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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
|---|
签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案5 H: T% d- y& E$ c! M
1.5请参照本章例题,编写一个C程序,输出以下信息:7 i1 R) N1 C# }$ x
main()
' k; ]* s. ^ d{, K( G" n8 ?. i' ~% _1 s9 S
printf(" ************ \n");
# r! p g" h. a% B7 k) p/ @5 \$ Mprintf("\n");) i6 P7 i, v# A, v& Q2 A7 \
printf(" Very Good! \n");* z) ?9 t. Y1 U, J0 Q
printf("\n");- y# [9 Y( j7 U
printf(" ************\n");1 T9 b9 @9 ^3 V5 `
}0 S6 q* |" Q9 V, Z5 N) N
1.6编写一个程序,输入a b c三个值,输出其中最大者。* h% \0 q+ h6 {9 n3 ^/ w1 z
解:main()
( E& L/ L( |7 i6 Y{int a,b,c,max;3 |4 W8 d7 J0 E4 q
printf("请输入三个数a,b,c:\n");
5 x( W& h5 t: rscanf("%d,%d,%d",&a,&b,&c);
% K% y L2 h& b, i, P6 fmax=a;9 c% @# h1 E7 L w
if(maxmax=b;
4 N4 R8 ?4 V3 k* Hif(maxmax=c;
2 b F; K6 r, h' Uprintf("最大数为:%d",max);1 O# |: F9 e# r5 e) E" [/ B
}! `$ I- u) K( M1 n
第三章# ~, K8 e( x3 Z9 O2 U' m
3.3 请将下面各数用八进制数和十六进制数表示:: L3 u# T& Y5 F* ^& W0 H8 i
(1)10 (2)32 (3)75 (4)-617& t3 m$ t d/ g) Q+ ^. q+ E+ }
(5)-111 (6)2483 (7)-28654 (8)21003# B; g- e' S, N! f/ Z
解:十 八 十六
9 Y: {5 Q9 M. |: I" C4 G7 C (10)=(12)=(a)" k3 T2 V L: V: P; o; ~' k
(32)=(40)=20
4 q6 p) y# j- S w (75)=(113)=4b
G( Q+ {6 O3 x9 \: K) t1 z6 E$ Y" i (-617)=(176627)=fd97
- J6 A. ~, {! ] -111=177621=ff91
$ N" N# _( z O 2483=4663=963
0 {7 I, w/ H& v; n -28654=110022=9012$ V( R1 w$ c& m- l7 {
21003=51013=520b7 o& I# ~8 u5 |7 s f. j
3.5字符常量与字符串常量有什么区别?
% {5 l& N+ }- V8 f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
" j- _; Y* p& A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'., B) i& i8 P& p' i
3.6写出以下程序的运行结果:5 G) G) L, a& k% S
#include
9 t. c0 A: O% m& wvoid main()
* ?, Z3 g8 b, H% |{
% D* v3 k5 E/ _- ]2 R8 o* W$ _- pchar c1='a',c2='b',c3='c',c4='\101',c5='\116';4 ]9 g3 c3 Q* r2 i" a" j$ N5 O
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);7 N) p! B7 ]/ z6 E8 [5 [% j* D
printf("\t\b%c %c\n",c4,c5);! X2 \/ a6 @8 D$ B
解:程序的运行结果为:
9 |, e- b: F8 p. _# L* taabb cc abc& y: l% e1 f! D2 u4 t/ |
A N% n2 x& |' b( `1 v( M
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,
3 z0 X+ }7 }* `& ~; n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".
) a6 }) b: X, u9 v% }请编一程序,用赋初值的议程使c1,c2,c3,c4,c5分别变成'G','1','m','r','e',并 Z! \6 l$ W; r7 t: i; r& [
输出.
