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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9 |, t; n; ~& z- J6 J
1.5请参照本章例题,编写一个C程序,输出以下信息:
9 z% z9 u: |! N* V2 Rmain()
5 ^! U5 L4 O! o, O{
, f/ T: p: A- n/ L5 {printf(" ************ \n");
0 V$ n6 E8 P8 R; O" |printf("\n");
3 T6 H/ j1 K' I) ?- Yprintf(" Very Good! \n");
* F5 X' _& B# [: u; m" b/ pprintf("\n");
" p% o8 g& F* P. Fprintf(" ************\n");, U, E8 T+ ]) i8 [! {2 X
}
; |" p" n: a, w5 R8 X0 }8 h1.6编写一个程序,输入a b c三个值,输出其中最大者。+ ^6 \; K% H0 h9 y1 V! Q- r. ]
解:main()4 j* ?" k0 }; }: D& D) B' N
{int a,b,c,max;* |1 o# q, B7 `2 N; N% r" n7 N7 p1 c" a
printf("请输入三个数a,b,c:\n");( v8 e( f: r( f3 ^: U
scanf("%d,%d,%d",&a,&b,&c);1 g5 B( S% [4 N! q
max=a;( }/ ]( i$ N. ?+ l
if(maxmax=b;; r! m3 {9 v& x3 B: d+ s/ Z
if(maxmax=c;! Z7 h1 C9 U6 f$ v2 d
printf("最大数为:%d",max);% F1 F" n0 E4 m; n
}
5 l4 G {% S, l# ]: N7 N第三章
, j: _- u2 d4 J3 i3.3 请将下面各数用八进制数和十六进制数表示:
; R: Q/ d9 X( P* [. g(1)10 (2)32 (3)75 (4)-617
4 _. X6 Q. H5 N: D# P- f(5)-111 (6)2483 (7)-28654 (8)21003) v1 M, x8 P2 W1 p* K
解:十 八 十六
1 ]8 `! j) D7 h6 |1 u1 U% @ (10)=(12)=(a)8 P3 X' c. Z2 Y% |$ C: I* }( I( ]
(32)=(40)=20% q3 s/ M5 V c# |9 ]- u
(75)=(113)=4b
- \- `: H/ L$ ~3 l (-617)=(176627)=fd973 R0 o' i. S) N/ ?% J0 w h
-111=177621=ff914 W( f" e) q) C# N
2483=4663=963
. B/ {3 B) h! V* i g8 F7 z2 d; Q -28654=110022=9012; ?5 l, h4 x0 K, Z
21003=51013=520b" m0 U$ s- U2 E$ U/ b; Z
3.5字符常量与字符串常量有什么区别?
4 u3 q7 Y4 Q* O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( t) B* k3 Q) F0 Z, @, _/ v* j/ B
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'.. r3 w) o. y- V
3.6写出以下程序的运行结果:
' Z% I3 f# X6 X6 K3 x/ a+ T#include
& q- m. P& i% A* [# \& q; D/ Avoid main()
! Z$ j( w6 M, l" x" e{
5 H; }8 ^. M, o# G* j7 y' T) Q: H0 uchar c1='a',c2='b',c3='c',c4='\101',c5='\116';2 K1 q7 O/ Y/ ?: G7 N6 u* Z1 F
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);; K( F- p# Y, i' g
printf("\t\b%c %c\n",c4,c5);
) G1 ` C' h9 } ~6 ^: A解:程序的运行结果为:4 O% L/ r# K, ^) }
aabb cc abc
9 c! z1 j A a; w# ?0 O! k A N% h K& t, s6 ?% b0 O
3.7将"China"译成密码.密码规律:用原来的字母后面第4个字母代替原来的字母,
) q! Z e' E6 s8 V$ Y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".' Y2 e; A0 H5 n# Z- `
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; n5 o1 n" c* A0 R/ q& [
输出.- I$ ]5 a! i$ j" T1 x
main()
0 t: a( ?+ h. O{char c1="C",c2="h",c3="i",c4='n',c5='a';' Z- ?2 h9 Z) ~6 i/ |: e
c1+=4;
* n6 A$ B/ Y; }+ O B) C" U! lc2+=4;
E& g3 e/ d# g" a+ [c3+=4;; U& _8 q* d: Y5 V" S
c4+=4;( H2 s. |, F! S4 ^2 M" V L( |
c5+=4;' m+ w/ `5 F4 `( _
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);4 g$ c- X N6 Y7 H, j
}; i5 M8 `5 k9 D2 F& @" `5 S7 f/ j
3.8例3.6能否改成如下:2 S& f1 H* g) \0 A3 q2 q- x5 i
#include' F& n# ~( _1 ]9 E) ?, C+ S$ Y
void main()
$ r8 K1 |2 o' \; O5 b3 @9 z{
7 |' j- K5 o' R5 i, V9 |int c1,c2;(原为 char c1,c2)$ w* G: f& w" ?( X. h) q0 b
c1=97;$ H* N0 q: b& d0 v& L2 w! X; ]
c2=98;
8 e4 h3 u9 T8 a5 K3 q% gprintf("%c%c\n",c1,c2);
( E, n: d5 a- X. m# v& jprintf("%d%d\n",c1,c2);
$ C+ D- d, w% Z: H}
; q0 C/ V* ]! ~. F8 N: s5 }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.
. t2 e( y1 C/ c9 W) b: `3.9求下面算术表达式的值.
( W; s- Q# g6 z# Y& J7 ?(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
" P2 z! Q2 C( u& C(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
! w' K# L3 O9 D5 v) D4 x3.10写出下面程序的运行结果:
, I+ N/ t* \6 r8 b8 S+ ^ Z0 E/ N* C#include) C5 c: ^+ i9 v4 i
void main()( @8 W4 C5 F+ M! k) E
{4 f+ \/ _: g' Q; K. T% Z
int i,j,m,n;$ { Z* x# s4 r% P& y$ z+ E, L
i=8;! r. D1 P+ p b' Y
j=10;4 D4 x( n; D0 Y3 k" O
m=++i;
4 h; b/ e7 g1 L3 ]1 Y8 in=j++;& k8 P$ j; p R; |8 ^2 H9 m/ ^! h
printf("%d,%d,%d,%d\n",i,j,m,n);4 M! Q5 ~5 ?6 m2 |9 |2 r. Z) z9 D% U
}
( ^' {* `/ m/ A解:结果: 9,11,9,10
$ r. W, k6 c8 M第4章
8 c9 ]. a5 E' u" U2 `. r4.4.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 I; i+ _4 g" h" v; s
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:9 q( Z8 v9 K. q
a= 3 b= 4 c= 5; E1 x7 `: [0 l( G% P' S
x=1.200000,y=2.400000,z=-3.600000
0 o; ]6 }* T) { M: J% b3 k+ bx+y= 3.60 y+z=-1.20 z+x=-2.402 T8 J+ _- T+ ?8 v: U1 N
u= 51274 n= 128765
& p p! [4 q e Y8 O$ W6 @. Lc1='a' or 97(ASCII)
" Y3 F. `% N1 y9 ?2 b8 R# z1 O5 m3 m4 Tc2='B' or 98(ASCII)
1 j7 W" `6 a6 ]% ?解:* c0 W5 Y; V8 E5 Z* p' U5 N
main()- r" @' I8 q$ S7 r( R
{
1 | s/ d9 X2 Q8 q4 }1 uint a,b,c;
' X, ^* f1 E+ y W) d$ tlong int u,n;
4 P, T: a, a) \/ Dfloat x,y,z;
6 y4 n5 |& }: `; w- achar c1,c2;: c1 W" y$ i1 I, k b
a=3;b=4;c=5;
# t* {4 B- q6 U( J! nx=1.2;y=2.4;z=-3.6;
0 y+ S' y4 l5 }: bu=51274;n=128765;
) Y7 m/ |; l# m1 d/ v1 ~4 ac1='a';c2='b';
- \2 g& x0 B8 `3 Iprintf("\n");
/ V4 R4 Z( H( Zprintf("a=%2d b=%2d c=%2d\n",a,b,c);' c, y; g1 c- x0 ]2 o7 `# z0 T
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);2 U7 y2 I8 ^. b' \
printf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);
" N. B- V9 a8 W6 p ]" i; C, [printf("u=%6ld n=%9ld\n",u,n);
- } ~/ ]9 V, u/ Tprintf("c1='%c' or %d(ASCII)\n",c1,c2);
! f1 O O! a8 m8 o+ Cprintf("c2='%c' or %d(ASCII)\n",c2,c2);* ]! h" ^6 R) l E4 f# X- `: Y
}! r P* W: Z. m0 L
4.5请写出下面程序的输出结果.
! W$ X8 P8 M4 {# K4 H3 A. F结果:
0 r- G( F/ K0 J& l& r n57) Z& a- l# e$ T9 f
5 7* j/ G6 I5 J+ n7 i$ N0 v
67.856400,-789.123962
2 x4 Y' J9 S- L! Q5 P' j2 Y67.856400 ,-789.123962( @% a( x' ? t5 A
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962
- @6 t' A- T6 [+ r6.785640e+001,-7.89e+002/ v* H, B- o3 b: }( V- N$ U% w
A,65,101,41
: S" G4 b0 X0 [3 S) C1234567,4553207,d687% ~2 x: |' p" x( R4 b0 K" _0 k
65535,17777,ffff,-18 K- Y& p4 S5 e/ M3 V M/ G) c* O
COMPUTER, COM* j; a( T @0 k% o: w7 F
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',
9 d1 g, h& }. [9 ?" }% X, d问在键盘上如何输入?7 b2 g7 Y' M. p% t. p$ E- \7 s
main()
- s- F! y$ S! I{/ }% [& j4 v8 _5 h. o2 [
int a,b;
3 I: d6 b( J& O, G7 Mfloat x,y;
' b/ m7 K: H+ lchar c1,c2;: H; ~9 i: s8 r$ N
scanf("a=%d b=%d,&a,&b);% n2 Z" W8 o1 ^6 n1 h
scanf(" x=%f y=%e",&x,&y);
q R6 }3 M5 u9 O+ [2 sscanf(" c1=%c c2=%c",&c1,&c2);$ u9 w2 W* g* y7 b! s
}. E, k% F3 I$ ]3 Q! o
解:可按如下方式在键盘上输入:) b. @ |, r6 W+ v" R
a=3 b=7
_* n- j' O- w. Ex=8.5 y=71.82
! D! K3 x) ~* Q( c( `) Cc1=A c2=a$ W( \" ^' B( f
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"被第二0 n) e& u; [0 D' F$ Q$ Q
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/ X# `+ B4 \" E/ ~7 ~
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一9 J2 ] o, {1 x, h3 O! q5 l" q
初步概念即可,以后再进一步深入理解.- w- S6 }8 g3 {$ N& k
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=-" c* z3 W/ d1 \ {, [
3.75,z=57.8,请问
$ ?! F. Y y2 z& H; S* a u在键盘上如何输入数据?7 p/ H6 o! [- R' t* G* S
scanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);3 m( S9 ?0 g( v4 N# m6 B
解:
$ N# x( j E- C( c* _4 W5 ]main(); P) O+ K+ V$ ^) u7 B; s; M
{9 q% G8 v5 x* w1 W3 N& z" {
int a,b;
5 m1 h3 P+ w, C% o8 N' sfloat x,y,z;
, r# g2 N' p6 o2 achar c1,c2;
6 I& T1 p' q+ }7 H8 }# Qscanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);+ @' a7 X' D# {$ B. ]1 Y4 a
}. @, d; o. a% F6 R# n( C- W4 U$ y
运行时输入:, M- ~% }3 D u+ n
10 20Aa1.5 -3.75 +1.5,67.8
( M+ W& f7 O0 m5 u7 K8 K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' t! N2 M7 `" e
是用来禁止赋值的。在输入时,对应于%*f的地方,随意打入了一个数1.5,该值不
+ C2 l% {7 N( A会赋给任何变量。
! c' H- l6 p/ C( @2 C% N5 S$ |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,
4 K4 W( x3 e/ W6 _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, p! p" h6 O2 @) y% [+ }- x
程.) e6 Y/ \& O. F9 u: ~- k
解:main(): U- E, [ h3 y" z& B, _# c
{3 l; \" n: Q, L+ ?
float pi,h,r,l,s,sq,vq,vz;
6 e/ X3 r) t! a# l, [7 ^pi=3.1415926;
3 d; I) B1 o) V' ^0 U1 D7 Dprintf("请输入圆半径r圆柱高h:\n");/ \% n0 J. T6 B3 k
scanf("%f,%f",&r,&h);
% s; j( o* y5 `, m" Z" Ll=2*pi*r;
/ q+ N. r( j3 I h% s$ s8 F2 d' |8 xs=r*r*pi;
; [( i; y$ U4 k! p- V- K$ ^+ d# Msq=4*pi*r*r; X7 {7 Y! k: {# [- j! ~
vq=4.0/3.0*pi*r*r*r;7 E8 S7 r7 x: L) y6 y$ f' r
vz=pi*r*r*h;/ K; p4 ]6 c* ]* T3 o5 ]
printf("圆周长为: =%6.2f\n",l);
1 m' O% K4 |! L; {' }5 Kprintf("圆面积为: =%6.2f\n",s); V5 e6 Y/ i' K6 {5 N
printf("圆球表面积为: =%6.2f\n",sq);: c- d8 C. r1 Y% _ U2 d `5 P
printf("圆球体积为: =%6.2f\n",vz);
) B" L- O& ~! y8 I h0 P8 \6 h}' s2 _3 K( e" t1 a2 G
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,! p# N8 A' o" v) w
取两位小数.
0 j: G( F8 D# X+ ^2 {2 v; @解: main(), a0 Q' e' O" t0 T) @3 Z8 T* Z
{
/ C s0 l: k: k- x) M9 Kfloat c,f;
' y2 G( f3 ~* r. lprintf("请输入一个华氏温度:\n");3 T1 ~( h& K# `6 c* _
scanf("%f",&f);
7 ?, [5 I, u1 O! M. T$ q' H5 I+ ec=(5.0/9.0)*(f-32);1 l" Z. G0 V9 d* j
printf("摄氏温度为:%5.2f\n",c);2 C, e4 o9 c& e2 v# i6 e
}' Q/ ]4 b& {6 u0 F( i7 [
第五章 逻辑运算和判断选取结构, Z) x& T4 b5 T4 o! \
5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.+ \$ I5 _0 z6 }; c& K+ O9 {2 K
main()
3 Q0 m8 B9 X! L6 p# `( a* s) W+ W{: F+ S" F. h& | K7 S. |# g
int a,b,c;4 C' A( ]* x% d% } y* {* m
printf("请输入三个数:");
5 n: o" z, N! Z' k0 n5 T Uscanf("%d,%d,%d",&a,&b,&c);
2 F8 W8 P' r @if(a if(b printf("max=%d\n",c);
' U( K7 E. M8 k+ S0 ^ else
, o* C9 ~6 }8 w: Z: t printf("max=%d\n",b);
$ }. u8 U3 i% Qelse if(a printf("max=%d\n",c);; F7 ^5 E/ S( V+ e1 h& k
else5 E, P6 ]; Z6 o) L) ?! L. i! `
printf("max-%d\n",a);
. X3 \4 Z' O, ~7 O% t}- m0 e3 _0 M. q; T7 V
方法2:使用条件表达式.
+ Z% ~% ?# D B- K* C& w1 Imain()+ G) H/ X6 l4 a1 n5 U/ L
{int a,b,c,termp,max;
3 W' I9 d7 b" b# P/ B2 f printf(" 请输入 A,B,C: ");
& l2 [" I1 Z! n+ [ scanf("%d,%d,%d",&a,&b,&c);5 Z+ o ~/ H+ v& s R0 r
printf("A=%d,B=%d,C=%d\n",a,b,c);4 y5 Y& _8 _+ e3 S2 V3 G# F2 s% Y
temp=(a>b)?a:b;3 h0 r% k2 c, \6 P. Y; I& ~! }
max=(temp>c)? temp:c;% Z c, N/ ]% ?/ p
printf(" A,B,C中最大数是%d,",max);/ P1 |, U2 j" N; r% m
}( h, p2 p+ {3 Z# q6 G: R
5.5 main()
" f; Q! z; E, C; _0 o{int x,y;9 z. L) B/ ]( |( O X! s
printf("输入x:");
: A0 C3 ]9 H- g$ H. ?6 F! Q* yscanf("%d",&x);' u2 i1 A# N! | w
if(x<1)9 L& i0 |; X- z d
{y=x;
, |' |. g. ]" _; p3 I n- m printf("X-%d,Y=X=%d \n",x,y);
0 h( W' `# K3 i- K }1 p5 m5 p [& |+ S" o; }- |
else if(x<10)
' D. t4 L$ ] R6 @/ I {y=2*x-1;1 m; d* N5 C5 l9 ^- k
printf(" X=%d, Y=2*X-1=%d\n",x,y);- M& X. g* Q- q% R; N/ s
}
# p' u+ C) t! h" Q" velse
! t; c: X; r; s( q {y=3*x-11;
. I0 \5 l- V" D! p printf("X=5d, Y=3*x-11=%d \n",x,y);
% D8 k7 h2 k' l }
* j: |8 [3 f* O6 d+ U* D, z3 y+ N( B}; v; B) K8 {$ j% E+ Y
(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! $ J0 J5 J% F+ n: K5 {; f* R& i
void main()/ J3 |0 p% S6 J! j" @' x
{
6 u: e* z, Z7 K9 b2 kfloat s,i;+ t- l$ Z& Y6 G" s; w/ J _( z
char a;
' k$ G2 k) A; f8 P. F" R& Xscanf("%f",&s);
5 T |+ V* n+ ~" M6 B% d. uwhile(s>100||s<0)( a" O7 @( K0 a& w0 \! t a
{' V6 |5 ]9 {# K7 [
printf("输入错误!error!");
/ a) O. Y4 m! M+ n4 ascanf("%f",&s);
0 t, E: x/ J& |, M' ], z( ^}$ m! s6 l8 ]1 l% W, R4 L; c3 p! h6 z
i=s/10;* H$ P* R6 J6 y; @
switch((int)i): w! \* H" u0 I3 t
{" U$ b- r! o5 }. d
case 10:
; `' _7 u4 ]7 `# Ucase 9: a='A';break;
+ x& {; N4 Y/ N2 \- A, o+ dcase 8: a='B';break;: t% X& U+ u" V# B% U A: N, _8 U
case 7: a='C';break;
; v2 l7 E: P, \' v. i( C) ocase 6: a='D';break;
" o, S. v% R* O# g0 ccase 5:8 _" w1 n; _+ [) z! X
case 4:; I! M* [8 u1 r0 f0 p$ M( m
case 2:
8 F; V4 x/ r, u; o' x3 C0 Zcase 1:
+ g! t3 \/ t. W% }1 pcase 0: a='E';0 [6 z v1 h: n# N- O6 {0 k
}
4 m# V. k3 e: X! c( r' q& M6 mprintf("%c",a);1 p: U0 e( I8 u) `' w
}9 i. a4 F' I8 Y6 H( X+ S& B4 g3 {; q
5.7给一个不多于5位的正整数,要求:1.求它是几位数2.分别打印出每一位数字3." |8 W7 h4 b! O3 w; D
按逆序打印出各位数字.例如原数为321,应输出123.) b# f8 @$ _# t
main()
/ [) S! q( b1 z7 v {1 p- B5 j' A+ e+ Y: J7 S
long int num;
6 l( ], v! b |7 E int indiv,ten,hundred,housand,tenthousand,place;" W: @! d5 o% }, W$ U% K0 @
printf("请输入一个整数(0-99999):");% n! h( O7 v8 G, W; H
scanf("%ld",&num);
$ q+ d9 Z1 g0 U1 n2 `( S$ q, D( I if(num>9999)
( Y. U9 F& R; h: C) U place=5;% ` k, @- P6 a/ Z
else if(num>999)
9 b! u: _6 C' R( S! q- W8 f- T! t place=4;! ]6 r3 G% c0 D f) X
else if(num>99)
5 F9 i2 ~2 n' W0 N place=3;
6 g/ Q- @4 F0 f9 Z+ D. s5 g5 Relse if(num>9). Y! W( @& G" r* @5 E) N# B
place=2;" h- f2 V* i2 K1 r' q/ M
else place=1;, l- K; y6 [6 [, W/ Y. u J1 a/ Y, _
printf("place=%d\n",place);' y2 b3 b; L$ v1 f I# e* {5 |
printf("每位数字为:");
; l' U7 ^! U$ Gten_thousand=num/10000;
: b6 |) a8 }3 j+ g" q, a* x# Kthousand=(num-tenthousand*10000)/1000;
5 J; W# m. y9 I# g s# jhundred=(num-tenthousand*10000-thousand*1000)/100;
( j" t! O& Y: U& Q$ Iten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
8 g' j8 n8 ^9 G. Sindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;
2 p" e# Y( M M _: L: ?0 jswitch(place)& D/ [: M/ v, D
{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
3 h! G& N6 |5 _- ?+ m printf("\n反序数字为:");7 }9 ]" s- G3 ~( X# h" m! a! _ L
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);& k9 K$ d) P5 t$ b- X1 O
break;2 a# S. L4 ]! `" i$ O
case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
+ M2 u* N+ X, Q$ L9 V9 }+ Z printf("\n反序数字为:");3 t# A, U2 K! t6 l' L' i! ^/ B
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);
* G$ d/ D7 z) O+ x# y$ a break;
8 A9 h9 C" M' M7 Ccase 3:printf("%d,%d,%d\n",hundred,ten,indiv);
5 w5 K! ]# W x# |. t printf("\n反序数字为:");0 ]; o7 v: S4 ]8 s6 Z
printf("%d%d%d\n",indiv,ten,hundred);
+ U' p6 ^& g- w6 S+ l( Fcase 2:printf("%d,%d\n",ten,indiv);
4 g; h6 O9 j' u! ^ printf("\n反序数字为:");
8 p6 K1 [1 @6 } printf("%d%d\n",indiv,ten);
2 G5 l* d3 h r5 a$ B9 ?, ucase 1:printf("%d\n",indiv);
. B; w$ e6 v: g, P printf("\n反序数字为:");
- N7 R. R& F) A; v( b# u printf("%d\n",indiv);! V. ]8 T. |$ r: E" ^* i/ r
}2 k0 }. {" y( p
}
7 a; l$ s/ k4 T. s9 T' i7 W1 {5.8; }2 ^: V- |6 G" ~8 h
1.if语句
+ _. a9 L/ b: T: r' lmain()+ L+ I4 M6 h) T4 A1 ^5 y
{long i;
4 b& j4 x+ f" r) K$ j0 l* } float bonus,bon1,bon2,bon4,bon6,bon10;
! i( ]" ]; [3 y# e3 Z: E8 a bon1=100000*0.1;
' \1 h1 A+ Y; E bon2=bon1+100000*0.075;
8 z4 B2 H u+ ^4 U: q/ M! E; L bon4=bon2+200000*0.05;
( f( b0 ^7 ?( M. @( u/ M bon6=bon4+200000*0.03;
) T$ s7 Z1 T0 K! S% _- a: D" L bon10=bon6+400000*0.015;- j9 U) u4 F9 ~6 R
scanf("%ld",&i);
R9 B' B, F, w' }- _5 [ if(i<=1e5)bonus=i*0.1;6 A5 ]: p- h$ x) b
else if(i<=2e5)bonus=bon1+(i-100000)*0.075;, X* X' G/ v& R, g) g& h% U% c
else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
' U( i @2 W% d else if(i<=6e5)bonus=bon4+(i-400000)*0.03;
; _% t- ?2 k* A. j: U& f2 N' M! z else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
! M: r/ F0 D% _2 G; H0 z% I6 ? else bonus=bon10+(i-1000000)*0.01;
: \$ T- }: b( W9 u% Y4 v/ J1 d m! v' } printf("bonus=%10.2f",bonus);/ {# q$ c+ ^7 Z5 L2 P J) G
}, A* t7 Y$ Z5 p# w+ k4 z
用switch语句编程序
5 V" ]- T1 t) X3 ~3 Hmain()
" g; C' `* W2 s{long i;
1 c5 ]; H6 }& R7 l1 p float bonus,bon1,bon2,bon4,bon6,bon10;
$ @: A* Z4 ]' `9 r- F7 @+ ^. U int branch;
$ [: |3 y& D; p; L: L) y8 V bon1=100000*0.1;
0 r; u0 i. \" s9 ?, S3 u bon2=bon1+100000*0.075;
" J3 @( G* P, w bon4=bon2+200000*0.05;5 s& P4 G' r9 L" Y; }
bon6=bon4+200000*0.03;7 c% F) A# ^3 r( }/ N
bon10=bon6+400000*0.015;0 @) M. I9 K& D
scanf("%ld",&i);! [# Y: k( z: [
branch=i/100000;
8 H- c) \; v1 O( p: F4 ?7 t if(branch>10)branch=10;
- |. A7 r2 D/ k. B4 @3 F switch(branch)
5 a/ z/ h4 L* q7 Y( @9 K3 p {case 0:bonus=i*0.1;break;$ D: U- I% [+ _+ H; t V
case 1:bonus=bon1+(i-100000)*0.075;break;! g1 j, |9 k1 [/ H
case 2:5 |3 t5 L# i' ^' ]( t `
case 3:bonus=bon2+(i-200000)*0.05;break;: p: f' j; |6 _) O1 h
case 4:. @, ~ u: B/ n+ `7 @5 ~" n* y
case 5:bonus=bon4+(i-400000)*0.03;break;
* c5 y& U5 q: Z, R. ? case 6:
. }0 ^, R4 u9 d7 X: r2 G0 P. I9 a case 7
* n- A' j1 f2 M2 ? case 8:9 t. g% U+ }! [, a' ^
case 9:bonus=bon6+(i-600000)*0.015;break;
/ m2 t* j8 s1 D! k- G5 u case 10:bonus=bon10+(i-1000000)*0.01;7 k9 e1 g k7 E2 b
}* O( N' [6 ^6 Q% {6 F
printf("bonus=%10.2f",bonus);9 ~' a' c% O" ]) R( @% a) O. }9 e
} 0 G) i7 W6 T9 `( R B
5.9 输入四个整数,按大小顺序输出.
