus=bon4+(i-400000)*0.03;break;
3 _) T |& U' x case 6:
0 z" U F$ A K* ]% J( D case 7
+ S' y$ f, J+ ^4 L case 8:8 l8 o7 i% d! R% u, a4 m; h6 d
case 9:bonus=bon6+(i-600000)*0.015;break;
! |9 E& R2 u C' \6 w case 10:bonus=bon10+(i-1000000)*0.01;" ^' a8 K$ b* R2 l# D/ }
}3 O/ W3 C3 H& D& ?4 [
printf("bonus=%10.2f",bonus);
: N+ s' q) r, o$ ?% P- T}
/ [, j/ D2 g& F2 @8 J: Y5.9 输入四个整数,按大小顺序输出.5 s( t9 }2 a3 N7 ^- q% k# C. Z
main(), ?7 \* k5 i" y8 x/ F, s2 V
{int t,a,b,c,d;5 |; {1 }2 e& W4 s$ X
printf("请输入四个数:");
* V k+ t( _% F. G) `& C4 P scanf("%d,%d,%d,%d",&a,&b,&c,&d);
4 T: I4 T' C7 T% }+ t- P7 C# b printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
5 M i' y. S1 p" D$ [) O if(a>b)9 `" ~) i9 D/ V7 q- D
{t=a;a=b;b=t;}% x5 G* c F) X/ Z
if(a>c)# A. S9 `1 n( q! r
{t=a;a=c;c=t;}
P/ ]7 y) Q- Z7 i% s; z6 q if(a>d)/ h/ @5 c+ S2 a. V( E2 L
{t=a;a=d;d=t;}' M/ P( v, i1 P
if(b>c)
2 L: v7 A7 `* z% p* y {t=b;b=c;c=t;}7 x9 i2 P7 f. J3 f& \- I
if(b>d)6 W5 x" k! s9 I. o+ e" Z, Q
{t=b;b=d;d=t;}5 q; `- U7 |1 P# S2 i+ b
if(c>d)9 i; B" X# O; U) Y, D# c, N
{t=c;c=d;d=t;}, ], M) }4 u7 R7 {
printf("\n 排序结果如下: \n");
- l, M( @$ K0 z3 I {printf(" %d %d %d %d \n",a,b,c,d);
: ?6 A- s3 G9 r9 H7 U}
2 G* Z8 D" P, B2 G5.10塔
* {1 f7 {! L1 ^ {! l5 x8 bmain()' }& V- V; g0 d& s! K% j
{" h3 n- m' w1 H
int h=10;
! n6 g" k9 k- Jfloat x,y,x0=2,y0=2,d1,d2,d3,d4;) Q3 A, q% I$ `4 B/ M7 {
printf("请输入一个点(x,y):");- U b4 ~* R( k( K. s( Z
scanf("%f,%f",&x,&y);9 _3 `& g- b0 g8 l. l
d1=(x-x0)*(x-x0)+(y-y0)(y-y0);! g3 r7 Z4 m3 \: m N$ G7 t
d2=(x-x0)*(x-x0)+(y+y0)(y+y0);' `8 n6 p. e7 u3 B
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);) m( R* S2 I4 A* Z
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);
! n1 h& f9 ]( o, M& s3 I4 g2 yif(d1>1 && d2>1 && d3>1 && d4>1)
& \" J8 A( V7 G5 gh=0;, E. e+ T5 A1 y% E& k3 i: x- i
printf("该点高度为%d",h);$ n9 x' b6 V! ]6 s) M- D: M
}0 A, _0 M6 S) _8 e2 a5 M0 M
第六章 循环语句
1 Q& R+ ~: X! |) P& F' p: Z7 Q6.1输入两个正数,求最大公约数最小公倍数.8 X- I' m8 y) A( x' H {* M3 f# `1 F
main(), B5 f7 [( \# X( p7 P' O# f
{
; t" K( i( c4 `' Hint a,b,num1,num2,temp;
( D+ \8 ~, P C$ s" p- Eprintf("请输入两个正整数:\n");) X: x8 W7 n' h" t
scanf("%d,%d",&num1,&num2);
* H2 y0 l* |( n9 F1 u/ @if(num1{
, h0 p( H3 F9 p7 y0 V( |; G. E7 {temp=num1;
; N4 K0 E# t D3 V8 Lnum1=num2;1 | o* J) Z9 P! E: Q8 ?; C; R
num2=temp;
4 f5 c+ I5 O, j( O u}
7 N$ }' ?, Z( ea=num1,b=num2;0 x; B1 ]# @6 q5 b2 J8 P+ {/ Z; Q. H
while(b!=0)
0 S7 M; g- V4 n" p6 p' e$ `3 e" f" s {
0 e' P6 y. u) u0 Y5 S temp=a%b;
3 P$ }3 b" G- r a=b;
/ a4 E3 ]; B. [( \9 x1 P( a b=temp;
7 R8 y/ F+ ~& p1 q$ Y. J }
5 p8 V' p1 e: U( f1 l2 n, gprintf("它们的最大公约数为:%d\n",a);7 l* h! W2 Q* s$ F6 j
printf("它们的最小公倍数为:%d\n",num1*num2/2);( y6 v8 n `, f1 U D: {
}: p5 P4 _' P% H, D- s
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.
. }2 s( ~& M3 U0 {+ |' m B解:* F1 Z7 a: q' u7 v0 L8 O
#include < >! W* i- t7 r+ a P$ ?1 U
main()6 E- A2 I1 k3 _9 R# z* R
{3 n& e, q2 b! n
char c;
0 x/ c# d2 L- d4 Wint letters=0,space=0,degit=0,other=0;
, y6 A/ w% m# b2 ]/ xprintf("请输入一行字符:\n");0 A+ B: z, y2 n/ Q' R
scanf("%c",&c);3 a) f% y8 j* R; [" j
while((c=getchar())!='\n')
8 r, h+ O, G# h: ~{% v+ R$ `6 o/ u5 i& y* ]' N& k# c4 J
if(c>='a'&&c<='z'||c>'A'&&c<='Z')
8 _+ A& G; G7 i- r1 Jletters++;
! |% l; j* `+ F$ x6 l& Zelse if(c==' ')& Y; i% }7 q/ I. g9 e/ F
space++;* p8 V. X3 m$ j. _4 L; P8 ?
else if(c>='0'&&c<='9'); @# V( ] P: M5 o
digit++;. U' e) |$ E6 E F) X5 f" ~
else
+ w! ?8 {& W1 E/ A- ]other++;
% m( |3 A0 o7 N% _+ P0 T}( C8 V; y y6 `4 ?* W3 k
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
& N1 f% U8 b6 P$ b6 dd\n",letters,space,
& {8 J- r9 Q7 U/ v% m7 e( T2 _6 Udigit,other);2 T: c4 B1 w( m/ ^$ Z1 i
}* a! n5 ]4 A, S( }9 K* q' Q
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.& Z& x- t4 x' h; w: b
解:. F& c* C2 R4 p0 S0 R
main()
$ z2 y2 I. [) o; h% p{2 y) } Y* q9 ^* |
int a,n,count=1,sn=0,tn=0;8 L# ?0 d: K, K0 |: S
printf("请输入a和n的值:\n");; y, j- j2 j: u% ^% K
scanf("%d,%d",&a,&n);
" i7 V7 ?& r6 h7 L/ Kprintf("a=%d n=%d \n",a,n);- ]( r" t0 Z9 h; B2 N/ H; \
while(count<=n)' `7 r) Q% E( {
{
7 i, P# n$ ]- p' o) m) K4 A8 z. |tn=tn+a;
4 f" ?. G. v# csn=sn+tn;) E4 D9 u3 J: k. Z; `
a=a*10;! m! }2 y2 B& N1 U& x: s- `
++count;
5 i, a) V+ S( y" L) ~}
: T1 }! f! }4 t( }7 Oprintf("a+aa+aaa+…=%d\n",sn);
- z- }! v0 B, v- d, H/ l6 p}3 ^" M, R- l+ {2 w2 S
6.4 求1+2!+3!+4!+…+20!.2 p- r3 ` x* F* j3 t
main()) Y3 d" a Z% w; K9 f
{ e4 p7 P( v" G# `" C
float n,s=0,t=1;
, D) L* d" v7 W9 w) W" d4 @3 Efor(n=1;n<=20;n++)- u# G6 G) u9 r7 {' L
{
! m' e) X" f$ ot=t*n;
- v/ ]/ X3 b; ]9 G- fs=s+t;. N% O. M# E. I7 I4 N. o6 \2 K
}
$ @# M1 J, k7 o9 _; W/ E0 Nprintf("1!+2!+…+20!=%e\n",s);/ q' H/ L+ a8 E% G
}) r. u4 z( @' G1 W
6.5 main()
' |) ?3 Y2 [% E# q$ a- l- |4 V{
' _% s2 Z% g6 \5 t% }int N1=100,N2=50,N3=10;4 j7 H7 F5 P+ @! z! T1 q+ V
float k;$ R2 j2 m U1 ~5 F3 n! E
float s1=0,s2=0,s3=0;; S5 L& S7 t. o& M
for(k=1;k<=N1;k++)
, y7 {8 Z+ F5 h{
9 ?, [1 n+ C, ^4 Z4 Y0 ls1=s1+k;, {2 e6 k( E( W, t1 b% h4 w
}" m' ~; r. r9 s
for(k=1;k<=N2;k++); e, Q* @8 } i/ E
{5 Y2 |& x/ t5 V
s2=s2+k*k;% q0 v/ e0 l, m( J, G
}
^$ A; V) D* nfor(k=1;k<=N3;k++) C5 X' I7 Y$ |" L( u4 n" }$ M
{3 n" `$ C+ [! M" |- X
s3=s3+1/k;, N( C& @7 O& }) z Y: P
}
. o* Y0 ~! `6 p9 o' X' cprintf("总和=%8.2f\n",s1+s2+s3);
: ^* Q0 W* G1 J4 E% e a! J4 ]* J}( P) f" N" ?# M( k; _! T
6.6水仙开花
# m! ^, H# J. x& I* C% omain()( j. T# V4 C I
{( j* \6 {; U$ W1 O
int i,j,k,n;4 s# \6 J& [$ Z) ~
printf(" '水仙花'数是:");
4 [7 m! w! u- i4 u! Q. bfor(n=100;n<1000;n++)
6 [# Q1 x9 Z0 _{
& Y/ }6 S# j% i5 k! c$ ki=n/100;
# N1 i D, r2 `4 `. t+ A9 {& d/ Ij=n/10-i*10;, p) @4 k/ N+ W+ m/ d' k
k=n%10;" I' F+ c* s" x: X& c
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
2 v% G" a& R( C- R{) x, e! b' j D$ o
printf("%d",n);
( E9 V$ S% j9 A( L: G# v; x7 x8 Y}4 Z: ]. `( f5 B p* T
}
5 a% s7 S6 H- ?printf("\n");
# x5 k) L, C5 l& B& d}
- Y: _" y) ]( l( s% R& N6.7完数6 n8 I, o* _! ]
main()2 k2 x" ~$ S$ ~( e
#include M 1000
/ G d6 O$ i( L& ~; fmain()
: z5 `$ D* ]0 d% Z7 s; ]( ?" t{9 r/ K- j- Q: S) G! A
int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;1 g$ V9 j/ j) I& t' H2 k
int i,j,n,s;
+ r& Y1 T; z, C5 [7 Zfor(j=2;j<=M;j++)
# V3 l7 N" r6 P/ m- _) }. v7 w& T5 b, B{
& J4 M; u1 E/ F9 i/ dn=0;% {& ?. p( S& f
s=j;
# x1 r9 a7 T2 b$ L: ifor(i=1;i {* [: K z/ V4 L1 K( Z: R
if((j%i)==0)
6 h1 j) v& E# S% i5 H {* G) E ~, a W) s; |( z
if((j%i)==0)
1 i0 F6 q+ N% M$ ^0 L: {" h8 V( V+ F {
) }& i- p8 Z! x9 T- c$ i$ H n++;
4 p. L6 R, [$ ^ s=s-i;
' D7 `' w8 g3 L+ W4 C! L" r switch(n)( e% m8 w9 w4 u9 ]9 b8 W
{
) \6 a5 \5 h9 Y/ o$ s I. U5 w* T) J% [ case 1:1 ]/ r% N1 l! D
k0=i;3 A+ z6 r" p$ A5 A& w
break;
$ k6 ~' O# i2 I/ L$ c2 \ case 2:
9 d" [; J& `. c8 U* g% A9 N k1=i;) @, u: X* ~) h U) n
break;3 `+ g# x* V$ W1 A0 B) J0 e# a
case 3:6 y, K) {+ L" D# K n
k2=i;8 c% @* Q; p7 r( u3 [7 d& l2 z$ q
break;; [3 ~0 A0 J I+ g/ J$ s: ?
case 4:1 r1 Z1 \+ O& X; D
k3=i;
% P$ ?0 Y+ t' H break;" x) V8 w: u& a/ i) }( y
case 5:
5 e: s) b- z% L0 a- z. F' \4 }5 e k4=i;7 A. L5 v, ^% c" X$ k
break;
! i- l9 ^7 Y/ E" y9 T case 6:
! H/ k9 W$ [1 y! y k5=i;
. }2 G# a0 a" H7 Q; p break;' x. y' K& r a6 j' t
case 7:
4 Z& J9 _. t9 n. g$ ?$ y/ C k6=i;
2 E3 T& Z2 O0 Z8 [8 x) M" L break;3 n, z" K. r( Z; @: x' x" P
case 8:
5 w5 f, l2 H* R k7=i;" E9 c' [2 m4 O+ h8 D
break;
4 ]9 U) ?* b( `) V+ t& c; L' @* C' S case 9:4 f, n: f j9 G. ~3 G" _: P+ c& i
k8=i; a/ x& f7 }0 z. @' ]- L
break;
; H1 r2 N5 z; ? case 10:
! r; i% ^- m! R5 s* _3 ?" C% { k9=i;6 e; I) h+ M- B
break;
" c& {: A) O: y& Y0 ]% c( T }
6 z. r5 o# S+ i. e8 F }
4 C, \6 D$ g1 \, G( } }" ` e) L' W9 P
if(s==0)8 @6 z. C3 O7 O3 k3 l3 e% M2 A6 x
{- i) T+ B7 a3 g) k1 C( x( E2 k/ T! {
printf("%d是一个‘完数’,它的因子是",j);; t s P& r; m4 S! I
if(n>1)
7 e {4 p* K# N- u! r9 V printf("%d,%d",k0,k1);# H4 `( `7 W# L( j5 h
if(n>2)
) O, o0 o! E+ L& ], i printf(",%d",k2);
, H/ B1 D9 a. A4 r* g% ]. [+ p! z& uif(n>3)
9 a% G/ A, r# n- J" B; {2 z printf(",%d",k3);
4 F* @3 \' [' _- x1 ?2 Vif(n>4)1 x* k" n' U- H- a
printf(",%d",k4);
( }, {6 X* h+ J5 | jif(n>5)
, L# v% t- ? Z printf(",%d",k5);
# n* G- ^. j) a# t9 c& p# l) E9 q( G/ Jif(n>6)
/ ?9 t, h9 P' Z' {* s. [' w: o# n: U. V printf(",%d",k6);
3 H# ?4 b$ P2 Z$ d/ qif(n>7)7 N1 D% W4 ^3 Q3 [
printf(",%d",k7);
$ W& @; \0 a2 c( X: @: ]if(n>8)
% s H1 a, e) L4 a printf(",%d",k8);+ D8 m$ _2 X3 Q4 `3 N
if(n>9)- q7 T; r+ ]+ L# V$ X
printf(",%d",k9);" O3 n7 W6 i, a4 A. Q
printf("\n");) Z& f# Y5 R; @# {- W
}. {, a' ]1 ^' D) T0 \* n( F* o
}; u1 _! ?- t, a' u( R& {- B$ C
方法二:此题用数组方法更为简单.7 T2 R2 Z; t8 R
main()
: P, M# v. J9 u, H* v{
- H1 u3 Z( b2 Z/ Ustatic int k[10];
) L7 q7 Y. C- jint i,j,n,s;
' V0 f1 t$ I/ y) E, k( Ofor(j=2;j<=1000;j++)
% w7 ^$ P# o8 _( m- U2 p" W{# z3 N* f( d. F2 f2 r
n=-1;/ w/ N- }5 C( X5 h* v
s=j;- @/ X# F5 n7 y- w! Q; |. A
for(i=1;i{
/ k( a9 G" q. u$ X0 X2 m% T' eif((j%i)==0)
( h( H, n2 w5 z n{
6 D$ ?. W9 G; W% c' G/ Jn++;
" l& p! z$ t. @; f+ ?; Ts=s-i;& Z1 P1 L( R V, M
k[n]=i;
% {! X; l" L2 I. b3 t! n( Y. j, Q }
* S; P: n0 r& ]8 l# `; D- ] }
8 P, X% I' z X5 vif(s==0)8 s% J5 C! y* \
{
- m' Y2 Y$ A: v; K# b1 ]printf("%d是一个完数,它的因子是:",j);2 q" t `7 \" o3 g3 N
for(i=0;iprintf("%d,",k[i]);
7 D1 s% b8 L5 v7 I7 p8 Tprintf("%d\n",k[n]);
1 \7 @# `/ k# M) l}
`% B' m' r! q5 j. @}
. y1 ]0 D* a; f% ?, ?' B l6.8 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.! }, }3 w# _3 T1 B! i P7 s
解: main()& n6 b6 m2 h: {' Z! n/ Y+ g
{
7 r, r( r- D9 Y; Qint n,t,number=20;
2 ^$ U& h/ c2 m$ lfloat a=2,b=1,s=0;9 X' S" ?2 Q5 F0 U
for(n=1;n<=number;n++)
' J0 e, @- `( b{1 {9 D) j# R8 S: e
s=s+a/b;$ n1 [' r* W/ T- q6 E* W
t=a,a=a+b,b=t;/ X( S: x. f* \* t8 q8 [2 c
}
" S3 ?9 B% M4 N- r* a7 Uprintf("总和=%9.6f\n",s);' e9 G& x8 {* U/ |; y5 g
}1 W; t1 _" y/ F+ n8 p0 [6 k
6.9球反弹问题
8 S2 w9 A) }) ?2 V zmain()
# f+ D' ?) {0 G3 d! N$ v{/ t8 H9 z# q" k8 Q% ^7 U( o. D$ `
float sn=100.0,hn=sn/2;
8 ]2 Z7 P2 I% N8 B$ l5 o- Eint n;
; L7 X2 {0 v; }+ P' x0 B% xfor(n=2;n<=10;n++), {# z' B$ V4 i1 u0 ~5 k7 z) |
{
# ], p! C- r( a$ E& E- K/ k+ tsn=sn+2*hn;
; j9 [+ R4 q9 n2 c7 }hn=hn/2;
" S* B. O2 Q# a$ C5 }}
s3 C2 |# Q% T. D: nprintf("第10次落地时共经过%f米 \n",sn);# a* G. a M. ^& a0 z% U
printf("第10次反弹%f米.\n",hn);/ M5 Y. W, u( o8 f' u, _ H6 J
}3 U4 G7 U* k5 s0 J! f1 j
6.10猴子吃桃( h" s5 W* C: ^# C
main()
. c W7 G1 h( i6 o6 ^) b0 N{! H6 M5 t* h0 D4 w9 z: s
int day,x1,x2;
& |" S( z5 g9 zday=9;
1 U7 J# F+ M2 d d/ {. \$ rx2=1;
* G0 l) f: L2 j& v/ f: v$ swhile(day>0)
0 |$ @- ~0 Q) S+ c2 E; M8 P6 ^{) z) }, c) E" q& o. \ o5 a+ h& T
x1=(x2+1)*2;& |5 q9 R; d/ v
x2=x1;
& A9 z$ h' c, U* \ R; Y& `$ B/ mday--;
, ^% D, w6 P, c}( x+ H% b; x( n# |2 F: i* \
printf("桃子总数=%d\n",x1);* n. f T( W1 j$ G7 [
}' }. c$ M) ]; l: d, d1 b
E1 w5 w. t' s% Z; |6.12
' K9 L, D0 U# L4 f3 @" M. z#include"math.h"
) n, b! u' W. e5 kmain()0 v/ l, C, t0 f- l3 O2 [# R
{float x,x0,f,f1;
4 X3 {4 r( y/ ?7 J. H9 g x=1.5;" q) `1 G7 d4 f; k$ B5 F$ ^
do
I M! b# N# v1 V) v {x0=x;1 C$ N# y/ w8 e8 o2 M( Y
f=((2*x0-4)*x0+3)*x0-6;
* d- }2 m. Y' G% i' A! k& x f1=(6*x0-8)*x0+3;
, R \9 i) f- ?3 p0 X x=x0-f/f1;
0 Q2 b. F/ f1 Z" i, i+ g }2 r2 v9 j9 W' n8 M
while(fabs(x-x0)>=1e-5);9 y- i+ h1 T6 Y7 q* x$ |
printf("x=%6.2f\n",x);" S- m# B; P Q: R
}
+ X# x @3 n) Y# }8 x8 A d' Q4 H" ]8 H9 P5 P
6.13: ?7 W! j; d+ ^( i9 ?