& s g% i0 ^! p1 b6 E) T8 nmain()/ k3 C; Y5 O! Q2 I
{char c1="C",c2="h",c3="i",c4='n',c5='a';
+ Z K6 O' v* K& [/ C+ {c1+=4;* h% W# m8 i) T8 j# _0 u
c2+=4;" E( A9 `- M( v! Z8 C! J( w
c3+=4;2 j, G, v: K h6 L. s1 U
c4+=4;7 k6 g; T, c2 q; L, t% p
c5+=4;
5 x: [% r& G0 X! T) h4 y% Qprintf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5); J, E* T' H& g8 k" v" j- D3 J
}# P0 n5 t! h8 s
3.8例3.6能否改成如下:0 G3 c7 h* b0 w, s
#include
& ~* ^& x8 C& K! l. S* avoid main()
- C& t- f) j& G{4 [$ n I+ A+ m5 p3 ^& M5 Q! k
int c1,c2;(原为 char c1,c2)
& _# C8 i' b i/ a9 ?% rc1=97;
+ M$ W, B3 T$ M2 \c2=98;/ f: }, f& o8 M
printf("%c%c\n",c1,c2);
B. G, g5 O v2 wprintf("%d%d\n",c1,c2);
! R6 R4 Q4 K* d4 j; B}; w* n6 ~" W' l/ L3 h8 \
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.
" `8 j( \: e4 V: \$ k3.9求下面算术表达式的值.4 j ` {7 e$ m3 I4 m
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
1 _+ w: [: z0 M6 f(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
; z6 W1 \2 F+ I$ @3.10写出下面程序的运行结果:. w( S! \( Y' K2 Z
#include% s1 R/ \7 S3 }; ]
void main()
3 ?# _. w) |0 d+ j, k{2 ^4 k" n& v+ z- I" ]7 V6 }9 X" c( H
int i,j,m,n;
# I; M/ o* q Qi=8;$ u, g4 g4 `: d0 e. v7 N- c
j=10;& V3 O+ `6 Y% f$ G. ?
m=++i;
. l5 S2 L% Z/ j; o1 U( Z5 qn=j++;7 Z% s6 F/ p; s1 g
printf("%d,%d,%d,%d\n",i,j,m,n);0 j N7 X# i( ~) J# e% ]! z! f
}, y9 h+ h; j F. s8 E1 s' I* @
解:结果: 9,11,9,10
+ b3 [! R3 ], Y第4章
& ]4 R6 E6 ]; h( W4.4.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4 i/ e9 b( ~: |% z9 Z: H( o
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:" Q2 b& V9 j( f. y& p3 c
a= 3 b= 4 c= 5
. K' D e' Q, c$ L2 d0 G) Tx=1.200000,y=2.400000,z=-3.600000
' h3 s% e4 H) x3 m, Ax+y= 3.60 y+z=-1.20 z+x=-2.40! l- W+ f6 p! x
u= 51274 n= 128765
! q& L2 c. ~: w% T, K: B' Vc1='a' or 97(ASCII)! K. f7 Z# |7 g# ~0 h5 F V9 O
c2='B' or 98(ASCII), J# q8 [; {4 [1 G
解:+ g l9 n7 w- M% G* @3 e# K7 s O
main()# s# b2 e( n- x" I
{+ k0 e4 L( n, i) M; G1 _$ @, M: K* d
int a,b,c;
0 Z% \$ M/ O2 n+ Nlong int u,n;! E' O# @( V" C' O$ }
float x,y,z;
& z4 f9 v) ~, l0 T: |char c1,c2;
/ W& \3 ^2 J7 V% sa=3;b=4;c=5;6 J; z# F! v5 s8 V. e4 d, d2 N
x=1.2;y=2.4;z=-3.6;5 q% A: w+ [6 g4 c
u=51274;n=128765;
+ l* e& i. z1 E* ec1='a';c2='b';
* G E+ W7 O. y- aprintf("\n");
% S8 @8 y# e0 R& e8 m' O% P1 b5 Mprintf("a=%2d b=%2d c=%2d\n",a,b,c);/ q2 l$ u2 w( O# a7 z
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);0 a/ \# q' b% U% [* g; P9 I
printf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);
6 |( w p( Y9 q' [9 E& T) w# ^printf("u=%6ld n=%9ld\n",u,n);
/ F) w3 ]) X( ^9 R9 V% ~# Kprintf("c1='%c' or %d(ASCII)\n",c1,c2);
$ `1 v+ U! V' s0 U+ i' |0 Q1 xprintf("c2='%c' or %d(ASCII)\n",c2,c2);, I# Q. @2 ]( m/ V4 ~2 X
}- _5 x2 T" x! C
4.5请写出下面程序的输出结果.