6 W/ w9 [# X9 B& ?0 j( p) Wmain()
" Z/ o! ]8 t' k! Y" }9 |& @ {int t,a,b,c,d;6 J3 O) o* z, V6 {
printf("请输入四个数:");# t6 q2 Y# r6 }5 A7 d! Q
scanf("%d,%d,%d,%d",&a,&b,&c,&d);
! S1 M+ q9 k% M printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
% J9 I9 Y3 R$ |6 k if(a>b)4 Y9 B- r- q) Z
{t=a;a=b;b=t;}! F% g( M8 v$ n! G( h1 q
if(a>c)
7 N0 F/ C' U3 A, ~' P2 F {t=a;a=c;c=t;}
/ K" t% p* g' o) y. @ f- P( P if(a>d)8 B* u L l% L$ L2 y( n
{t=a;a=d;d=t;}) U/ h$ W7 r/ s
if(b>c)( b7 L: K7 }1 f, A; P; H
{t=b;b=c;c=t;}
+ g) q8 c+ ?" y+ e9 y9 G) \* [ if(b>d)
' i% C V( y' c& j0 `! q2 U6 q3 ~3 F {t=b;b=d;d=t;}6 V& m5 a1 w* y) k
if(c>d)* y) ~% E+ R. n+ R3 A l4 d" L
{t=c;c=d;d=t;}
) Z3 L$ K3 A' R5 Xprintf("\n 排序结果如下: \n");
0 r# q7 R3 C) u( p" {' qprintf(" %d %d %d %d \n",a,b,c,d);
9 E# B) E, b' R6 U& T/ b) k7 u0 w}
" |* g& W1 [! O$ k3 V5.10塔( s, J1 B! ]! c
main()% }. R4 J" K7 H- C s
{4 f- f H6 C% o$ H- k- n
int h=10;
! i$ [7 R( @2 Y- y' n( d+ |4 Hfloat x,y,x0=2,y0=2,d1,d2,d3,d4;" v8 a1 X! O1 e$ Y1 c2 F
printf("请输入一个点(x,y):");5 L* F0 i- [- D, X/ W$ z
scanf("%f,%f",&x,&y);
4 o( Y' ^' w4 F# I% O: ^/ O+ L; V1 }d1=(x-x0)*(x-x0)+(y-y0)(y-y0);* } x) A" |8 \+ C% ]+ b: K
d2=(x-x0)*(x-x0)+(y+y0)(y+y0);4 H6 K2 [5 A/ B; |0 g4 E+ g
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);0 z# P% q! [3 n `* M2 K4 Y) r1 q
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);3 i! e' v8 }2 `; L6 P
if(d1>1 && d2>1 && d3>1 && d4>1)
Y X5 o6 z( m9 f; ph=0;0 E; U9 s2 K( r
printf("该点高度为%d",h);
4 ?# Y% s. B: N: I* j}$ l( i+ J; |; F0 G
第六章 循环语句( s( t. n6 ` Q9 |
6.1输入两个正数,求最大公约数最小公倍数.3 D9 c7 O" _% P5 q( M, P# V) _# L/ G
main()
6 S6 A7 P) E f# r. x& V{# j# P; R( v! O% Z* C
int a,b,num1,num2,temp;
( ?0 F0 v0 z/ iprintf("请输入两个正整数:\n");$ j( a' h$ {* U6 K
scanf("%d,%d",&num1,&num2);; Q N1 ]0 J1 Q+ L7 h. B
if(num1{
* _/ I4 b. ?$ ytemp=num1;
9 i( k8 l. d9 {) T0 mnum1=num2;$ v; t( q6 ^9 V9 I; d( G
num2=temp;$ j7 e: B/ U6 |
}
# \5 Q) I Q/ A% k1 h: y: [a=num1,b=num2;) ]! Q2 @6 n' i- K( w1 c
while(b!=0)
0 ~" J# o. K7 m {
; E1 o+ O* |9 P0 z: R5 V. D& }1 ` temp=a%b; {' W& e7 I3 l7 u7 g7 K2 S# S8 N
a=b;
9 d' L7 {# e; p0 Q+ L* B* b; A b=temp;
0 a" T0 `# v) [+ F' y# |8 C8 P8 M }! _9 u7 Q6 W& \7 Q" v& Z
printf("它们的最大公约数为:%d\n",a);
Y6 O7 d4 q! M9 Mprintf("它们的最小公倍数为:%d\n",num1*num2/2);
4 n5 a. {* _' M) x: H5 f$ S+ G4 Y} i3 _) x2 s) e) o2 J9 g
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.
( c% B, E2 z; v$ d4 u7 X解:( n: W: _5 o( m
#include < > Z4 C `1 V; Y; g9 Z
main()
# Y$ ^9 a" x. i{7 \$ l& t% c( V3 ?/ O( ^
char c;3 }5 \+ Z6 G! k M7 I7 ]
int letters=0,space=0,degit=0,other=0;
8 W( L3 V0 @5 w8 Iprintf("请输入一行字符:\n");
+ {! s M0 y0 s9 N' pscanf("%c",&c);$ S a1 Z9 E' r9 h" P7 F5 Z
while((c=getchar())!='\n')1 C1 U2 x2 @* B
{2 p* [4 I; d5 @* ^& Z
if(c>='a'&&c<='z'||c>'A'&&c<='Z')
+ P( o9 y* q4 |- Gletters++;6 {: Y O3 V$ m1 t
else if(c==' ')
: d2 `; {& p! O: A/ o/ {space++;0 k) l8 q3 P8 N% _8 d
else if(c>='0'&&c<='9')' ]' @, Z; @" W: O
digit++;9 c- s2 D9 q" I9 }
else# k$ j% K7 G% M6 Z
other++;
0 I, ?6 r8 J& z: I9 d}# G8 G; L: T6 G' u" o- |
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%% [/ i# B& u7 j( t! i( G
d\n",letters,space,
; I. v3 J% {+ d( z5 P$ c, t- ~- W% Adigit,other);
- c7 {" J! y, ~% t% `" ~}+ {0 i+ I- @! u
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.6 @" m% x. d# R& @
解:
3 H5 e9 W2 g: o$ d5 q& B& bmain()! e+ q9 w8 O4 g1 j0 Y! T
{
) H& S# t8 Q+ Eint a,n,count=1,sn=0,tn=0;
" Q. N9 g" S, ^printf("请输入a和n的值:\n");
1 x6 D" G2 d' L. ~8 pscanf("%d,%d",&a,&n);4 }6 T W3 @; P7 r4 O! M
printf("a=%d n=%d \n",a,n);
" I' B6 H( q: y8 K- e8 Owhile(count<=n)( r6 Q8 S; r9 o$ R4 z5 Z
{
. h/ Y6 j! P) U6 R3 ]2 n: n4 Itn=tn+a;- I ? w" _) w; y% k t/ m9 ?, Q% i# ?+ D" x
sn=sn+tn;
: d: V# W# u8 Q# X3 K( sa=a*10;, b$ W4 v. v( x" [
++count;
5 x4 {$ U1 @4 x9 Y' h1 `: j8 Q}/ [- @: s. |' u! _/ c7 {6 f' L4 H
printf("a+aa+aaa+…=%d\n",sn);+ E8 q, M7 O- j
}5 S* B$ \' J5 g' j3 m+ Z
6.4 求1+2!+3!+4!+…+20!.
V0 \ \! m9 _+ N5 K" E omain()+ J$ A0 S, d; a- C
{
% R% O. M8 U& J: hfloat n,s=0,t=1;
}8 Y1 d5 ~' V* H5 b" efor(n=1;n<=20;n++)
! Q: B2 c/ J! A) {/ T{
- A |" `% n' S! t! d5 jt=t*n;
u9 G/ b# b: K R, M. ys=s+t;
& F( r3 s2 ^# J; d4 ~: @2 p}; o& ~* f, A" ]4 v; B- Z$ C
printf("1!+2!+…+20!=%e\n",s);
! x. n9 E: X, G3 c) ?}) Y7 f3 L% t1 I1 J* ^: g
6.5 main()7 T4 N" {: v& K& L& I5 G( H! D* H
{, @/ Y1 s: b! T0 s3 F, |) ?
int N1=100,N2=50,N3=10;
8 G( ~0 J- _: o0 R" vfloat k;
( G3 k; A' y; m6 G8 efloat s1=0,s2=0,s3=0;
( t2 J/ I" o" A- y- r A" J7 m5 Wfor(k=1;k<=N1;k++)3 M4 s3 B+ X3 v+ {
{
0 W8 l/ t+ n. q* ms1=s1+k;+ }# ?; H S; M) c5 r2 c/ `$ _
}
$ ]. P$ G% s& P" s+ b6 kfor(k=1;k<=N2;k++), d" k8 q9 B0 o
{
7 B, m) Y3 a0 m- |8 Ss2=s2+k*k;. {6 J0 z. Q) T* `. N2 R6 k" Z
}: f2 t7 k+ [4 z
for(k=1;k<=N3;k++): k$ K2 B6 s0 f, p! z
{6 z: z b9 V# w5 I' E; J3 }' {
s3=s3+1/k;
( ~, ]! W- \4 X+ I1 K& [5 A5 B}% ?% w5 f# v- J: e* ]
printf("总和=%8.2f\n",s1+s2+s3);/ M: j1 `5 L0 l8 [$ j9 e
}
+ f s+ I; q; L) Z) }% _! i# n* l8 S6.6水仙开花' [4 [" J0 E1 a( l- x; B
main(), B* F9 c; G) ?' o
{
( r+ f$ [, o" h4 V8 a; j: ~int i,j,k,n;2 y2 X5 k7 H: ~6 m5 B. N0 }
printf(" '水仙花'数是:");
8 t( k$ O. Y/ f, ^for(n=100;n<1000;n++)# i9 F4 i$ Q4 [6 G$ ?
{
8 d; K) O3 [( T" Pi=n/100;
6 G# ?/ j, c/ o* r+ i, n2 rj=n/10-i*10;
7 W+ ?5 T! S( I+ ek=n%10;
3 c0 p+ W" I" ^$ E3 L# u& k) f( x% yif(i*100+j*10+k==i*i*i+j*j*j+k*k*k)8 }% I" A2 @7 o1 h4 z0 a: a% C: f
{; y% J$ l! y! h/ f( x+ A& @$ C
printf("%d",n);
9 V; O7 A3 |4 Q2 |8 z, m}# s0 v: `. ]8 Z$ [' S
}/ u) B0 X2 q) m5 E; u
printf("\n");4 D. K% R8 K! E
} ]1 Q2 i d# D: Q
6.7完数. k j* d/ b- @: V8 j! w8 u
main()% c; c- r% c, g* ~
#include M 1000
7 |, b, B, P6 l3 u) {& C! m8 pmain()( u: F+ M8 W6 R3 a! ?
{
& ~) C E# Y' \1 I6 oint k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;7 E+ F5 K+ Z# h* j% |+ Z: u
int i,j,n,s; A) x- |2 p1 V7 [ c
for(j=2;j<=M;j++)
/ w! @2 A5 f- S! @0 k( n{9 c0 T7 |, X' _0 D4 W* k0 b
n=0;
/ K1 e6 H8 b' `" c7 i* ls=j;/ f' s0 U+ J( q
for(i=1;i {3 k. K8 O( t J1 ^% j& |9 C9 e
if((j%i)==0)/ S: J9 k9 W1 V
{7 y: h; z/ k$ C% u
if((j%i)==0)
1 j" F" H! w- i1 F { l5 M! w' J4 H! z+ J! U
n++;
- Z/ q8 Z7 b) u k$ R1 L s=s-i;
, h! Y$ t9 ~7 n v+ r0 } switch(n)
. a( I2 |% K" C: O! n- c- | {
3 z1 x; Y, W' X! X' W" T) P case 1:5 S$ o# L* B9 {
k0=i;
$ ], \; C0 l) \! A! J6 G. \ break;
) v* @- o4 U- ~) C# c. _% `% s7 X' E case 2:
# ^0 t+ V. p) I, e8 I- K/ f k1=i;
5 W8 j& M' @ K! y* M: ^6 E E break;
0 r V1 `# e- @1 T case 3:
9 a1 Z" j- n1 y9 ^ k2=i;
' W1 e v$ \6 i0 K5 D break;
, \* Q0 Z* b. b# l# e7 h; Y; b, ? case 4:; B e) d( ]4 Q
k3=i;% ]) B5 Y7 B9 @% @" y% j2 Z
break;6 c* I0 J3 N ^7 ?
case 5:
5 s: ^8 k" X8 c" d k4=i;
5 p, J# ] z1 ?$ Q( k& [, k break;
9 ]" E1 I/ ~0 L3 z0 ~ case 6:
: `: z' P. V$ V, \ k5=i;
* Q7 b0 }/ `) k( W+ J; A- Z break;7 D1 p4 ~: C+ e$ A1 m' {
case 7:$ I3 d0 O3 R1 X9 y0 Q! {+ H
k6=i;- g6 ?4 }; b# `/ S; x* X
break;
5 v8 e m6 Y2 R( l; Y. c case 8:
, H7 Q) j8 X5 p# V0 E' ]9 t k7=i;
& ]9 N" i; M5 `3 }6 M; `3 k break;
) p! _8 Q; T0 d) Y1 Z& U/ ~! J case 9:
; {3 t5 v0 R6 v9 r; q: y6 b k8=i;
. Y: M) A5 ]2 ^% ]5 q break;" y- n( O9 U d/ d1 o6 q
case 10:
1 Z7 V' C8 R( Q) p* T k9=i;
3 e+ U$ m# m. H$ Q- A break;
- A/ w5 i& ]: j! Z0 ^" m }. b- b3 ~- {7 @# U( b0 j
}/ W) h6 C( \. G# V
}
9 H* S' f( q }2 k; s5 \if(s==0)2 ~6 y* Q/ G K" F
{
9 s! _0 W( R/ r) C1 Z: Cprintf("%d是一个‘完数’,它的因子是",j);
8 R+ U8 f7 @6 N7 C4 cif(n>1)
4 Z& L4 u& V) I% q printf("%d,%d",k0,k1);
- k. X: C; g' o9 [" A2 u& ^ uif(n>2)
$ {4 z$ o/ _& v, r; M9 | printf(",%d",k2);
/ r' O, D3 o6 a3 C9 B0 bif(n>3)% m8 U; Q( x( X' m) l, y
printf(",%d",k3);
$ r& ?. ?* Y& I2 \9 H+ Bif(n>4)( x" y9 V [. q) r5 p" ?
printf(",%d",k4);" r* D$ f* O, \# b
if(n>5)' T) X" D6 \ R2 c: b# \) T
printf(",%d",k5);# {+ B/ N) z& O' }
if(n>6)/ q9 O0 j" c" G8 m: C. _
printf(",%d",k6);
/ n" `1 E8 R4 L1 l% X' U5 J; Iif(n>7)" \% _' M7 c% n+ y: y
printf(",%d",k7);
$ K& M* |* n" f; b9 l' aif(n>8)
7 p7 m3 ^" }% M C6 t printf(",%d",k8);1 v: g Y) ?& @+ x
if(n>9)' ^3 g l6 O! V% i/ h; r2 G
printf(",%d",k9);" Q1 ^; _3 U ~5 d& K! R* S
printf("\n");
. w8 c" t$ i/ }! i; o6 | }
* @* m& |8 g+ I \% l}$ c% X8 ~* u, J7 @2 c# x
方法二:此题用数组方法更为简单.