#include"math.h"
: l8 |' I4 w H# X( c8 Q9 Umain()3 j% v% n# j2 M# E1 a9 x
{float x0,x1,x2,fx0,fx1,fx2;5 _" M1 z9 P0 U
do6 `& s4 X2 B4 r" g6 \
{scanf("%f,%f",&x1,&x2);+ Z" j$ E' J7 a6 V: i* i
fx1=x1*((2*x1-4)*x1+3)-6;
( Q' o0 } O2 ?! d8 k% R- O1 p9 U+ o fx2=x2*((2*x2-4)*x2+3)-6;
& f0 P# J: W9 n6 Q+ ? }
5 d ^$ M$ T+ w while(fx1*fx2>0);" r4 w% X5 [' P6 y; c) P" t- N( m0 j
do% n$ ~+ D8 b" ]8 r4 k
{x0=(x1+x2)/2;
. R6 c4 Y' j/ }' _ fx0=x0*((2*x0-4)*x0+3)-6;
5 e8 ?; g( ~7 e. y if((fx0*fx1)<0)
/ c* T' t# e% I. v, r7 ~6 j {x2=x0;
T( A" }( N, X6 n fx2=fx0;3 K y1 t8 f' d' T" @9 [" C; }
}2 v: z% g+ } } l1 ~; |% f0 o! D
else. O3 [( t2 {2 z+ s6 i# H4 R; @
{x1=x0;& D/ B0 W3 r5 v3 V
fx1=fx0;
# d% q: ^" s& d }. l* L# ?2 `5 _
}& E& A1 s" S8 c3 r& z
while(fabs(fx0)>=1e-5);: o% J, k3 `( R3 z% @2 Y2 J
printf("x0=%6.2f\n",x0);, G; Q" y5 a; c" _' R" _
}
* [, w4 N0 w! H9 B6.14打印图案! o7 {$ B( T: p, p0 R6 R
main()8 s; G9 w! b6 y0 ]
{int i,j,k;
, w4 c$ r+ u- l# \) |+ e: x* T for(i=0;i<=3;i++)
" E" v* y0 G' v2 H4 `4 D" S F {for(j=0;j<=2-i;j++)
6 ?7 ?! Z/ I \' }" h printf(" ");
$ G7 i x2 i6 P for(k=0;k<=2*i;k++)
- r& r( I, n! o printf("*");
' {' {2 w" @2 P printf("\n");8 `4 H' k7 k7 A8 B: O
}
1 X" Y- n8 `$ u9 ^' M) c j for(i=0;i<=2;i++): o# B* F# u2 A0 }0 O
{for(j=0;j<=i;j++)# f% o5 W% n& Q: P4 D2 T8 s
printf(" ");
t2 u) K' G3 ?2 T for(k=0;k<=4-2*i;k++)
4 o% _. `+ X' j printf("*");
% \! K/ S2 @7 @. p& P2 p# Z printf("\n");
* F! ]. d2 G5 q2 [, Y) e2 g }
& G& {/ o; q# a0 o' P$ _8 u}
. G% N' V' Y; C* b* z% T. W) j6.15乒乓比赛, g6 R, v, n' r( K r! I& Q
main()
" m2 c9 p5 y+ j, j5 a( N# b{
0 K7 Q- L8 w" a: tchar i,j,k;2 S5 n' o3 ~ y. M$ W9 _1 P! S
for(i='x';i<='z';i++)& l& I, u' L) v0 z) t7 l
for(j='x';j<='z';j++)$ V# }' X+ U; E0 C
{
; }- A8 t' h0 Y9 ?% \if(i!=j)1 u4 }) o1 o2 X4 L. P
for(k='x';k<='z';k++)
$ T! O& q5 d) ]1 G( Z% N {
0 \% h2 _# ?2 M G2 j; Z4 Zif(i!=k&&j!=k)
) u0 [, ]' ~. D( D' H3 z( J {if(i!='x' && k!='x' && k! ='z')
% m( s8 ^# U0 _# a2 _- V- Iprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);9 q" \" i5 S, ^* k/ ~* I6 J* F
}
+ A" B* c7 z% q K3 n; r" ^ }
+ X3 W1 k; ]) I6 I! K1 P }8 A2 F0 h; j' W& ?: H
}; Q9 A; V5 }8 Y
C语言设计谭浩强第三版的课后习题答案. \$ W1 y+ I3 y: t
7.1用筛选法求100之内的素数.: ?; W, p, `: N4 j! J, }; {
#include
+ z A r: n6 J: U7 |#define N 101
6 n& E( [0 c% Bmain()" G. z) _- W. E
{int i,j,line,a[N];4 s! [7 W/ S T6 y1 I
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)6 z& g8 p( h8 O! c9 k" ~( x9 t. N
if(a[j]%a[i]==0)
% M2 o9 Y7 A: o a[j]=0;- r3 e$ s7 c8 f
printf("\n");
+ K- ?9 W1 I+ S: _1 Gfor(i=2,line=0;i{ if(a[i]!=0)! n4 |' f v' D" \
{printf("%5d",a[i]);' \/ h0 [; I( J5 _8 W' g
line++;
1 M9 |; i; `# z2 p7 D- Q if(line==10)
8 T5 A3 M8 v7 }. `* | {printf("\n");4 G- {1 I# F0 |" n/ e8 s2 W- g3 i
line=0;}! h9 I: ~! Y* T' a; ^9 Y3 s
}- ]. @6 _9 A6 g" Z4 [
}
0 _ [) b6 i& |+ v0 c7 m7.2用选择法对10个数排序.5 L. c) e( G5 W! R: M" w, H9 h
#define N 10
0 s0 U; Z! W* [1 z* @2 Vmain()
& O( r& Y$ K1 p0 [2 C{ int i,j,min,temp,a[N];
* W7 B3 n; p. ^% q/ d3 Uprintf("请输入十个数:\n");8 \3 X+ e7 J$ X/ M
for (i=0;i{ printf("a[%d]=",i);/ u& B- y8 D' q' S& f; D2 y
scanf("%d",&a[i]);: B( H9 v0 g- l$ X' S% }
}2 f# L( V+ _& R
printf("\n");- _2 q5 u2 P. ]
for(i=0;i printf("%5d",a[i]);9 `$ {, r& }, h4 X+ Q5 z
printf("\n");3 s5 Y5 W' M( u
for (i=0;i{ min=i;
7 H3 H7 R3 G/ ~% b( l for(j=i+1;j if(a[min]>a[j]) min=j;( m$ |0 `: z. W
temp=a[i];
_/ ^% l6 J3 U" C% j e3 j0 l a[i]=a[min];2 Y7 F7 Z( E4 Z+ h* q
a[min]=temp;
$ G. Q. S& H. K# P- B}
9 _3 B) \: h/ m Nprintf("\n排序结果如下:\n");. G2 a" w5 e0 K; A% j/ q5 F+ ]+ ^
for(i=0;iprintf("%5d",a[i]);5 M% u0 m% Z; h9 B' G
}
. U1 i: P; |/ W) w) U1 B5 N9 @7.3对角线和:% [: Z5 f" ~; K2 p% S, Y, ^3 f
main()
( w" l$ R. a3 L0 l; t' f5 h8 k{
7 ~, E, W) p8 E$ w4 |" g Wfloat a[3][3],sum=0;! L5 d: U5 C% o
int i,j;3 l8 k `, l4 V; Q2 W# [
printf("请输入矩阵元素:\n");
' n1 L+ k/ j D) u% j7 Kfor(i=0;i<3;i++)1 j: Q2 A( t! G' U; X7 r3 w
for(j=0;j<3;j++)
* C6 c L7 ^: `; Q scanf("%f",&a[i][j]);
1 j: y( G8 _8 Y X3 Z* M for(i=0;i<3;i++)8 H5 n3 v9 m! J, U. O; H
sum=sum+a[i][i];
0 Q3 T- m1 _2 g printf("对角元素之和=6.2f",sum);
* C+ L0 _+ m. P; l7 F}
1 c. E9 R2 x/ o7.4插入数据到数组4 U2 O% A5 \& n. S
main()
* r4 a3 h: Y" u$ O T- A- Z. V9 L{int a[11]={1,4,6,9,13,16,19,28,40,100};; w1 T( z5 [/ V( O* V
int temp1,temp2,number,end,i,j;
" f4 P D- m# f2 Kprintf("初始数组如下:");
9 }* s7 }- \( o- Q1 rfor (i=0;i<10;i++)8 n7 s, x" H" K$ z! O5 k1 y5 {
printf("%5d",a[i]);
9 ^/ f6 v& ?. u1 Nprintf("\n");6 q4 i. q& m- ^
printf("输入插入数据:");% M6 }9 ^$ P! k; o" [5 w
scanf("%d",&number);- U( O& V- j/ E5 c' r8 P; h8 m
end=a[9];
- {) e/ {; ]5 o4 o3 U$ i5 h! j7 L" Bif(number>end)
! M8 U: ]2 ~" Z$ F* e0 q# |a[10]=number;6 g% h2 e* j5 Y, o$ V
else
8 `2 d# |( W7 X6 t$ M' i+ S, M+ G. S2 b {for(i=0;i<10;i++)
/ C0 q! w5 ?! i! T6 v { if(a[i]>number)
- ^0 C. @4 o6 ]8 M% _: j {temp1=a[i];4 }% M0 t* w* a1 }/ N! {
a[i]=number;
c& ]) J3 t" L8 _5 n for(j=i+1;j<11;j++)
* g( ~$ F: Q$ S+ H( h, ?- C7 o {temp2=a[j];
4 |7 L: ]$ h- p# [" ~8 ^ a[j]=temp1;# z+ ^' V2 y3 D- o1 \+ q
temp1=temp2;2 l. L4 n8 I9 I7 B+ B; E
}
, d3 L7 z E( l6 L4 O break;
* P" j/ k7 u& [/ d; A- G5 `( l }- `4 Q/ _" G9 Y
}
7 S8 |$ }9 e. O; [" l }2 A" v2 R4 i& o4 o6 D* ?
for(i=0;j<11;i++)
/ o6 F8 l. B6 D printf("a%6d",a[i]);4 E- R2 z; r0 s8 _- ]2 W+ N W0 D! o
}% M# @8 j: ~& `# ]% G# a
7.5将一个数组逆序存放。
# X6 U! i. ?& I. ~7 \#define N 5
8 A+ S' ^2 {3 |; |. Ymain()5 j8 g$ `+ q, M% D
{ int a[N]={8,6,5,4,1},i,temp;+ q# ]6 s: k5 }+ D7 J6 _
printf("\n 初始数组:\n");
! Z/ k& c8 c0 c- nfor(i=0;iprintf("%4d",a[i]);
9 t0 _! v* f1 L- N% ~: dfor(i=0;i{ temp=a[i];5 R! W+ }0 |& l; }
a[i]=a[N-i-1];
7 [- {0 |+ z1 O9 X+ r3 \3 o a[N-i-1]=temp;3 B! k( t& g. ?6 ^$ p7 O
}1 y4 {2 Y, \1 q W" Z9 @
printf("\n 交换后的数组:\n");
& U6 j0 _; f: H; Q, B+ @for(i=0;i printf("%4d",a[i]);
6 j+ P0 K9 `; S}
8 ?; ` I' d7 C) P3 s- g! O7.6杨辉三角
- W1 S: K7 {+ r; G+ j# k#define N 11
3 r0 D) B9 c; N8 N) l, ?& Lmain()
: B( h( t$ n2 K5 b Q% K9 Z% {{ int i,j,a[N][N];9 }% X& [+ S6 y& Z; o9 W6 g3 j+ K
for(i=1;i {a[i][i]=1;& M* ?' k6 W1 q; l% y9 M
a[i][1]=1;. l8 A F9 y/ S0 F
}( l* y3 U4 f( v. s0 {9 c7 s
for(i=3;i for(j=2;j<=i-1;j++)
% E$ g" E9 |9 h5 p$ V a[i][j]=a[i01][j-1]+a[i-1][j];6 y& g! Z9 z$ |! Y
for(i=1;i { for(j=1;j<=i;j++) X: P0 M/ @$ r% B! m. ]
printf("%6d",a[i][j];
8 B: f/ v2 x/ P' A) D, ?+ ] printf("\n");$ f, }% G, s0 W0 A8 V+ M
}9 q& i& d3 U3 \1 C
printf("\n");. K- `% m @ V
}0 d& l% d7 P$ J: u
7.8鞍点. d7 ^# C* w: q0 f* a' ^! r% T2 s! e, i
#define N 10
1 z, ^* H. d; g! S#define M 104 X* }, n( K. L Q# a
main() I/ _* n: x6 D
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;
- l$ x4 ]8 V/ ^ p5 v( ^ printf("\n输入行数n:");3 o% O/ K& }* y% |/ ?4 i
scanf("%d",&n);
8 ^ y" i3 k" ~: U) p9 A printf("\n输入列数m:");
$ R4 A: S0 I- x6 ]/ \ A+ F Z scanf("%d",&m);9 a1 M$ |+ Z, z7 b7 [
" y( S8 ]. h+ ?, T5 ]8 l4 L' E: y
for(i=0;i { printf("第%d行?\n",i);
% S- H9 w% w" | for(j=0;j scanf("%d",&a[i][j];0 c& n" l* Z, i: e; h
}
+ `- m; H: H8 ~( R+ L- k7 c; x for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
% v1 z' [5 H7 N- ~- @/ y3 A6 G pritf("\n");+ a2 k( b H* I
}9 a' u; h+ A6 s+ [8 Z' i9 @- A
flag2=0; D9 A; U7 V# ]# t! \: j$ ]
for(i=0;i { max=a[i][0];
! s( I7 O, |9 @. e W for(j=0;j if(a[i][j]>max)
2 ]0 e$ a8 |% d3 }; d7 L i( A { max=a[i][j];
0 i8 R7 W2 t, e3 l1 ~+ U( l# X maxj=j;9 Z2 A! a7 s. ~9 o& y! @2 Z F
}
& N" _ y* ^# p9 ]6 m2 l for (k=0,flag1=1;k if(max>a[k][max])
% ~# Y2 `( S, A flag1=0;( U4 s2 ^. A! O( W
if(flag1). F8 a0 J+ T/ Z( ?3 S
{ printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);/ N5 a9 {! d% @5 @) u$ b" U
flag2=1;5 `% P' w2 f( c* D
}7 [1 a( M8 H) _0 V5 {
}' K; W# H# m+ r
if(!flag2)6 o3 a8 o8 C4 [9 g
printf("\n 矩阵中无鞍点! \n");8 f* N' m# l8 r0 a/ E5 x$ |; ]
}4 {* m9 y7 u6 ^ F- @1 j7 ?
# W6 {. W* j- N$ y8 l- s9 v5 P
7.9变量说明:top,bott:查找区间两端点的下标;loca:查找成功与否的开关变量.