: ?1 v( V+ Z$ O! ]+ _! t结果:
2 Y& p1 j$ s) |" F579 i' D9 i* h6 m- X. S
5 74 v& D" H2 R1 A7 L C* c, B
67.856400,-789.123962
5 w+ J& h3 y8 d. p67.856400 ,-789.123962; X2 y1 P1 I$ W; f5 O7 @8 P; i
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962" M& Z! h( G; U
6.785640e+001,-7.89e+0020 `5 j# E6 E, P- ] F% C
A,65,101,41$ K6 }! P- u4 P1 m' t5 V
1234567,4553207,d687
1 d9 h$ b3 G/ A7 n65535,17777,ffff,-1" x: q2 W# @7 K- x X, H/ _
COMPUTER, COM
" z3 E' y# D r, x* P) s4.6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',
' S: [3 Y, K8 U1 P问在键盘上如何输入?
( p2 R! ~5 I( ?! Y! y W2 i1 W, v) kmain()
/ B0 M( x; @* d9 ]- n{
! x( L7 i$ G# {/ E5 [9 o8 mint a,b;
2 \2 z4 `8 s( [6 \float x,y;
7 i" J3 I0 a: f: j& L) P4 ]char c1,c2; G3 }) Z9 o& S. j# D) i8 n
scanf("a=%d b=%d,&a,&b); R8 B+ x% J5 s+ }% E4 ]; \! d
scanf(" x=%f y=%e",&x,&y);9 A" D% r9 g, q4 P& B5 j4 o/ ?
scanf(" c1=%c c2=%c",&c1,&c2);6 }" g! g$ b8 Q, v. Q4 @; s+ g
}
4 q7 C0 i- k( G1 C$ Y解:可按如下方式在键盘上输入:
" Y6 |) u: C2 {5 X/ k2 ~/ i! A: ua=3 b=7
/ n' {! J, [/ ^8 y! R& |x=8.5 y=71.828 h& l9 K6 Z1 Y, y( p# H4 ^0 u
c1=A c2=a
% H1 i, k0 c7 ^$ e说明:在边疆使用一个或多个scnaf函数时,第一个输入行末尾输入的"回车"被第二- r( u: x; O1 b5 L6 h7 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
! }4 Z" i' c+ R6 f9 E0 n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
4 ]2 m8 E: N& I) f% l7 L初步概念即可,以后再进一步深入理解.
+ x% K5 d. `3 C4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-; k3 z, l" R$ ~4 V; z4 \
3.75,z=57.8,请问" t4 l, Z, Z1 `. x% ? \ l9 ^2 K
在键盘上如何输入数据?
3 k8 E6 t) Q! d/ F6 }" x; F; Rscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);
/ g4 i& F& a* p* G1 d解:
* ]. m+ q6 J8 C( A0 h# o0 P# `. vmain()
2 r; G3 ?" }6 y8 S1 U: O{, F" ?+ t3 `- r4 A/ N
int a,b;
1 L. N) {/ C$ V; N1 l' }1 ]float x,y,z;
$ v, a0 B; ]7 ^char c1,c2;
. a, D! c% J( w% C& H7 |2 F9 Tscanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
3 i# e" t* h3 Z3 u}6 t0 D J; x- X. w8 I `- A
运行时输入:( K: B8 b Q# n- u6 t. O' E! _" L# I
10 20Aa1.5 -3.75 +1.5,67.8' f1 `, V- i" z9 [7 ]
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- ]; U# a' g9 J
是用来禁止赋值的。在输入时,对应于%*f的地方,随意打入了一个数1.5,该值不
8 [$ c' S! T! D会赋给任何变量。
" E/ @/ d5 r% D0 H4 B4.8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,
. v6 ^" c+ g% B, a0 m' N# b) |8 C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
$ @. \+ v3 u" v程.