+ g7 |) g' a% f# `7 L9 a4 `1 _main()6 A4 B' ^5 {$ u2 Q3 i: g
{
4 [7 t3 k5 L8 F- [1 I) b" cstatic int k[10];
* V1 ~1 |2 S8 g% k+ Yint i,j,n,s;3 m% ^0 ]) g' J7 `$ F! o$ r
for(j=2;j<=1000;j++)$ w1 e+ ]3 c9 f! y8 K4 B
{) x6 u! p+ P1 C% Y
n=-1;
9 D; I* Z4 [, a% {. Us=j;: y a) @; _0 i, x1 J9 \
for(i=1;i{
3 M3 K, G$ Q( X0 z5 i3 P6 Lif((j%i)==0)! g+ `6 N5 Q+ @- Q+ h" C! G
{/ p! g& v( Y4 R# Q! x
n++;
z: X6 C# T, Ls=s-i;
" V. P3 d- Q, T, W: W1 ok[n]=i;7 f5 [( N0 T% R3 E
}7 @; |8 G. g. ~2 ? `2 i
}( W, S+ R2 z7 ^! f& p
if(s==0)
9 W0 c3 P; P! F" r3 k{$ e8 X; ^3 j x s$ b
printf("%d是一个完数,它的因子是:",j);3 r. f9 m9 R% P U8 V
for(i=0;iprintf("%d,",k[i]);
! c% n4 V5 w9 B( o! p7 Y# Kprintf("%d\n",k[n]);
* T R4 E" H" d; s}# `3 A6 e3 p& @0 o, H9 F; [
}6 ^0 V! M3 j0 _ H
6.8 有一个分数序列:2/1,3/2,5/3,8/5……求出这个数列的前20项之和.* D- N; V9 `6 E) y1 G+ d% @8 h" l
解: main()
# \5 s5 {* R5 L3 s, r7 I{: r6 T6 L' W; l$ H
int n,t,number=20;
% Z# f3 \( ~# f! M% @float a=2,b=1,s=0;
B; r( H! a- E' B- ]for(n=1;n<=number;n++)
: L. T$ S3 @) O* v @2 ?1 n) s{
8 U/ T! B) ?8 ?" xs=s+a/b;$ Z. G/ `, C+ [: F) L" Z
t=a,a=a+b,b=t;* }+ s; ], d+ t s
}+ f+ \7 c; G4 x" g* M
printf("总和=%9.6f\n",s);: O5 Q9 D/ Z% v* T3 [
}
; K' ?: L0 g4 B6.9球反弹问题% e. J( x0 F# X+ d0 O% U. u
main()/ c* A, N+ f* i% b, [
{. ]( Z+ i( e- F3 Y& T( |2 C+ e
float sn=100.0,hn=sn/2;
/ R: W9 y- v' L- c# H" mint n;' b. }6 x0 j: b2 L
for(n=2;n<=10;n++)
+ k( `1 p$ u7 ?1 `- u8 W{
, G) t+ @" r1 @4 q/ g' qsn=sn+2*hn;6 \: t8 r7 M0 T, }
hn=hn/2;, @, e4 X8 K2 T2 @
}8 y: N; Z# C6 a
printf("第10次落地时共经过%f米 \n",sn);/ S Z, B4 r$ a4 t
printf("第10次反弹%f米.\n",hn);
# V% A( A7 V1 N$ R$ s}4 x1 F/ V. Z3 D: [" L. g( x7 Q- o
6.10猴子吃桃
0 F3 w3 L4 p1 a# V- K* v; j' Kmain()
% c$ g9 C3 t- K+ K* ~: C{
8 ]' }6 q! v1 y. m! P& e- Xint day,x1,x2;5 @" T" ~+ o7 b0 q' b
day=9;6 M6 F. S2 r4 v) H2 C. s! o
x2=1;
+ `7 E: x% Y6 ` W; z0 Zwhile(day>0)
% D% X S. ]- X6 t. U4 Q{
6 B. f+ L/ g. J+ f2 ], |x1=(x2+1)*2;4 c# S. Y, G6 K" s; m* A
x2=x1;
: G* v4 V2 h b1 y- X- N- Hday--;
3 B1 o! s2 m# ]8 W q; T}2 n- i. M/ Z w' g
printf("桃子总数=%d\n",x1);
9 E& X- Z9 u3 G4 i$ e0 j3 T; O}
, G9 D9 v# ?) p: z- x
" w# f$ K" i8 ]) O7 l1 M0 [* N6.122 A% D$ A- e& r0 S
#include"math.h"
' ^/ t6 l/ V/ T. Qmain()
" m& v! F: b4 n6 C{float x,x0,f,f1;' X8 X8 D1 k' |1 O! B. S1 Y
x=1.5;
8 P2 C, D0 Q/ q6 V- \ do, j! N# Z+ A5 n' ] N c# b7 Z
{x0=x;
9 o$ a9 s! L5 C f=((2*x0-4)*x0+3)*x0-6;5 h/ o/ |4 ~6 V8 T# ~6 [* u5 p
f1=(6*x0-8)*x0+3;' W1 O" m$ L+ R
x=x0-f/f1;1 R- }) S' I$ U5 h* d* R
}: f7 t( w3 x+ R: U* P, V9 K$ o* L" B: f
while(fabs(x-x0)>=1e-5);# K: V( w" h5 U- f/ j& I- _4 t* |
printf("x=%6.2f\n",x); x1 o( V8 X- a4 N0 O
}& @$ }! ]/ U$ h$ u
U1 X: Y( {8 [( C! \
6.137 E+ Q, V8 r9 H/ b5 `
#include"math.h"- x7 S) G3 k. g; E4 X( }
main()' A0 g( W. w' ]( X P$ Q
{float x0,x1,x2,fx0,fx1,fx2;
2 y- ? H* B7 X P p do6 _+ U$ k! L" n/ F% ^
{scanf("%f,%f",&x1,&x2);/ J) g7 `7 g. w$ k
fx1=x1*((2*x1-4)*x1+3)-6;+ m& R) t4 @$ W9 [8 M5 I
fx2=x2*((2*x2-4)*x2+3)-6;
S/ C$ ]7 U- G# n7 D* C }
$ f+ J" L1 E# Z+ \2 }: `) r while(fx1*fx2>0);
8 S9 D8 |! x% m3 m3 W9 Y6 y# c do: D( w# }4 X8 q2 ~1 j$ G
{x0=(x1+x2)/2;
; n6 p4 l& c, S& S fx0=x0*((2*x0-4)*x0+3)-6;
1 l. p* ~1 i& P2 X if((fx0*fx1)<0)
6 r# T; b7 D4 M8 K2 g {x2=x0;
, {9 a, {2 `! ? T/ Q, D fx2=fx0;
9 e/ C( w! Z. V7 Y# _! w7 B9 h4 h/ { }
; H- o3 a9 g1 e9 o% j; N else
: y* V- _) H; R. d* y {x1=x0;, N# t3 E% J. ^! O
fx1=fx0;
1 K0 ]# q, Q! e6 y: l& k5 W1 f; S }. h: a* V7 [* ^) k/ A9 \! s. R9 l
}
' a$ N; e7 L% C' E while(fabs(fx0)>=1e-5);
7 A5 ~6 s2 k3 E" y8 V printf("x0=%6.2f\n",x0);
' b6 S6 @0 J/ n/ v" K& m}
4 E8 R E& a# a5 Q9 `) a- _6.14打印图案
2 z6 M0 p9 U' kmain(), E0 N1 w3 u% w6 r3 V4 ^- L& c
{int i,j,k;) B, T- D& I; F0 F
for(i=0;i<=3;i++) T$ g2 v$ a. ~+ V y, p. O& Q, s
{for(j=0;j<=2-i;j++)- c% d: M% {9 y9 B+ d5 z
printf(" ");: H( [; g9 C0 t/ n
for(k=0;k<=2*i;k++)6 L2 `* R" t1 E
printf("*");
5 E) G2 ?0 X( W( b printf("\n");& H" l5 Y" Z& c, W- W# `
}
$ h: T+ Q9 X. u$ s' _0 x for(i=0;i<=2;i++). ]. c& T. k, T6 K+ g
{for(j=0;j<=i;j++)
3 d! f, v8 m3 f- J& R/ ] printf(" ");3 q3 e# h! P% A; N7 u" l
for(k=0;k<=4-2*i;k++)& j# @ r; u, B! B
printf("*"); s3 y; `2 g1 m- ?& p: E
printf("\n");# x! H, ?. ^0 Q9 }9 G1 ?8 ]: n
}0 h0 @ a0 a/ ]2 \
}
5 O; l0 j2 h3 H' R1 s! j# R6.15乒乓比赛
% @# n3 h9 y8 ?' q% o% Dmain()$ ]4 a# j$ g/ J* g3 C( @; s
{" U5 _' f4 B( u
char i,j,k;
W9 F' P0 r) A# Afor(i='x';i<='z';i++)) w2 Q+ n) Z- [$ l. m9 f9 i8 l# I
for(j='x';j<='z';j++)
. B/ s9 s+ U/ x1 x {
" h5 u! |# x+ Q8 m+ Y7 _9 _if(i!=j)
, J9 s8 _2 m, L5 l U8 @3 ^5 sfor(k='x';k<='z';k++)) ?/ G9 {7 l2 B/ k; M: ^: J
{& Z% t( A! ^$ j Q4 r
if(i!=k&&j!=k)0 k% \! k! ?! R- X& H( s: V
{if(i!='x' && k!='x' && k! ='z')
B+ \. X W8 p U. mprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);9 H* S/ j& C7 U g; a Z @
}: e! ?( k) _+ F2 A. E- |9 V3 H
}
: [* C, d# d6 E( n0 a% u, r }
/ |0 w$ M+ q; z; j# [& Z9 f" g}
3 ]. D! |3 h' } `C语言设计谭浩强第三版的课后习题答案
\' h8 }6 @' _6 X4 x1 X0 B3 A7.1用筛选法求100之内的素数.
0 y5 j" e/ k4 z A* `: S#include: Q8 I# k3 V% l3 n% u; g1 ~8 J( c
#define N 101
& V" V t# |9 j; e5 X _+ ]5 Fmain()6 I6 f( K% ^ V' q1 i% `$ c8 O0 Q
{int i,j,line,a[N];9 d; a6 t4 b* ^+ K: F1 x
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0), `9 @% \+ n @& g6 d6 z
if(a[j]%a[i]==0)0 x& m6 w( g2 g+ b" e. [
a[j]=0;
: h! l* s# i9 g' Uprintf("\n");- Z2 ^( {# M6 D
for(i=2,line=0;i{ if(a[i]!=0)( F9 W& L) {/ o1 Y0 Q0 u
{printf("%5d",a[i]);
3 o1 O7 O& J- G- v0 D line++;
( r$ Y5 a6 A1 ], Q4 N* ^ if(line==10)6 J# V4 E$ f1 h2 @/ b( A3 J& ^
{printf("\n");4 X! |( W0 M2 N
line=0;}
3 M; M( }! O$ p% A* Z0 |+ v# g }
# x. I6 M6 W7 D8 E( Z& M}: x% M5 A, m5 E2 f
7.2用选择法对10个数排序.; P- N7 z- i% T2 Y
#define N 107 ^* K. p1 Q$ Z C% R& n
main()
2 i1 D, w M& a2 q( q{ int i,j,min,temp,a[N];' d8 j4 n# T0 U) N/ Q
printf("请输入十个数:\n");
1 a1 z) O& ^3 K. B2 dfor (i=0;i{ printf("a[%d]=",i);
; V: ~* w2 \& e1 \ scanf("%d",&a[i]);
$ \" @. w9 c; }$ J9 s}
8 Y/ m' X t. Zprintf("\n");$ [/ J; ]; O; I+ s
for(i=0;i printf("%5d",a[i]);, F; N2 x/ ^3 B/ }6 x2 W
printf("\n");
, H" r& ^* b2 ?' }6 O* Jfor (i=0;i{ min=i;0 N, ^. n: K* X
for(j=i+1;j if(a[min]>a[j]) min=j;1 h( S. c5 U, g. v
temp=a[i];
& |, C0 a+ b& k8 j) \ a[i]=a[min];) r/ X# a6 E7 Z4 B0 b
a[min]=temp;/ O- H1 I; d" ]' t6 w
}) U0 h5 e2 O9 w4 p6 V: _% J" W
printf("\n排序结果如下:\n");
" ^+ | q6 C, _# o" Nfor(i=0;iprintf("%5d",a[i]);
0 ^! J' F& N3 B1 w3 A) m; b}
! p3 x) a# }3 L4 a& H7.3对角线和:) h0 z8 ^2 z0 c
main()
( v8 x% ~4 {( f# a{. u+ q7 X6 z7 p6 G, }+ @
float a[3][3],sum=0;
; z+ t+ u9 u6 d; U4 f. ?int i,j;
- a9 }8 x" D1 p Dprintf("请输入矩阵元素:\n");9 h4 f0 T$ G2 b7 s* I: Z
for(i=0;i<3;i++)
+ B' w) |+ n9 ^* i2 R: I) m for(j=0;j<3;j++)
6 @* R) M5 N, R, F8 S2 N" u scanf("%f",&a[i][j]);
1 f) b: \1 Z& V for(i=0;i<3;i++)
/ G$ z# | b5 a* Q$ z sum=sum+a[i][i];4 }+ W4 ]. v# |4 n" W$ ^
printf("对角元素之和=6.2f",sum);
( j. X$ F0 X7 D, i1 V7 P/ D}! h/ f: L6 W- u0 Q1 q* z! }
7.4插入数据到数组6 C2 G2 J7 Z# B* X, j6 S
main()7 _& \5 w. W& w; Z4 `8 o
{int a[11]={1,4,6,9,13,16,19,28,40,100};
, M6 [. |% ], ^: w) Sint temp1,temp2,number,end,i,j;8 K1 k) {2 b5 m) J1 R! }$ x g
printf("初始数组如下:");
/ s! S3 t% O5 M( Kfor (i=0;i<10;i++)0 I. @. T/ [# F0 D! ` o
printf("%5d",a[i]);. ?# K# T, y# [6 ^5 O8 Z
printf("\n");
1 |, I {9 K3 Fprintf("输入插入数据:");! c' E% L! }1 z8 j; Y. s! D- H
scanf("%d",&number);
l# b# g, T' O7 p$ M2 fend=a[9];: M9 B/ o; g; {7 @% \/ P, A7 i
if(number>end)0 O% e0 r8 N$ P" \& w# m, g
a[10]=number;4 s7 x) u. X& b3 n1 b0 _
else
1 Y( G3 X; |, W) X' O; ~7 v6 u {for(i=0;i<10;i++)9 p8 O' k3 t* A; W3 f9 u% T
{ if(a[i]>number), E) x, O2 \0 ?' p& q/ B& T9 d( C
{temp1=a[i];9 @. c( G. A5 h) ]5 N
a[i]=number;
# S/ `5 P9 N/ g/ h6 b% n2 A# c: t for(j=i+1;j<11;j++)) g& L$ ]# K! d1 V0 Q' ?7 A
{temp2=a[j];" k# c- w6 c4 h P; [
a[j]=temp1;0 k% ~ L5 q! I8 W1 Q
temp1=temp2;
' z3 S, W# B% f7 v0 L }9 }, b7 t4 ]2 c* F! T
break;& a+ |; b0 l0 m2 J* f
}) B( q+ T, |1 \' V7 K+ o
}
8 f) K5 ?( u/ A( d. _' N9 U+ q }# U. Z6 n& L2 q8 Y' e
for(i=0;j<11;i++)( O! a1 n) N/ v2 N6 d4 ?# r
printf("a%6d",a[i]);* ]" ^: [3 S! s8 ]7 b- U* t
}3 X) s/ M9 ~2 b2 {" E; q$ J
7.5将一个数组逆序存放。# R- W0 F- _6 s: J9 u5 V
#define N 5
; o( K. Q; E2 \' g6 dmain()- g) E' Q- U* J: U6 e
{ int a[N]={8,6,5,4,1},i,temp;
# q/ i x% q/ }- b7 ]# uprintf("\n 初始数组:\n");. K. ~2 b2 S4 ]# n( v
for(i=0;iprintf("%4d",a[i]);* D. ^! T5 l4 @
for(i=0;i{ temp=a[i];
. w( H: N% a2 [- L: t+ I# i" Q a[i]=a[N-i-1];
5 K8 @4 A5 o8 f; A8 g a[N-i-1]=temp;
3 K, F! `/ z$ i* L+ X* V}) N: c# \ }' [# n, g$ E$ L0 n
printf("\n 交换后的数组:\n");" Y- q* |' ~$ Q8 g5 Q2 u1 Z& j
for(i=0;i printf("%4d",a[i]);
0 y7 J( x( [+ q}$ {. w! d- T, X0 `! c1 \/ q
7.6杨辉三角, q! ~2 E5 L. T5 i. q8 B# `
#define N 11
6 e( l6 i( N/ v4 ^8 w0 G4 umain()
. x' T6 j S0 y7 f0 C{ int i,j,a[N][N];
- k3 T2 e& N5 l. m6 ]) o, | for(i=1;i {a[i][i]=1;8 a _# K, n- n% Z
a[i][1]=1;; G' ?$ l; K& d+ F. \' ~$ N
}6 e1 J8 ?$ v. O3 |; w
for(i=3;i for(j=2;j<=i-1;j++)
2 N5 U% a) X0 o& I: \ a[i][j]=a[i01][j-1]+a[i-1][j];7 z9 ^; } s- \
for(i=1;i { for(j=1;j<=i;j++)& w O) \5 L+ B$ u3 j
printf("%6d",a[i][j];
& k ^) A, v- U' {$ z/ { printf("\n");1 }8 ~ L$ F! ^3 l/ w- l C, w
}% [" q! |3 Z. |& X ]5 e+ r: F
printf("\n");
" x; z2 g+ Y C}' Z9 ~/ P3 g+ r ^3 y* N
7.8鞍点& w" }& P7 M% ~# a/ g
#define N 101 j; I1 F, r, P# Q9 p' o9 P. v
#define M 10
, U: J& q; M; Y8 Y: M( X: Z. s5 h! Nmain()4 ]1 ~9 l `7 q; A# x
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;
' ]* F' j- Q' r# C printf("\n输入行数n:");; \# G6 ]' I- `/ _
scanf("%d",&n);
5 k0 p. `1 x- ? printf("\n输入列数m:");
9 q& h% x2 G& J& M scanf("%d",&m);$ ?( M" v6 M# W# T3 b5 ^# t( ^
! ?, s% [6 a& c' j( }4 a5 K/ x for(i=0;i { printf("第%d行?\n",i);
) A( I4 b7 m0 W" {1 h8 { for(j=0;j scanf("%d",&a[i][j];/ W9 x) ~3 h' n1 m
}
# D W9 F. I( v for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
2 y5 _; K0 @# V( \' {+ ]4 R pritf("\n");9 q2 [, F0 L% N$ o
}
, W& ?8 j1 A- x: F9 X flag2=0;
& y0 u1 t4 Q- D& \ for(i=0;i { max=a[i][0];
" R7 [3 ^: \% t0 {+ S for(j=0;j if(a[i][j]>max)
) n j1 T) ]; X) p- {. z3 K { max=a[i][j]; Z% | L, O0 N* ~ ?* r8 B
maxj=j;8 }' Q& s5 l' k* i/ H! L* ^* Y1 I
}
! f$ u$ L* f( g0 H for (k=0,flag1=1;k if(max>a[k][max])
* e' @( Z7 T; C0 N- f7 d% i- Q: F+ G flag1=0;
# u$ k: q2 }' E$ k- N8 v if(flag1)
: \; c, v/ M" A5 x" q: R { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);, u) `' P( `7 e4 q) x' R! b+ O
flag2=1;
$ G1 y- M" w; i( T- W; L( n- I }
! X4 p# A3 W3 h}, V) H, F ~# o5 p. K: g
if(!flag2)/ t1 y7 Q' ^' R3 e" h! C
printf("\n 矩阵中无鞍点! \n");
. t: V+ ?9 _9 Z}: T- ^& [8 w7 U( ~6 x8 p3 b
; U2 f: J# P( U: U5 L* R2 j6 f( A7.9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.( T5 F9 Q0 Z1 H& f6 U" x
#include' f/ X, y6 n$ q
#define N 15- w/ A% J* m$ L6 W
main()
9 s1 D9 |1 V( J{ int i,j,number,top,bott,min,loca,a[N],flag;) @4 v$ P! t9 E
char c;; K1 [1 }# N7 z9 _; n/ Z/ ?