0 e6 @( P, {4 `1 ^6 P6 m( j$ q. d#include
2 n' [5 i" h6 O+ E: I#define N 15
\; t* @4 D9 R' amain()4 Q/ d/ {/ ~/ @! D5 g$ L
{ int i,j,number,top,bott,min,loca,a[N],flag;
7 L6 u8 u5 N0 W! a: c* F char c;
2 `: Q, j. J7 I# M$ k printf("输入15个数(a[i]>[i-1])\n);3 A# U' H; ?# ?2 A" f
scanf("%d",&a[0]);
8 z; T7 j0 I& d2 S+ q; J i=1;
O' Y D3 v0 V while(i { scanf("%d",&a[i]); d' V/ U. G5 x2 q
if(a[i]>=a[i-1]), P4 F2 r. p6 w( S9 S K
i++;
3 Y: T( y9 w: I- n9 v, O6 G esle; D% o ?3 p! X& ~- g" s
{printf("请重输入a[i]");
' h) |; K+ ~) O! i. s1 h printf("必须大于%d\n",a[i-1]);
0 B: |4 V4 {8 W& b; F }
& j$ a7 T+ A; M6 W8 e! t: c' ^ }
! X5 E9 [2 w! x% F printf("\n");
G! c3 d: h3 E l5 p$ y# v for(i=0;i printf("%4d",a[i]);
" d6 J5 }; E7 ^6 X/ m" W# Q1 O2 r printf("\n");
4 C2 B0 A6 M Z
# _7 S6 J6 q! E9 i8 Y- U! b/ | flag=1;
2 P# H+ ^; C9 j. M+ r" \ while(flag)5 A. p% B M# C! m6 E, `% @7 ~
{1 S) Y, \- V1 B+ T
printf("请输入查找数据:");$ v+ m$ e- n& t: v. i% H4 r9 o6 ?
scanf("%d",&number);
; K* p- u# F6 w loca=0;- s3 [: I: N, j1 n, n
top=0;
" x) q3 ? M9 h% K# q3 s7 ~3 v bott=N-1;
! Y1 k& c8 E3 C" C, A6 G if((numbera[N-1]))
1 x$ L& { j+ v( m/ h9 L j, v loca=-1;
! A ~$ b7 R" k. L while((loca==0)&&(top<=bott)) I4 A- u1 w6 f/ }5 U
{ min=(bott+top)/2;" J1 W3 V a, L& b3 S Q4 a
if(number==a[min])
: o. c& p7 K, ?. R) T; e; A { loca=min;0 b' `2 ^2 R6 U# ?5 L H a8 L
printf("%d位于表中第%d个数\n",number,loca+1);' p R$ C6 ~- V0 Q4 Y, l- B& `
}2 S9 |$ _ w7 j: R5 ~
else if(number bott=min-1;7 a8 g1 }' i) u9 p% q
else
+ n. @& [5 J- |! C top=min+1;+ r# b2 c, _ \# U* f
}
) s3 T1 N2 r0 j9 E4 p, E( r if(loca==0||loca==-1)
5 e" }6 @1 s, p! C# s6 G printf("%d不在表中\n",number);3 N% Z9 ~% \' d; m, J3 e+ r
printf("是否继续查找?Y/N!\n");; M% k6 g. c( G' C/ L8 S3 g
c=getchar();; L( h1 z9 w" Z
if(c=='N'||c=='n')6 `4 c" ]- L6 k5 j4 Y6 D
flag=0;, k9 f G; e' `7 A9 b
}
. f1 N+ _% |- F& T- F" N}
% g |% J& x8 \" r 5 Z' J! V2 L5 J: R" }; ?( A& P
7.10
, ^7 T* B& K/ P# H3 M/ `- Hmain()
- |' c+ G. Z1 j$ x; g{ int i,j,uppn,lown,dign,span,othn;
8 \/ A# V/ s% h" \ char text[3][80];8 `2 I, a& s0 E+ q8 U; y, J
uppn=lown=dign=span=othn=0;
0 h3 J( R8 P" ^5 x! F for(i=0;i<3;i++)
3 a5 |- P: @* S4 j9 B { printf("\n请输入第%d行:\n",i);
& y2 ^) G# P4 ` gets(text[i]);( ~9 d& X$ J! o7 ]# i1 E
for(j=0;j<80 && text[i][j]!='\0';j++)
- R' ?3 f) j: ?4 l r) Z/ d {if(text[i][j]>='A' && text[i][j]<='Z'), ^9 N, l/ M7 U i
uppn+=1;8 h2 ]1 O3 X- ~; C6 B# b
else if(text[i][j]>='a' && text[i][j]<='z'), |) o* n. ]9 V8 U5 i
lown+=1;
" v6 a/ m; V7 o! t2 I# d else if(text[i][j]>='1' && text[i][j]<='9')' |) x- @; s$ \; q& Y, y/ Z% ]
dign+=1;3 i6 R/ m" z j7 `4 \+ S/ z3 F
else if(text[i][j]=' ')
9 n# F1 g6 {# r6 a span+=1;" p5 _# v6 n. t. F
else
/ ]2 y% w3 b, v' y0 r othn+=1;3 K+ q6 l$ D' e6 C3 a6 e
}3 o: v& D, ^8 d7 g
}+ K2 V' S! p& d5 |
for(i=0;i<3;i++)+ P" Y- E4 g6 H! d+ P
printf("%s=n",text[i]);
" @3 R% ~, h$ {, z) J# x9 h printf("大写字母数:%d\n",uppn);
' D) E2 q1 ~: u4 i! d, b printf("小写字母数:%d\n",lown);
8 G8 p( V& O" A6 r2 l printf("数字个数:%d\n",dign);
+ V: S4 Y3 a' @8 z, v+ G printf("空格个数:%d\n",span);
5 h. J' Q' v1 p* R$ I2 \3 \( W3 E! \ printf("其它字符:%d\n",othn);
+ X$ _" O" g$ `5 U$ C, [}
; ]4 k7 U3 e* J- D# r3 e4 A/ W% H* q" J8 h8 r7 B
( ?7 x; I9 @% ?
7.11% u" w7 n H ` X5 W
main(): |2 c. X: U. A& ^& P7 [: J+ D
{static char a[5]={'*','*','*','*','*'};
6 b' G9 z3 ^$ c9 {$ f/ C* _0 C' o: C7 } int i,j,k;
" v8 A; ~. m6 l, R; S. M! T: \ char space=' ';
- M; k, X% K: z1 c& [9 ]7 v& F for(i=0;i<=5;i++)
2 S) H, c7 Z8 z) Y; c {printf("\n");
. l. ~$ e! f) D5 j2 P5 |; _ for(j=1;j<=3*i;j++)
% k: U* f# y0 g9 d- J" I printf("%lc",space);/ A. z2 `& |, L7 B" q
for(k=0;k<=5;k++)* K% Q9 Z; W5 X4 M t) _! Y8 C+ s
printf("%3c",a[k];
* J8 i: ~4 V7 t* H7 {% e4 |% f. J }+ y7 f r: p7 [3 J- Q, n" g1 J0 c5 l
}
) G5 ]$ }; h$ J4 x6 w$ o( n7.12' F. _+ g/ r, h
#include
, a g% t0 I$ Nmain()* u8 r e6 V9 v; V
{int i,n;
: {2 a8 I. d8 d, ]: g- m- w) i* ~ char ch[80],tran[80];
7 {& v% E' H+ F6 v printf("请输入字符:");7 x1 G. a2 _0 ^$ \
gets(ch);1 D" `4 |5 [- Z
printf("\n密码是%c",ch);8 N% `4 W1 n7 D6 Z
i=0;
5 g P- {, M) }2 nwhile(ch[i]!='\0')0 F* i) d6 N" {% `: L
{if((ch[i]>='A')&&(ch[i]<='Z'))9 R* u- g6 H3 L3 A/ B
tran[i]=26+64-ch[i]+1+64; } \" L5 Z2 R. i0 m X& H0 X/ l( F
else if((ch[i]>='a')&&(ch[i]<='z'))% W }- M& R6 B: d0 ]: D: ]5 `
tran[i]=26+96-ch[i]+1+96; P/ Z7 s( g1 G; E/ s6 M$ g) m
else
, P' i+ k5 M! w& f0 c tran[i]=ch[i];
8 b1 |3 F$ q8 M+ Y$ [. ]# [ i++;. r/ b7 m0 e X; i' X
}2 n" A1 C# S$ |, v7 ]: }
n=i;
# c" u, F# C2 B1 q* i8 I9 s/ Cprintf("\n原文是:");* ] c6 {; |/ t1 @+ \$ c
for(i=0;iputchar(tran[i]);
) Y& E! L3 I" t* Q6 j9 {% l}
9 G. E$ Q' o8 _( V l7.13
9 c- t5 H1 d# E8 @main()
$ }# Z0 C h9 a- h. T+ { {7 G* o7 [4 ?* |4 A
char s1[80],s2[40];! D3 f7 |9 j& I7 v0 V& L4 ]
int i=0,j=0;
( q6 K* }$ y6 f* D0 F printf("\n请输入字符串1:");
6 x, s" _" v/ J" B% k# h2 K) p5 b scanf("%s",s1);1 l% C. y! M' D3 j2 ~1 m$ V
printf("\n请输入字符串2:");! L* d& {7 [( B9 B
scanf("%s",s2);) Z$ B& \( K2 ~- U9 N6 V, i& C
while(s1[i]!='\0')
4 x' L, q6 o: ^" e# l. d n5 U i++;
4 X7 T: \; i# E. B. j9 Twhile(s2[j]!='\0')
+ y6 R" `0 V; Y9 Y s1[i++]=s2[j++];4 R, V: O7 Z) ^# D: c5 c* J
s1[i]='\0';7 X# y) V5 a$ l6 Q$ U( l2 n
printf("\n连接后字符串为:%s",s1);0 K* L: V& z, Z
}
2 l2 @( M" Z0 _, T% ]: j3 `
% n4 g9 t5 b& w! H! A
0 H# m! x( H" F% A/ e7.14% M6 q/ S* t' V/ J! q
#include! l( O4 S7 x- B
main()
: p+ C5 }: U" R& d3 R- o$ ?{int i,resu;* ]/ E; d# A% j' g8 {0 ?- b' C
char s1[100],s2[100];
/ J* @) m$ e1 s( w$ Q g printf("请输入字符串1:\n");1 ]' r. v- Y5 n
gets(s1);, `" r; u' z- \2 L4 Q+ }% f! \
printf("\n 请输入字符串2:\n");
5 M. p! K: N. y% w) l# m gets(s2);! R! i. u Q! n% N+ w% O8 u
i=0;
, A* [7 T2 m8 I while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;5 o% O5 V6 I3 D/ m' X+ U
if(s1[i]=='\0' && s2[i]=='\0')resu=0;% M1 V' E9 P5 v- b c0 r5 u0 U6 Q
else0 Q8 `& M9 J9 ]" u
resu=s1[i]-s2[i];
0 _4 J) g' T g3 Y printf(" %s与%s比较结果是%d",s1,s2,resu);* I# Y5 S2 U4 Q* X% F6 Y
}1 B' E4 R0 f# Q
7.153 _1 [/ j+ f$ J: l. u* G+ ?& L
#include: M6 E/ p% T: ^- i9 S4 P( R
main()
7 a' v2 Y. v& d: c {
6 J. j% y3 M6 W9 T. h char from[80],to[80];
. Y9 t% P/ r/ R/ H' C int i;
3 |" Z+ ]. l" Z! _* o; ]1 [ printf("请输入字符串");
2 y* F$ }- h& q scanf("%s",from);" x/ i3 U" n7 I e
for(i=0;i<=strlen(from);i++)+ h2 z2 ^$ G' k y* v0 P* M! W
to[i]=from[i];
1 @6 j' G$ c2 v/ @' n printf("复制字符串为:%s\n",to);4 ]/ F! u0 A7 I! V0 }+ v
}
2 h6 l5 W0 R6 Q! T4 ]4 `0 |9 ^; z; j* ?9 N/ O- ~. S6 a
/ U8 x- y2 z- _8 e第八章 函数3 P3 a$ q8 r# x! h. q
8.1(最小公倍数=u*v/最大公约数.)
7 l7 d" f' a ohcf(u,v)
* \& h. c( ?/ M* Yint u,v; ]9 p( Y6 c4 s; X( F: B
(int a,b,t,r;
' r+ \) X: u' ^" j8 z, I if(u>v)
) X1 i: l s2 D8 i5 p6 A, L& O! T {t=u;u=v;v=t;}9 \6 _3 T* l3 P4 q7 b& P7 N
a=u;b=v;5 \7 R, |' s; p. b+ X, `
while((r=b%a)!=0)9 K- I. g) T1 c a# U
{b=a;a=r;}
j. A+ t8 @" D- {- o O return(a);
4 b! u" j+ S# t9 A }
+ ~& Y/ _+ A! S% G' R$ z3 U ? lcd(u,v,h)
% G1 W) q8 g) f int u,v,h;
& d1 n) ]& v+ W6 c g {int u,v,h,l;% z2 c& y$ T# g
scanf("%d,%d",&u,&v);
6 g& J3 V3 x/ V+ o$ k( K h=hcf(u,v);
& x0 F. Y8 p% o* o" D( ?" E printf("H.C.F=%d\n",h);; c* y% @8 u- j, W
l=lcd(u,v,h);8 f/ v; D! E% h, J/ r
printf("L.C.d=%d\n",l);( q1 S+ u$ @9 }. R7 E6 O
}. P" l# W/ ?8 {/ y! U/ W
{return(u*v/h);}" ^6 p: _3 V" w6 c( m' j! `
main()
& e$ z, s% Z. q {int u,v,h,l;7 }9 C2 x( F- j, ^3 ]7 D, [5 o, S2 n3 E
scanf("%d,%d",&u,&v);
4 y" R, r* `+ u5 W2 j/ n4 r h=hcf(u,v);6 O! ~6 w6 n$ S- S3 @
printf("H.C.F=%d\n",h);: H; o0 b% a, ]7 Y# b# C; G
l=lcd(u,v,h);
8 A* ^, J1 M: n; g; V' Q6 N/ p printf("L.C.D=%d\n",l);
, f# |. ^: b1 ?; G4 |7 A }8 ]9 ?4 a, H, _" V* H
, t/ p1 c& f) v2 p8 |
/ m8 k+ j2 H) a* T. t& K- m1 M
0 `# z3 V, ~ m* k* H8.2求方程根$ I3 [7 B( i- ?# {2 z. A: p
#include
7 X3 F+ h8 p9 j0 W3 Y+ V6 G2 H: q5 a0 ^float x1,x2,disc,p,q;
3 Y0 R( ?1 u& a9 q+ ^- C/ \greater_than_zero(a,b)
) l+ T& f9 y8 T# w n3 c" n" Nfloat a,b;
3 P( _" W: l1 ]+ y8 Y1 e9 t/ ~4 r{4 k% E% T' `* _5 T8 q8 M6 c
x1=(-b+sqrt(disc))/(2*a);
% ?3 d# T' q bx2=(-b-sqrt(disc))/(2*a);( f# p5 _8 q% W: E" v
}
- S- _6 M( ?/ t G& |8 i8 R! d Vequal_to_zero(a,b)
, g) S$ `3 C$ B: N, \float a,b;
& Y' q$ s" C8 |: |6 ~8 Z5 p{x1=x2=(-b)/(2*a);}; H+ [0 x p8 r3 W
smaller_than_zero(a,b)
2 K6 c1 P: z2 a$ \. ~$ z; sfloat a,b;6 |& y' l+ X; _2 N
{p=-b/(2*a);% h& E4 x; c l' {& K; V
q=sqrt(disc)/(2*a);, o7 T! G( N1 F* D
}) q2 F7 @/ E* P) V/ @6 r5 y
main()7 p- m. j) ~5 {6 K3 a
{
9 o; ?" y+ m6 h) M. sfloat a,b,c;. V4 x& a6 A3 \# v7 j
printf("\n输入方程的系数a,b,c:\n");
* [& j% \" A* C, j; {: Sscanf("%f,%f,%f",&a,&b,&c);3 V% L6 U) P' J& @$ b- M x
printf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);' z# O) A+ E3 c+ f! I4 Q/ V( D( T
disc=b*b-4*a*c;8 P3 f2 j# }8 v- l/ K+ ~4 S
printf("方程的解是:\n");
& \. i9 F; ~2 f% M$ Z) P0 L1 kif(disc>0)
* y6 d* Q _, S/ A% e+ U{great_than_zero(a,b);% ^4 m& j9 C' u6 R7 k" r
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);- W C* h* ~/ \8 H \
}9 {# M7 F, M+ b1 p9 l
else if(disc==0)
/ W8 a5 F' P9 I: d9 Z, [8 W: W {' j4 w8 H! {( E, N$ X% m
zero(a,b);3 A' W, o8 {+ k
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
3 O5 {! M6 J# | C }
+ o( C5 `- N7 uelse
$ w& C; t/ M( b: M0 e7 U {
~& J J( z3 ]6 o& l6 r small_than_zero(a,b,c);
* W7 C `& _7 @1 ]+ q, G8 b, q printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);+ C9 r# N1 r3 e- G, h
}. p3 U3 r. {/ P. H; S6 l/ @# X
}
' f# z' D$ x1 L& q$ T; @5 C, Y8.3素数
, t* `* t7 ], p7 i8 [+ Q#include"math.h"
# F k$ M9 w4 Hmain()/ ~- i0 ? F4 V# R4 s6 L
{int number;
+ y e [! W5 }& Z+ Q+ h) R W scanf("%d",&number);
; S" ?" n9 ]3 S1 z/ J/ w& |" G* W% H3 Z if(prime(number))$ _4 d: A2 I9 u& ?( f
printf("yes");4 {' @. ~& w: K: F
else+ @5 e5 _ u" O2 D- x# i+ i
printf("no");6 i y( P# Z3 ^* \
}
5 Z' q' ~/ x0 e) _int prime(number)) b. p+ ^7 m2 ~; c, E8 c% {
int number;- Y) M. k# C7 h+ P' N
{int flag=1,n;
( z* ^/ \. W4 ~8 } for(n=2;n if(number%n==0). y) c" A) J" O$ v2 i
flag=0;! I/ B, I" w/ t& w
return(flag);
6 d! p* Q: p$ g" A6 v! k: ^}# x, R0 E, Y7 {! M" T4 z4 S9 Q0 I
5 H) R- r6 l8 U2 a
h3 t" S5 X2 O
: a4 l1 O. s( K" N8.4" N! X, u& b4 H* W. K1 A# s
#define N 3+ T) C( a6 L' @
int array[N][N];
$ p$ ^5 g" K# J" ?convert(array)( p' v$ [+ R- ]0 M/ K
int array[3][3];
- {+ F# R5 f+ @% n. ?3 ] { int i,j,t;
& l$ [9 U/ u2 A# s% s, _8 q% C for(i=0;i for(j=i+1;j { t=array[i][j];, ^* M& j2 g% z' ~
array[i][j]=array[j][i];
! c7 R# p: c+ ?1 U array[j][i]=t;: U7 `! j5 l& k; Q& W, f( [3 D
}9 C& B! M, p. [) q) l
}4 D- M A- T& H( F/ y8 C
main()
' R3 i. I& h0 Y$ x: N8 }2 n: E: P{
7 w7 i) n% R6 v. ]+ Z0 b int i,j;: I* F! [2 Y- z0 l2 p
printf("输入数组元素:\n");
' w( O0 t: P3 t) O) {& V% t7 ~# \& P0 ` for(i=0;i for(j=0;j scanf("%d",&array[i][j];( y2 r7 i) [: ~0 G8 w7 O
printf("\n数组是:\n");$ ?4 j- Y) j9 u- g# W
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);# O( Q% ~! v* O6 E* K( m4 I# z
printf("\n");1 N' v9 Z2 y1 c! Z
}* R- @; W. l' w) R6 J5 D( r
convert(array);- N2 h* _# V7 h
printf("转置数组是:\n");
! o. m% Z6 w+ b6 Q* Z: j- q for(i=0;i { for(j=0;j printf("%5d",array[i][j]);; Z4 q9 ?# y8 w& O* o i3 ]6 ]: D- Y
printf("\n");9 M5 r3 i' d3 {$ L5 b* P
}
. K( | z: t, ~9 R2 R) g' e) S, g* h}
- _3 J/ T$ b% J* C! u5 h2 v2 q
7 O8 b* F! H2 n) O7 ]9 [5 y1 \7 P: v. q3 F7 c
! m. M5 {9 S. n0 g* n4 Q
8.5 U" ~/ m& L, {% V3 H: D
main()
8 a3 v. i5 c( j( F{, u( t4 D! ^$ s' I$ i/ v
char str[100];
% k' f1 p7 @9 }8 {* A% P printf("输入字符串:\n");7 ^8 ]( {! l7 O9 S8 ^! `0 e
scanf("%s",str);