; r+ A, A# u' e0 q# f4 g解:main()9 ]/ R) S7 W6 W# f$ j4 o
{
: R: u k q2 X5 p, {float pi,h,r,l,s,sq,vq,vz;) W f, A, D$ r7 d2 l
pi=3.1415926;4 m6 i7 F: ?. p4 y6 m& k
printf("请输入圆半径r圆柱高h:\n");
; u, X# J( Z0 e' s) mscanf("%f,%f",&r,&h);5 \5 P) T v1 o; P) Z6 K
l=2*pi*r;
; |7 z8 v0 o. [2 P; e: ~/ j: ys=r*r*pi; C( E; `2 K) ?* W
sq=4*pi*r*r;
, [) ^5 I& h! x" A& ivq=4.0/3.0*pi*r*r*r;6 C! a: V2 P% A& l$ X5 E# p
vz=pi*r*r*h;
( O9 |6 [' O7 d; i3 l: f. H6 J# eprintf("圆周长为: =%6.2f\n",l);
4 ~+ k/ l# r0 H# m/ ]) uprintf("圆面积为: =%6.2f\n",s);
! X1 [# m+ H a3 P A. e0 Fprintf("圆球表面积为: =%6.2f\n",sq); ]6 \5 T l* o7 m* A
printf("圆球体积为: =%6.2f\n",vz);
! U$ d. V1 s5 W5 ?+ W: u: I}% O3 D# ?# s& b, ?
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,0 z0 F7 A! X3 e+ g2 L
取两位小数.7 y/ o x9 b! F8 g! w4 ^
解: main()' w: w$ |; k# K% s T; R
{
5 P$ F( A' `3 P: L! qfloat c,f;& x! h) N0 I8 P
printf("请输入一个华氏温度:\n");, ^5 w& H2 U6 J8 u# B0 X
scanf("%f",&f);
+ c i$ @+ {6 w4 X% z0 Lc=(5.0/9.0)*(f-32);; q% L6 R6 z9 A: P4 S
printf("摄氏温度为:%5.2f\n",c);% D4 i1 n. m+ ]
}0 U' t9 h `1 S7 t+ R1 ^/ r4 H
第五章 逻辑运算和判断选取结构
- C! b; v' \2 {3 U6 {. }0 }1 V5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.3 o2 S; K: t: {6 J8 X. p" j% D
main()
% l- ]8 Y! Q0 @5 ?{
7 `7 [) `, H1 m) s0 L: P+ [! B7 l" sint a,b,c;$ K4 @9 R8 |! e7 k
printf("请输入三个数:");' \1 p7 b# Z2 C' t1 f5 Q$ f
scanf("%d,%d,%d",&a,&b,&c);/ T, Q4 S- c$ m( m5 Y! I7 w( j
if(a if(b printf("max=%d\n",c);: Y. y2 a5 e9 c( P& I
else( |5 p7 G5 \7 L8 H0 e
printf("max=%d\n",b);5 t2 ~7 R S1 n2 x4 w
else if(a printf("max=%d\n",c);# b' W3 l$ B. V, D
else
) v- E( k7 R) A. I printf("max-%d\n",a);
/ ?6 k6 J% Q- w) c+ J}$ @6 ~6 z; m* Z# F' u1 f
方法2:使用条件表达式.