printf("输入15个数(a[i]>[i-1])\n);
/ c, y2 C9 v2 {! a0 P1 h8 j scanf("%d",&a[0]);4 @1 U3 I5 U X8 _
i=1;
7 H( `9 z( l! d' P0 ~ while(i { scanf("%d",&a[i]);
- u. L- w4 n, G+ K% t if(a[i]>=a[i-1])
' Z; Y8 g0 l4 B1 y+ K5 K% J# t i++;/ U$ ^, s7 l+ Q/ `9 E5 l1 H
esle$ |& g* `+ }7 W3 s
{printf("请重输入a[i]");/ ?0 E6 l5 n: I0 s- Y4 j
printf("必须大于%d\n",a[i-1]);
5 C/ h, I9 O9 O+ ?8 B& f }6 j% g" M2 \/ b: A
}1 \$ X: d9 I% C' E, _
printf("\n");
r @% }, c* h for(i=0;i printf("%4d",a[i]);
( S- O( f' K6 g6 I1 T4 x printf("\n");
2 u0 [6 l1 L+ O+ l, H& f" C % B m$ L% M! ?% l+ v6 D! k; @
flag=1;8 P3 w3 B0 y3 [0 V/ j
while(flag): @. ^; G9 C, W* N
{; F/ e* o) e( d- J" q% r" ~0 m A) C2 N
printf("请输入查找数据:");% M1 R0 z% t( E
scanf("%d",&number);
0 G1 }( a- _8 a3 G; J. E* T3 _ loca=0;6 r s( Z" g, m
top=0;0 K! M: u4 ^( O+ E% q+ U0 R: p
bott=N-1;
; w/ Y( g0 I6 ^$ W* A+ @ if((numbera[N-1]))
- f2 A$ X6 | \: H1 K3 k- D8 p/ V' R2 ] loca=-1;8 W+ W: K/ r" @
while((loca==0)&&(top<=bott))
& v3 s0 m) W. T5 |. N- | { min=(bott+top)/2;
. [3 X7 t% A3 ` if(number==a[min])4 V; d2 a9 i. r* ^( j, n
{ loca=min;& h- a4 W: E% x* A
printf("%d位于表中第%d个数\n",number,loca+1);9 D c- }; r4 ^" D4 i# F
}
, o( i2 f6 O( A2 n$ E' w6 c6 O else if(number bott=min-1;7 _) D- d6 ]6 s4 W% [1 ]* K6 a/ g
else
+ b6 I, C/ p8 |! k% ` top=min+1;" T+ S, x/ M- D$ H
}) i: \& {8 o8 z1 e2 a9 |8 u! e% e' ]
if(loca==0||loca==-1); ^# `6 t4 {6 P- E9 x3 U
printf("%d不在表中\n",number);
7 U/ n7 O$ V G) N printf("是否继续查找?Y/N!\n");
+ d- n2 @$ F! H. Q* f4 q% ]* ? c=getchar();" G: G" ?& o5 I
if(c=='N'||c=='n')8 F5 d9 K% z/ N' w+ A
flag=0;4 R( e. i/ x+ p1 O! Z( w" E, j
}
1 A) x* v# ~, i5 u! V* X; ?# g, R}/ y# j0 k' M2 x8 D" _/ o' f
" v& r; R% a4 v. M a+ J+ M7.101 P/ ] d/ K- Y5 w) B. x5 q/ P' v: _
main()8 i+ ^! c( m8 ~% V* f
{ int i,j,uppn,lown,dign,span,othn;
& ~7 V. `- D& ]+ Y1 { char text[3][80];
" E$ t! k8 ?- D& m1 f# C2 y& d uppn=lown=dign=span=othn=0;% J8 g& `0 a) F7 n5 r; A0 _
for(i=0;i<3;i++)( B; q; U; G" N0 J# H, ^3 @
{ printf("\n请输入第%d行:\n",i);
1 A0 K \4 u- U( _, s gets(text[i]);
! a: W( l9 ?% r9 ^ for(j=0;j<80 && text[i][j]!='\0';j++)# ~6 |' g. G- O1 [& l3 G5 \& J
{if(text[i][j]>='A' && text[i][j]<='Z')( |) F6 T! i8 V) ?7 N9 P
uppn+=1;+ l8 P, Z2 _/ m6 }8 f
else if(text[i][j]>='a' && text[i][j]<='z')
! O2 j1 ?% g2 X* g; B) H g2 G# w8 | lown+=1;& ~- k& O1 c, h% }
else if(text[i][j]>='1' && text[i][j]<='9')
. n( ]5 p9 B. d4 z2 A$ u dign+=1;
7 u# J4 R Z6 F1 x( H4 ` else if(text[i][j]=' ')
( q6 ?# B& i; X$ _( V* l& I span+=1;
4 Z. J( J6 N; U* W else
# N' A" ~5 a0 N3 b7 [0 T othn+=1;
* c! D% n, b E }
! {. \2 z& p: e }
9 y, V- U m, f! B% k: b& k for(i=0;i<3;i++) A2 a# ]% W; g; V+ L' s
printf("%s=n",text[i]);5 J2 m7 ]* M K% t. b0 H" q
printf("大写字母数:%d\n",uppn);
4 b' K) R- Z4 H printf("小写字母数:%d\n",lown);; V8 c* i" O, U, X8 |
printf("数字个数:%d\n",dign);
% P* m7 n; Y ]' z printf("空格个数:%d\n",span);; k) n. O; s1 @2 H$ R) I L
printf("其它字符:%d\n",othn);
( Q7 s }; }. Q, N; q B3 L5 t}$ z! j: T* Z1 r: o2 z5 v+ p) U
) p Q& h1 M# C+ ?8 o8 v
7 y( i. C% p$ O: a" Y
7.119 m6 a L% i6 _
main()+ O2 C# n7 E% ^, e& h" j3 B
{static char a[5]={'*','*','*','*','*'};; D" I2 z& y8 y' N+ X
int i,j,k;
: s$ o X+ @- F* O( E/ c% c5 S3 ? char space=' ';# ^9 x; w/ h# C
for(i=0;i<=5;i++), c; H6 `; }& Q' `3 O
{printf("\n");+ i' C: ]0 j' \3 Y1 p: i
for(j=1;j<=3*i;j++)# E" s0 V5 J$ A f
printf("%lc",space);: F7 O) L" [7 }
for(k=0;k<=5;k++)
1 |) z* |% m' a5 ?% I printf("%3c",a[k];5 H/ |8 L, E/ b, `/ e
}
/ M# x4 z; K% _0 Y}" {5 d) K% k; l3 ~+ `4 l2 s
7.12* f9 E& s/ v! z0 Q! d, s* l
#include
8 ?' \! a: r; A. z! ]main()
0 Z* F3 b4 w9 ]2 \- O: v4 h{int i,n;
# d2 E. g, |! z, I. @: p+ | char ch[80],tran[80];
; f( P4 ^* d$ Q/ Y& F7 M! j% T printf("请输入字符:");
% e( \: O+ e7 K: F gets(ch); g2 r. u8 v, i. h n4 w( \8 b
printf("\n密码是%c",ch);8 r! ]! U4 m# \2 y. @
i=0;/ b& v- j; v v$ c/ n- K. o3 H: m
while(ch[i]!='\0')9 U0 M. A# p5 F% Y& e% G* [
{if((ch[i]>='A')&&(ch[i]<='Z'))% _; `1 b( Z) b+ S
tran[i]=26+64-ch[i]+1+64;* z/ a5 Q" Z( G& R( L
else if((ch[i]>='a')&&(ch[i]<='z'))
0 f' G( z: m; X Y" p& h E! X5 R tran[i]=26+96-ch[i]+1+96;; L! C0 L9 N3 c2 t/ [
else6 M6 g% ~( w; c1 P
tran[i]=ch[i];
& e% [& i( ?8 x' F i++;7 }( Z* W! T! ^- W- k
}
# B$ }- n) A' m ~; b3 kn=i;" f8 I: M. {: i6 ~& `+ R; ?! ~
printf("\n原文是:");8 ^, ~( C v& k/ N+ [
for(i=0;iputchar(tran[i]);& G2 p; L7 V* U9 ~ J
}
$ I- K6 X1 g0 y3 j, g/ W7.13
4 V0 N0 a0 r e* ~/ y wmain()' \- K6 d# ~0 U
{: E" y6 q, v+ Q; p
char s1[80],s2[40];8 _" e0 ^4 W% \; z
int i=0,j=0;' ^* z% _" w0 j) n3 Y
printf("\n请输入字符串1:");
' @6 e3 k+ V! G0 Y+ \9 E" U a scanf("%s",s1);
' O! m0 a7 c& N3 m; R- ^( z7 h printf("\n请输入字符串2:"); T; b6 y& x( m
scanf("%s",s2);
" ]* p9 N X# D3 _# |1 d while(s1[i]!='\0')0 a- A) M" d+ W- F# w
i++;
$ X7 w( e) u1 Mwhile(s2[j]!='\0')
8 x f3 r! n) z s1[i++]=s2[j++];4 K! N; Z" b1 {- X1 v6 F) j* h
s1[i]='\0';2 q% t7 V' X- }8 {& W
printf("\n连接后字符串为:%s",s1);
6 K( R9 Z) f m) w }; w6 A. W3 ?( e
8 f7 _3 Y" g; D# \- m# [
6 f+ y6 v0 _. x! M2 Y) x/ }7.14
8 j; V M1 T( W1 K# b; h#include. t: w& h' ^& c, R" |
main()
% _; u# E4 H+ S# C5 w% D1 h2 V' l{int i,resu;
( w! \! k# ?3 a char s1[100],s2[100];9 P7 m9 \! \+ Y6 d+ ]: D$ T1 v
printf("请输入字符串1:\n");
& B# h- a+ B" i# H2 ]9 K* l gets(s1);* z! G' N3 T/ F
printf("\n 请输入字符串2:\n");
) v# g) _' [9 X7 F- O gets(s2);
- ^0 A) K% J! r4 X! |. W3 q5 z i=0;0 p8 [! Y5 `9 \
while((s1[i]==s2[i]) && (s1[i]!='\0'))i++; @2 T; F: @9 c7 f% y1 i
if(s1[i]=='\0' && s2[i]=='\0')resu=0;! x* C, M' e# ?
else6 O" j# ^4 t) a$ G8 {0 G
resu=s1[i]-s2[i];
, l- n2 o( }/ L3 K* _: G. x/ s# t printf(" %s与%s比较结果是%d",s1,s2,resu);- f/ Q9 }: Z. _5 {
}/ e$ Q4 ^8 k' S# ^* A: g3 P- b
7.15
9 ^& A- E# y x$ m' T2 z#include
, t V# `# w$ J4 T1 ~5 E$ ^main()! A9 ^: |) i$ `9 G2 @9 c
{
: K) ]) v9 q: T3 D; ] char from[80],to[80];3 J9 m& \ b+ @. m$ T
int i;' z, _" m4 v p! R
printf("请输入字符串");
! c$ ~" G6 L1 Y$ k scanf("%s",from);4 g6 O; z/ ~; _3 P1 `9 M3 L: k
for(i=0;i<=strlen(from);i++)
) R2 O) K0 Z1 M' F0 ~' G7 h7 R7 }7 l to[i]=from[i];' L1 X! h, y; H) _. J
printf("复制字符串为:%s\n",to);
2 U5 i. _( j4 @) t, [; F }. l% R; n3 X6 ]
# ?6 Q6 G+ C" W& T1 ^
& m: ^: R3 W5 H9 u0 w3 q) Q8 u3 \第八章 函数6 R* M+ G) f3 n: a
8.1(最小公倍数=u*v/最大公约数.)
. W( f, ]7 G W$ b) c6 Nhcf(u,v)
( ~! @( m6 V! A" Y6 u" l! dint u,v;
0 `* D; H7 D: h8 H( s" b(int a,b,t,r;
9 X9 \; ~, `; k1 ? A% f if(u>v)$ Q% x( C! s8 U- C+ K+ }$ H
{t=u;u=v;v=t;}" H' o, {, q8 O
a=u;b=v;
[1 Q1 l; m3 C8 S while((r=b%a)!=0)7 m4 i& S$ h/ R3 I9 L" n0 e
{b=a;a=r;}
, t! i7 j% I9 U H return(a);
4 _$ O) R( S* D' M t/ P0 N }
0 c- C+ x8 S) E: w9 | lcd(u,v,h)2 }" V+ E8 t+ o, L: j1 Y( n+ C
int u,v,h;9 k: \. `; H; x/ s( p6 v2 y) }
{int u,v,h,l;+ P- a4 T$ f3 p: q
scanf("%d,%d",&u,&v);/ F6 @. I4 R2 l- q! }
h=hcf(u,v);
, r" T; \, {; E) g$ g printf("H.C.F=%d\n",h);
$ C: u: v* p- ~. X, \9 l l=lcd(u,v,h);
% A# l5 a& _1 {8 E; t) p- j printf("L.C.d=%d\n",l);
' A- ~+ p1 B1 D- Z$ | }
4 |, s3 L8 I, r7 D. t {return(u*v/h);}
N# X6 b$ l7 {0 A% y7 ^ main()
% ^2 v6 ^, D* V) c2 | {int u,v,h,l;
; o' l+ v% c5 m$ h scanf("%d,%d",&u,&v);
" w3 w( \2 L6 H- u* e h=hcf(u,v);. l. L1 p! s6 Y( B
printf("H.C.F=%d\n",h);3 P9 b: u3 J: h
l=lcd(u,v,h);' F$ F R/ @% m7 ]; S
printf("L.C.D=%d\n",l);$ O, n6 E9 `& Z1 y* s# ^4 C
}
( l( a! G8 X* z4 T+ _% H0 T8 z
; B: U( F+ L* d) i8 c+ U7 |: ]1 G+ ^- K" t
2 }! }- r3 V' Y5 K& U9 F
8.2求方程根
: Q7 L+ Q( d" s. X' G, }, { O0 B- Y) i#include, k, y+ j* h+ S) P
float x1,x2,disc,p,q;
+ r5 e5 ~* Q( S% J8 [& I, ngreater_than_zero(a,b)
Z* N. m. h& j9 O7 C- x& g& Gfloat a,b;* ?9 V! h/ X2 }/ N: ? a1 k# L
{
, O9 s! t' X9 W0 Z( r' w4 N* xx1=(-b+sqrt(disc))/(2*a);0 n# ?6 H% `0 T, U$ \
x2=(-b-sqrt(disc))/(2*a);' G0 A Y; n# m4 t3 A# Y
}
% D% |2 s- h* w R# w3 h+ Q7 Q: T+ wequal_to_zero(a,b)
* t; c* u9 P, x0 bfloat a,b;
( A& c' x+ {+ m2 p( F" h% H{x1=x2=(-b)/(2*a);}
9 ?( Q, }+ C0 S- Psmaller_than_zero(a,b)8 G! P3 ^5 V- `7 k( b# ^7 m
float a,b;
, f1 j" Q2 @% n{p=-b/(2*a);) U! c3 Q b- ?: P) G
q=sqrt(disc)/(2*a);$ z7 j: V, h9 c. ]4 t. @9 i
}
# Y: Z3 j+ R* \$ C. a0 Lmain()
2 J8 x( s5 O' I: r$ k{
% ]+ H- I5 n% ffloat a,b,c;
, q6 h' Z1 b9 L: n6 Zprintf("\n输入方程的系数a,b,c:\n");
5 q- r1 m# F4 fscanf("%f,%f,%f",&a,&b,&c);
7 A5 ~3 I) M9 v3 xprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);
8 q. C" J, k- j [$ T3 ] A% Vdisc=b*b-4*a*c;8 \7 S& T4 Z2 y0 x
printf("方程的解是:\n");! u& }4 ~0 c+ s4 Q+ C k& X
if(disc>0)! m1 {1 j: m3 } m2 X
{great_than_zero(a,b);
/ I% t k4 w. N0 u, k* uprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
6 ^4 }4 W0 C j. \* U) ]}
! T5 ~+ r& T1 o" V: p! t6 n8 F' h$ `else if(disc==0)9 p4 N4 W; o q, i
{
0 d6 h0 j. |* u1 Q- V( }zero(a,b);- S+ l6 U d3 @* R- L
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
: j) ]. O! Z1 ?; B7 |. B" e6 E }
& b& @/ z' |0 ]8 G3 P: f' Xelse
: U* y4 a0 q! d+ |/ ?/ ]; E+ o {3 _9 p& v' r$ f! j7 w: e
small_than_zero(a,b,c);
: ^. Q, y" J* o, N printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);
0 W& p F2 o; O0 f5 Z }2 m) Q# w* j% j3 o7 j1 H* }6 u
}
& E7 R% R' D" `' R- @8.3素数1 |4 K; A6 R2 v5 H" ^* J! q
#include"math.h"5 ^4 x3 @8 X; ^. t# c# S! m$ A
main()
/ w! _: Z3 i. Y& E! ^' O; l: u{int number;7 h0 E% H& \0 ?; ~) b5 \/ |: l) l: U
scanf("%d",&number);
! R+ E: y& Q" V% ~# }8 g if(prime(number)); p. S k' }( T
printf("yes");, o# `' y! M2 D- L! m$ W) }4 E
else8 p( B, ?6 G+ I! @5 z9 v, O
printf("no");
9 y2 M" i6 c( g& w* U}: p& q, o F9 p
int prime(number)
4 h5 \" w. }- U' r: e2 |0 ^int number;
/ M+ Q J- m- W+ q{int flag=1,n;
, M b, K' o$ V0 D; G for(n=2;n if(number%n==0)
4 [& m* X3 n. {" B" ~ flag=0;8 }# E- D0 |" H1 f8 U: b
return(flag);
+ K$ j; |8 q3 p V}
0 T; o2 r( P1 G* ~1 ]; i, B% K! d; D9 q: \
* f% W7 C( b' J6 I' ?
1 d5 B! z0 u7 F# q0 N
8.4
/ x, a; m8 n& y6 C e* b#define N 3
: o: R" G- C' yint array[N][N];+ j' u7 m, W% E/ b
convert(array)
+ {$ N5 g" t+ o) ~9 q. d P! V: }int array[3][3];% C/ L: `, F3 ~# t& J9 @
{ int i,j,t;
, a4 L7 l. |9 K* n for(i=0;i for(j=i+1;j { t=array[i][j];2 `+ P, h5 s2 [- H
array[i][j]=array[j][i];
# |# O, F. J5 v3 r; B. Q array[j][i]=t;: n* Q- O+ E. r) ]
}
. K) Z8 S+ _ f/ g3 Q9 ]6 c }) c. A! ]! z- u# @, f( O
main()
; h; z; o, Y [4 V2 P' B; \{
- f, B# i' P% G, i1 E4 J* b9 d int i,j;+ h, X: z3 U2 V$ g" t+ H
printf("输入数组元素:\n");
, m. k; q7 A( } I7 M$ x/ E5 o# X, I for(i=0;i for(j=0;j scanf("%d",&array[i][j];$ H8 N9 r: d6 N, I* }
printf("\n数组是:\n");$ p) ?1 W: o+ Z1 e7 R; _& z
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);% \/ y( z s+ a
printf("\n");, k0 v) X# J: q5 \/ m- E' |1 [
}9 N+ O, B) A2 ]" z! }1 `( e0 s
convert(array);8 N- ]8 [* T! ~5 |) ]; x$ F5 h
printf("转置数组是:\n");
; l7 \, u* B& i- G2 l8 j* X for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
$ N* g9 Y( h! U printf("\n");% K7 Z4 z3 ]2 E+ N) ~3 P
}
- W; E) F; c. y# B}$ g1 _2 d6 L8 h( K0 g5 ?1 K9 R
7 @$ C r, x. H: G. L' g
! G! \. b+ j% i% |2 f- z7 X% T9 ^: @2 _+ L; Q" {1 a6 Z
8.5) a. S: Y& B) w+ q' U3 {8 D! r J
main(). M, `6 v1 ]/ N" z* a' [8 T
{
; w- ~( v2 T# X4 @$ S0 j. r' D7 pchar str[100];
4 ]. ?9 V: C) V7 Q, J! F d printf("输入字符串:\n");
+ A- V" t% p, e scanf("%s",str);
' A& Y( f7 ^; a! @1 A inverse(str);
5 O9 J& h4 R. o' `! r' F* l printf("转换后的字符串是: %s\n",str);8 x6 F$ R1 @# ~0 b3 G# L
}
: f" U# ^1 h8 D3 F( \inverse(str)
" ^7 y7 [' u, U3 Cchar str[];' ^1 r5 S3 w' s9 x
{
( \& U6 \4 n9 a9 Q0 A- q char t;# E6 `2 U$ E, z! z: n* r9 {# X
int i,j;
/ w/ l" D x: h$ Y4 n: i% j for(i=0,j=strlen(str);i {
# F4 X; X- u( g t=str[i];
; j f, ?1 C: D0 V! L str[i]=str[i-1];8 h/ \7 v8 C; I0 S; Q3 }7 F
str[i-1]=t;: i. ~& G0 u) _' X# C
}
/ d( g+ o( m8 N$ F I}
) y# }2 g! Z& t) V* H1 v
! k2 T9 D" \7 Y O0 H' b6 k, B) m0 ~$ q5 ?1 s6 U% x+ |( F
9 [! y6 n5 l- m9 L8.6) [, o! ^# m5 ~+ `: Q7 X+ [
char concatenate(string1,string2,string);7 `) _6 M; W+ T5 j
char string1[],string2[],string[];
) E: I3 X) Z/ B0 H3 l. y ]) Z" M{, V% H) a4 q' @) R- }- o v
int i,j;
( q! k$ P5 D/ w' o) i; E1 wfor(i=0;string1[i]!='\0';i++)8 [& s# L5 d0 l: f5 Z
string[i]=string1[i];
' A# T5 V6 [: l& H9 B7 i0 kfor(j=0;string2[j]!='\0';j++)
. m2 F5 L7 n# o% M string[i+j]=string2[j];1 h/ L: b [8 @$ S" a/ J
string[i+j]='\0';
2 v. z( P/ L" F9 n7 i/ Y$ k}3 {- s, `0 z* h* u) l* V+ K/ ]
main()6 Q5 L0 g9 K, l# j) C
{; [) Y9 G i# Q& Y' ~1 S
char s1[100],s2[100],s[100];3 {, Q" [ U# g! ?6 @. X
printf("\n输入字符串1:\n");& ^- @! C. K/ N! v, l/ `
scanf("%s",s1);
5 H5 `2 a9 x2 J p) q printf("输入字符串2:\n"); f* v# a' k: n$ S
scanf("%s",s2);
( z$ p7 L& G4 ~' y5 l" Y concatenate(s1,s2,s);
3 \, R- N g5 g# f- c6 | printf("连接后的字符串:%s\n",s);. n2 c, h1 S/ f# e) O7 M4 X
}& d6 M8 K5 |, M+ H; C p
4 U$ `* ~( | x
2 j" N% i* [0 ]6 W* E3 P8.81 A' H7 w1 D( e" a5 R1 K
main()! G3 w F4 j# s* O. D6 _( `& S
{
4 i' W" x+ x; _# J char str[80];
1 w9 X$ j% t, B& U5 _& q) E3 i printf("请输入含有四个数字的字符串:\n");# f% X/ T+ B# z! Y* {; s
scanf("%s",str);
& b# [, w, m, o8 w/ V8 m insert(str);
* f9 K1 ^) d/ g' q0 m/ E/ T}6 ^- ~" g5 u ^/ D+ N
insert(str)- G0 F, G5 b$ \
char str[];2 o) s- t8 i* v/ o; n+ d6 ?