' g |/ m5 C n. D7 l& I) k inverse(str);
' O; s/ U' {8 K; _. i. `; q1 W# L5 j printf("转换后的字符串是: %s\n",str);
; Y- Z- v; ~% m9 ?) ?) ~2 K}( ]2 f! f( \( u5 a. Y! y; C
inverse(str)
5 ?$ M1 q! g. T' u. @char str[];
/ p6 q2 W* B# C5 a! ?{
- B) w- J! T) ]6 z' x$ e char t;
3 U% U- E, w7 s2 M5 [+ e int i,j;
1 o. F4 F7 E2 c- h for(i=0,j=strlen(str);i {
- W/ \' C. a/ q: u t=str[i];
) J; a I# a8 f8 B str[i]=str[i-1];/ b" ~8 A1 Z) n* w
str[i-1]=t;
4 J& D* z! ^4 b }
# A4 f1 B0 i0 r7 X( T6 W7 t4 e}
' q* X" X9 n& u5 L9 q
& Y/ U% E8 I3 |* P' x7 m( c0 j% h' t" `" l
" h* W& G6 I M, W5 x8.6
$ ]! b8 j9 B. s$ V0 uchar concatenate(string1,string2,string);
8 U1 P" c' i9 L, w$ n% g7 R6 Gchar string1[],string2[],string[];
+ Y2 W5 X% O/ r0 d8 }" x. G{
- q f, i5 G! [, J3 d8 k$ bint i,j;6 f6 |. B' u" D) M" P' |
for(i=0;string1[i]!='\0';i++)& j8 }- A8 m1 c
string[i]=string1[i];: {, h- b7 q. B, ~3 B
for(j=0;string2[j]!='\0';j++). q0 x; w+ V. d* u$ r' S
string[i+j]=string2[j];
7 K, S" _$ F& [. E* f string[i+j]='\0';# q# j3 F3 B4 t4 T) N2 V
}
3 g5 c9 R' t W( ?# \main(): Z8 o$ R$ `/ p6 i0 c, c4 Q
{
2 |' U8 C. }0 q) b; l char s1[100],s2[100],s[100];
- S, \/ v; H2 ~ printf("\n输入字符串1:\n");" c( o( E3 ^& U/ k
scanf("%s",s1);/ u( v* Z6 e2 A8 y H& }
printf("输入字符串2:\n");
) W; e7 g( s+ V6 ^8 S" I scanf("%s",s2);) r6 B4 q: j3 x( |4 i3 A# ^
concatenate(s1,s2,s);* Y1 _$ w. W2 k% o7 b% q
printf("连接后的字符串:%s\n",s);" x5 @% i; E" q
}
3 F1 Z& I* s' O# _! b: k/ I# w6 t4 G" h
/ _1 E) s; m% O+ e1 b
8.8
+ O' i9 L8 L2 q, ^1 fmain()
* e& T, Z' h! D p& u! K{0 D: X2 w* \7 U# K" ?: X" }, `
char str[80];
" Z/ w7 W8 C" f- u. a( Q3 O. D: ]1 \ printf("请输入含有四个数字的字符串:\n");" J; f4 i1 c( w' R5 R3 n
scanf("%s",str);
. S! u7 V. B. {7 g2 ^0 d insert(str);
) Z: e$ O3 J5 U6 A c}4 c L/ F) b8 |; p' n! e
insert(str)2 w* @8 O0 \ t
char str[];( u* l4 e% Z# R0 ~# s% H- x( L
{
0 {1 _& S1 s9 U+ [# P& A int i;
8 T u/ x$ |! T. p% e for(i=strlen(str);i>0;i--)
" c$ \% J2 D1 d6 Z { str[2*i]=str[i];" R& S$ j0 b) t; u% ^3 [* i
str[2*i-1]=' ';
1 j& S" d4 {# g$ S) f. O }
4 }/ j1 P( w) [% I7 @, x* z9 o printf("\n 结果是:\n %s",str);' [; \: ?" C- i g# X$ D
}, @/ F- l* B) f( m8 W+ i
% q6 `3 g" C+ j! m% n2 u! _
0 x7 t! Y& v& l7 W
3 N, u9 z* h& {! Z& I8.9) G6 d# F* ~3 \4 T7 E+ a
#include"math.h"9 f! f% p$ G/ C7 i: g: L
int alph,digit,space,others;
: {; Q4 N' N; D2 J1 ]main(). d# }4 J1 ]* N: h, F7 C' v
{char text[80];
; c( N, k) @! u. C" H( |& L% X gets(text);- t* K$ ^+ S( z9 j% D- {
alph=0,digit=0,space=0,others=0;
0 x9 V2 z9 R) U. ^7 H count(text);
m9 R$ d( N6 v9 @$ S% y) f printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);
2 d# a9 D+ I% x}7 b8 P5 V, M' A6 x
count(str)
0 g! c o7 ^% qchar str[];. x8 S0 Q" u4 n k, l# ]
{int i;
, Y. J( J3 R$ w" t6 i* \+ N1 m: Q for(i=0;str[i]!='\0';i++)
u4 ~0 J8 ^$ d3 S if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
" Z0 ]! I" X1 |, j. H3 y alph++;
6 m) A8 X I& y+ p- Q+ l/ \' Y else if(str[i]>='0'&&str[i]<='9')3 X% ~, j2 y2 V0 I6 a. _9 W" ]
digit++;8 }6 q1 ]% f3 c; f+ C
else if(strcmp(str[i],' ')==0)
9 J) G' a0 e8 W3 B6 @& Q& R0 _ space++;
1 x7 K! R" s2 t* h5 h else
; H/ k1 R6 k$ A1 M% D$ C others++;' B( U* u, m6 F+ s2 F I$ [
}1 o; }: L1 o2 U0 l& w/ g
( Q) A1 w2 M9 c& q j: K
2 d8 g' C- J# B5 m' I* ]
8.10
) K+ K6 ]6 o) Q# e4 l9 r' ^int alphabetic(c);
* m2 G9 _# w/ s1 ] Xchar c;# N3 T4 y! W" @ S! \+ g
{
, W# e9 ~( T5 x- @ if((c>='a' && c<='z'||(c>='A' && c<='Z'))$ w/ B) r' G& h4 H- T* E
return(1);
! Z8 y8 M6 I% O else
- F: m9 B6 v' W return(0);3 N- z# Q6 l2 [. N
} a/ N: Z4 f- S! {5 Q0 A
# b! |& ?( b& Q% Q3 f# t% w* Kint longest (string)
( I0 }/ l! x) ]8 B& i Vchar string[];' o5 _0 _. I+ c$ _
{* {2 d9 l( h7 o
int len=0,i,length=0,flag=1,place,point;( ^8 ~7 O ~4 O7 e( Q! T
for(i=0;i<=strlen(string);i++)
/ G [8 m, {; A9 x9 H: c% |& o if(alphabctic(string[i]))
' b) s1 G {+ s% c8 I) ` if(flag)2 M8 p% \% a/ o& X7 w
{+ a# q+ I+ @4 W: [1 F- m6 v- L
point=i;; q- T8 G$ ?) |# ^" @+ ~1 }
flag=0;( t$ R. }( A8 @" B9 q8 i
}
8 H& j6 D- m4 P j else
' n+ L8 i! T4 o7 `) |8 F len++;, [7 `4 u) X- }3 K
else# W; r$ t p: V, H
{ flag=1;
" t% V5 R( L0 O$ u2 q" }7 K6 e# P if len>length)6 @4 x* j' }9 ~; m, X6 }* X% f
{length=len;" u/ r' h& [& v5 k4 ^# w1 J, a
place=point;
5 d: I4 ` f$ s7 Y H% @ len=0;- D/ k& _- ?: ? l/ }- A
}" }: w7 V0 S; v& q6 l/ E
}
# U1 y$ f7 l& I! P6 g. X return(place);/ j* z$ A& I9 M+ p
}
" a. c# ^6 y+ lmain()( s- q6 B$ V! v$ h [1 t T
{
( Q, A" [1 O$ }) `int i;/ T) M1 ]) q) i" ]" }, L3 k$ [
char line[100];
8 S* P5 _, ?% X1 Fprintf("输入一行文本\n");2 L+ R, _- q. X/ `
gets(line);
" l, i! e( ?* g; i, _8 _! n( I: Zprintf("\n最长的单词是:");
- g+ I; M1 M% m$ ~0 e1 Kfor(i=longest(line);alphabctic(line[i]);i++)& T" S' o: G/ |+ ~5 @' Q! }$ I
printf("%c",line[i];
7 Q0 V+ ]" r6 y: V+ _printf("\n");& t* J) @, K5 z) Z
}, M2 K3 f% a- Z8 j1 D' m+ o
0 u- X# F5 r) A. Y. a) Q6 M2 ?/ i$ [; ]
?2 Z9 e2 i+ L3 F6 B% e# e
) h8 _* d3 t5 {- f6 Y/ R# O8.11 m8 m$ L* I, T4 Q/ V+ U4 s
#include) p) Z: E) D9 s7 [1 S& _* H# e* @) R
6 N" [# G8 O, ?4 u# W0 m {
#define N 10" [) ~- f0 p! V$ A4 }, y
char str[N];
4 |/ x+ `% T7 j7 T: C* Ymain()4 T2 C/ D5 l, I- |- s5 r" A
{5 g2 q' a, |/ E! Z2 i$ k
int i,flag;
# R" B4 G7 s6 }for(flag=1;flag==1;)+ Q8 G; B- P0 [& {
{
: J2 I5 t3 O: y9 w printf("\n输入字符串,长度为10:\n");
- A2 }3 B7 ~& N. K6 L6 t scanf("%s",&str);
7 g% K% ~0 {) L2 L' K if(strlen(str)>N)
4 I+ ?; p; a* D# x printf("超过长度,请重输!");
# F5 A8 j; P1 ~. g& l3 B else! K( a" T7 y7 |. \, p! c3 m
flag=0;
' W2 B4 G( F1 i+ d! ~! o6 C1 \}& z% Y; f, W7 p
sort(str);0 K' U! l/ F/ ]) L- x+ T& d% ^6 r
printf("\n 排序结果:");0 t( [: N+ U% s1 F+ F$ V
for(i=0;i printf("%c",str[i]);2 O8 }7 \. s {% U+ s% E
}: E6 @. U: f& e( _
sort(str)
" \, c" e. B* x" r; p* w* _; Gchar str[N];
% }4 O* k R! \- P{
: p5 z, z. I+ G, k: K0 `int i,j;
, @" G( B5 J$ h7 vchar t;
+ U( ~7 n6 p/ Y F3 o- ffor(j=1;j for(i=0;(i if(str[i]>str[i+1])" m3 Q* r4 c5 _5 I- [
{ t=str[i];. @ O+ h+ W6 P
str[i]=str[i+1];. W- u: h% F6 c, r8 W
str[i+1]=t;
2 W, L! l' `4 n }
; x1 t- f$ W' l9 N! J) G}: h& l( ]. B5 K! K7 K6 t
8.12! |# _. f# O7 ~* x
#include/ C; }% k$ H. o! n! p
#include2 o) P( K7 @4 W' T) H' w/ C
float solut(a,b,c,d)& X7 W4 o5 v- e! E% R
float a,b,c,d;% M' @0 A: x1 y. z% C" Z6 o
{float x=1,x0,f,f1;$ @9 z1 j5 J" Y9 g# @
do
% d+ a8 ?% }. B" W# G {x0=x;! l( Y% k) c6 j
f=((a*x0+b)*x0+c)*x0+d;
5 ^$ m! h' ^- J# H f1=(3*a*x0+2*b)*x0+c;
. B- L' L" j& ] x=x0-f/f1;
) u( I( S/ _- Q b1 I }$ S' W8 [" w1 F( W- |9 K
while(fabs(x-x0)>=1e-5);# x) n/ e- A5 \3 F; M1 h
return(x);
5 r# m' r+ y" k}; V/ a2 d- L% e9 K
main()( V; T- M: T7 }5 G7 W" d
{float a,b,c,d;, d* @1 ^, n: ^1 s
scanf("%f,%f,%f,%f",&a,&b,&c,&d); a1 f% |$ R3 ?4 F( q( G/ n
printf("x=%10.7f\n",solut(a,b,c,d));
' g! R/ c( V, d7 J6 \( r}# i: u6 h% Q* b
8.13
! ?3 n" `8 R$ T/ U3 a! U#include
( g8 S- C; z+ j d7 U2 Xmain()
% R a4 y5 D8 O m{int x,n;
: `7 J! z( c6 \) `2 U+ _6 ]6 e5 O$ ? float p();) F2 k* {* n ^# e% G( B5 U8 b
scanf("%d,%d",&n,&x);
9 M/ m! f& Z" k6 B% a8 p; w printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
# q: l2 c o; H) h7 Y9 z: n2 j}
6 P% e- }, P1 l) mfloat p(tn,tx)+ H0 d5 [% H, D
int tn,tx;8 @4 M* L' T) r2 d" ^
{if(tn==0)+ v0 y W3 P2 m; o# `0 l& A3 J
return(1);) I& D5 Q4 W( ^6 j; t3 {
else if(tn==1) ~: K& }# r5 V+ H7 o5 y5 E
return(tx);
C) T1 j1 X2 V: d y else& f/ m7 u# b9 \7 L# N0 c4 r: I- u' w9 f
return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);/ i+ r2 m* I1 V8 g+ J
}2 s" Z: u$ W4 b0 Z: e0 V! m/ i
8.140 n$ a0 ^6 K: X8 x- Z
#include "stdio.h", c( D% a# W- N& _# `
#define N 10
|. v6 p3 `# C2 `#define M 5
6 n3 Z! E: C: [1 b Y' @4 `2 |float score[N][M];$ N' y2 A% Q8 g1 q4 P, E
float a_stu[N],a_cor[M];
e. C& M' K5 S: m0 q% dmain()0 y1 k& v" R- R) j
{int i,j,r,c;9 v( l8 ]9 R0 Z- p6 [0 \0 a
float h;7 u$ H2 [6 U8 g9 v1 c2 r4 J
float s_diff();+ ^! _" W0 ?. {. [- V' V
float highest();
- |: {! Q$ k* M _! w& P; Y r=0;0 U- o. r; U: P) x4 a
c=1;5 A) K$ d' L3 i, d% Z/ d$ n
input_stu();
! C0 q3 S* m [1 W! ? avr_stu();
3 E6 U0 R+ G6 v. J; D# [, { Q avr_cor();
- {8 O/ P# k7 b' O: X+ o! T8 m printf("\n number class 1 2 3 4 5 avr");2 v; [$ ^8 F/ p% U
for(i=0;i {printf("\nNO%2d",i+1);
1 n* H( f/ R" Y; A6 U for(j=0;j printf("%8.2f",score[i][j]);
9 C4 E- d8 J& T* H7 x9 f printf("%8.2f",a_stu[i]);; Y. [8 I8 \; A; p8 S5 F4 L
}. k' d$ ^% X! e; y9 I
printf("\nclassavr");' E# p9 D1 W) M1 r
for(j=0;j printf("%8.2f",a_cor[j]);+ b2 m, ]: G6 H" j+ b( Q
h=highest(&r,&c);
: \1 w! y/ f0 K8 l; ?4 v printf("\n\n%8.2f %d %d\n",h,r,c);
+ B8 ^8 a- e) b( b printf("\n %8.2f\n",s_diff());6 w/ f9 U% w, E/ u; ]5 W
}5 p8 |/ g$ h6 O" g) C
input_stu()3 ^- E2 U: S& x6 i6 |3 @
{int i,j; ]7 X3 d6 M" s( z9 c( G/ {
float x;3 l& e: o8 e4 {
for(i=0;i {for(j=0;j {scanf("%f",&x);
3 `) w j v' i; ^% ~ score[i][j]=x;7 o. p9 V, ~* R g5 K+ b
}5 g! p! j8 R8 M
}
* m i, X$ m Y8 q& s}1 w* o2 N) C$ R0 b" G
avr_stu()1 P, w5 m% F- A* k/ g' S
{int i,j;
* ^8 |. c6 q! L, e1 J float s;6 N" B2 O0 I( S3 s4 b* O3 l
for(i=0;i {for(j=0,s=0;j s+=score[i][j];
4 |6 X% b$ x; H( G+ h a_stu[i]=s/5.0;
+ I4 w! G5 m# i2 `9 [/ c9 J& ] }: ~- U) O# x/ A
}! `3 l2 O8 u7 p9 s
avr_cor()/ H" S! K' j# h1 ?* ?6 _
{int i,j;7 I7 F; x& e% n. } q/ ~1 y7 V
float s;
, Z3 s2 ]9 j, g+ N! P for(j=0;j {for(i=0,s=0;i s+=score[i][j];4 Y7 g9 A- T/ s/ W' X1 _" A7 O
a_cor[j]=s/(float)N; M$ V- ~ E8 R2 e* _! x
}/ `2 y9 Z( ?& H8 e4 O
}* K' E+ w" H- Y @' d
float highest(r,c). `4 z1 P: ]5 A6 a" j
int *r,*c;4 ?* u% N8 D( m$ C$ ]
{float high;
, S; ]# ^& K5 Q' {5 E" b) C5 w* ]7 W int i,j;: p& v- Q# [2 U! p
high=score[0][0];
$ X6 {# V% ], x! U6 U for(i=0;i for(j=0;j if(score[i][j]>high)
+ P+ F8 G; Y# x; u# V3 u {high=score[i][j];
7 n( Y( H- |3 _& J6 M: n *r=i+1;2 W4 r1 |" d/ b/ ?* y% c2 |
*c=j+1;( [" B- L s; ]+ F: j
}; B9 b6 F8 @* j' p5 ~% _
return(high);
: g3 q1 v9 ^ Q2 w- F, }6 t}
9 d) l9 g: M/ g+ ~float s_diff()
% h0 p4 _7 O6 p6 Z$ i( p0 e8 _{int i,j;
3 D8 |. W+ q( V; N$ _" c# Q float sumx=0.0,sumxn=0.0;3 V, A7 G" S' q( Z: w/ ~2 d M
for(i=0;i {sumx+=a_stu[i]*a_stu[i];
" s; k- n) }4 M; \ sumxn+=a_stu[i];* U! O/ U' D2 V1 _. j1 p
}1 Z' D' p- l% {+ i2 A1 ^" A1 J7 }
return(sumx/N-(sumxn/N)*(sumxn/N));
5 j: E: L1 A1 o& f" A, {}/ T) M8 w' W2 i5 k7 s
8.154 ?& b- ^$ W- L3 v% u3 K
#include* J: _& n. z! ]8 P
#define N 10
0 G1 H6 i1 w6 \+ G5 P. o% i: Hvoid input_e(num,name)2 R( Z( O& K9 N, t1 c) P
int num[];2 }2 Q% }" r1 ~( `( U0 @7 Y4 K7 I
char name[N][8];
* @ q1 C) Y0 N& h# E* Y{int i;
" f: q) W8 X6 I% V6 R, r for(i=0;i {scanf("%d",&num[i]);% L/ T) j. `' V) s! f# O: ]
gets(name[i]);
; M0 e; T4 O+ O4 z, N }9 O% A( q) n9 _6 S% W( S+ }
}5 l; }! z0 v- _. |% y4 Y2 |* j
void sort(num,name)
: X/ o/ }& \$ h" V7 D; Dint num[];
9 t( Y" D* i; Gchar name[N][8];3 p9 F+ T G, c* {
{int i,j,min,temp1;9 u+ E# E' }: y4 G1 _+ R; c
char temp2[8];" q6 I% G! D/ C; E5 L- v3 E4 _: I' |
for(i=0;i {min=i;
7 F+ F& {+ H2 z( t$ D" N6 K2 r for(j=i;j if(num[min]>num[j])min=j;+ K! I v; ]- K8 ~2 Z Q# h1 e: a
temp1=num[i];
; }; F; p6 S- d5 k( g num[i]=num[min];. z2 T2 H& V4 K/ ?; p' B9 b
num[min]=temp1;
3 k M& r) S9 C) r; m strcpy(temp2,name[i]);
8 x0 D- s. z' t- h: l- F strcpy(name[i],name[min]);
# \! Q! k/ O4 J+ q8 N8 C2 O. ~ strcpy(name[min],temp2);9 W2 E5 m9 N8 o; f/ K
}
) y2 ^* S5 T# v4 Q" j! A for(i=0;i printf("\n%5d%10s",num[i],name[i]);
) q& X% z8 h4 i8 x}0 s9 _" {/ L G( a' k v- o4 L: N2 }% q
void search(n,num,name)1 `( e: G) P2 ?