% W6 A6 c% \2 W S; P* Umain(): C! t! y: H9 E# F* c. d' z2 U1 }! E
{int a,b,c,termp,max;
3 y4 Q+ f' v( s) _. z$ K printf(" 请输入 A,B,C: ");
$ K& u- H% K, n) x scanf("%d,%d,%d",&a,&b,&c);0 D3 f) ^: \8 H9 N8 R
printf("A=%d,B=%d,C=%d\n",a,b,c);
" }. i) P7 Z7 ^: L) R/ ]( t _: O2 [8 k temp=(a>b)?a:b;
3 ?4 N B4 X% j u. X6 q- a5 U: } max=(temp>c)? temp:c;' [ o- |# r, Q$ e1 ^ `
printf(" A,B,C中最大数是%d,",max);/ y: K- r6 B% E6 z; G# ^5 \! Q7 B1 F# S
}) x/ n; Q0 o3 V3 Y& e" }
5.5 main()8 a* Y1 \# w5 j1 t; Y
{int x,y;: R$ j$ W: S2 J; p' u( ^9 X# O
printf("输入x:");% @! w- n1 @( J7 D
scanf("%d",&x);0 X" R5 g1 M Y! ?7 w
if(x<1); L% {3 ?; R3 e9 z5 ]$ D2 R; Z
{y=x;
5 q7 M7 O% i3 O- g2 l2 S printf("X-%d,Y=X=%d \n",x,y);$ Y) V5 R- i! O$ t ^- b; U: Y
}
5 B; z. U- R2 _9 j& _0 xelse if(x<10)
) i+ ]! Y! N e& w' h1 G U+ | {y=2*x-1;( N/ t4 \: Y3 v. [
printf(" X=%d, Y=2*X-1=%d\n",x,y);- x- B% S5 p/ ~. c. a/ Y
}
0 V4 i* A" w6 f" j6 |else
# K6 s4 g- T8 e3 o- a {y=3*x-11;7 z0 Q6 V; @' V* X; g. \
printf("X=5d, Y=3*x-11=%d \n",x,y);
- k6 m! Z: t1 l0 d8 F }
7 i) G& L5 `; q6 \}
( a: _1 h }; T- T(习题5-6:)自己写的已经运行成功!不同的人有不同的算法,这些答案仅供参考! : U! C. S/ \* `& I9 m5 c" z
void main()- u" r0 \1 R& H O& G' g
{
`4 w' j/ L" o4 Q# n+ efloat s,i;
6 o& l8 q. K% d; n0 F. {; X( Cchar a;" p0 r3 w. S- V( D- C( |; J
scanf("%f",&s);5 K M6 y- j6 N1 f
while(s>100||s<0)9 D( z' C. m$ u" N
{( k' X: y, ?! u7 [ @
printf("输入错误!error!");* k2 a4 ^7 Y+ f
scanf("%f",&s);
% G' a/ `% Q e2 ^}
( F [% _5 c/ @- |+ K4 Ti=s/10;9 }. q Z' ]% T1 L* f1 X9 `
switch((int)i)
* r$ p. P; O; x c' c{
/ e7 ?+ Z K: q. [$ b8 b2 m3 zcase 10:
: ~* I9 X3 h* n* a% }% \case 9: a='A';break;
; l6 r# q" P5 [/ N& H+ Bcase 8: a='B';break;
# P) d* Q' N5 ?' \9 Ccase 7: a='C';break;: c0 ]+ l8 z- C! b. y% O
case 6: a='D';break;' @1 b3 ^2 T' \+ Q3 {2 J
case 5:
- N: g5 H' A6 g7 S3 t" d* |case 4:
+ A6 N% u7 T" k3 @case 2:
/ ?0 @" ?3 g H+ Ucase 1:
7 ?" M! |5 d. ], B! Wcase 0: a='E';8 _7 V2 l1 ?* s0 X3 s
}' o1 e; @) o9 e2 N
printf("%c",a);
! {0 `. B7 [( b3 b t& w; Q D}
9 ^4 K. g; }& ?4 j" T9 z# D! F- w5.7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.