{/ t! v0 T( q+ v, s3 n7 C# R+ d
int i;
8 u9 x; j- C- s2 d( g for(i=strlen(str);i>0;i--)
* p1 v* [/ \! {" } { str[2*i]=str[i];
5 T' B3 R: X. ^5 K/ [- g6 T str[2*i-1]=' ';8 h! Y& h0 X- _
}
7 \' _0 |3 g( A; X! E printf("\n 结果是:\n %s",str);
1 g a+ V/ x5 n$ r }
2 _. x6 C/ o$ O2 m# }0 j7 h" ]; c0 g6 O
1 X* z) }, l0 I) N) |8 Z/ z6 M1 I+ \6 y" K3 ]
8.98 G" R9 c b- G; n7 n* Z
#include"math.h"4 l6 V# w9 `: A1 m5 E( c/ `
int alph,digit,space,others;) m b) t* h/ w* v1 @" j7 n
main()
, u# O5 T) l3 A{char text[80];
, A. A7 ~; f+ F: b4 A, G' i( b gets(text);
& Z, W/ |% ^ B! q alph=0,digit=0,space=0,others=0;9 d) h" g! v; t0 O; \
count(text);
' I( s6 D$ S/ }; d- f printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);" _; T: X5 Z) }
}
! C/ m( N. n& {count(str)8 p! ~" H) M. R- K5 r5 X5 M9 M
char str[];* I1 w; A; A+ ?! r% j7 \; k4 Z$ Y/ |1 }9 V$ [
{int i;% U, {. e2 u9 P7 f: `! {
for(i=0;str[i]!='\0';i++)/ a5 r) L' c r
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
; o2 h* N; }7 _* D! G: { alph++;
! C- c8 L) a1 H1 M2 y: c" V else if(str[i]>='0'&&str[i]<='9')8 F* T, [6 T# s% u" I' |: _' U0 J
digit++;
8 j8 e/ x# w3 m8 I" W3 f6 V else if(strcmp(str[i],' ')==0)
2 k8 f$ s/ W- g( L. u space++;% W' t3 S8 @: O; q
else
8 y1 N% E; y+ f0 `: Z+ k others++;/ w# v- ]5 l$ s4 G
}" T: B6 p" Q+ C' f1 d' |
, J$ Y& {; j2 H! ~5 [
" n/ i' N0 g' @, V2 X8 E C2 T8.10
& D4 u3 Z9 l7 m; P) `9 B. nint alphabetic(c);
! G# {) s6 [5 A1 ^ e) L" X( U' Kchar c;
, k. k* _; X. Z{/ X% }1 f7 s- ~
if((c>='a' && c<='z'||(c>='A' && c<='Z'))
/ m4 g# r& t6 I! I6 a return(1);
) \$ E9 H7 z# M else
! u3 {' ?# B: p' V! |# ?* Q return(0);
& N/ D, H. h) N5 l}2 y8 H' E" ~3 f8 T8 z" y
$ T, F( E0 v: | g. @# i. B+ ^7 w" r
int longest (string)# C1 `+ N- f% s' w
char string[];
4 n/ y% m8 b+ K{/ Q' B9 K5 u4 x7 R+ G; d( S
int len=0,i,length=0,flag=1,place,point;
7 l" y9 J7 P5 G# T) k for(i=0;i<=strlen(string);i++)
4 y r$ o' T! Z B r if(alphabctic(string[i]))
4 h/ q. J) ^8 ? o8 s if(flag)
' z. j3 |/ R- }7 l/ R( o" K' D {
6 f, Q% N! K9 i8 L' t8 U0 {) ] point=i;
$ h$ S( U. ~; U8 y flag=0;2 Z5 S/ n+ F5 v- C( D; E
}- w5 \7 ~+ d3 @3 v' p% w
else
S! M6 X( y5 c& _4 N len++;
$ d0 y8 q$ `. O9 H5 q7 z' x" t else
9 y* d4 Y0 n$ [3 {5 M { flag=1;
) b, E" I+ }$ w) l7 t if len>length)% x* p4 ^3 |2 Z: a) C8 E
{length=len;8 ]3 F3 U- Y% R6 T. h4 i
place=point;7 c2 x8 O& z! _6 C* C
len=0;
% w% u: k, D: x! D( l$ ?& C$ B }
6 i+ T% v0 e9 Z }
6 Y0 `: x0 C. o return(place);
+ K5 v. L+ D5 g }) i- r, G# Q- C
main()
2 W' T2 I: |; Z; l7 @{
- P( P1 W/ N' x' [int i;* \7 D! ~( J: {! V+ {4 b' {9 k
char line[100];; _% y4 G/ _, l8 b# u
printf("输入一行文本\n");9 C) z. j( F ]/ v& w9 k
gets(line);. Y) k& a& g3 G ^! A |& V3 r3 p' ^! d
printf("\n最长的单词是:");% z( G. N& g9 f5 |7 H) }
for(i=longest(line);alphabctic(line[i]);i++)
; t- J! U0 m) o1 Z( e printf("%c",line[i];
C; a% z+ V& ~5 lprintf("\n");. S/ x) {5 w# v3 [- ]
}
! f0 X! P( n& n, }7 K0 G
- x) \& q' w K/ S% }' R# j6 m3 w/ T1 {" B) ^
' g' b l6 b3 N5 _' S' c; v2 v8.11( q1 Y3 U0 m* G. K$ z- n/ u
#include
0 \. \ R2 B: W% q" c* \3 a! u3 O- |; c$ W: [
#define N 109 m1 C, F* @' p+ r
char str[N];
% w8 \1 O! b# B" R: s4 q# L! ^2 X5 Omain()/ u! F* O) a; b; X0 d. l4 o$ l f
{0 z8 X, C1 V: j+ I$ q
int i,flag;) b6 ~0 X" o& C) W1 j
for(flag=1;flag==1;)1 l! o% H" O- S8 W1 a7 B
{' ?) Q8 ~ r/ w
printf("\n输入字符串,长度为10:\n");
% O, Y3 J& e w& x8 X5 H6 u0 n9 a scanf("%s",&str);
+ f9 R- A# h6 i, u8 V" B7 q if(strlen(str)>N)
. z4 U# a/ P& ]( B1 ~! @0 s g! M printf("超过长度,请重输!");% q1 T' T6 O! F3 v
else8 Z0 w" m' e% f1 i' V0 Y3 P) ^4 d* m
flag=0;6 p( n, t7 ^; I9 m0 D, N
}
% U7 L% y. e8 Usort(str);
4 F% H. D( J1 x5 V" Qprintf("\n 排序结果:");
* H% B3 b% j9 m+ b" n# Nfor(i=0;i printf("%c",str[i]);8 D- E0 A H/ l* u# z5 e X8 c
}; N$ A1 b7 ~# x3 e
sort(str)$ k8 g! K# W# g. g" D
char str[N]; X/ p; z4 i' L9 J2 \ S; P: U
{
3 W! Q$ w8 b: T& w' c2 R6 ~% g) u" r! kint i,j;
# w5 J5 N7 ^$ g# t3 i4 M( y) }+ [1 Echar t;
( `0 t% C; e$ R$ s* {for(j=1;j for(i=0;(i if(str[i]>str[i+1])1 k; r, v) {" ^; k0 U
{ t=str[i];
2 h+ u% ^( l8 V5 r) N: U str[i]=str[i+1];# i4 @8 N; }, w" B
str[i+1]=t;
|' H4 I' k$ a7 N3 b# `# W* r }
" s2 v" f# ~- r: ^}$ {4 A- }" t& g( \9 o" m( ?% ?
8.129 i3 T1 g7 N& |1 n- Z" I5 A
#include7 T- {' }! G4 ]6 e p* ^
#include
# I, T, |8 u& w% m7 u$ x1 wfloat solut(a,b,c,d)' l8 ^! Q" n% w* O
float a,b,c,d;
/ ~2 o6 t! ]8 I9 C: _{float x=1,x0,f,f1;/ H. O( c! ~* h8 E+ f1 z
do
9 w* Y0 @9 J3 \: V* d& C {x0=x;5 Q! e6 y: L# \5 j+ ?3 n
f=((a*x0+b)*x0+c)*x0+d;* ?8 y; E, V8 ?- c! t3 b" R
f1=(3*a*x0+2*b)*x0+c;' K& V. Y* {! P5 e. o( z: D6 f8 M
x=x0-f/f1;) \( p& a' z* f2 ?- @ ~5 a
}$ A1 F- o) k" `; f
while(fabs(x-x0)>=1e-5);
2 s6 r& p% \# l return(x);0 @% ~4 d x" t
}% D8 ~: x2 e9 e+ ?
main()
; c' D: Y& t, G* t9 ?" i6 ~1 J/ k{float a,b,c,d;
. @. }6 Y" U% K+ @, B" Z4 n scanf("%f,%f,%f,%f",&a,&b,&c,&d);
) r5 a9 h( u0 \- I printf("x=%10.7f\n",solut(a,b,c,d));6 w6 u: J, M& b0 A* {
}# e4 w; D [( w% K
8.13# `1 o7 Q$ B* s( K* L1 ~" b
#include1 X) g; l0 U$ H9 y) Y
main()
3 A( w/ S4 ]& U{int x,n;
0 y Z. m$ |9 n4 C0 M float p();( D. U" e) U J' k
scanf("%d,%d",&n,&x);! ?/ P1 Q7 | k, }( n
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
/ O' j7 ] b, S1 e) \}4 y4 W1 a( ?( }" Q7 q
float p(tn,tx)( v$ k8 C0 k, q N' P; r
int tn,tx;
/ T: a9 j. }& d2 E{if(tn==0)
1 {* r5 }0 o* N" e return(1);: p5 t( B; {5 m, x, O8 v* E
else if(tn==1)
9 ^0 p! U3 C0 A0 P; D$ }- j return(tx);2 v/ d! V; A( S$ u
else
' y- P- m {! L return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
; a/ V: @8 R* z* i4 ?% u2 e9 L}
5 q2 r1 D- B0 j, m2 n8.14
/ r& O$ ?- H' y4 j#include "stdio.h"& g: z' t" F5 a, Y; X7 L
#define N 10' o: s3 L! R6 M( N* t
#define M 50 z; L3 K9 m5 A: U5 M/ T9 G
float score[N][M];
8 |; B" d, P& L9 U7 i; b0 rfloat a_stu[N],a_cor[M];
2 r! F" d: \2 ]7 Imain()! f! o4 C$ Z, P; Q5 |
{int i,j,r,c;
. j' `* n: P: ~8 g+ W' M float h;
4 V0 [ ]# `+ m/ ^" u- |( ] float s_diff();
" X% l7 T4 C6 U/ f: k. k( K6 D float highest();
; x- c4 R& x3 R4 [) Q5 X7 o/ x b r=0;3 C% s6 d% N/ @- }* ?
c=1; w1 E4 P8 e# A8 J! v2 O
input_stu();6 @; b6 Y) l3 g; d( l! @9 d
avr_stu();
8 C4 R$ K9 l( T; W6 A7 S" c0 G avr_cor();
0 H3 p* {" K q u: z printf("\n number class 1 2 3 4 5 avr");
8 R$ R$ |8 j* Y" K for(i=0;i {printf("\nNO%2d",i+1);0 I% Q+ y/ }0 C. m, N
for(j=0;j printf("%8.2f",score[i][j]);
0 U. I# D) V3 Q+ F printf("%8.2f",a_stu[i]);8 P" R) S* M7 }7 ]7 Q0 Y
}5 `" B3 w, G, S; c( G3 E
printf("\nclassavr");8 s% {) z3 a7 R+ S% m
for(j=0;j printf("%8.2f",a_cor[j]);
, L$ q. f/ Z! E1 w; p h=highest(&r,&c);
+ i5 N; a2 [- L printf("\n\n%8.2f %d %d\n",h,r,c);3 A7 b- i6 f7 i" u& D6 T
printf("\n %8.2f\n",s_diff());2 D5 P* R. b* J p
}! ^$ k* y% `/ F
input_stu()
6 X! }2 ~8 W9 R# N7 ]" ?{int i,j;
. ]; a G+ Q0 y) q7 G' @8 h float x;
, m; O5 R2 l0 c* X+ | for(i=0;i {for(j=0;j {scanf("%f",&x);+ @+ k+ H' F/ K1 r
score[i][j]=x;
& Q: ?; m7 T6 J& Z) |$ G/ ]9 } }8 b% u- r! d" M
}! m) H) {" m' f2 x. ]
} T' Y% E# P2 P! Q0 {/ I4 d! ]
avr_stu()! D( s2 Y0 f7 u$ c/ r& ?8 I
{int i,j;
: U* n& g7 O+ h float s;
/ n7 ?7 K. V: i4 {/ r# C5 E for(i=0;i {for(j=0,s=0;j s+=score[i][j];
# A4 ]1 f1 x% Q a_stu[i]=s/5.0;0 M: v. e6 g- m! F! g( d q/ N
}
- z _) T. I. ~5 l" q1 U}
. Y4 v- R: M) q1 `, S, f3 h/ }avr_cor()- l c# F5 _% {* h/ C( Z* {
{int i,j;
3 I4 l J6 N8 ~" V, K float s;
1 t2 T. S# K [2 G for(j=0;j {for(i=0,s=0;i s+=score[i][j];
1 e+ g* r$ u- X% w7 J) K9 B a_cor[j]=s/(float)N;
& a1 w, u" t7 a3 a. F$ }$ U }- z$ C. ]) S" n. }" K z
}! K& S/ _+ D6 `5 E. g
float highest(r,c)
( `4 g! J! B; f$ ~int *r,*c;) u7 k" h r6 F' P5 R
{float high;8 x/ j H: q" S+ ^! I% _. a
int i,j;
9 D& _1 ?# {, N high=score[0][0];" V7 t- t. V9 H
for(i=0;i for(j=0;j if(score[i][j]>high)
! {$ K' }) _# A {high=score[i][j];$ \, q) Z# U" @% Z9 w$ B& t2 b0 N/ T
*r=i+1;
/ F; I& V- x' k2 D- c l. } *c=j+1;
0 Y/ ] C9 _) P. _3 Q }: K2 h" T3 P8 z) H1 Z; X/ D
return(high);
& I2 k p2 f; [}
- w! @. n- m; G/ R% y" k) Lfloat s_diff()
! Y$ t7 T7 l6 s! x{int i,j;
5 l1 L5 q" H) m8 {) J( ^6 a float sumx=0.0,sumxn=0.0;
3 y" F2 [( @& L2 r1 }' J for(i=0;i {sumx+=a_stu[i]*a_stu[i];
3 _5 ]0 [% d j3 P; h) a$ d0 ? sumxn+=a_stu[i];! ?! `' ~4 P7 a( ~6 S
}+ `4 \" ]) C1 E. n ?
return(sumx/N-(sumxn/N)*(sumxn/N));) d, C% {/ H8 ^) v; b; G/ E
}5 o- R1 Z* G% W' p
8.158 E& M, x; }7 o
#include
1 l- U3 r5 f# ?) P* w$ l. x3 ]1 m' P#define N 100 M6 }/ `: D: Y
void input_e(num,name)( B8 V! c& e8 w# u! D
int num[];
/ D% |* W' V" Y& Lchar name[N][8];
0 v; @. E5 F9 c{int i;- L6 E/ J t6 z7 z7 H
for(i=0;i {scanf("%d",&num[i]);
9 P, n% |* J1 `: \0 m gets(name[i]);
. J) A& I9 j0 d3 p% ]2 Y8 y4 x }
, V3 f$ M: H) c: r}
# }2 F$ {+ O; ^1 s# j# Zvoid sort(num,name). A7 U8 D9 G* ?7 [9 C6 S5 g. e! R
int num[];5 R+ M$ W* F" N
char name[N][8];
( Z9 w# l; a( V$ \) n{int i,j,min,temp1;
! P1 i) I6 ?% h! n4 d' H9 m) Q char temp2[8];) t0 I. Y9 i0 z" v; N8 i4 u8 m, i% i
for(i=0;i {min=i;
& F! O( ?: [7 k for(j=i;j if(num[min]>num[j])min=j;- m% ?# O0 G4 ?
temp1=num[i];1 o; S) Z. `; j
num[i]=num[min];( ?' H# R; @1 x
num[min]=temp1;
. t/ o0 A J3 P1 U+ Y, q: |8 H strcpy(temp2,name[i]);! ?2 L; P& b* s g; z
strcpy(name[i],name[min]);
+ q% u/ q. }% p/ G strcpy(name[min],temp2); S+ N, d0 @% X8 v( Z
} T( I1 P: ^$ g W
for(i=0;i printf("\n%5d%10s",num[i],name[i]);3 l8 l+ i. J: I
}4 v* k' Y6 k: G1 O" e, {+ e
void search(n,num,name)
5 V' _" o! Z' |( Bint n,num[];
; J4 v( b9 o" G" {0 T. @' ^5 ?char name[N][8]; j5 |* f* k* u% N! i1 u0 v
{int top,bott,min,loca;
/ }" ?0 D! m* C) ~ loca=0;; n# `5 D5 I* E: @1 X% f" p- b
top=0;
0 @# G( o t- t8 ^ bott=N-1;$ B) Y/ r9 @: V
if((nnum[N-1]))# R: n1 e6 p5 W" u. J2 H; d/ [7 V
loca=-1;
* C. g% R. a$ e# i; x while((loca==0)&&(top<=bott))
/ O! u5 g/ x/ T8 Q1 w Y ]9 p( W {min=(bott+top)/2;# {6 J1 T/ C: Z) L$ v8 {
if(n==num[min])3 }% k" N1 N5 Z
{loca=min;) k* d p+ N% n4 c6 ^
printf("number=%d,name=%s\n",n,name[loca]);
! J# f& `* v6 {) G% V0 U }2 @2 X u; f3 ^, W; C
else if(n bott=min-1;% N" F N# l, i( n$ Y. c' r! E7 F7 N
else8 S- |' } {! b$ m; f
top=min+1;
* `' L5 F9 _* j9 N& Q }# F* J' Z/ d; M- G
if(loca==0||loca==-1)
9 e7 a+ t! |9 v5 }* y( Q printf("number=%d is not in table\n",n);
0 w& n. { Y/ U0 o; A}0 o( v, L1 _! |6 X
main()
/ ~2 G% m. l) E2 L7 F- \( j5 e7 O- }{int num[N],number,flag,c,n;* n7 P9 N9 `9 p) ?2 m; G+ T" R
char name[N][8];% S1 z% I3 q8 Z }
input_e(num,name);
1 |! ?, J+ B+ g0 ]$ B+ Y& Q2 }+ T sort(num,name);0 ^9 J7 Z2 Z% A
for(flag=1;flag;)$ O& t& ^( M7 Q! ]
{scanf("%d",&number);
- O F8 N& x1 P) M9 c search(number,num,name);9 }; U: M- y3 g0 t- T
printf("continue?Y/N!");
( s# ` z3 j' b0 G) r8 } c=getchar();
& B {4 u6 g) y' G! f9 Q if(c=='N'||c=='n')
- w4 l& e. O/ `8 {9 t' _ flag=0;
3 ^& M9 y8 h( |! { }
$ A6 v- K( L3 r. k$ T( S}
6 w o" X5 l k! v5 _; G6 K8 V% d. S' A8 T
8.16/ G+ S( a' ^; o
#include; E( W" }9 W8 O" X
#define MAX 1000
# i. b8 }2 y. Y$ K+ Nmain()8 N( R$ V" h# i1 |
{ int c,i,flag,flag1;
4 H t3 D! i$ K B char t[MAX];6 H; j4 V/ H' a, z) y' ?
i=0;
4 E/ r+ w2 m3 d3 u9 G flag=0;/ j; ~! O. y# X" N7 W7 W+ I4 W9 Q, q, s! d8 j
flag1=1;7 V9 h; Q: z! Y( Y
printf("\n输入十六进制数:");
O- i& B. d* A0 g0 w2 t while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
, r- |: E* \, `: V {flag=1;* L3 l8 I/ e* I: ?