int n,num[];
4 _1 X ^7 k2 ychar name[N][8];/ w; L; P' z1 @1 |# L
{int top,bott,min,loca;
6 P/ G7 W) y% \7 ~( g loca=0;
4 F) N2 y7 u7 X$ R4 U top=0;
; N/ U! r& X" k' N bott=N-1;
- P3 N1 v- X. l! V; c/ ?% _8 Y if((nnum[N-1]))
, u& u) p( l8 ^: z1 {; a" E loca=-1;
/ @- }6 i+ N% s5 R6 v' e) P4 O7 m while((loca==0)&&(top<=bott))
/ _4 w: I3 d8 l* Q& x2 p* X {min=(bott+top)/2;4 w+ {! j% |0 K3 F$ g! v& f
if(n==num[min])8 @5 W3 r+ w7 ?3 e
{loca=min;
5 ~- @2 t; _, Z1 J; Q3 D printf("number=%d,name=%s\n",n,name[loca]);
( s" e) s' g f }
+ {3 [0 J/ V% z/ n4 H' B% U else if(n bott=min-1;# a, O/ ^" z* y: v" l# f5 N& G
else
: K2 W* }3 a( d top=min+1;
* _; j9 J' O, i }& r( E* i x4 F1 ~% w- x/ T+ j
if(loca==0||loca==-1), k1 X# m$ U: k$ L2 u0 V3 E; P# F
printf("number=%d is not in table\n",n);6 b( S" w, d: _, I
}
7 N- p4 ?" I! P9 G$ j: D& _& emain()
0 q) B0 ?5 Z, Y- ]0 }$ J' @) N) s{int num[N],number,flag,c,n;3 V+ Z$ g1 z( x& d2 w0 O9 R. B
char name[N][8];
: w/ Z, V+ G/ \2 p# m, _- r7 R input_e(num,name);% q1 `3 k. R+ ]# Z" K" O
sort(num,name);2 m1 r2 I4 f/ Z1 `1 N
for(flag=1;flag;)* a! L( k2 N. P! S& H, y' M( ?
{scanf("%d",&number);
0 f1 s, v t% ` J' i search(number,num,name);2 w1 v$ H) _% o; x
printf("continue?Y/N!");
9 ?3 `' p: v) o9 M* y c=getchar();
5 b1 ]$ Z1 N4 n2 B; g4 ^" g& _+ ?7 } if(c=='N'||c=='n')
7 U0 ]9 b; ]1 c* H1 I/ P! e flag=0;
/ w" Q! ?- r0 H! ]& v5 y. ]7 r }
) Y' I0 s) S$ K2 V' l2 c5 D}# W& @. P4 L8 M- ~. t, X( v2 a3 ]+ f/ X( z
# X5 J+ y% A9 h9 @6 p
8.16
7 i9 x" y0 }' W6 r, [" x#include+ k' w2 o9 A% o+ ]
#define MAX 1000
- V! F- F/ e- ?3 U' Y. `! _, gmain()
' J* O" K; `; ~6 |{ int c,i,flag,flag1;
8 I# C. p' O! h! Z5 l4 V5 M5 [ char t[MAX];
- y* u( y. p0 M3 ?+ {) q w i=0;" G1 o7 M9 K2 [; s4 L/ w4 X
flag=0;1 {! }5 E( _2 Y4 X' c
flag1=1;
) s t" B3 Z" L) S0 m; L printf("\n输入十六进制数:");: n" H7 ^7 Y8 O7 K) L8 B6 k
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
. f4 h. L5 D4 c5 s2 k [ {flag=1;
1 O4 q ^$ }$ H" o4 `& @7 Y s t[i++]=c;" n2 H% \5 }) Y6 b% T
}' W+ N, w1 Y- m$ J! N
else if(flag)) f6 c9 K- i& u5 R+ Z: P
{
) W( X: p9 N! I" o# ?# V t[i]='\0';
& W9 k. H- \ g printf("\n 十进制数%d\n",htoi(t));! H7 s; k1 {7 }6 u5 H- j u
printf("继续吗?");8 @. o! j: A( d0 \, \7 S8 R
c=getchar();
- \4 a6 h+ t: f1 _( U! Z; z- P if(c=='N'||c=='n')- J5 l8 x5 d7 V8 L0 F. R2 z
flag1=0;+ Z3 q9 m/ {+ |4 Z2 q; o8 I. a4 n
else
$ ]: M2 ^; H V7 a {flag=0;
8 E# }7 @8 z! ~: K8 c2 p i=0;) \: ]$ W7 |& o
printf("\n 输入十六进制数:");
- j! `: h5 |6 B4 C7 P$ U }7 Q, z# D3 e A; \9 d: |: \
}
/ }. j5 g: y7 i; M5 `* X0 o7 m0 b& ]}' _+ ^9 f8 S8 s, P' }) F, P
}& z, ^, O- _* F( h( k3 ]1 l' G
htoi(s)' @% |3 o+ f: X8 ^2 q1 W( ^; [1 M" `3 S* s
char s[];
- e+ Z: x) D/ Y- S/ Y1 d6 S{ int i,n;
+ S8 a9 w& O8 y' J7 |* e1 S n=0;
8 Q- \. h6 H6 ] for(i=0;s[i]!='\0';i++)
( r- V! b6 R. }4 X# e j {if(s[i]>='0'&&s[i]<='9')
8 I+ ~4 \! Y; M) u n=n*16+s[i]-'0'; @3 I5 l4 u+ w. B1 ~$ r
if(s[i]>='a'&&s[i]<='f')
8 c2 q% A; r( t: A) h n=n*16+s[i]-'a'+10;
$ R5 u' t L4 `* P* [2 M if(s[i]>='A'&&s[i]<='F')
' R- I+ i+ @6 L8 k- R) z+ {/ b n=n*16+s[i]-'A'+10;
7 l: n0 c; H1 N3 a# a7 Y( k }
- l# C, e; F7 Y' j return(n);1 q1 q/ J* n- J# t! _% u7 x
}: c9 E6 n0 i5 t
: B$ ?0 D* `2 G% J8 x4 H! e B, I+ [4 x. C3 I8 X3 ~9 Y+ \6 G4 N" O0 X
4 T* x7 ]8 e1 @" E
8.17
) g; a' z; k/ x+ h#include/ s3 g: Q: \ }6 f
void counvert(n)
$ s& U" H1 w" L: O; g1 u# y9 Vint n;
( |$ i1 B( E+ d/ e3 `" f* K{ int i;
2 l; W5 z8 h |, N( I3 }* J if((i=n/10)!=0)7 c" n# N/ m d8 K( }' F
convert(i);" k( _( V( H! Q* ^2 R+ j9 E
putchar(n%10+'0');
- e8 ~& i1 Y0 p+ U0 j. `4 `3 d}
# k, i3 n0 m& _/ q, xmain()
2 q1 I0 y# h4 I3 X) h8 T{ int number;* p0 Z$ ^7 f9 I/ R0 z2 S4 c
printf("\n 输入整数:");" X _/ o& t7 q0 q. C
scanf("%d",&number);
; f) c0 f8 z, P, o7 P printf("\n 输出是: ");
; I7 P1 C& @/ E. k9 }3 ?; x* _: W if(number<0)5 E: ~7 L+ J# z; a9 q% O
{ putchar('-');4 g- B5 V7 h5 T
number=-number;
* G7 A. H# K4 H& x7 f' n7 K }
p* ^$ g2 g0 @- D# Iconvert(number);
' E; q4 z/ X+ c1 z} ^# ~5 r* }2 ^; X+ a. J0 [/ X- C
* p# b+ u: y4 q4 ?$ F: l
% X+ e( ~ n# a, g$ q) U3 J: L7 e
# E) e$ E# e: G( [, D8.18
) G- O" ~* H9 ~! j2 }3 Tmain()- S9 [8 X1 B; C) T
{
, Q" w& p: Z0 G# h& _ int year,month,day;
9 }7 a# `9 r4 O3 o7 `! P int days;( q1 b6 |' A, A
printf("\n 请输入日期(年,月,日)\n");% ~" W* Y* x- N& L- x3 ~* U! Q" ?