: G0 A% M4 J' U6 r3 T2 \' y按逆序打印出各位数字.例如原数为321,应输出123.+ f2 I" T7 I$ {0 c! g& ?) M E
main()
) [1 H5 i2 o0 Z {6 _2 i% a2 W" ]$ A
long int num;
. b7 V& G; y2 m int indiv,ten,hundred,housand,tenthousand,place;# s8 z8 h' W2 i. u
printf("请输入一个整数(0-99999):");& \9 A3 N: X3 _* V/ y
scanf("%ld",&num);; `9 h6 F0 J2 e6 Q6 I
if(num>9999)
8 l8 r! r* |2 ^0 g0 j: U& B place=5;" {! g) O e) t9 X8 O
else if(num>999)
0 Y8 M2 ]+ n' b! Y& y) o" a6 k place=4;6 _: q3 J' ^# ^1 A
else if(num>99)" K3 C7 W4 K8 Q* v$ m8 h, ^) q h. V
place=3;
( t. ^* { t$ ?else if(num>9)
* _, ^% m; G- G/ D1 C7 F place=2;, x6 b, t3 L# s4 |) l; G; o
else place=1;8 \4 j# G: b! T" t) ]1 P2 K
printf("place=%d\n",place);/ z3 w- B3 f5 d, ?2 ^3 H
printf("每位数字为:");
% g. A) U; y3 w5 }6 |ten_thousand=num/10000;
: O" G( ?: `* Jthousand=(num-tenthousand*10000)/1000;6 [9 G- I9 t+ }
hundred=(num-tenthousand*10000-thousand*1000)/100;
( _! R$ @- r w( N: w& iten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
, y% M! j. U6 [; h, \/ n6 hindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;6 l; X( l7 ]9 s1 w) N" _/ _
switch(place)4 d! p, E' I- y0 T1 l/ L5 A
{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
& s7 H7 w/ L( p3 v' K8 J U! |* h printf("\n反序数字为:");( p; _. y5 e, q0 _: R& |
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);
' S8 P) K$ j- d* w v break;
# A& m. P, H6 p4 `case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
9 K$ b3 r2 A" z8 X$ s printf("\n反序数字为:");: A o2 m3 V5 X; i6 s# S3 T
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);# V7 `' q! B2 O" ?/ f0 l
break;7 w. d4 Y7 G; x) A" j6 c
case 3:printf("%d,%d,%d\n",hundred,ten,indiv); }8 q* e2 @/ H/ t" P& X, i4 X9 K7 c
printf("\n反序数字为:");
# b# j, c4 _# s4 w printf("%d%d%d\n",indiv,ten,hundred);
7 J- C& L! w7 Q: Z( dcase 2:printf("%d,%d\n",ten,indiv);
6 P& [+ n) s) o) o- a4 u+ U% ] printf("\n反序数字为:");
$ D7 p) a- W7 ]# r printf("%d%d\n",indiv,ten);
" m6 q/ X3 A8 Wcase 1:printf("%d\n",indiv);6 _- d% M. E4 |& m7 G3 E) a
printf("\n反序数字为:");
/ }) Q* O5 k$ I v! G% }2 b+ h printf("%d\n",indiv);, J3 X u2 ~# _, a5 P5 w
}( l1 m0 A4 `( s1 t. S: n
}( o! h% v% S; G* x6 d( g
5.8 X+ C5 a' y# N5 z% P
1.if语句+ v, M! k9 \. [9 ]) W. M
main()
4 B; i* f4 }4 C8 P7 {$ v! |! J3 w{long i;
( i& Y6 y }" M# }9 {! ]9 H! D; B float bonus,bon1,bon2,bon4,bon6,bon10;
Z$ E5 L" a9 b6 ]$ e0 B5 c- b v bon1=100000*0.1;9 ?9 B; s h; W7 {- R; x5 \
bon2=bon1+100000*0.075;
/ K3 m3 p+ {# Z8 o) e7 |( k; M bon4=bon2+200000*0.05;& Z) b% |6 f. }
bon6=bon4+200000*0.03;! F+ ~, g* P. [* g3 Q$ ^3 E! H% h
bon10=bon6+400000*0.015;2 q% E( x$ \+ w" A/ {7 v! ]1 {
scanf("%ld",&i);( O' z" U ]- |8 g B
if(i<=1e5)bonus=i*0.1;
5 \6 A. [$ N/ S% d# \ else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
' E8 t& p# c& w, G7 R5 k- G! x else if(i<=4e5)bonus=bon2+(i-200000)*0.05;2 t0 [' _) U1 g& V2 g' B
else if(i<=6e5)bonus=bon4+(i-400000)*0.03;
3 \% W! L% {9 ]' Z; |( u else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