t[i++]=c;% w% Z9 @ A' W) x- }" `
}1 T3 L$ V1 ~6 Z4 l. |% ^7 W
else if(flag)
' o4 b$ A9 Y6 X# d4 R& @1 n n/ y7 Z {
3 C! [; |/ R `% Z! I; q0 @9 l t[i]='\0';
! @+ i+ t6 f/ Z; h- \& `: O* G% D printf("\n 十进制数%d\n",htoi(t));
6 i8 X5 t2 h' `0 j printf("继续吗?");* Y8 e" J1 k7 {: l% d
c=getchar();! @* Y. T3 l' K& c% T, [
if(c=='N'||c=='n')5 u- S# m; k9 f' }
flag1=0;
2 j9 m' T4 @# Z, ]+ } else8 _: b+ F' C9 F
{flag=0;- h- p* q. ~2 D% S
i=0;
6 Y% Z$ i9 D1 K; k! ?% t printf("\n 输入十六进制数:");% ]6 f+ r5 I" A! {9 ?' I, x0 T
}
" z+ \4 [6 }+ \0 ?1 ^( i3 N; h9 [}
4 }: A$ u0 Z+ Y4 C}
! F7 s0 i" h5 U; J# D5 _- B6 o}
1 H% s0 ^ l* _4 I2 ^( Yhtoi(s)
: G! V8 E; C. s9 q: Cchar s[];
( t' q4 j% e$ ]! S9 v! ^{ int i,n;
d" f1 K0 I J; ^' Y n=0;" ?" ~, c. F3 I% @# n8 J
for(i=0;s[i]!='\0';i++)
8 h u0 k# r6 \ H' Y$ I {if(s[i]>='0'&&s[i]<='9')% k0 ]/ K! ]( g, g* N7 e
n=n*16+s[i]-'0';* z4 _$ o* q _$ g7 u
if(s[i]>='a'&&s[i]<='f')' z* q4 }9 {& Y4 t k5 w3 ~+ h
n=n*16+s[i]-'a'+10;
3 x5 Z ~( ~" x9 W: Q! l$ _# F if(s[i]>='A'&&s[i]<='F')* l" t2 I- _; D% w+ e" ~$ b
n=n*16+s[i]-'A'+10;$ N/ M( X6 l! O5 N% I: b. Q
}
' f7 p5 E3 k$ O. \& ~ return(n);1 D. V/ v# e) u* T9 {
}2 K" e3 D) K7 n2 y3 h9 c
* v& ~+ X+ B; n/ V
9 p$ y: T% h$ b, \
: Q9 z2 @7 p: x8 M
8.17* Q+ P! ~/ i* _* R
#include+ o8 h* b5 F; u1 C+ {
void counvert(n). j( R6 o6 B6 `" t! k
int n;
5 I- N( \) k0 R5 K{ int i;
% Q$ o8 p) p: Y4 a I: D if((i=n/10)!=0)' z. T4 _$ A5 H9 l5 \+ ]
convert(i);
5 H9 o! s C2 Y" D; Z( I putchar(n%10+'0');2 g0 Z& U. [ J, o S) N8 B
}! V) ]# k% M( I) U0 u# Y5 q
main() E( v: g9 \ T+ U1 m
{ int number;
6 P' ?, R1 a/ h6 o printf("\n 输入整数:");& H* P+ B7 N: ~& }. b/ a5 D
scanf("%d",&number);
0 j7 j/ D- ~) T/ l8 g+ {4 ]; v printf("\n 输出是: ");
/ ]4 g& g5 }7 r7 k, b! Z if(number<0)
# d, H ?; {) W$ W/ e0 n& _ { putchar('-');
$ }1 w5 V3 h: x8 Z2 h+ X number=-number;
- r5 a: ?" M* N) @/ i6 l+ C+ p }2 K9 u2 j' Y Y2 G4 y
convert(number);
. O" L1 g y0 v! w# Q3 f4 y}
* K1 l0 i+ e- ]- i& D4 F9 H# I" R: |+ n' u
" S- ]: Y8 H Z. u4 ^
! l4 \1 b; Q) Z0 A9 Y1 n/ x8.18
; e4 K# O3 }1 bmain()
" _/ Q# k8 J) u3 h9 I{0 S2 A) j2 A# t/ m% `# f; M
int year,month,day;- B- x& l' n* h, l! r
int days;$ y5 h8 Q* [7 f) }) ^/ V
printf("\n 请输入日期(年,月,日)\n");1 f8 s1 z6 U5 j P$ b
scanf("%d,%d,%d",&year,&month,&day);
- t! j% u; i9 @" F: n! g printf("\n %d年%d月%d日",year,month,day);; r8 \4 f* P! H8 V
days=sum_day(month,day);
% w( T! }8 a+ x7 O" z, l6 h if(leap(year)&&month>=3)
N9 \9 r. d7 F% t. ?& H8 ~( i- } days=days+1;
7 i8 U- T9 f' d# H printf("是该年的%d天.\n",days);0 s9 u) R7 f/ h6 n, o
}
3 K f0 G, a, Y8 W static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
" J2 o2 o! n6 [0 B: e( m int(sum_day(month,day)- r9 f- s$ h% S6 \. ]* t# V
int month,day;
6 t; ~# z- k/ O+ u2 ^! O8 w, { {
- Z: T7 ]% [+ f0 v" D) N6 g5 S int i;
0 _* j3 D" A! O; R0 s& \ for(i=1;i day+=day_tab[i];7 J+ F% p) p" I4 ?
return(day);
$ H" e7 L$ ]5 ^5 X }
7 E/ F0 d! R- F9 u2 g+ o5 I int leap(year)1 W0 Y' t. q3 y# J) k+ N" s
int year;
( P* v2 B. V3 u: k( w7 [; ^1 Q {
1 `9 P( \+ O* w7 M' H3 z1 D1 N int leap;" t* Y& O% l3 l
leap=year%4==0&&year%100!=0||year%400==0;
( W. K4 h' s3 j+ k; D) v return(leap);% b& \5 w# G9 X
}
2 A" B3 O9 Y; d. P第九章 编译预处理
% y! |% Q' F6 |- j; B9.1
1 c2 U# K) p" b+ I7 H+ B#define SWAP(a,b) t=b;b=a;a=t
) }/ J& k( g( t& N. Z8 \, e/ umain()
3 B( M$ R+ }: w( F{
, P: ~% s5 n. m3 q/ _int a,b,t;
8 i' c0 @# e _+ I& Zprintf("请输入两个整数 a,b:");
' {" ?! B% m" M3 |. Yscanf("%d,%d",&a,&b);# ]2 X3 ~9 H" Z/ \; i0 b
SWAP(a,b);
8 L7 O0 F3 m3 q0 C% ?4 V# E( nprintf("交换结果为:a=%d,b=%d\n",a,b);, v2 t6 d. f' f' T- y- C
} 5 ~0 m" T3 `) n
% I7 b2 @% c: C. q/ E t h* P
2 c. S) t8 w8 r( b+ u$ |
9.2/ s7 |: ~& s4 O( r4 S/ O# r
#define SURPLUS(a,b) ((a)%(b))
7 |0 Q" `/ k5 @) B( c4 K$ tmain()$ m& U% g. K% m9 I
{% V; H# L/ X ?7 a; N0 Y# ^' I
int a,b;0 ^+ S% q$ Q \& m5 d% e
printf(" 请输入两个整数 a,b:");! o% ]$ X8 i3 x" d& I% O4 D
scanf("%d,%d",&a,&b);
1 l. Y1 Q4 C" ^) }printf("a,b相除的余数为:%d\n",SURPLUS(a,b));2 [0 b( N1 T, h f/ m5 [
}
4 z- l+ |! _9 D0 h3 B; m% y7 E+ e5 ~+ ~% u C2 i# K# A
# g* q9 E0 Y3 T7 b" u! D. ^
9.3$ _5 w! e/ g$ i* r7 f3 C( M
#include9 u; ?/ M: c( R* X) E- {$ ^0 x
#defin S(a,b,c) ((a+b+c)/2)
3 L& b& A" w9 c- ]#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-
3 ]; h4 h9 G# @- h# B) Xc)))1 s0 g7 i+ f" M; e- N
main()
& S6 i) e$ {& `' _ {$ i' H; N. G+ a- n6 F9 ]3 R
float a,b,c;3 z% X* g( g9 k
printf("请输入三角形的三条边:");* L" K+ t) o- n, h o* q# `
scanf("%f,%f,%f",&a,&b,&c);
" ^/ z; g( n1 T) A# v2 E8 p if(a+b>c && a+c>b && b+c>a)8 c! a4 {% t; a' O% r* ]3 W) [' Z
printf("其面积为:%8.2f.\n",AREA(a,b,c));
6 M& h- J( Y% G, y2 k else
9 m' ]8 v/ O4 o2 o4 h" W printf("不能构成三角形!");
h# F7 ]* c% Z( g }+ d$ q$ B! v; t6 S2 l5 r1 ~
# X2 g, m' @5 h% W9 Q
* G$ B' n3 W% {) M$ p. \, H. b; A
9.4
/ \5 {7 |6 l3 K#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)
& S% o$ `; S/ I6 f' ymain()
/ a% ? n6 t. ?" L+ C% I; U {
$ i& c8 }' }) K3 m: b D* Q$ h' z int year;& \% a0 |; K2 [! {1 ]
printf("\n请输入某一年:");5 @1 C: _( Y/ o$ T* f
scanf("%d",&year);
7 w. n2 z4 [' Q" Z) i if(LEAP_YEAR(year)). T4 c. R: e8 T7 Q7 j4 x( @
printf("%d 是闰年.\n",year);9 Z& B+ P2 k9 z0 c8 A9 L
else
! c3 C/ i/ A' F; y printf("%d 不是闰年.\n",year);
8 f7 ~* Z$ B( v- C }
X* C# ]8 |9 d+ X- ^* s6 e* j& J6 |3 r# L
" g+ u6 L1 Z, D
* I9 t- k2 R. {) q
9.5解:展开后:% ^: i6 r; C- Z2 }% G; M
printf("&#118alue=%format\t",x);( k* H6 S# \6 t, E$ y
printf("&#118alue=%format\t",x);putchar('\n');
4 U8 C8 Q+ g$ s- g' |printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');
' x6 F- H4 ]/ q6 Z9 b; c输出结果:
1 z+ |" m* I( j4 v' q3 F. T&#118alue=5.000000ormat &#118alue=5.000000ormat
! Y( ^7 Q4 S$ y% O3 Z7 H. f3 [&#118alue=3.000000ormat &#118alue=8.000000ormat
# p/ S8 n- s) \
% o% e- r6 e# p$ k' w1 @( L
* }) l1 D" L4 U1 a( @8 x* h& d& |9.80 E1 ?/ a, f. N7 L7 s. Z4 F' ~
main()# d& B) p" a p) n; i! `
{$ r& q+ X0 C& r6 ^
int a,b,c;
/ ]" L0 C; Y# t, |1 M0 }# y% ` printf("请输入三个整数:");
3 H* k3 x/ ^2 l& B scanf("%d,%d,%d",&a,&b,&c);5 x8 f, _! M! ~! r# y0 M" W& \
printf("三个之中最大值为:%d\n",max(a,b,c));
) t" i/ B1 s: U" }1 A6 t }* [7 ?; T$ U8 U k" U
max(x,y,z)3 G3 g5 o# G2 Y. n4 p& l( Y+ H
int x,y,z;$ ^( k5 q* Z3 x! j( y8 ] n2 F
{; M2 ]9 A& |4 `1 @
int t;
& n7 @6 \$ d8 W6 G% U" v3 j t=(x>y? x:y);; _9 I! o1 p |; P" P0 ]
return(t>z? t:z);' o4 d& s, w0 ?/ f/ ~
}* ?8 [7 X. K/ M, p
5 c/ T& t2 ]9 H6 `% c( Z) H
$ M+ ]! i4 l2 b) F2 \# t" K
, n. d+ m" s I _9.107 U- H1 H( e: C5 L2 u& E
#include
% a- R% p2 \( v* B: Z8 p. ^#define MAX 805 @1 G9 a* |' j& c
#define CHANGE 1( F. ?' w+ z8 S) s, q+ [; {
main()
7 P% m# i0 b: `+ O{/ K# j# t- Z! S
char str[MAX];
1 h. `+ } p! N( |3 r9 z int i;7 [ a# q4 W' h, O+ x$ C
printf("请输入文本行:\n");
" `) ?+ t8 n7 T8 `* A5 g6 t" C scanf("%s",str);; f0 Y0 X( E& k& q' W
#if(CHANGE)
1 g3 D6 t; o8 R& I% V0 q6 G, M# ] o; g {$ I$ D7 B2 w& [3 |4 H0 r; S O0 M
for (i=0;i {
* g/ l- M) f2 M9 Q" L) @0 ~ n) s if(str[i]!='\0'
& v+ C: J* g0 l7 K9 w if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
: d" r G2 l4 w7 B- b( F2 S5 H+ p str[i]+=1;
6 n2 }9 D3 @" a* X Z2 N else if(str[i]=='z' || str[i]=='Z')
2 F* q2 n7 `% V5 f str[i]-=25;# R0 ~/ @* ?* K5 A2 d
}
, x& i3 S; H& h5 A$ r E- v3 Y}$ o: E7 I x7 D$ ~- Z9 L
#endif
" M1 ~/ o$ i' L$ Bprintf("输出电码为:\n%s",str);# M- ^- R1 r- W3 v5 Y
}
) k- {( ^* N) R" m0 V) L9 R) \; ?第十章 指针- T& x! e, H2 B7 e! Z: `
10.1
' a$ \4 g+ w4 C3 Ymain()
) G9 D( @- ~0 s4 T1 {{int n1,n2,n3;# j8 m* _9 [# Y2 b J
int *p1,*p2,*p3;
, s2 E" s8 \% x* F* j scanf("%d,%d,%d",&n1,&n2,&n3);1 F/ G& R8 s1 \0 O4 B/ _7 i. y
p1=&n1;
# O6 O$ u" a5 N" E( L p2=&n2;
3 x% {) K% H$ I& b p3=&n3;; o% c3 D ]1 {6 l
if(n1>n2)swap(p1,p2);5 F W. M9 y+ s4 O
if(n1>n3)swap(p1,p3);
2 p5 R+ b0 j6 ]+ Y( G* U4 A/ j3 c if(n2>n3)swap(p2,p3);
( q) y Y, R4 T printf("%d,%d,%d\n",n1,n2,n3);
# ?$ I/ ~) @" e, p% O8 I: R}5 \* X/ a* S* h/ H+ x: c
swap(p1,p2)
, I; L+ n# I( E! gint *p1,*p2;
7 u7 m' U! o$ o& Z8 K; O- `{int p;
1 K2 b7 }/ C( @; ^/ `% m8 y/ j p=*p1;*p1=*p2;*p2=p;( k8 g6 |8 p" E$ E
}
2 S7 | a$ Y" [, J10.2
" ]9 O6 u# _5 G3 z$ G6 ]main()% K" R/ ?5 h6 z6 n
{char *str1[20],*str2[20],*str3[20];
& `0 f. W) s" l$ p y char swap();
7 \& x6 {5 Y0 {3 J9 [ scanf("%s",str1);' R) E6 o d3 _4 T4 ^9 A& y
scanf("%s",str2);
8 I7 ?5 S! y. {7 N6 Y; N- Z: i4 L6 i scanf("%s",str3);7 j) M, T# m6 X, q- O
if(strcmp(str1,str2)>0)swap(str1,str2);+ v. Q; s2 ~5 Z& m, h
if(strcmp(str1,str3)>0)swap(str1,str3);
# a6 y3 y" _. D" P- K if(strcmp(str2,str3)>0)swap(str2,str3);2 y6 C$ M# G- K
printf("%s\n%s\n%s\n",str1,str2,str3);2 {! n! Q S2 }" C0 v
}2 S: f% C& i3 U- H
char swap(p1,p2)
& {7 w7 l8 B8 i# B$ h! M* Rchar *p1,*p2;
5 F$ l% g1 i! e* V2 d+ M{char *p[20];/ o" D5 b9 K" v: S1 P/ \
strcpy(p,p1);
4 g3 T* _2 _4 d1 a" {2 a strcpy(p1,p2);: a; W" ]- z0 i# W5 S" z- |& \: m# Z
strcpy(p2,p);' j8 Q G5 u7 X' W# i
}
* Y6 |5 r9 L9 X c. n+ q8 z7 O10.3; P5 q2 @7 j ~9 Q: X( y; h/ D
main()
# l, K, N) N: t" ^+ b{int number[10];; I; u0 @& F" g4 X
input(number);
4 |3 ]' u j+ [( F max_min_&#118alue(number);
3 q5 P8 K) o- y7 T; x. Q L0 d8 M output(number);
2 u8 p5 a( ~& r0 B, { }( Y- r5 q8 H0 q/ r8 Z* s0 M) x
input(number)% i9 L* N/ t5 v, S
int number[10];. i. N; a# ^+ _" B' t( F4 J/ `
{int i;
. D' C3 |% i3 Z# E; P for(i=0;i<10;i++)0 k5 ~! Z) X' Q% E
scanf("%d",&number[i]);2 C ~* k7 L, I4 Y4 z5 J/ Y
}
0 L6 B* n# G+ ]; [7 E) M0 s9 O' a+ [max_min_&#118alue(number)
" Z: o9 y1 P" A. U% p1 s2 {int number[10];9 C- Y: g0 Y; V; ^* a" p, m$ y
{int *max,*min;
! L* { K. ]# l+ J% N int *p,*end;
\5 `3 u/ |- U9 B, T+ x. |8 _ end=number+10;: Q2 g$ R8 s- z( N* k* K$ x7 v
max=min=number;3 y+ A, C7 ^! E$ v
for(p=number+1;p if(*p>*max)max=p;
c: K. f) u( b% I8 L else if(*p<*min)min=p;
/ u8 t/ C: }9 c! i *p=number[0];. \4 |% W4 d; S; Z
number[0]=*min;6 B. A8 `/ o! I- a
*min=*p;% \9 s2 F7 L/ [6 h2 Z/ k/ B
*p=number[9];9 e, |9 R) _# ]% n! n
number[9]=*max;( k) s1 _/ I. h/ W& z2 ]
*max=*p;
2 @- Y' M! G0 m return;
& v1 Q7 T: q7 K}# [- Z5 T- H9 C! [/ q& l! \
output(number)
' n5 Q+ s" f0 q0 k" h5 F& qint number[10];
2 l5 H! h# H6 k& Z2 U: C( r, `{int *p;
" d2 I6 d, {; l& w for(p=number;p printf("%d,",*p);
* e4 m" |$ K, v. G% [& }9 `4 S4 y printf("%d\n",*p);+ q/ Q7 M6 g* ~1 T
}
& S% n* z" H1 B) k1 D& V10.4
. I. S1 k1 g' d8 m, N( J1 pmain()4 T% p5 ~1 l& @! {9 K8 @
{int number[20],n,m,i;
& M+ H# B+ U& \) r3 }0 c scanf("%d",&n);
4 ~9 W2 j( g7 u2 j/ t, ~ scanf("%d",&m);
* b7 R" t% q: H for(i=0;i scanf("%d",&number[i]);2 b# e, |) x, H& S4 D9 |" \
move(number,n,m);5 u* Y- I! B" Y4 L
for(i=0;i printf("%8d",number[i]);
0 B9 Y u0 U9 D6 h0 d& ^/ e; v2 a0 V}
& y; B, S* ~6 `3 Y' emove(array,n,m)/ Z% C2 k- O ]- P3 N5 i# |2 E0 T
int array[20],n,m;4 l p: X: i! t- B+ r6 v/ h; l1 Z
{int *p,end;
9 b4 r5 p A; {* u& j0 P end=*(array+n-1);: c% `, r$ x' T& x C
for(p=array+n-1;p>array;p--)/ w$ ?' [$ r. X
*p=*(p-1);* l! M+ P2 E: G' `
*array=end;
. F* o/ |& U5 C6 \5 V& X m--;
& F3 {+ ?/ d, W3 W if(m>0)move(array,n,m);5 P" _: U. X8 j4 R
}