scanf("%d,%d,%d",&year,&month,&day);
6 d b) Z7 e& L; x+ j0 {7 e: j printf("\n %d年%d月%d日",year,month,day);* c- l$ B/ U& Z2 W" W4 I; b) ?* B
days=sum_day(month,day);
. _+ p O4 H( L% G7 d if(leap(year)&&month>=3)
, S A, [; c4 ?1 M days=days+1;
$ ~: g- p3 t2 ]( R6 s printf("是该年的%d天.\n",days);
' X/ X- l( L w* \! C }
2 `/ ?! k7 m5 v" `) I1 c" Y static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}- o* B9 H. I( P W
int(sum_day(month,day)
2 y) ?% n" h; W int month,day;
7 u1 b( v" x: [% c9 I6 {$ B1 @ {
5 L2 n1 f9 ^/ ], E4 |- T! U9 i int i;6 v7 Y: K e8 h( s) `- _
for(i=1;i day+=day_tab[i];7 I8 p" ~$ r( e4 T; Y! ~
return(day);. Y; F) s3 b2 E! K. k
}0 A6 O/ _$ ]: Q- S5 P% a
int leap(year)
. |' S, J3 O4 l. L z+ v* L, e6 J8 v int year;$ W/ p; W# n/ Z
{
1 U: {" U: f( A5 j* x int leap;
( u( Z+ |' D. C# C0 A D; v4 i# J leap=year%4==0&&year%100!=0||year%400==0;& }( Y8 \! J" X4 b2 N$ z
return(leap);/ `8 i: k! C8 u2 D5 F$ V) o) @$ M. A8 G
}: G5 a! v: i1 B2 v6 }" p
第九章 编译预处理
5 f! ~7 e! A& w% \* r9.1
2 I+ G* o" ? S6 H' T" k% E& Q- C#define SWAP(a,b) t=b;b=a;a=t
. V2 |% Y8 U; E* g9 F- o8 [% Cmain()! |3 b! K" ~( q6 \
{
X8 Q! k: d+ A8 pint a,b,t;2 @9 x+ P, {, B# i! u- `* p! r
printf("请输入两个整数 a,b:");6 `1 W4 f& G( x
scanf("%d,%d",&a,&b);1 p8 h' a" z2 y6 o {' g* F$ B
SWAP(a,b);
. v/ \6 _6 _( V5 G- y, ?printf("交换结果为:a=%d,b=%d\n",a,b);8 S c; A' ?& q1 D$ ^0 `* M
} . u( Q. I% k& _& `
0 d c& G4 E7 ~6 j
) U& p8 C7 t/ E4 L9.2, l7 N/ }3 p9 w- \1 m
#define SURPLUS(a,b) ((a)%(b))
" J% [0 G" \8 O* \( {4 r& @main()
" D# r, M& X3 A+ l& t) s {
% G# `# G# a0 w+ H! U int a,b;
" J( ^3 a' T/ f7 B2 s: A printf(" 请输入两个整数 a,b:");
- r& E/ g+ X. \! @$ H* f; B- X scanf("%d,%d",&a,&b);
7 w% D6 B( I# }1 }printf("a,b相除的余数为:%d\n",SURPLUS(a,b));
, a; p6 x' N( R3 K3 ]4 t; X: j }
" {3 N4 L- f7 u5 n7 Y9 r0 \& g0 f* D; W/ t- p% [
2 x r( x9 S/ Q! t4 O4 U9.3
) c K% c4 X: ^0 t#include1 f2 N! |) x6 ]' u; G/ Y
#defin S(a,b,c) ((a+b+c)/2)
" ` s( c- k$ I+ y$ \, w; @#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-2 K, |8 T' Q7 U. r N
c)))
& A3 Q8 i# Q3 J/ ~ x* [main()6 P& e+ S, o' u3 N- w' ]
{
0 W0 Q7 q* ?, N2 y float a,b,c;
/ _$ ?/ \- n& t printf("请输入三角形的三条边:");
. Q9 v- u/ w3 [0 W# j scanf("%f,%f,%f",&a,&b,&c);
/ J" I( c. V! A# u' V if(a+b>c && a+c>b && b+c>a)
u" L+ @& Q3 D) B printf("其面积为:%8.2f.\n",AREA(a,b,c));0 z& V7 i' F# _0 p q3 v2 R; ]
else' C4 `' j- v2 s1 m2 X6 \' i# _: i
printf("不能构成三角形!");6 Y; j }: x2 G2 v$ G5 y
}+ i- Q5 o' ]# J5 X1 ~. C: E
6 y5 C# C% Q2 M3 c$ S
1 H4 u0 j0 s( p7 m$ E8 i0 p
, i# s1 i' B, n7 P) R1 |5 C9.4
3 {. U4 k) \5 v: j U& h1 O1 V* C/ O#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0). C4 V: [ `. k' J2 ~9 `
main()2 I2 b" O- _; k" ~( `0 o; h
{
1 |" L w% D$ z, E int year;* M+ Q$ z& D8 m) ?! x
printf("\n请输入某一年:");
3 W4 _8 w. ?9 P6 |" a) }* e scanf("%d",&year);
4 O- U% r& G- O, X- O$ u- U if(LEAP_YEAR(year))
; e, W/ n8 s/ i f printf("%d 是闰年.\n",year);
* d7 J4 Z: N7 k0 D0 G else
4 `" Y+ }; s7 L2 F% U- x printf("%d 不是闰年.\n",year);
9 m. @3 {4 a7 j6 e5 H& P0 d( M }
- a, W, J4 M0 w! f) D# y) Y! A; z$ O3 D/ D
' J( z' g# ?' I |* [$ N
: ]9 M* l( L) [, g7 F0 I3 r. E0 v9.5解:展开后:
. A* F H* U) b8 z% |, sprintf("&#118alue=%format\t",x);! U+ f* f2 m7 C8 }6 ]
printf("&#118alue=%format\t",x);putchar('\n');# S n" v/ G1 R" A$ @7 s
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');
" `/ M: l4 X7 ~输出结果:
. m& D$ F. |, }% o5 S&#118alue=5.000000ormat &#118alue=5.000000ormat
' h6 c/ q2 F8 f- t6 W&#118alue=3.000000ormat &#118alue=8.000000ormat
- [9 L0 N2 J( ~: P* z* v. T+ b3 `/ q8 g
( Z# u: A$ a, ^" D" M+ B2 c8 O
9.8
5 [; p: R6 F% C; A7 S P4 C5 Qmain()
8 Q9 r0 a) o k1 Y0 ^9 ? {: i/ h3 S) V" u0 F' h' J8 k
int a,b,c;
" ^$ b- t* z$ p$ E2 W printf("请输入三个整数:");
: K5 ^( f$ {+ q5 O scanf("%d,%d,%d",&a,&b,&c);
/ G8 w% a1 W6 } printf("三个之中最大值为:%d\n",max(a,b,c));0 |6 |. V( w' ]6 V) O' }
}% J- y" I. `- |, A0 M8 \
max(x,y,z)
7 J; d+ P5 y V9 `0 P, E- M/ w: Q int x,y,z;
K F6 _2 ]' C7 x3 F i {
) i) P9 J) R/ v; r( P int t;
# S" a9 A7 P- z, G& g% U t=(x>y? x:y);4 |1 P. y8 P" n9 w3 [" g
return(t>z? t:z);
) x! p D" o$ G' h' o$ Q }2 j; B7 G2 d# i9 P+ M4 Z2 Z
8 v) z1 ~ ]( H) O
, n6 s9 Y3 d& m$ c6 U: O
* d: O, S- h# V$ C* G$ a, V9.10' ]; @' z+ T' W+ L
#include/ X1 r% O5 }' F( [
#define MAX 803 v7 D$ G9 M/ B/ a3 X, j
#define CHANGE 1
0 a2 L' ~- ]# l1 K) kmain()
7 _ ^# L8 O7 D6 ]{( R5 C% Z* o% e5 ]! o) Y" k% n
char str[MAX];
, l" _) I0 A1 j3 O, Z int i;
: \: l7 p M* Y' j& w printf("请输入文本行:\n");
- P) ^+ K9 }2 |* _, T6 @, W* ~, A scanf("%s",str);
3 T, \" g' ]8 D #if(CHANGE)3 ?# x6 s- s7 p/ f
{* S- k6 {7 [% f3 h& @
for (i=0;i {+ K/ F8 D' ?, ]1 G
if(str[i]!='\0'
: |5 L* K/ q" C- f2 v5 i. t/ ?# M if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
9 t* u9 D8 m- g# T6 ] str[i]+=1;7 y" ?. T) Z$ ?) y/ d7 K
else if(str[i]=='z' || str[i]=='Z')
2 W# n1 R% B9 U* L& U6 k str[i]-=25;2 i ^+ v0 \( u2 V; L, @5 G
}% ]$ S' n) K/ g' ^2 h
}2 h5 I! w, n5 a0 o8 { Q
#endif
, _0 T1 X- R! N: i; J/ Y2 ^9 x% [printf("输出电码为:\n%s",str);
8 c J' S+ d9 @0 I& u+ M}* q8 u; `0 z( r) Q M
第十章 指针# T+ N+ N" C/ l8 t' N
10.1
* F5 ?$ f( {/ ?; i# [5 v8 r' Cmain()
# Z$ Q0 A4 e0 S9 ~) }4 Y: X" q+ a! q{int n1,n2,n3;" z, h1 }6 A% }$ b7 X. {
int *p1,*p2,*p3;' y! Z) a& b4 y+ I
scanf("%d,%d,%d",&n1,&n2,&n3);
4 {) R& b! L) S p1=&n1;" @* ]7 K- f* o0 [; v; G T
p2=&n2;3 B2 ~4 ]! k: L! ^, n5 `! B2 S! h; y
p3=&n3;
. D! E$ b2 E! I* |( S! ` if(n1>n2)swap(p1,p2);0 n% {8 C$ N5 Z+ z w
if(n1>n3)swap(p1,p3);0 b. p/ V8 J2 O) M$ y' l
if(n2>n3)swap(p2,p3);
/ r( d0 h5 ^/ s. A! h, U' f printf("%d,%d,%d\n",n1,n2,n3);
/ h6 j% K' _8 i/ k# P" o# M}9 F* S! m7 \& {% b3 @8 ~
swap(p1,p2)
+ ^5 v6 _' T: C/ H# z0 {int *p1,*p2;
5 S v! y2 Y4 k9 D. g; V{int p;: i8 t) A6 `; m. E6 x5 r; p
p=*p1;*p1=*p2;*p2=p;
$ r+ N3 n8 {" U}4 K5 M2 }8 v2 I( B7 t& c
10.2
4 X. p. e: w' Bmain()
. o2 m4 p" D2 S- i{char *str1[20],*str2[20],*str3[20];
6 `8 f: n* U, j+ ?; p- o8 \ char swap();( A; a0 C) ^+ V5 g! t
scanf("%s",str1);0 k2 x7 ]' \$ H( F: Z
scanf("%s",str2);
* d1 t" b% \* Q& z/ i, p) \# e R scanf("%s",str3);
- ?2 X% s6 B2 U7 I" ~# q# K, ?: w7 W( q if(strcmp(str1,str2)>0)swap(str1,str2);2 k i" g4 X0 H, B7 B, d; Y7 F
if(strcmp(str1,str3)>0)swap(str1,str3);
; [2 B9 ^' s& x if(strcmp(str2,str3)>0)swap(str2,str3);3 q0 {$ T8 V" e4 c$ A; W# b
printf("%s\n%s\n%s\n",str1,str2,str3);
" G" ^) y3 F! {5 D: k}4 C4 g% F2 G) c0 r6 g
char swap(p1,p2): m h0 H* o: \: B
char *p1,*p2;
% o3 I% }$ o- z{char *p[20];# j- P+ q8 R4 o# C
strcpy(p,p1);
4 g o8 R0 l9 D+ r9 j( J% q strcpy(p1,p2);
% Y; |. I# ^. ~+ y; S$ N8 m2 | strcpy(p2,p);) T% R' W& B% F! o( v( F
}# t, T b- S: q: x/ U. R
10.3 q' ]! k1 y" n# ?7 x" T
main()
. v! @+ m' n3 D, F2 \{int number[10];
, q8 _/ M4 A ^4 @+ W input(number);
+ s% K ^1 c H9 e+ s! }! P max_min_&#118alue(number);
2 i; A0 {3 y7 T0 v' r output(number);2 J# N3 g: n+ v/ ^- d
}
, R- V J' x+ Y+ [, j einput(number)
+ I/ Q; c4 U- }9 |. z. y4 I& j( Mint number[10];
D2 T6 _$ |* E& Q$ ]8 F' r! [+ O{int i;' v+ N' A* t+ I P8 {! |* D6 s
for(i=0;i<10;i++). i; y1 ~6 E+ f
scanf("%d",&number[i]);
6 r* L5 Q [, g2 l( p( n}
9 p( U/ c( o4 ]# Omax_min_&#118alue(number)
/ x& l7 c$ i' @1 |1 A ~/ @int number[10];; z. ]; l8 {0 @! Y) T
{int *max,*min;( A9 f2 Z+ [1 }) [9 p0 x
int *p,*end;
8 T& `4 Q3 B2 _ A+ T! e end=number+10;+ X, H! C; \1 K0 v" o9 J# z8 a6 \
max=min=number;
2 {$ Q( t h- Z, k. H for(p=number+1;p if(*p>*max)max=p;2 N7 V( L/ Q0 V& e' D5 Q
else if(*p<*min)min=p;6 d: T7 _1 W" G4 G
*p=number[0];2 `1 w$ D% W, L% F! j$ u8 u" _6 _& A
number[0]=*min;1 J3 E9 n6 _0 H6 @ E
*min=*p;
' } H7 z- h# ?0 T. ^; p h *p=number[9];& F& s7 b; S* R6 `: r w
number[9]=*max;
- H' y+ S0 F4 Q# \) ~8 X! y0 F *max=*p;( S0 W! h9 t4 n* ]& _1 o; N( Q- k/ H
return;
2 \9 [( Q4 W: n1 I}' f/ s6 ]0 C6 K& n
output(number)
( i) Y$ ?2 d2 Dint number[10];
- b- t4 I/ x) ] s{int *p;$ R+ O/ e- r3 Z7 w; L# K
for(p=number;p printf("%d,",*p);
! _$ c# }6 f. _- d0 q printf("%d\n",*p);
! y- }- b8 O! ~8 r1 I. T' O) H}
' @9 \3 ?* Y8 g10.47 U* V: B3 D+ U- W7 k, R8 S) e
main()* I2 M& c) I2 U8 D% o- q8 d# n
{int number[20],n,m,i;! ^" i) Z E+ Y! v- S- f% w
scanf("%d",&n);! n o9 e4 A9 N$ X
scanf("%d",&m);
$ z: {0 H4 X3 G* `/ q! f& s for(i=0;i scanf("%d",&number[i]);
& i' I& r6 j- |: |5 k$ N move(number,n,m);
+ }6 ?% r* C5 d! X9 o for(i=0;i printf("%8d",number[i]);
' D4 E4 P5 v5 h8 i0 N: D}
9 j, n x3 m7 v( { V3 p% e! Omove(array,n,m)8 d9 I3 h* j" S1 n
int array[20],n,m;) t" J6 s* K) ~, J# K
{int *p,end;+ h+ y. Q$ R& p5 M! o, d
end=*(array+n-1);
, n) a( M$ ?# i o6 I6 w5 n7 x for(p=array+n-1;p>array;p--); z1 L# N( Y! _2 l
*p=*(p-1);( P9 A& d) x1 J- l a! ^0 j7 {
*array=end;/ }" R" i, x( V- e
m--;# r: p- b5 W& _
if(m>0)move(array,n,m);
' `5 n- I8 z4 ^' N! ?5 X}
5 j* M# `0 Q8 J$ Y0 }10.5
" ^. u; m' \% ^3 k7 N#define nmax 50$ g, C1 |4 K3 b6 N* K' t
main()8 ?/ G* K4 _( ^9 Q( {
{int i,k,m,n,num[nmax],*p;
_6 U8 a) ]. z0 k1 K scanf("%d",&n);1 Y0 p& p. i j) m
p=num;: B- d H/ s* D5 X
for(i=0;i *(p+i)=i+1;
5 W: A, q8 t, d* O; V( v i=k=m=0;
7 t7 `; V7 ^, n while(m {if(*(p+i)!=0)k++;
) H: D( M D' i$ e% | if(k==3)
2 a+ F; H8 y. ^* a {*(p+i)=0;! x! W9 n' P0 R* W
k=0;; y' m% A8 A! r* X6 d: H9 {, e
m++;
: E8 O0 i9 Q. g0 p9 [ }
0 K3 U& k# l8 j ]3 g i++;' b2 y) M: G" w8 X( `
if(i==n)i=0;
/ x) F/ }( V& \+ |# B$ x }6 | z- r. ~ F& z, o
while(*p==0)p++;
% K8 n! \& b# p6 y4 i+ | printf("%d",*p);4 w2 @6 |1 O* j9 S, z: ~6 A( q) s
}
" b' X& \6 U# C6 g6 X7 M3 @+ o! z10.6
7 `( Q/ S" K9 h J& t4 kmain()
- a _0 S2 K3 C7 O# @5 j7 M( E3 Z& N$ H{int len;
- X1 ~, Y0 H* G" j1 J" o char *str[20];
9 b9 J3 b8 q1 O r scanf("%s",str);" ^9 D7 H5 I0 y* W& B7 l- p
len=length(str);* U2 _1 S! k$ B
printf("\nlen=%d\n",len); M/ _* l! i' J: s& k2 A+ p M
}
& ]( y1 S# h! C+ qlength(p)$ u. d2 h2 E# Q0 y3 y: z6 w
char *p;- W0 f, A6 _* f
{int n=0;
+ o' o! c; Q, k. m! P while(*p!='\0')+ ?% r% f, U7 u) ?
{n++;p++;}
- g: d9 A# W! \2 u% } return(n);
( n, ^1 X! v9 D9 Q+ m# e( O5 Q1 z/ t}' V+ [" w6 }6 P- d
10.7' E* [/ m6 D$ V; W) _ Q
main(). r* j7 p) h. o; y3 c; W& ]/ z
{int m;4 K1 b& i: e0 A! h
char *str1[20],*str2[20];
3 V( s9 ?2 c9 @4 B scanf("%s",str1);
* i4 f: u+ P* V& E scanf("%d",&m);
- v5 Q8 G# p% Q) M6 M if(strlen(str1) printf("error");
@# i4 l9 a# i, V9 w" m else" J+ R8 j" ]1 W. Z( n
{copystr(str1,str2,m);* b d4 i6 ~: C4 }
printf("%s",str2);
. Y: S I0 K P) f& Z/ L }
5 @( K! g( b( {8 k5 h: d5 I}
9 t* G i' y5 [9 O! Hcopystr(p1,p2,m)
$ k* l2 q1 o7 Xchar *p1,*p2;% ?) }$ ?7 W5 P* F
int m;0 I5 ]- ?- ]/ T: M7 a
{int n=0;
% p3 I3 A+ C- g. q. s5 f4 q7 z while(n {n++;p1++;}
* M1 f9 e, p! [5 Z/ F* z while(*p1!='\0')
8 X$ d0 R6 i w3 N( N {*p2=*p1;8 G% W2 J* L( @ _
p1++;2 a' T1 K, ~6 o
p2++;5 S! t- Y: V6 b4 k Q' U/ A' [
}. a$ d4 Q& G* v# E- k
*p2='\0';7 ~9 [4 t% E" W r
}
! k) a' L% _% T8 ?' d5 n10.8
: ~# i# f9 \+ V/ n* R; h9 Z#include"stdio.h"& h$ ]0 ^ K! g# J$ }
main()
5 d* i- l! R( ]/ d& [7 c{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
0 Z0 F7 X( J0 [( L+ P& J# ]/ C char *p,s[20];
) X O: m9 X9 L! ] for(i=0;i<20;i++)s[i]=0;
$ W1 W- P8 d2 O i=0;7 A4 C$ L3 C1 N& N: ~ X
while((s[i]=getchar())!='\n')i++;9 D# o+ W* c& F7 F f9 |
p=s;7 g% Y. h! j- M
while(*p!='\n'); q! E1 k4 ~% y; e% b
{if(*p>='a'&&*p<='z')# I4 i L' z# |3 Y6 B, I: P
++sle;% v7 h" ^: D! x7 `
else if(*p>='A'&&*p<='Z')/ {$ z' h" P6 f0 t
++cle;9 e1 f; E B( K, Q6 t0 s
else if(*p==' ')7 T5 S" p& m' O) S
++wsp; f# A* E1 V# h8 [
else if(*p>='0'&&*p<='9')
; F2 }% J3 u! R C- D. E ++di;
" v5 v& @3 e5 q, I6 z' j% b( F6 X; h else
6 H. l$ [; [! z* \# y ++ot;
4 V2 L* ~0 O' w) {& d8 M p++;
1 x$ n1 W7 a' P) B1 x8 { }
" [- w$ p" {6 o, { printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);- z" R) ?: _% C5 N& Z. g* W
}
) X3 ]" r k4 K4 l' ~10.9
) A I# |! ]( Z9 Y. W$ Xmain()& v/ W+ x. i( X' F+ D
{int a[3][3],*p,i;+ d+ z' \ t7 t B/ y
for(i=0;i<3;i++)
! O& M8 q% r9 | scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);
+ `& b' j: C# }8 Q: N$ v% E% Q) e3 a p=a;2 ^8 Z+ o" ?& v% N; s
move(p);* o; _& B8 q; w4 o% u+ [! e
for(i=0;i<3;i++)' V; M7 f+ S3 y+ V1 D5 }
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);
5 [, N' [' j2 p' |2 A1 w! u}$ Z( p1 X; K4 f% e) P$ K) O! A$ I
move(pointer)/ k, U% ?7 f3 @5 F3 J* E2 m# `
int *pointer;6 N2 P5 w# w5 V6 u, c' A
{int i,j,t;
3 s e# b0 j& {. { for(i=0;i<2;i++)
, L$ E% I r+ o/ ^ for(j=i+1;j<3;j++)2 H! I. x' S h4 ^5 M; t/ k2 S
{t=*(pointer+3*i+j);' F* W8 O* u% J$ A$ q, ]4 X
*(pointer+3*i+j)=*(pointer+3*j+i);
5 n! z& P4 x9 q6 O *(pointer+3*j+i)=t;) W C5 n) L, V/ o1 C
}
& |* Q/ v' G# S6 ^* V. p( B, ~}6 K" i! B& R+ \' C% |
10.102 k. T q3 @% }! ]
main()7 A Y, n9 R# H) T8 S
{int a[5][5],*p,i,j;* T C/ {- V/ C* a
for(i=0;i<5;i++)
: o9 ~3 ], X4 f t4 ` for(j=0;j<5;j++)
4 a1 p& c- ~6 V2 s6 t scanf("%d",&a[i][j]);
1 [6 X& R- W- w* l p=a;: D5 r' p% a/ x# o
change(p);/ X. m: N x% I( R" q2 Z
for(i=0;i<5;i++)* C" x8 p/ q9 H
{printf("\n");
* S4 R7 r. t: C for(j=0;j<5;j++)
2 c4 y, S$ L" _4 e; }8 Y printf("%8d",a[i][j]);
2 \1 e* Y. j/ Q# h u }2 A5 x; Z# l C! \, G8 `
}: o, Z6 e' U/ i
change(p)- T5 \2 q& d' w
int *p;4 P9 X8 ^1 A0 @# h, ~ L
{int i,j,change;
: Y) \0 d; U: k4 s8 b int *pmax,*pmin;
* G4 N# b9 g* B g# s pmax=p;
& m# O/ ]. L6 h, S& L pmin=p;9 I& |' Q! e, F& k4 o+ w5 h
for(i=0;i<5;i++)
: l. A p7 U0 {/ l. B$ b for(j=0;j<5;j++)% z! ]4 s% l; o( N
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;
6 N4 M! F! {3 N9 ~$ K$ ` if(*pmin>*(p+5*i+j))pmin=p+5*i+j;% x K% r3 [& O7 I
}
; ?8 A/ m! {2 @$ I L7 p1 j change=*(p+12);! _& T2 C! p: t8 d4 {* F$ ~
*(p+12)=*pmax;- Z, Y( \- r5 a t# Z8 w! c
*pmax=change;
2 w# g0 B4 W! L0 s% f. ~ change=*p;
9 a# I! |3 l/ `; w1 P8 e *p=*pmin;
; V. A5 Z9 A( X *pmin=change;0 v$ w1 R, p* o/ e- s# ?5 V* s
pmin=p+1;
; f1 j( T% \6 H for(i=0;i<5;i++)
/ `6 U/ P8 c+ t, f) Q: r for(j=0;j<5;j++)
& o! o; E; H. W l& E if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;# w Q. n% q2 P4 W P6 T3 M
change=*(p+4);, m& Y# E% K' e% f- e
*(p+4)=*pmin;
6 O3 R9 W' X! o* Y0 b5 g9 a/ \ *pmin=change;6 H$ q0 O$ E" Z. M, L' Q4 A
pmin=p+1;
+ h: z- t/ _! X. s- A for(i=0;i<5;i++)% a( r4 z+ e9 ~7 a' {; F7 m" D9 L% k
for(j=0;j<5;j++)4 M% x# u; f7 Q2 D! ]
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))) z3 [4 w; j+ O$ [% C0 ]6 l
pmin=p+5*i+j;
) L( g& l( N7 w' v9 c+ T change=*(p+20);% q$ r1 d- e) p+ G2 F
*(p+20)=*pmin;
* c" t2 e1 z" ]# P0 k *pmin=change;
; [( \; x' p$ y6 r# [, V/ t0 r. o pmin=p+1;$ h% l% I3 [* Y! X! p- X+ o
for(i=0;i<5;i++)3 G- P) W0 W* z4 e4 N
for(j=0;j<5;j++)
& Y9 {: J( y* k5 [1 T if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
$ ~+ M+ O" u$ k" v' q2 [! X0 l &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
4 F) m I8 T2 q) K/ p6 { change=*(p+24);
# K6 |" ^# ]$ ]: x/ h! q *(p+24)=*pmin;, `) S4 ]: P- v/ _$ N, U+ ^
*pmin=change;
7 G9 E* H; M' P3 k}
. A n+ ]: G3 B) c8 t+ m5 _4 W10.11
) T- F' h$ v! a6 i9 J; fmain(), ~ W$ ^9 c3 d
{int i;
5 y" t0 Z! ~, c! g& \ char *p,str[10][10];* ~: ^, P; Z9 q0 L+ _: D/ |! G
for(i=0;i<10;i++)5 i5 s8 E: E8 @0 o+ U+ v
scanf("%s",str[i]);
9 f8 t4 o/ ], a( I/ w p=str;2 m: t% Q* u( Z1 G& _
sort(p);
, G. ~+ L% z3 s9 M, ~, [ for(i=0;i<10;i++)
8 o2 S6 V( L& r2 A$ B0 o6 _ printf("%s\n",str[i]);