2 f$ L) i4 f/ x Q else bonus=bon10+(i-1000000)*0.01;- E# M ?; b% n
printf("bonus=%10.2f",bonus);
0 ?' V+ b8 I! ^8 s& I) v- J7 t}
O8 r4 U4 n' t# ]( G( M0 N3 O用switch语句编程序
3 y4 l( h* w# D) h: Mmain()( j8 N) E* y0 U4 X
{long i;
! l |" i- ?. G$ l' y/ D float bonus,bon1,bon2,bon4,bon6,bon10;
4 I9 U3 n b2 E+ ^: I& ~0 d+ S int branch; T7 k& C A5 n
bon1=100000*0.1;- D* u5 b. L# ^% X7 D2 f
bon2=bon1+100000*0.075;4 Q( d3 _* q5 w+ B+ J4 u: K
bon4=bon2+200000*0.05;
7 `; `% k( _) }& @$ a' w1 G bon6=bon4+200000*0.03;
0 y" o+ T9 r6 W bon10=bon6+400000*0.015;, s0 s n# @* G1 l' T. P" X; {
scanf("%ld",&i);
" ^# N; q# ?( P2 k& R7 [, V branch=i/100000;
4 i/ N( A+ b1 w: e1 H if(branch>10)branch=10;
& J( E# d$ y% S: ] g4 c. y1 |$ k) K switch(branch)
7 { @& \2 n( v2 F+ T {case 0:bonus=i*0.1;break;
1 a8 q- `; q2 i' L! b case 1:bonus=bon1+(i-100000)*0.075;break;
; I* @9 k/ X( {% {" e case 2:
7 B, J3 W- S7 s- [4 U case 3:bonus=bon2+(i-200000)*0.05;break;
D: r1 m! p+ `* F- z! W& w case 4:
0 M. C x. U% Q1 P. e; s case 5:bonus=bon4+(i-400000)*0.03;break;
* T( u7 l. t1 C U: t case 6:% _% E6 H5 C. c4 A$ T% C1 N/ K
case 7# L. _- m7 j% c( C1 t
case 8:
0 Y& O; f$ X: ~& F' }1 f3 \ case 9:bonus=bon6+(i-600000)*0.015;break;/ T8 ?3 s4 d4 ]/ P4 O: f
case 10:bonus=bon10+(i-1000000)*0.01;+ P6 q1 U, F5 E5 M& |; y
}& L7 I# q o# ?- \6 H
printf("bonus=%10.2f",bonus);# G! V0 [5 k. j; V
}
: |. y2 F7 m, ~9 A- g0 L5.9 输入四个整数,按大小顺序输出.# H* ?& D" N% t( l/ B8 f R" I) k
main()" T9 `; Y9 n& f b5 C7 h
{int t,a,b,c,d;
, _* u2 x- v' @+ P1 _+ {4 { printf("请输入四个数:");
2 G; B: D+ m9 e' t scanf("%d,%d,%d,%d",&a,&b,&c,&d);
o- I; u/ B1 a9 O) H3 a printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);0 l, h0 S- l/ @
if(a>b)
/ E; R, N f( E- i+ s9 K! } ?: X3 H {t=a;a=b;b=t;}" K' _4 C: R) b2 \) ~+ |" P
if(a>c)
7 a @9 E6 {3 u1 I( m6 a1 ^0 `& ] {t=a;a=c;c=t;}" p8 B2 q& X0 l' T' ]2 M( y4 V
if(a>d)
! C+ e* I4 ]( n3 u& e {t=a;a=d;d=t;}
! W$ K; N0 W( n/ _4 m( F: S1 _ if(b>c)( S! Z! G& F2 Z0 h
{t=b;b=c;c=t;}
. G0 u2 U9 x0 d3 L6 { if(b>d)3 A! h' |3 q) q, ^' P6 k& ~
{t=b;b=d;d=t;}
0 D4 K2 @" B8 `: H+ l if(c>d)
2 \4 E; g7 R0 i: y* l9 F {t=c;c=d;d=t;}
& w) r, I2 P Eprintf("\n 排序结果如下: \n");2 P$ |: U% m9 I8 _" g
printf(" %d %d %d %d \n",a,b,c,d);
) f# g) M$ X% c# x}6 z9 N8 W& i N8 ]& X; _* u
5.10塔
s6 [3 K5 o3 w( J3 r# W( Imain()1 a' U. V! D. h2 P* y, E T
{
( h- r9 f+ t1 f; d! [int h=10;# B( `* q1 w& z9 \
float x,y,x0=2,y0=2,d1,d2,d3,d4;% W \0 U7 N5 X' F
printf("请输入一个点(x,y):");
! c$ }/ L3 X1 G; bscanf("%f,%f",&x,&y);; n0 {9 R$ K$ c4 ]
d1=(x-x0)*(x-x0)+(y-y0)(y-y0);
/ Q6 g; i1 h6 v- l" w: s! d0 s0 ed2=(x-x0)*(x-x0)+(y+y0)(y+y0);1 l" M& r# `$ ]% \" w7 S
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
; ~. u/ V2 n# k, nd4=(x+x0)*(x+x0)+(y+y0)*(y+y0);# T6 d2 a3 u& k' d" m1 F
if(d1>1 && d2>1 && d3>1 && d4>1)
. N" l8 U I/ O* Fh=0;: q' O V. ~2 E, ~4 _9 u$ S% Z
printf("该点高度为%d",h);3 b( G. S: ]8 P% U+ o
}3 G, T. T. P9 X l
第六章 循环语句
4 |7 [) _2 M" c4 W6.1输入两个正数,求最大公约数最小公倍数.