' b5 p6 n) P, y, G3 i* x+ ]7 t10.5
* M6 R% _0 k0 x/ ~; c#define nmax 50$ j) r- }! @$ F( ~7 r
main()% g9 v; e, Y# g& y$ q+ `/ i
{int i,k,m,n,num[nmax],*p;: Q. p- {% R$ [& Z6 V
scanf("%d",&n);
8 p k6 v# T/ X5 _0 x6 t ^7 H p=num;3 P' p7 r' p9 m" m) P
for(i=0;i *(p+i)=i+1;
) e5 w6 k4 L% h0 }8 q. E- [7 s$ s i=k=m=0;% c, S7 v* W2 g6 y- z
while(m {if(*(p+i)!=0)k++;+ }+ ?5 G/ o; O. F1 C" B- v
if(k==3)
+ P3 w8 P, s( T9 D" t {*(p+i)=0;
; z5 X6 ^( P, Y& m0 Q* h k=0;; J' \4 Q6 a8 I3 H9 @. q; M5 i+ o
m++;* ?$ A# H6 e# v4 a i4 w' n7 d8 J0 q
}
) F4 l+ i; @& I- Q5 s5 S i++;* D P1 S o, C$ x! _; y& h: t" F
if(i==n)i=0;) F2 T2 ~7 {" c2 Y( e% Z1 z5 f
}: p# e* ]: C/ L7 ?9 P
while(*p==0)p++;
+ k+ x/ n% s/ Q% f printf("%d",*p);
8 x! Y# N' g r2 |3 G; p1 [}
, Z {5 [6 h; j; \' V10.6
- o3 v9 O* ~5 ]- J) ?2 K$ a* pmain()
0 p% w/ n0 U" o& [" r{int len;. X5 _- J6 L5 { D9 }( H' L) Q
char *str[20];
& D% ~+ h2 j2 B: _ scanf("%s",str);- T8 X) G8 G: S& Z; u6 K
len=length(str);
: ?: l r5 N/ \ printf("\nlen=%d\n",len);; r0 \1 c% j# ^3 n) Z0 C
}
+ Z) m8 M p2 Llength(p)
8 Q* ]3 h8 `3 s& f kchar *p;
6 w5 c& r: U( t{int n=0;+ y H9 j- J# |' M# j8 X2 N
while(*p!='\0')2 g& [1 m# w5 t& B0 t* ?- ^
{n++;p++;}
1 E0 V" t% ~! u+ l, T$ X1 s return(n); s& z+ b% u1 H3 v. N
}6 e. M8 y8 z3 L, R
10.7) [$ w* x/ I, G6 }
main()2 a- d) y: a' m
{int m;; W6 U1 K* g' {# y
char *str1[20],*str2[20];9 w! {/ Y6 `! T3 }5 U3 r& P1 g1 s
scanf("%s",str1);
+ ]2 `3 G6 w7 v) p3 n# j0 M' e scanf("%d",&m);" X' c2 ~" W! h; z% s$ @/ i
if(strlen(str1) printf("error");
( T, F5 x+ k) T/ L else7 F' {+ E9 y% p, O
{copystr(str1,str2,m);
, ?! L" ~( V& f* g" u9 U2 d printf("%s",str2);
) _# L2 l/ Q1 s. n) v2 \; M6 Z }3 ]+ T; F2 |# T
}
' W8 j2 O" b. ]( dcopystr(p1,p2,m)4 i {4 T- p0 B- l
char *p1,*p2;- g9 k% F' ]% c, U& p3 x9 `
int m;( C9 m, |- b# W- Y
{int n=0;
+ j0 Y4 }2 [# w/ w9 D) h7 H while(n {n++;p1++;}+ j- e( [ w, N# _+ R
while(*p1!='\0')
# i- ]! ]' |% `# o' X {*p2=*p1;
4 U" l$ x* B; m) \" V p1++;0 M) \6 u3 C* x' R8 h" j
p2++;
4 a' T# T5 t: j4 p7 o }
7 V0 m+ C0 ^# P" R: ?- N( o/ ^ *p2='\0';
5 P! L! }: V% k}
$ g1 j5 I6 r* ^, l% W2 h3 R$ W4 x10.8
e# e& U, S; O+ w- a# s#include"stdio.h"% C- X/ U2 ]9 B
main()" n3 m& X5 R+ ?- N7 V" T/ U& e
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
; @4 v' u8 v' W* g+ o char *p,s[20];
2 T/ }* j' O4 N# l for(i=0;i<20;i++)s[i]=0;
7 g0 I9 `' R0 a' N: k& l6 P8 y* T# d. r i=0;
8 T4 Z* q6 g# d, d+ ?6 [ while((s[i]=getchar())!='\n')i++;. Z @1 U3 M) {# _! a
p=s;
$ _3 b8 J2 E+ \- ` while(*p!='\n') ]/ P0 k" k/ k' y/ l) ^" J( [
{if(*p>='a'&&*p<='z')# U; W) X/ ]; x7 ~% H3 q$ z! T
++sle;# \: l9 V' I- @) D& W+ l! I( Q4 o4 r7 l
else if(*p>='A'&&*p<='Z')
* v- D- n! f A! v& m2 z$ I6 C/ x ++cle;& i) u: ~0 V; f* O! N4 {( H# c. B/ G% `
else if(*p==' ')
7 e+ T, m2 `' c' ]& A* _5 ~ ++wsp;( K( X, H* S8 X5 y" R
else if(*p>='0'&&*p<='9')3 C* K. D K/ F$ @
++di;) E, ~& S* c$ W4 O' z# \: A8 y2 D, ? R
else
; [$ O. x/ m; e; H ++ot;
( ]1 H2 x( U' b0 | p++;
; g* h* I. T3 G }' w+ d( D' c3 n$ `/ j) W, Q
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
; Q7 d' ]- j# q( C}0 U! _- y) i! e
10.9
- P8 Q- L8 J9 b8 ~" p0 K/ w3 ?main()
; |) q% V7 [% m' \% l. C+ ?{int a[3][3],*p,i;
2 I- j6 f4 j. }8 Q ] for(i=0;i<3;i++)
, ~! f9 Y6 C" t5 H! ~$ a scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);# j5 I& ^1 r, V# m
p=a;# i' d/ T2 } v( D5 A& y8 N
move(p);, g- l: L0 n- }. k) H8 j$ c
for(i=0;i<3;i++) g' _' w C* ~5 r' o& x: A
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);0 G$ R; j- m% P
}
# Z. a3 O# `' m1 u7 ~move(pointer)! l) y/ L4 t4 E! L Q
int *pointer;
+ d; N. _4 [0 y7 _5 M{int i,j,t;# X0 d7 ?7 P; O4 m2 H, z; z
for(i=0;i<2;i++)
% x6 i* Z6 ]# Z6 D4 ]! I3 D6 G for(j=i+1;j<3;j++); M) b. c% m/ g) ]( e6 ` o
{t=*(pointer+3*i+j);! b# |# K. L# | K
*(pointer+3*i+j)=*(pointer+3*j+i);7 a" p" ^$ \ ]
*(pointer+3*j+i)=t;
b3 k& F' k+ @/ @9 Z3 c }
2 D% Z( [0 ~* _4 k}9 P# X5 C9 ?; ]9 ? |1 |* c5 I t
10.10 K7 a4 Q6 ~- M/ ~% T
main(). Z7 I' o' _: {! Q y
{int a[5][5],*p,i,j;: ^" S0 h$ @, X' l9 y1 i
for(i=0;i<5;i++)
% x- ~7 D l& k* D9 k$ V x3 S/ I* M& B for(j=0;j<5;j++)2 f0 ~- Z d, E( S. Z
scanf("%d",&a[i][j]);
1 L& A& B% c* v' V p=a;% w* J. e) ?% W7 C- {; u5 z, r/ q4 B
change(p);
1 w$ t h- T# f, X" s7 [ for(i=0;i<5;i++)
. N* X+ I& A w6 @ {printf("\n");
( A1 P, L. T( C; h0 B: E# c for(j=0;j<5;j++)
$ B4 t$ z# x. W# X% U, z printf("%8d",a[i][j]);7 ^' o: Z6 S) {, \/ K
}5 V$ E+ o1 V4 I3 ~" h- t
}8 M- R8 E+ P6 ~/ A$ [
change(p)' W8 E: n2 H1 H6 p
int *p;" U9 X( J" w/ {6 h, u/ }
{int i,j,change;4 W$ c: ?1 s4 C' a4 {2 j
int *pmax,*pmin;5 k% c4 C# Z& n
pmax=p; m V" @' c. d# L4 x
pmin=p;
6 A$ u# Z, ]6 k+ s7 ]: a5 L% w# a, ~4 N8 A: R for(i=0;i<5;i++)4 X5 y1 z7 R: B
for(j=0;j<5;j++)% d6 N& A: v/ a: l& U7 ~ Y
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;5 B s2 D4 P8 Z: `1 j
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;7 N) o' K% S0 L7 U4 ~$ m( N4 ^
}
' W% g2 V! d4 x [% w change=*(p+12);, A8 Z0 k% I4 N( {" j
*(p+12)=*pmax;
/ T3 @4 L% D* N" V$ h; M *pmax=change;' J* P1 U0 u( q, R. B. ^
change=*p;6 E; `1 B n! r: G& F
*p=*pmin;
# H6 l' a8 R# D, z# L7 ?2 X *pmin=change;
3 b; o% S5 L! I ~8 u; C( c pmin=p+1;$ ]1 U S u( ?$ L
for(i=0;i<5;i++)8 m! ~: e8 ^5 f2 w5 y1 ?. c: o
for(j=0;j<5;j++)
: K" g. Q# s# `9 Z S/ D$ @1 Y if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
! R# n# Z- M8 l; l change=*(p+4);* M4 H! l. U+ m J
*(p+4)=*pmin;
) s5 B6 x ~; F [9 F; V- c *pmin=change;$ ]- l. H+ g- r( y
pmin=p+1;: |9 ^. I! S( z: r: v- O5 o- q
for(i=0;i<5;i++)( b/ @. J9 v: W0 b: v- ^
for(j=0;j<5;j++). k3 Q5 C8 B+ t$ @( s- k+ N
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))% Q2 K% Y* e. P0 d' ~
pmin=p+5*i+j;
/ v1 V9 h; z2 b3 Z r J change=*(p+20);5 G. J1 m. k9 N5 {
*(p+20)=*pmin;0 }& E1 B4 Z- R% K$ N" \4 z
*pmin=change;
& w2 w4 [+ T; h8 z2 A O+ [* F pmin=p+1;; ^) O. _5 B1 h; O" ?
for(i=0;i<5;i++). c7 r+ X9 v9 K2 L3 a( M
for(j=0;j<5;j++)) k5 ^$ S( p! p: U4 @5 c- `
if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
& C# N- e( L- ]( z4 w &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;* Y: K. q" |4 B c: ~ E1 C! E: Y
change=*(p+24);
4 t! y: ^* G' e7 r. j% q *(p+24)=*pmin;
: w6 @& h2 b8 J4 Z9 o8 R) |# x *pmin=change;
e0 C( {! B5 A9 T, V' s}
7 Z/ ]; z k# t$ {10.11
' J2 `+ a% ~$ o# qmain()! U* h& z1 W! R. q+ z
{int i;
/ {; W2 J6 F; M% T$ l5 ] char *p,str[10][10];
# h8 Q: C, A9 k for(i=0;i<10;i++)
; n. f! o0 Y7 o* D. _/ { scanf("%s",str[i]);
6 D5 l: W; u/ W p=str;6 t: f. J9 u5 b4 ?
sort(p);
! j; _" X- z' p8 O for(i=0;i<10;i++)
" c {. B8 J! b* f3 [ h printf("%s\n",str[i]);
0 B. B9 n P/ F0 [6 m# h c; \}6 M7 y; w" W6 w; H. I; ]/ v
sort(p)$ ]9 L" J0 R* ~- g$ N0 {" U8 W* ~
char *p;. G6 y$ S7 k4 b' ]5 q- ?
{int i,j;
- W" x" T% r9 }; [ char s[10],*smax,*smin;1 {* i/ q, A$ ^' Y' t. w
for(i=0;i<10;i++)
" l; N, |1 H. a# b! W {smax=p+10*i;, V1 S9 {* k/ V
for(j=i+1;j<10;j++)7 |9 w* u, {- L5 S% \6 Y% C
{smin=p+10*j;, C" a8 l% U# ~. q! c! M3 f
if(strcmp(smax,smin)>0)
0 K/ P" l% F* D. `8 g, l {strcpy(s,smin);+ G& V8 P( M& n( W g' k3 p
strcpy(smin,smax);3 w5 Z& H+ s- Y% H% z
strcpy(smax,s);1 ^7 b( f6 b/ {9 U* s* X" c
}
- Z; r& q6 V0 f. r6 a* y/ } }
$ N+ w9 ^6 F7 {- c- }$ R- F3 t }
" @1 {3 b; b- x}
8 X% W1 F5 \! ?1 @; z8 E10.12* y7 S7 E# L/ D9 \# j3 {
#define MAX 207 E/ a7 Y6 l$ z: R
main()9 M/ T% i- [' v4 ~ G0 b# F
{int i;. t4 b K: k, y8 A
char *pstr[10],str[10][MAX];( E, P. a, v7 m
for(i=0;i<10;i++)! B R( r; K f: l1 g
pstr[i]=str[i];
5 m6 y, f6 R2 P7 q n$ l for(i=0;i<10;i++)
- i& _2 A1 E7 Z z& R; o& J scanf("%s",pstr[i]);
- P; G; \/ ? j8 v0 m sort(pstr);* F z2 f4 k; i& h3 l0 @( i: N( a
for(i=0;i<10;i++)$ N* r, i' F2 o: q
printf("%s\n",pstr[i]);
3 K: c! r# @0 [( R: N/ M0 c}
$ V) U. o) f7 F1 x y- wsort(pstr)4 ]" U5 r( B R8 Q' w- q" x5 @
char *pstr[10];- H# l. r5 k7 R+ P& ~
{int i,j;
8 E/ ~/ d i% W* X char *p;; `9 [& ?( Z4 k# g( v8 `* w) }
for(i=0;i<10;i++)# p& s8 J8 k. r* b) L9 |
{for(j=i+1;j<10;j++)
4 }1 O1 y4 V0 \/ S" G {if(strcmp(*(pstr+i),*(pstr+j))>0)# q6 V# V5 J7 w* a3 v* @1 _, R
{p=*(pstr+i);
" `6 U) ]: W1 |, @; l1 e1 E *(pstr+i)=*(pstr+j);+ k" G% {6 i! ?- B+ f1 A+ l, x
*(pstr+j)=p;' I7 ~ J! K) d4 }" Z0 _& c
}
/ e g! i$ B X: H9 Z- z4 o }
, @ j2 B' G) w. E. }0 Y: t' | }5 v! w% n# P7 R" ~1 u
}
, ], E) r! x2 W* O% C, I( C10.134 Z! Z$ [, }2 i, j0 P
#include"math.h" G1 D- H* }4 [4 ^9 j% \
main()) K2 T; T8 a2 R3 d8 ^ `
{int n=20;5 O$ [6 G4 z6 q$ H+ `
float a,b,a1,b1,a2,b2,c,(*p)(),jiff();* N; Q4 l0 Q& ~6 \7 j+ E9 ^
scanf("%f,%f",&a,&b);+ O" D; ~, H( e
scanf("%f,%f",&a1,&b1);) c/ G3 W; h) Y
scanf("%f,%f",&a2,&b2);
8 Z* _. o B8 I7 w$ M p=sin;
# u/ K( N. c) q+ U6 [; ^ c=jiff(a,b,n,p);7 z& F* r" C" p% i+ O' z
printf("sin=%f\n",c);; ]( g+ H! w+ C( D. @0 ?8 i' s
p=cos;$ z c7 }+ j9 s2 i
c=jiff(a1,b1,n,p);
$ Y( n* b1 m- v( r printf("cos=%f\n",c);+ Y4 Y. H2 }/ ]
p=exp;
1 R5 g$ _; i( E% y D$ n' H* e, W c=jiff(a2,b2,n,p);
1 H/ z8 \& M0 | printf("exp=%f\n",c);
) Y$ h) L' q% y/ H8 \}
$ t# Q7 ~4 J$ K% N# I* {" q2 ufloat jiff(a,b,n,p)
, a- V8 e2 S& @8 ~float a,b,(*p)();
X. n; g3 F9 q4 p, O* U6 g$ b0 K# Xint n;) |. M. X b; I) x1 `4 J' x n
{int i;
% @5 S$ l* S" T! T% c7 f$ ^+ |% h5 L float x,f,h,area;
. F9 _, [, h2 {8 V4 C; ]0 C h=(b-a)/n;7 X- g+ [" ?$ b* W' ^& l/ p
x=a;
; B w! S+ Z$ x, H area=0;
# X7 T; `0 S3 d }6 M for(i=1;i<=n;i++)
; P- j! g+ u+ ^' l7 b/ }! ] {x=x+h;8 ?9 y/ x6 g- P* R, q
area=area+(*p)(x)*h;
% f) r+ G1 W$ Z, T2 j }9 J& {3 d1 h/ q) x- q# P
return(area);* _$ g1 J( \, B1 u8 W+ _! U" w
}
3 c4 Y" Q7 L9 H7 Z10.146 Y7 m# |4 f3 \: t, J3 B, e
main()1 b" J& u2 R8 d* l6 T% ]
{int i,n,num[20];4 N& w. K# ~$ X( a! Y4 c! Q- Z
char *p;; V2 Q/ ?" ]" g, i+ F
scanf("%d",&n);, f4 l3 T% I6 l! k% }
for(i=0;i scanf("%d",&num[i]);9 T \7 C$ d) p- y# o$ x
p=num;. Y2 R6 r. k* d: ~: o& ]. x
sort(p,n);
& X" c& E9 A1 z6 j5 F' Z# p# A! n for(i=0;i printf("%8d",num[i]);
0 e( X0 Y8 p, z}" z) |% R0 ]3 I% ~/ r; F
sort(p,m)2 W* L# h7 }9 i
char *p;
! w: A5 I! n+ S' _. |int m;
: ?$ N3 C$ f& x- n' n$ S$ c) s, n{int i;
2 ?, m! \2 R- W& \. Z' p. O, d- l) K char change,*p1,*p2;- _4 k) A$ g& q f
for(i=0;i {p1=p+i;- h1 }1 G* P4 l; s9 X1 u
p2=p+(m-1-i);
, k5 b2 i- k0 m. w) l change=*p1;3 d- E# h: |% s2 I3 S5 s
*p1=*p2;
, a& D# P, O: p0 g *p2=change;
2 Q4 |/ l7 U' d0 ^: m }2 n9 Y0 O6 C1 r1 F8 e
}5 X1 Q# }! {4 t! C" b3 f' u' [
10.15- ~1 A7 u0 h) U, H
main()
1 X& d; P8 ^. Y6 b7 F* p* b{int i,j,*pnum,num[4];# z" \# V" x, o$ ~
float score[4][5],aver[4],*psco,*pave;
& Y2 n+ w0 L$ E, ~' c% ]0 ~ char course[5][10],*pcou;1 a: R& L" O0 e! y N' A
pcou=course[0];! e" ?9 r/ w* B: J# B
for(i=0;i<5;i++)& I. Q( D( A! U) }
scanf("%s",pcou+10*i);
2 V$ `" I2 p( R1 y printf("number");
) k0 J: d' h6 \7 l h* S for(i=0;i<5;i++)) [( \( C9 o) {* p. s5 Q
printf(",%s",pcou+10*i);
8 v$ Z+ s6 r4 s' g I6 J3 C. n printf("\n");
n( j9 ]$ _2 [ psco=score;9 W7 n8 K) p+ l3 c* z* q- k! }
pnum=num;
5 `1 O3 [9 [& l! ?6 M4 S/ i for(i=0;i<4;i++)3 M& k, B" @& ~+ S( M6 ?