9 W9 ?+ p8 b/ g}5 E2 Q7 v5 |8 P5 ?
sort(p)' E4 F+ M* {6 d
char *p;, ^8 k$ J7 ?5 y5 [% D
{int i,j;3 U( \6 I$ G+ ~+ X: e. S
char s[10],*smax,*smin;: S# Y0 @5 f' n: y9 t( w& i3 F/ A2 K
for(i=0;i<10;i++)3 }- c6 L/ p. E" e
{smax=p+10*i;
# q$ ?3 J5 o4 W. ~4 O for(j=i+1;j<10;j++)
2 ?( k5 ]) i9 t: @( T {smin=p+10*j;8 `) J! g9 r1 v5 V2 V$ t
if(strcmp(smax,smin)>0)
: b- b* x$ ]7 T0 V: @) L8 M {strcpy(s,smin);
$ l& W5 J* f: t( y. X# ^ strcpy(smin,smax);
; c2 F6 }/ K4 r& C strcpy(smax,s);9 q" z6 ]8 a' d! g& H" p1 t
}
4 A" u f0 K$ M7 e6 S: b: b) d }4 K$ G% R- x- i* D; o H
}
/ }: \6 _& ]7 P8 h, ~" u}
8 _& g1 k: g$ \: k! T10.12* S$ Q9 \! x2 }+ @
#define MAX 20: |: z M) _4 K( b. k" {7 }5 j6 o
main()
& n+ n% d D" C" q n# Y/ N{int i;
( o6 k2 l& u O* o1 [ char *pstr[10],str[10][MAX];
5 d: V9 Y% F( G8 ?' _ for(i=0;i<10;i++)
# ` @7 G. o' n pstr[i]=str[i];1 r2 P1 G8 k5 d8 A" ?! N- k
for(i=0;i<10;i++)' C( K2 r% v; L, t5 p! i$ n# `) ?" Z
scanf("%s",pstr[i]);
5 W9 E4 o1 I% }1 @6 t/ h& l; k, O sort(pstr);
7 ^1 M( h' u) k' z6 D3 G3 [, G for(i=0;i<10;i++)6 k) H- q* D0 l8 U4 q8 x; y; t) v
printf("%s\n",pstr[i]);! ]& J |; y( }- b
}
: e, t" x4 [. z3 E$ G w. Csort(pstr)
/ h5 T0 I6 j2 M; @char *pstr[10];
& U& c! Y8 ?7 B{int i,j;1 B! i' R) z3 K. J# F+ h1 Q& o
char *p;
. _; @8 _6 v) s% T h5 T+ U for(i=0;i<10;i++)
T3 _+ L! A$ E6 Z$ l {for(j=i+1;j<10;j++)
7 l6 x d* y$ }2 a$ i. } {if(strcmp(*(pstr+i),*(pstr+j))>0)
1 F* V" L# o$ R7 A {p=*(pstr+i);, E" u' }5 D2 c, a. ]+ q" r; Q
*(pstr+i)=*(pstr+j); R/ W C2 Y& x( F2 n: S
*(pstr+j)=p;
+ y, M$ u- U+ h; Z6 Y3 i6 F- C/ w }, P, u d! a0 K$ B
}8 T3 B; r% {2 E
}
7 h5 s: W1 J0 n}0 k% |; z/ `* O- B
10.13
( }6 i1 _% i; |( L; Q2 O& t#include"math.h"+ v7 U" P" S7 ?) N5 ^- y
main()
5 B7 |/ q4 f2 C, U) T{int n=20;
% W) p1 a. C% `" G float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
: [- M ~& C, R+ ` t/ Q6 b$ j: \* s% \ scanf("%f,%f",&a,&b);
! R* ^; L6 O) k1 D# N scanf("%f,%f",&a1,&b1);- P# ]# }7 r- Z# l
scanf("%f,%f",&a2,&b2);/ s% m3 z, y) T2 S( r2 d: ^
p=sin;4 j# j! _5 @- K& p% e
c=jiff(a,b,n,p);; n/ { H3 W$ D0 E$ ^" E
printf("sin=%f\n",c);2 U' T9 D! I% I0 E3 N
p=cos;
: g1 R. k1 Q6 k; H& L3 {5 y7 P c=jiff(a1,b1,n,p);
& a& c& \+ K2 M- N9 p printf("cos=%f\n",c);! y& X6 v! S3 `8 l3 ^! H( c* g
p=exp;
! r6 k- S/ o- e9 d$ y c=jiff(a2,b2,n,p);
9 k+ b+ P0 C$ h" J: z. _ printf("exp=%f\n",c);# z1 D3 D* u4 k$ @4 M
}
& T# B1 z3 N0 F! Efloat jiff(a,b,n,p)
5 h: b: ]- ^5 sfloat a,b,(*p)();
( ?0 E$ @0 ]: ^0 r! oint n;
) d( L9 d$ w6 @& |. P{int i;, l# p* C" m) Q& g, \/ _8 d
float x,f,h,area;
- y% M+ _/ J4 o1 c h=(b-a)/n;% X+ L* @) H9 s+ z" B% v+ U% Y7 H b
x=a;1 a0 n/ h; v, c# G: c+ z1 w/ Q0 l
area=0;0 O+ v0 q# Q6 q& y1 @- Z F4 y/ M8 I
for(i=1;i<=n;i++)1 T3 T/ P) t! Y/ N2 h
{x=x+h;
2 l" h3 o. q/ k, } x; ?! ~ area=area+(*p)(x)*h;5 R6 ~' e9 T9 ?" Y7 L f5 m
}
" P- K; Q6 Q+ \* H- Z return(area);
: m6 u; G8 z0 H( S4 h5 C}% w, Q9 r7 }* B
10.14* v, d& F' r1 D/ b2 J
main()
* b0 J+ i2 H" e! E3 g2 p{int i,n,num[20];7 |/ ?* t9 b+ p+ X6 j4 X: J
char *p;/ H* T3 u5 A- C# F6 p
scanf("%d",&n);
4 V' m! Z; J" [ for(i=0;i scanf("%d",&num[i]);2 t% f; H* a2 {
p=num;
- t. L/ R0 ^- F- k4 k; ~8 o! S sort(p,n);" N( s* X$ d+ v2 y0 p! F$ S( k
for(i=0;i printf("%8d",num[i]);
9 ^" P' M0 _# ?5 @ r1 w8 d}
6 L3 T/ B* q. I5 `& m$ X4 vsort(p,m)
" H) w' o: u- L1 |: W2 ochar *p;9 j) l# ^2 ~/ H) q6 j
int m;
$ A( U$ c& o; O{int i;0 b; \9 I- s# m6 q2 @: A; {( l
char change,*p1,*p2;5 X* G9 l+ s& U) e; ^2 c1 W
for(i=0;i {p1=p+i;
6 |1 y) ?* S9 F% i; Q p2=p+(m-1-i);6 w8 K8 A" O+ y5 C- L3 o- a, ^1 W
change=*p1;# q5 N" ?+ f. i) I! j6 K9 x
*p1=*p2;
7 `3 e" N5 x& q( z: f *p2=change;7 S/ |) C* q8 G
}" h$ V( M& M( e# t, q
}. o% o$ t \" e8 X
10.151 [0 F' @5 H* o/ W
main()
0 _% R. f* m* X1 r{int i,j,*pnum,num[4];
/ w* G6 h- v, T: ] float score[4][5],aver[4],*psco,*pave;/ }7 ~ s4 R& n6 M3 k4 z& X
char course[5][10],*pcou;) h. j$ `9 e3 h8 y/ T3 |% @
pcou=course[0];. `% H$ M7 h" V; Y9 N9 ~4 E
for(i=0;i<5;i++)
7 D8 S1 ~. R4 f scanf("%s",pcou+10*i);, }- k; ], V+ E9 ^1 D
printf("number");
7 I/ P% F+ ], g0 ~ for(i=0;i<5;i++), d9 f, e; z7 o5 C) O9 z+ o
printf(",%s",pcou+10*i);: H5 z- U$ e' W' ]9 A
printf("\n");
- N2 H8 p x' y$ O: P9 n5 V psco=score;/ J" M; j( f+ q' Q
pnum=num;
8 Y1 Y3 d! m7 u+ O% } for(i=0;i<4;i++)
5 V$ [% l/ |6 D {scanf("%d",pnum+i);
! n- E" X3 o- Y T h- F( c) ^ for(j=0;j<5;j++)
O& g, P; @5 p scanf(",%f",psco+5*i+j);6 y" A. z: I1 ~5 a$ Q+ h8 W
}
! D; g. y$ E, B- R1 T/ `! ?- B pave=aver;
4 t" Y& }% `. Z4 E* |: Q" f printf("\n");4 L, o% s. P: U& ^* \, T+ a
avsco(psco,pave);6 T8 M) a& X$ p! t% ]1 |* H' o
avcour1(pcou,psco);; Q2 M5 K/ S2 V% v* f
printf("\n");# i0 L0 k" G3 p. r# q
fali2(pcou,pnum,psco,pave);$ }' B$ Q/ s6 U# Y
printf("\n");% V" a& ]; [" Y& x5 c* W5 b
good(pcou,pnum,psco,pave);% E0 b; ^; ^: t
}
& _1 S0 r1 S7 Vavsco(psco,pave)
) u" I. R0 ]! G8 }$ n9 efloat *psco,*pave; d1 \6 @! U, I0 F
{int i,j;
$ } s# x; v5 X2 D, ^/ L float sum,average;
" ~, K; n* P) [% r; L for(i=0;i<4;i++)& @: D6 K! H( r2 b# e4 H% f
{sum=0;
) E! t3 }2 }( i% m1 z& \ for(j=0;j<5;j+)
* Y$ ~9 _% H: k; _' c sum+=(*(psco+5*i+j));
8 W6 a- b. g, x: c4 K( z average=sum/5;& h+ A5 {0 t/ b$ ~
*(pave+i)=average;. V Y0 T5 {1 `+ e; x1 u
}% I8 J& ]7 d! x* z1 U! E, Q- K
}
6 d5 P1 Y( L. d& s" Davcour1(pcou,psco)! Z* J8 _3 u! s- o+ K
char *pcou;
8 w# ]/ k7 C2 v xfloat *psco;0 K8 v0 b$ ~ w; a; }1 ?4 R" E
{int i;
5 Y' l; ~: Z' C, R! Z8 h# ^% m& H float sum,average1;
( }7 a3 q% o( X) S2 Q# F7 w8 q sum=0;
4 \7 Y6 e- j" ~6 ~# d* p for(i=0;i<4;i++)1 i4 { F4 }5 J$ P+ z( O4 v
sum+=(*(psco+5*i))
6 a1 R& E. Q4 r, C average1=sum/4;8 x+ [/ n e9 H( ^! R
printf("%s %5.2f\n",pcou,average1);
7 ]5 @- j& G% d}
% G, F1 e- w: F, ]fali2(pcou,pnum,psco,pave)
/ P% T* m7 A! cchar *pcou;
5 R% Y) s0 X& ]3 qint *pnum;! [% ]5 M* b" x9 u, ^
float *psco,*pave;, E8 ]4 G% x* S3 H) B% t* B
{int i,j,k,label;
D& x5 t: X8 G1 T1 y! I printf("\nnumber\n");1 b) T1 [, v: a- v% Z
for(i=0;i<5;i++)
3 b% D- a0 c8 O3 ~* h printf("%-8s",pcou+10*i);
2 O5 n- q5 t: P- i printf("\naverage\n");) w3 {1 `; M9 ?7 E6 ^
for(i=0;i<4;i++)
/ Z7 o8 r/ g6 D {label=0;
- t9 p) y" ^$ U: d; { for(j=0;j<5;j++)
/ h8 Z0 g# V" U& ]* H4 y; a if(*(psco+5*i+j)<60.0)label++;. n R" ]* J! x! U% `2 @! a" I
if(label>=2)
6 f8 K/ Q! {, x) ?4 n- [. j {printf("%-8d",*(pnum+i));
) L3 f+ S6 ^) u0 N* ^+ t for(k=0;k<5;k++)% ]/ p, ]4 q8 b$ H
printf("%-8.2f",*(psco+5*i+k));
5 P9 ]3 D/ M) d( c printf("%-8.2f",*(pave+i));
' U; Q. E( f1 n/ m f1 q! O* |& u7 O }0 y, E' N9 A3 |4 Z- z0 A
}! ~# |( b0 ]: d% O. Q/ [
}
* ]" o0 L4 \1 o3 `, z% C bgood(pcou,pnum,psco,pave)
{7 Y' z2 {) F2 i D* tchar *pcou;
2 O" s- [4 _8 D. z& o0 E0 tint *pnum;
7 j* `) G0 h9 {. |: M+ @* b8 Bfloat *psco,*pave;
* X( G I" H& U- }$ K; N{int i,j,k,label;8 H) D7 [6 x8 O$ J
printf("number");
# b( F8 g* }- ?' s6 @2 n for(i=0;i<5;i++)- k' ~5 q; ?; b- Z1 }
printf("%-8s",pcou+10*i);
5 x( N5 a2 l& s4 k3 e" {! X printf("average");, i! i1 ~3 w0 H; i0 l2 c: V
for(i=0;i<4;i++)( ?: b2 B5 q) Q) j# A* T
{label=0;
" W6 c i N' B* E0 l( V for(j=0;j<5;j++)
" y2 \ ~2 N! D3 U if(*(psco+5*i+j)>=85.0)label++;' K9 J* e. q8 Y& O3 |% s+ `7 I) N
if((label>=5)||(*(pave+i)>=90))
|9 T' h: `$ }3 y- M6 q {printf("%-8d",*(pnum+i));
- b! U8 R4 r& i0 ~9 a( Z" n3 g( q; T for(k=0;k<5;k++)
, E5 f$ V! Z( O+ }( O" M printf("%-8.2f",*(psco+5*i+k));0 W* T. a" N& y1 f+ U Y& j7 \- O
printf("%-8.2f",*(pave+i));
! i5 t3 i3 K2 J8 [; }: s2 v }( H1 D2 K$ z+ B
}8 G4 i8 @0 n7 h; c
}
6 j8 q3 A. ^0 H$ p. F( R9 O& `' f3 X# J10.16( e$ v c* R2 k
#include"stdio.h"# ]& M% L0 S" v
main()+ e/ d. b& }( h0 g
{char str[50],*pstr;3 x g* [: j: u# ~3 r
int i,j,k,m,e10,digit,ndigit,a[10],*pa;
* ]8 _( X l5 l- \& x. B2 C2 u gets(str);& ]& o- I2 _: d3 T4 J
pstr=str;$ n( P! j5 B4 y; t; ]2 h
pa=a;
5 d. Q7 }+ W( f ndigit=0;
# S6 I, [' r2 `3 o2 Q( ?% g' m N3 ^ i=j=0;
, d: K8 v6 p# u( I& M" n while(*(pstr+i)!='\0')
' q0 G; j6 u4 L3 Z {if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))
* k7 @6 D# ?4 }5 P& o j++;5 A1 _' F- j+ Y, F, o2 J; L5 H
else
( u3 B% k) ~- L8 K/ ^9 @ {if(j>0)
I6 A4 K' J( n8 E- Q$ r/ y {digit=*(pstr+i-1)-48;0 t; m& t; s P1 F
k=1;' A& D7 q f. ?9 S3 X& o
while(k {e10=1;
. O0 ^+ o, a0 J* e9 Z: y% \ for(m=1;m<=k;m++)
! b4 Y! A9 j6 O* Q! p* W+ U0 ~0 Z e10=e10*10;
0 N; E2 H$ i$ {3 |2 W6 z/ a+ } digit+=(*(pstr+i-1-k)-48)*e10;
, S7 Q' ]5 P6 L& g: N9 T$ n k++;
; H( w! e/ P1 H. W: b1 f! D( o* b }& K9 x K0 G. U) Q
*pa=digit;
6 h6 `1 S2 J; T ndigit++;% T/ A5 d) j" ]
pa++;: l+ P8 N: ]7 w: u4 W! k# Y
j=0;- d. L3 {1 ~7 u, \) F
}
0 \0 ^, P# c* Q! w }5 ~$ m( S( h5 c# W+ i
i++;9 s4 }) }' N; S5 ^) w
}- ^ N* q' _9 h0 n2 O
if(j>0)3 W3 X9 L- a: O8 b7 |! K
{digit=*(pstr+i-1)-48;
3 {3 ~+ ^0 p7 u2 {2 @ k=1;- l X( A P- [4 e) g; P. S
while(k {e10=1;
) J7 u, X6 V B% A for(m=1;m<=k;m++): D) L# H( T/ F C* s
e10=e10*10;
2 ?. @. T- Y/ a8 h digit+=(*(pstr+i-1-k)-48)*e10;
0 w* W O$ m1 @+ e k++;8 B! c1 Y: H" l4 k/ T
}
1 ]! L1 |4 [8 |& o# y4 H *pa=digit;
0 ^. k9 C+ B* U# N. }5 F. L ndigit++;
f; J2 B8 J5 W/ C4 q8 | j=0;
& p2 u) M4 l# `; E" r9 f C Z } ) R5 h+ r5 {# T/ i8 B! j
printf("ndigit=%d\n",ndigit);
& H0 C$ h8 I+ @1 h# m j=0; n. w, Q$ W" X0 ]( F1 ]% ~
pa=a;: K* A; f' ]3 K: `8 w
for(j=0;j printf("%d",*(pa+j));
3 V5 V1 }1 P0 S. w( P2 D}
1 s3 E/ @! O2 P10.17
5 W( |0 c9 O# [9 B; Kmain()
3 {& o# t3 h- U X{int m;; y$ J7 t+ t$ @( n/ p2 O, B
char str1[20],str2[20],*p1,*p2;4 H8 P4 K ]2 g T5 L: c
scanf("%s",str1);
4 R) T( S) f3 f$ I, E: ^$ D6 Y scanf("%s",str2);- W* F/ M" l. X* k5 N, D& d
p1=str1;1 e5 f4 {' R/ K' m5 X7 i. W
p2=str2;
; i3 v7 d i* ?! a8 n& O m=strcmp(p1,p2);
7 ?* A8 g5 [, h: f4 E* j% u printf("%d\n",m);
2 k0 p5 p0 S7 t0 z/ c}, z! a/ H g, N, W9 {# `. h7 s
strcmp(p1,p2)
/ J) @# j7 ~3 \" g& F/ ?1 c2 u. Ichar *p1,*p2;8 g: H9 T0 }9 \/ w7 s
{int i=0;8 y4 p( U, ` q$ U" p& d
while(*(p1+i)==*(p2+i))
0 `" j# @2 X" G- s3 Y if(*(p+i++)=='\0')return(0);( G+ F8 L/ x2 o7 o+ u! q
return(*(p1+i)-*(p2+i));" m+ r9 W! F/ V9 N: O5 K+ p
}7 o% G. C) w" y e) m* D: o
10.18
; Y* r% o, ]8 U9 K1 }+ H. Imain()
, M! x1 e$ ^/ b1 e* a& \{static char *mname[13]={"illeagl","January","February","March",
" c$ L, I% W# y( i+ b; q* b* x9 d" Q "April","May","June","July","August","September","October",* t/ k: P) w3 r5 h" l
"November","December"};5 ]. z7 @! f* c' `" J* r: W4 i0 H
int n;
1 t8 b5 t8 Z* \3 j* U2 U) ] scanf("%d",&n);
7 H9 v( l3 b) _$ s; D4 Z if((n>=1)&&(n<=12))
6 a h. b7 c, ?+ w printf("%s\n",*(mname+n));2 p. I! m k, e/ L, V" _
else. d( f& z1 }8 j8 e
printf("error");6 q# x: t3 y) ?# r B- A% @
}
9 I% u5 ]% P3 P- y10.20- s: O$ b3 T- I3 Q; }" M
main()
6 i1 R5 u7 L1 P1 j; w{int i;3 f8 P* I5 @, h7 F
char **p,*pstr[5],str[5][10];
) d- ^+ G8 Q) `0 ~/ m0 D for(i=0;i<5;i++)
; i/ z6 U! F7 i+ D4 j5 @ pstr[i]=str[i];
. z6 q' E+ ?5 m/ `- L2 Q for(i=0;i<5;i++)/ T3 b/ B) H) f0 w
scanf("%s",pstr[i]);
, _' i( R+ r$ ~! X9 ~+ P4 i p=pstr;
* Z* W: A0 b( ?9 [3 T/ `8 L( x0 f# J sort(p);4 e! w: t: v4 t {4 i
for(i=0;i<5;i++)( ~& E' d& ^9 @ f6 n% u3 x0 A7 d+ R
printf("%s\n",pstr[i]);
& l& u5 ?4 a. z$ f& I$ r4 x}, y @% V. w) I( T
sort(p)
$ @, s7 `7 R! S( Vchar **P;
; j6 G; L' o/ l5 x{int i,j;
, N2 o- Y4 z! ~* D" ^+ t. | char *pchange;- s1 O' |5 L; {# C- |
for(i=0;i<5;i++)
5 u! y; d7 g. {# _6 o8 N( M! R {for(j=i+1;j<5;j++) W, \9 t' L% ^0 X
{if(strcmp(*(p+i),*(p+j))>0)
" V& i+ \+ p I( M( ], q2 O" Y {pchange=*(p+i);4 H& K. j% C5 L" W; G2 t+ B9 |
*(p+i)=*(p+j);7 ^: m+ ~2 _% H# ~
*(p+j)=pchange;' w% B$ b* r6 k: `% v' |7 ~
}
% k' f. U% Z+ }+ ]. R }( p9 q; H9 B: J9 t' y, @0 l7 y
} d" m8 c( C" w% J# ^
}" N; m9 }1 x/ p1 l0 M1 [
10.21
Q+ e j1 H4 R' \8 i. ?main()' j% Y1 q$ d/ |8 a) M
{int i,n,digit[20],**p,*pstr[20];
T6 I/ [, @; X8 i scanf("%d",&n);9 ^2 N1 {0 H6 ]- J$ F8 ^5 v
for(i=0;i pstr[i]=&digit[i];; h' c- U* d2 c% \' b
for(i=0;i scanf("%d",pstr[i]);# q8 O# ? P/ o) Q6 n
p=pstr;" O: _8 n" m3 u0 y0 I$ `0 r4 h
sort(p,n);
9 ]8 e% c4 N& V1 [ D for(i=0;i printf("%d ",*pstr[i]);* r* J0 u* `$ a- g% f) P8 Q F
}3 \4 {& x) w* Z) O0 W* z8 g
sort(p,n)4 H0 G7 l8 D( O. H8 g( {# q) e' S
int **p,n;: l' I& g$ r% T% U2 |1 z% \% g
{int i,j,*pchange;2 e" z$ d% M8 L7 [" Q. B
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))5 {+ H1 V2 d5 J
{pchange=*(p+i);
* \+ v9 c5 H( {4 k- k2 Q$ q *(p+i)=*(p+j);
) ?7 o2 h; h$ T9 O* f *(p+j)=pchange;( p* v) o7 x: ?0 V0 ]' [
}; B) H4 z+ y2 S1 ^% o/ J
}6 @( w# g& b( |& H! Q! _/ c
}) C* G9 S3 B6 L: {) r- U0 r2 [
}
3 u- Z; H5 H" t; p0 G1 }1 g) v5 J第十一章 结构体与共用体
) ~! a5 V* A3 G2 | D+ U& C11.1. V$ I8 I, n6 i) `8 u! x* n$ |) v
struct7 {6 P0 d3 ^5 t+ A! \0 R
{int year;- J! Z; f" H* t: C; x" c
int month;
/ `5 T/ ~ [) l int day;
[( N+ i7 E- a: U6 ` }date;
1 |' E# U, f, d# S Fmain()
: N, A( k! V: Y: h4 o/ o0 ~( S) R{int days;" q; h; M! i ^' V0 g& ^+ N8 e5 y
scanf("%d,%d,%d",&date.year,&date.month,&date.day);
+ E6 X, |, [1 H, v switch(date.month)' J5 I; a- J+ K/ J) Q) ^
{case 1:days=date.day;break;
8 m+ E8 [/ o' |6 z case 2:days=date.day+31;break;
% D5 _! C- F1 [# z3 b4 U" [ case 3:days=date.day+59;break;
+ n! V# J7 G$ o3 M/ K& b case 4:days=date.day+90;break;
, v/ Q( k* V# M) n. i# B9 ]/ } case 5:days=date.day+120;break;- q. {# N; O7 h% H7 e
case 6:days=date.day+151;break;* V* @, r3 l( s" w5 ^$ f( |
case 7:days=date.day+181;break;" `, |: ]8 X; y" Y! M) Q( Z
case 8:days=date.day+212;break;2 H$ E4 c; Q) K1 B' I `! g
case 9:days=date.day+243;break;
F! P! x, v- V case 10:days=date.day+273;break;4 d; O4 T# t+ w2 c! @1 y
case 11:days=date.day+304;break;
% w2 r6 [/ R3 Y case 12:days=date.day+334;break;
) @& C$ ], w/ f }5 N: K! D N' H/ V+ y
if((date.year%4==0&&date.year%100!=0||date.year%400==0)( ~3 I6 B8 O5 b+ v, W
&&date.month>=3)' _1 g g1 R$ D8 z3 z: k3 X$ D
days+=1;
+ Y d2 C; j4 S4 B. y& C+ r1 a printf("days=%d\n",days);, L) z. }7 k. w0 n6 Q0 p
}! \! u4 K8 f2 ^
11.2
. n0 l1 P. @4 N% o; kstruct dt
: l; o- s8 V0 \4 d2 v3 j {int year;
2 d( N1 `9 M( F! O) s int month;
$ l3 E- L- _! Y int day;
$ D% Z* H" d. T" F2 Z; [$ X H# o }date;- P; _/ e! f% X7 A" ?% e& j& a
main()# s2 N$ p4 C, g
{8 P4 ?4 T; N! b+ V) F- D) Y1 z' _$ A- J
scanf("%d,%d,%d",&date.year,&date.month,&date.day);) P) a4 @: C! A6 ?1 ~( ~8 {
printf("\n%d\n",days(date.year,date.month,date.day));
7 `% Q- {$ ~) C9 F$ |: G- x}
5 L0 C0 `9 h/ s8 sdays(year,month,day)
& C" e! n# I ]$ aint year,month,day;
" C* h! @. I7 I" a& ^8 W8 W{int daysum=0,i;( A i- }0 W& l$ g% j1 G
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
6 n) m0 T' K0 M. x for(i=1;i daysum+=daytab[i];/ W0 G8 @9 N* r5 t9 n& E' \
daysum+=day;
" v/ L5 H2 Z/ D& U; j* w4 P' _ if((year%4==0&&year%100!=0||year%400==0)&&month>=3)
# X0 a& v0 z' h% v* z daysum+=1;8 O6 O- u8 G/ ~2 K1 ~6 ?
return(daysum);
! Y1 D+ F' O0 G7 i}
, h# T5 E4 B9 n: O11.3" J# X' H2 v3 Q
11.47 ^5 [ _3 `$ `/ B, o; V3 L, w
#define N 5: L$ Q- q% E) a0 b: O! m5 v' |
struct student' U, ]) n( i- }2 b7 P
{char num[6];$ \; Z" L) N5 u/ @9 S" o2 t3 |
char name[8];
4 w; l. B7 z, c- B6 f- j int score[4];
2 W* k/ x. y& |7 f) M( r }stu[N];
3 b7 I7 M$ V, F) cmain()
7 ?) [4 q- X) W' a" p0 c7 B8 |{% B% U4 q" M, l( i" ]
input(stu);
" x6 b3 N% G8 f7 h2 A print(stu);7 {) N5 }# E' ?( s
}1 }- t7 z; ], {. H5 ^1 X% b
input(stu)
! `( B! d0 O# Z3 M8 j) D9 Z1 hstruct student stu[];3 B. d5 Q0 i5 q6 \
{int i,j;
8 P" W9 _8 w2 S! d d4 t for(i=0;i {printf("number");
8 R& i( L; V" E2 A3 q( y0 i9 i# E scanf("%s",stu[i].num);
% ]$ q6 F1 R9 S9 x8 V! p printf("name");
1 {5 |# i( g; F+ J scanf("%s",stu[i].name);+ G" j' E) p" t. w0 {2 D
for(j=0;j<3;j++), w0 K: k3 |# t( S( q0 P
{printf("\nscore\n");
( l! z' @2 s, N scanf("%d",&stu[i].score[j]);" B3 y5 O& K9 o8 y
} B" V& K: A" r3 ~4 }$ x0 I
printf("\n");
, `8 O/ n! Z1 P+ A. M }8 o% Y7 R8 H( H$ @0 A. e( B
}
u* [. b+ V6 H1 ]1 Z8 |print(stu)
Z- A$ z# M. W- H2 O3 jstruct student stu[];; K$ |5 S m m6 Y
{int i,j;
2 R; Q+ f5 O* w+ P8 j6 Y printf("\nnumber name score1 score2 score3 \n");
4 Z6 b3 j4 e7 f2 ^4 q+ O; I; O for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
- g) g$ k9 y9 R* a3 D1 S for(j=0;j<3;j++)* `+ l4 `& L0 c# C- h6 x
printf("%7d",stu[i].score[j]);) b5 X% n4 Q! V1 ?0 i$ [& @' W
printf("\n");
6 T5 \ y1 ]: N8 a$ s# h }
8 L- w( R/ K% _1 S4 G9 y}2 q" x2 l9 u& y2 n" p: }3 _1 {
11.59 B% v6 f: P) C/ N, S+ U
struct student
2 t% Z9 \) w! \& m' _ {char num[6];$ z S& J2 u, z- F
char name[8];% u) i; P5 m3 F. b6 p# d, [
int score[4];
5 q5 {) w6 L. M, _$ x4 `! @7 s0 m float avr;2 `+ l+ u' D- [
}stu[5];. g7 A' a1 O$ \
main()
+ u( N3 A* _+ [4 M! k5 o{int i,j,max,maxi,sum;5 n9 @2 F0 \: d& F2 A% H6 o
float average;
0 g1 Y5 @' {5 k' E- x7 j6 q for(i=0;i<5;i++)
8 x' I% A& y+ P8 ` {printf("number");
M: t0 b+ E3 ]; _ r8 m6 w8 a9 N scanf("%s",stu[i].num);
u* T7 e; g; N H4 |8 }% @; L; x printf("name");
2 p) _& d, d% r) q0 ~ scanf("%s",stu[i].name);( C O! U2 v% u! F) i% M
for(j=0;j<3;j++)% T, T3 C c- w
{printf("\nscore\n");
3 T* U8 A# A; x5 k& M% b! _" l& @( { scanf("%d",&stu[i].score[j]);
7 M" ~" u F9 C. t5 W) U0 {; V+ P }
2 S( P4 K/ \& d$ u- n }
( r+ U, l7 N; s) D5 u average=0;7 N! U- P' X3 S, E
max=0;* c s' l+ @+ u
maxi=0;* H! C5 @1 b; z* y
for(i=0;i<5;i++)
, O8 E4 P. b# f8 w1 t3 Y5 ] {sum=0;
$ `- ~, X& ~, J8 B for(j=0;j<3;j++)$ n) F1 g( F' H0 K' @
sum+=stu[i].score[j];
8 b7 X% L4 z, ^8 h stu[i].avr=sum/3.0;1 X2 T0 p* q8 m* c7 _
average+=stu[i].avr;
! V: ]$ @5 q- ]' A2 ?- i if(sum>max)+ U; ]1 y( M4 v4 c- L. Y% ?8 B
{max=sum;* ^4 r, L4 h B5 l
maxi=i;% C i; T0 w* F% I5 c# W
}& {+ d+ d+ P& L; i) k# I
}
- B: G9 [0 f- C5 m average/=5;
8 n! X7 d |% U4 w printf("number name score1 score2 score3 average\n");
6 H/ ?3 F }# o0 O8 |% e% G for(i=0;i<5;i++)
7 \/ F! _$ b* u# _2 d1 J {printf("%8s%10s",stu[i].num,stu[i].name);: `6 \. Q4 F" \% {7 z0 \3 g
for(j=0;j<3;j++)! S4 x* @ a% ]+ Z
printf("%7d",stu[i].score[j]);- z }' A' B3 k* j0 J- w/ X/ ?
printf("%6.2f\n",stu[i].avr);9 i. E$ H, W+ i+ ~' c2 i
}5 h% H* g; P* _. u
printf("average=%5.2f\n",average);# P; q6 K5 O" w n
printf("The best student is %s,sum=%d\n",stu[maxi].name,max);7 N3 ]. K2 U; p0 }/ U7 F( N+ d
}
! g, ~" W/ U6 a/ @1 p& R6 L
8 F, `2 L! n& ~6 ]1 O |
zan
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