{scanf("%d",pnum+i);* b8 u6 P* G; D* _7 Z! ~! W* k
for(j=0;j<5;j++)( O3 s: ]% R$ R I: V' d& \
scanf(",%f",psco+5*i+j);
0 m2 {. E0 N% N; h$ G }
, ^/ v: n/ L7 u1 Z( l pave=aver;6 p, |+ A. n; {" Z; s: y
printf("\n");5 t1 |; `0 m) P5 [" n( H
avsco(psco,pave);! K9 r3 \1 ?: ^8 e8 K
avcour1(pcou,psco);: l3 t% f+ F; J3 l% T0 n& j; {
printf("\n");* g0 B. Q1 m; C5 f
fali2(pcou,pnum,psco,pave);
* F2 `( F, R' l- r8 y; \! Y printf("\n");
. ?7 r5 E7 }' i6 |+ H6 N) [9 B5 } good(pcou,pnum,psco,pave);
& F( P# v1 V* X7 Y% g, U' N}
" G! [$ U! i0 U( v" F1 davsco(psco,pave)' E: H6 j/ F1 z
float *psco,*pave;
4 \" q% q; ?0 T$ w# ~: w{int i,j;
8 b! m. c& M! ]7 _( f/ t- M- \ float sum,average;; G" p8 Z9 g( ]8 }# q; `2 [) r/ R
for(i=0;i<4;i++)
" C6 [" _, k. H2 t7 M& K! F' ^3 O {sum=0;
5 ^3 O9 {8 N, p, _* e# c for(j=0;j<5;j+)
- V2 L4 G8 r0 h& U sum+=(*(psco+5*i+j));
3 L! h0 o/ N8 x8 b- _6 J8 @/ Z average=sum/5;- W& q g% A" b" v: @) H# p
*(pave+i)=average;& f' j- w$ P1 q( }9 w9 S3 v
}
|% k& h0 A* C* J}
@8 p$ h9 \* ^& Favcour1(pcou,psco)
+ l; A) m7 i, ]; ]% w8 Ochar *pcou;
/ n- G1 p S8 k$ P: d' ]' o: _float *psco;$ D6 u, y, C6 X: i, I4 X) v
{int i; q0 s) P( d1 Q
float sum,average1; N8 h4 o3 S4 Q) G
sum=0;. }+ Z5 }7 l+ h0 z: R2 p/ u
for(i=0;i<4;i++)
7 R, C `* F8 \' O0 t sum+=(*(psco+5*i))
, U! Y6 p/ O4 Y; Q w: H average1=sum/4;# d- U/ ^1 N2 H- \' |
printf("%s %5.2f\n",pcou,average1);
6 |* u: l% o8 V3 m9 p}" _/ E, O" X; J$ l( E6 `6 S/ [
fali2(pcou,pnum,psco,pave)
) _! ?+ @2 R2 {char *pcou;5 E$ R# c" ]9 {) W
int *pnum;
) c3 y7 k g9 S- }" V5 nfloat *psco,*pave;
% ~" [ K# e/ d [& _# B3 X{int i,j,k,label;
# R, @7 A! N, o9 E printf("\nnumber\n");; j; q, h$ r) m0 Y: i
for(i=0;i<5;i++). K" o) F; A2 l8 Q& ~4 E- Z {
printf("%-8s",pcou+10*i);& d- Q1 \ b/ f) z, a
printf("\naverage\n");( X* y9 w' b0 J
for(i=0;i<4;i++)
1 k* C9 j) x' E8 k {label=0;! w+ P; w# k3 [5 k
for(j=0;j<5;j++)
: E, C+ s. G# E1 a if(*(psco+5*i+j)<60.0)label++;7 K" M7 t: z0 M* a
if(label>=2)
- R7 G9 W, ^9 h. J9 u8 M3 i {printf("%-8d",*(pnum+i));2 w. _% p; V/ N2 z' l. v' n1 n+ S& c
for(k=0;k<5;k++)2 V9 S4 J0 q& M1 e
printf("%-8.2f",*(psco+5*i+k));
, n) n) O" n; D4 U( I printf("%-8.2f",*(pave+i));
$ i" T. M# v2 n; X }
! S6 B; a2 e t( L- c }" l& X; T$ w9 M3 F0 p
}
+ O* ~+ ^5 I) n# Agood(pcou,pnum,psco,pave)0 q7 G# N; G) s+ g
char *pcou;% T. e. u! q S$ q8 r
int *pnum;
* @0 Z9 B) _4 Q$ C+ P0 wfloat *psco,*pave;
1 m7 Z& T5 V3 @+ {; K2 M{int i,j,k,label;
# y# o, c& Z' S, E; U5 y( z printf("number");) Y" p3 E8 P1 z# d; P' Z4 T' o
for(i=0;i<5;i++)0 h# Y" \. o" B0 {0 I
printf("%-8s",pcou+10*i);
$ D# m& E; H' B9 i3 k$ } printf("average");. H, p, ^% @. F- f# K) U
for(i=0;i<4;i++)6 r, ]: A" I" L1 [
{label=0;
# a1 ? Z2 W: N for(j=0;j<5;j++)
. F1 q. O: ]' ^3 y7 Q if(*(psco+5*i+j)>=85.0)label++;
+ T5 B& P4 s; Q ^1 i, n+ x if((label>=5)||(*(pave+i)>=90))" v, d1 q. y+ `' A' i
{printf("%-8d",*(pnum+i));) ~, e5 ^( `+ H/ `5 M! ~
for(k=0;k<5;k++)
# Y" f! `# `' U6 U' m0 i% w printf("%-8.2f",*(psco+5*i+k));4 w& d1 \) n' N& y0 K! n1 O5 d
printf("%-8.2f",*(pave+i));
4 \7 w+ d+ z5 j2 X( v/ o, A }7 C& B( }9 O) R d+ ]0 m8 i
}9 C0 {( ~4 z6 S% n
}2 d) @; ?1 x/ m% U
10.16
) I2 x' a* Z* C8 M#include"stdio.h"
- I! T! B, I" Imain() A4 {+ x# w) Y8 U
{char str[50],*pstr;1 T. m. u g9 K* N! L2 y
int i,j,k,m,e10,digit,ndigit,a[10],*pa;
/ o) g5 j) L5 l* _8 k5 n, r gets(str);0 I1 i1 u$ y- R8 Y2 E W& S }
pstr=str;
/ ]! o' |' b9 ^# a8 A ^ pa=a;4 P& N' ~# G# h. p o9 S! |
ndigit=0;
6 G) ]& T8 K0 w C6 y/ T: k u i=j=0;
8 T, a& X( x6 e V6 w9 n while(*(pstr+i)!='\0'); j3 i5 f7 \8 ^+ ]" g6 d
{if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))
' K: X7 ] ?7 N! h j++;2 Q" J8 O7 R, j& w, x+ S
else$ B% o; x: ~! ^9 S" ^6 {
{if(j>0)
0 o' `( `1 `8 E8 u# X$ Z& E9 | {digit=*(pstr+i-1)-48;
0 d$ x2 ~& J h k=1; C; z. o) I+ a6 k4 h6 {
while(k {e10=1;
5 Q. F3 j! ~1 X4 X for(m=1;m<=k;m++)2 c X5 d( c4 B
e10=e10*10;- ], t4 j# I& }3 _4 Y4 G3 }
digit+=(*(pstr+i-1-k)-48)*e10;
! T& E/ J$ \0 s# W k++;
+ H2 j7 B8 |, w( N }- A8 y1 @& h B9 R+ a
*pa=digit;
* E w5 Y* G" G3 z: n8 l ndigit++;( k2 @0 g( M) n# Q: W
pa++;
( w8 y X- Z: C+ x {* W' K \ j=0;' i" S4 `4 _; Z% c
}
- y) D& x5 _& g1 X0 x3 p- k' N }
# `' O! }$ r' A% i: w i++;; k( v6 e2 b8 o% O3 C- z, C; [
}7 N& m) q; `' ^9 ^
if(j>0)0 _1 J9 \. p6 r. J5 z
{digit=*(pstr+i-1)-48;0 u; K% E% D6 d$ z- S) K: F* H
k=1;, J* z* ~3 D3 _( b+ P
while(k {e10=1;
7 p1 g" R7 Y" B* ~6 T for(m=1;m<=k;m++)7 }7 o5 t2 {. X! y
e10=e10*10;7 f" b! o/ Q$ H0 Y0 ~4 c
digit+=(*(pstr+i-1-k)-48)*e10;
* C/ I: I7 L4 W, H$ B M( I3 i$ f k++;- `7 q/ m( d& O9 K9 }
}3 U% z6 k4 g: f+ F# T
*pa=digit;8 L( F- }- k9 X5 y
ndigit++;: ^ \0 x) X, L$ W M( x6 d; ^
j=0;
4 D8 j) G/ f/ ?& x$ |# ~ }
$ t; R# P, T5 A: R% {( P0 N printf("ndigit=%d\n",ndigit);
! k( e% E& _, n j=0;
i/ c) U, B/ [5 I: @ pa=a;
% c- {5 j/ R; S1 r; \6 H for(j=0;j printf("%d",*(pa+j));
$ Y) p! A* c8 ?4 z- r. s- X$ `}( I% T6 f+ E( G1 T
10.17- z+ V1 V3 ?( G0 r" [3 T* u# V
main()
' Y' F6 H" M. _{int m;3 E& n8 R! C0 v( T
char str1[20],str2[20],*p1,*p2;
# }7 [4 \( U$ j2 z. i' O/ B scanf("%s",str1);
0 g$ N# Z- F+ Z scanf("%s",str2);
( u: U- D! D v' i ^( B( ]. z p1=str1;
& M# h) O0 I7 T% R p2=str2;& ]3 o. R$ E n- \6 s, u
m=strcmp(p1,p2);
& O6 l- Z: x2 m4 h! m3 @5 ]- f printf("%d\n",m);: Y2 V5 G3 X+ {8 o. j6 N
}; v) `4 Z: V5 j- P- |
strcmp(p1,p2)$ a0 R: ^, t) x" o2 N- i
char *p1,*p2;; X J/ T$ ^' y; |: I3 ]6 o9 {% Q
{int i=0;
7 E* w+ V9 n- y4 |1 ]* U" h while(*(p1+i)==*(p2+i))
1 x% H _0 ?! q# F% c if(*(p+i++)=='\0')return(0);
# m+ g( q+ V6 A' r( _ return(*(p1+i)-*(p2+i));
* |5 r' m$ Q4 L4 z9 J}8 b4 Z' ~' k' {! Z
10.183 s% J( y' p7 P3 ~( F; N
main()
" \% n' V/ U$ B/ X: y9 K4 f2 y{static char *mname[13]={"illeagl","January","February","March",
/ F8 w* q+ H. z "April","May","June","July","August","September","October",: Z& `/ D2 f3 L9 W3 I
"November","December"};. t8 \: E [% I/ k2 `6 i& M/ f7 u; n
int n;, h( a% ]/ [& _9 a4 X1 J* M! f# o
scanf("%d",&n);( ` ?' {# \7 Z8 j- Z" N# c4 r, P
if((n>=1)&&(n<=12))" d( P; C% F0 @: ?
printf("%s\n",*(mname+n));
6 ?( \' g, p; r0 N else1 l' O# b8 b4 a/ \* m( z& Q7 `
printf("error");
, ^) f2 x3 C' d" }8 {}' M7 p# j! @: _9 J% ?
10.20
& S0 V4 |$ J# _ amain()
6 _: [ S u. C3 h5 R: P: z{int i;; f: O/ I$ S; o+ U: n" D) O/ O
char **p,*pstr[5],str[5][10];, K% d7 c% @( ], b: t2 c
for(i=0;i<5;i++)8 n2 N. `9 ?" o9 B9 ~: \, B
pstr[i]=str[i];
8 J2 L1 b* o/ h: v. t0 Y- q for(i=0;i<5;i++)6 H- Q. `) |/ m# h
scanf("%s",pstr[i]);6 q8 N1 |" l, G7 C
p=pstr;
5 ], y4 N: N h Y2 S6 } sort(p);- u z% T# e# {" I+ Q' A1 P4 S5 t
for(i=0;i<5;i++). i. w0 o! `( n! h6 J8 @
printf("%s\n",pstr[i]);4 f' C$ W" J0 I, l
}
, y+ L. O, k. _: P. Y. ysort(p)
; E2 n+ ^5 n4 X% Achar **P;
1 w* T' A$ v9 g2 |7 o6 P{int i,j;" A: ?$ x% m- ]/ a0 C
char *pchange;1 I2 P1 k3 o2 H5 J5 B
for(i=0;i<5;i++)
# U% Y) r, C, q, Y5 g" l, }% f5 b {for(j=i+1;j<5;j++)
$ Y4 \5 @6 v- d [( t3 D# `& e {if(strcmp(*(p+i),*(p+j))>0)" M- F$ c$ j4 W( Z5 l/ b
{pchange=*(p+i);
% ?# T. p! n4 S' `# H *(p+i)=*(p+j);& S. e$ f. N+ K
*(p+j)=pchange;
( h! s, v8 s% E, C5 Q2 n9 T4 V }7 f5 P3 _2 |8 U& h/ D8 W' x
}
- I6 u. k* K( F. [2 }4 r }
6 H9 y9 p6 h- q: N0 h } o" Q1 q8 m}
4 E2 S+ w" L" ?0 _' Y/ K10.21( ^, j. R* v$ w9 @# t( ]$ X
main()! N( t8 N& G9 F" f4 X! ]$ j
{int i,n,digit[20],**p,*pstr[20];, o: [; W" G v% Z- C- C) [1 A
scanf("%d",&n);
, X/ o* N0 q; t# |4 V for(i=0;i pstr[i]=&digit[i];
+ I- D7 b8 l7 T4 D1 c+ z$ e for(i=0;i scanf("%d",pstr[i]);
+ \( x, U @7 R p=pstr;( y/ F$ ]5 A5 c. \9 f
sort(p,n);! t5 R7 T+ r# i# @ q% J
for(i=0;i printf("%d ",*pstr[i]);( M% b* J' \: z( `) T
}
$ W+ j8 m; P, M) s7 y; j6 @sort(p,n)
I. [5 J. ^( f; H8 L: Dint **p,n;& B+ Y& Q& H; g# x& `6 m* F
{int i,j,*pchange;
3 b( i4 T6 I5 U% W; c/ @* z! [ for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
) L4 k! [ q( S' w, c, _- w( k {pchange=*(p+i);% i1 \7 N9 E0 [' O& p( u
*(p+i)=*(p+j);* A+ B) B. i7 b7 l/ h3 f* O$ \
*(p+j)=pchange;8 e* n0 h! m4 q, D+ g6 M' U
}
) U% z- K+ M1 e- T$ u9 p. s1 b }
3 j) d M, ]) J4 F }2 S* p$ i D! C: o7 y" Y7 |
}
: G' z3 S; I- T x0 r第十一章 结构体与共用体. G4 k: z$ W5 k& N) t- U3 U; w
11.18 q0 v. m) O' e K E; h& C
struct
/ N4 E/ N3 v2 e: ]/ _: Q' F- x {int year;
9 F/ P6 H. w" @6 t int month;
6 u' d. t5 s0 O. D$ d, l int day;# u" C$ w& d9 q; m
}date;
1 C" X' K1 d; L+ v5 emain()( z+ F$ s! f2 y! [
{int days;
, e2 I7 [% J2 ]4 a scanf("%d,%d,%d",&date.year,&date.month,&date.day);, I/ s8 e, H* y4 ?
switch(date.month)
, w' u1 ~8 A: M0 q {case 1:days=date.day;break;
/ z, a3 O0 W. ~6 E A case 2:days=date.day+31;break;
4 Y, `6 a( l, J3 i5 j" O; `7 T case 3:days=date.day+59;break;
0 H" e: B/ S& v! O ` case 4:days=date.day+90;break;
+ R% X1 t9 }( H& D2 ^ Q4 B case 5:days=date.day+120;break;
$ B. _3 y6 t4 F1 [ case 6:days=date.day+151;break;, `% g/ `) [: v1 O) P4 z s
case 7:days=date.day+181;break;
4 E7 z: R$ i- U7 T. i R" Z case 8:days=date.day+212;break;
# W$ d! j7 P# D2 G; D case 9:days=date.day+243;break;
0 y1 C* q2 p. ^8 V) u case 10:days=date.day+273;break;
9 J. g4 q" ]8 {' f case 11:days=date.day+304;break;
0 n0 |0 B$ B% p5 f" e case 12:days=date.day+334;break;$ r% k; v3 Q3 v' J# \: H' Y
}
) y/ ]% P5 W/ r5 O! G3 X+ I if((date.year%4==0&&date.year%100!=0||date.year%400==0)6 C4 T! h f2 @ H
&&date.month>=3). `+ ^( ]) @$ f3 j2 M9 j
days+=1;/ H' Q& y9 \) j: T/ n# P
printf("days=%d\n",days);
8 u: N0 Q2 e8 d. V: z) y& l}
& e8 |2 L2 w+ M2 a11.2
3 i/ ]/ S6 V- Qstruct dt: E# y0 `! W0 \
{int year;
* U J6 g! h# y+ O! [4 @" C8 k: E) E int month;
' j$ `+ a" t1 T9 Y i% i# x int day;$ a0 i) k A p5 F9 e. a
}date;
- c+ S1 l+ i8 Z$ Hmain()6 s6 R3 N$ f: p1 Y) v7 }7 t, q" n
{4 H t1 M7 u& X3 z! `
scanf("%d,%d,%d",&date.year,&date.month,&date.day);
4 B0 A) N3 \: e& Q printf("\n%d\n",days(date.year,date.month,date.day));$ x \6 t* ~% b! R+ z/ D- F
}
% z' B: q2 `* I. }/ Fdays(year,month,day)
" Y/ B; C2 ~# R' [1 Z- Lint year,month,day;
% q/ j V. h7 o5 i M' F$ J{int daysum=0,i;% R' A& @5 c+ v5 i1 _& P
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
* C- U- H) V9 l$ x. |1 E for(i=1;i daysum+=daytab[i];- R3 C" I0 p% i- m1 A3 s
daysum+=day;6 E0 y( F7 p5 `# a4 C9 c k: W
if((year%4==0&&year%100!=0||year%400==0)&&month>=3)7 Z' S" R+ D; K% J5 `! C# P$ O
daysum+=1;
2 ~7 [+ E# I/ b8 a7 d return(daysum);7 n4 q8 \6 A1 E( ] \
}
& M+ {: o, o M11.3
* u- k M9 V. f. \. |6 [2 a$ ]' ^11.4& j' J4 K( G- T, \8 _; Y
#define N 55 m& j. t) V! P7 J% Y
struct student
( |! V( j9 [7 m9 S/ {, @ {char num[6];
$ }# X6 U) F- ] char name[8];
& h# D( E6 Q; h8 O" V7 | int score[4];
" ~1 T- A; V; g1 Y& A/ s5 g }stu[N];
8 `2 r0 T) [- H& g7 r' _7 m: Kmain()
5 @7 K# I! |8 [& v# m{: Q! N+ y6 ?- R
input(stu);6 m0 Q$ R! N1 Y4 [+ B
print(stu);9 D5 ^+ r* R" z6 g' k
}
7 ^- C& N" P+ Y+ Q. ^# U: tinput(stu)" j. v" H, b; O2 B3 D' Z# q
struct student stu[];
% W& d& S/ c) C( h& o+ H; w{int i,j;% D3 d- V. J( t# Y v/ k
for(i=0;i {printf("number");) L1 [* v8 K, C
scanf("%s",stu[i].num);2 t" q! G# ?$ ?1 n! P: ^- O
printf("name");
( z. x3 S. f: r. l! z3 Z scanf("%s",stu[i].name);% m) F1 o9 K* A: P! q
for(j=0;j<3;j++)1 p: d2 j/ {( C! Q, A% J6 ~
{printf("\nscore\n");9 _% Z- R3 G, S7 g7 [: t% V
scanf("%d",&stu[i].score[j]);
+ D( D" |* u2 u8 s! r2 e }2 F5 _" d7 F% u
printf("\n");
5 ?6 ]6 Q. V* X6 R6 T! |% b }- |1 W4 g& X O9 v( W6 _+ C; ?( h
}
8 h, f- Y6 D7 m- xprint(stu)4 Z/ h. s; x6 F+ \+ w; X
struct student stu[];! H% i" U' y: A9 W. g
{int i,j;
3 _; F. X7 |2 h0 N printf("\nnumber name score1 score2 score3 \n");2 W& Z6 Q. v5 ?, j; E) h! [# `
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
, r h+ c `( S/ v# {. `( A for(j=0;j<3;j++)6 l; e) m" C3 j5 F% u# K& n
printf("%7d",stu[i].score[j]);
; K" j5 |$ s! Y [4 v& x# z printf("\n");
l- v0 M; v" @# ^ }4 `+ p( X' V. W
}. S$ {% U5 o1 v- B j
11.5: K2 F* o4 ?/ m* x# t3 w8 I
struct student6 B0 j B, S1 v5 h- P% @- M% G+ g
{char num[6];
& Z, c; M1 E' G* W+ Z- I char name[8];
# V- P6 C+ F/ ^+ w int score[4];
/ a& L" @! v$ S. V3 X; S* n; u float avr;4 d0 I V/ R9 s1 S% Y6 \( J
}stu[5];: [+ I+ z0 i5 x5 Y$ s0 I/ o
main()" V) u0 G; J( y4 n4 `( o1 M* p0 ^9 N
{int i,j,max,maxi,sum;
4 G8 N+ c' y: i( O6 d4 |, z# ^ float average;! ]* Z( `1 ?- _6 u2 d
for(i=0;i<5;i++)
2 e, K0 _ N3 X {printf("number");
4 k0 F( t7 G" s9 [, n4 b+ ` scanf("%s",stu[i].num);
$ G$ z2 Y: y# Q printf("name");. f& M6 h1 y/ q: G/ U3 b) e
scanf("%s",stu[i].name);
0 X4 F k. \" \: w0 j( }% V3 _( u for(j=0;j<3;j++)
( p2 ^ \" X5 p# Y3 Y {printf("\nscore\n");" U; g: d8 c+ |' ^, y3 } Q% S
scanf("%d",&stu[i].score[j]);
! O9 N- M6 k1 m o }5 L0 G- `+ V" n" @7 J
}8 ~' w; i4 Z( d: z/ m) g- ]
average=0;) o, i2 A) R# j$ f: a) l9 |- P
max=0;
0 z; g8 Q% W7 |8 h2 ~0 H+ U. `2 Y maxi=0;
, C( n2 a4 Z" ?2 {6 o for(i=0;i<5;i++)
3 W* v1 {) Y) y1 ^- t- U& v {sum=0;4 V- C @* }) i6 S& ]: P8 e
for(j=0;j<3;j++)6 H- {! O3 x6 K: A
sum+=stu[i].score[j];6 L# M! _# \& k& }) f( P# ?9 G
stu[i].avr=sum/3.0;! k) {0 M, @& Y; I: b
average+=stu[i].avr;
* M2 G! g( K( G# q$ q- f& l1 S7 s if(sum>max)
. D3 T& M6 P( H {max=sum;4 B3 x% E6 i- _) L
maxi=i;+ M# z/ x/ p, G7 d4 C6 C
}: `7 X9 ^0 Y: O: u% n
}1 Q2 S. c/ ^6 ^( C0 l; f" F
average/=5;
1 L# x8 a7 `( q) F printf("number name score1 score2 score3 average\n");
' p! _5 @1 a/ u" A5 s0 {$ J for(i=0;i<5;i++): d' r1 p3 w$ p" \& {5 p
{printf("%8s%10s",stu[i].num,stu[i].name);
, R3 F6 o1 k c- v: [. t6 k for(j=0;j<3;j++)8 i* u& v5 d/ I6 a; {6 R( E0 v7 z; r
printf("%7d",stu[i].score[j]);
2 h. Q- E$ R$ _" i4 U, W printf("%6.2f\n",stu[i].avr);
e* k8 Q% O* |- s; u }
3 x" L5 O4 V1 Y6 n8 r; q# h/ r printf("average=%5.2f\n",average);
- i4 c9 {# H9 u( X3 w$ H printf("The best student is %s,sum=%d\n",stu[maxi].name,max);8 T3 w! I' p+ L* U* n
}
o* A- b- k7 R; H* ]# Y0 F, x4 c1 W' T* }$ j2 ]
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