7 B& @3 c- f( z$ m/ f |: `main()" c: P8 k: x" v
{# P( c2 c) N1 X. Q1 o
int a,b,num1,num2,temp;
6 e$ n! e- O7 mprintf("请输入两个正整数:\n");# P) _; k: b( q2 G4 G' v# h& O
scanf("%d,%d",&num1,&num2);! x0 W; v+ Z' ?6 v: }
if(num1{
* [- E" E; l; atemp=num1; u, u$ N) \# b8 _( i& p( y
num1=num2;# t# D& P" R1 T, ?$ g. v
num2=temp;$ p4 K% Z7 S- W9 d9 S- N% f
}
2 J5 h8 a$ o) j qa=num1,b=num2;
% J$ d& [0 u/ f6 m$ O) K- Nwhile(b!=0)* r! \1 x# Y5 n3 N$ R
{
2 K* J, D# [4 v: M$ R9 F+ j temp=a%b;
% K$ ?0 p( N. s) ` a=b;
. R( Q' M) {: J/ T# w b=temp;
* @7 d$ m& S% P- c3 c' [5 G5 j, S }5 A* J/ E' m0 m
printf("它们的最大公约数为:%d\n",a);- v& O9 ]3 ^& B! d4 h( k& c: N4 K2 N
printf("它们的最小公倍数为:%d\n",num1*num2/2);4 f. @. y/ } _2 G% k! N
}; ^) y; g5 ^8 @/ g/ m9 m: p L7 v
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.
, r8 s; D9 C* m9 z1 ?解:
, [8 H6 E9 _8 K: S( {% `#include < >
) T4 j( C, u3 C2 [' Fmain()
) s* c* r8 z6 Y* Y) `7 I* p{
* w. h! h/ d; |1 jchar c;
8 b7 D3 r/ d- J! Bint letters=0,space=0,degit=0,other=0;/ M n+ @) i* |4 Z/ p( e+ {2 O3 ]
printf("请输入一行字符:\n");# [7 T# A! `: x8 ^3 p' v
scanf("%c",&c);
# h- z0 y9 ]* g2 j2 s% Pwhile((c=getchar())!='\n')
5 v% P; S7 b3 h5 }# h7 [9 }{$ ~% k! l C) F/ B
if(c>='a'&&c<='z'||c>'A'&&c<='Z')
; D$ @5 M0 i. Y u8 d% y0 t) e, Cletters++;
+ Y b" W2 H' v% qelse if(c==' ')
1 ~' T2 v6 u: U! Q& mspace++;
1 e5 q/ u4 H# b6 telse if(c>='0'&&c<='9')6 t" m% s+ x" G( ?% n# T
digit++;
6 V3 J; S7 b6 `! yelse
5 `6 ?& l7 t; T# ]other++;( [& {& K1 G4 ^+ \* t
}
3 W* i" i- v4 @6 a( Vprintf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
; h+ C6 f# Q. R- ud\n",letters,space,
! F6 | i" ?% pdigit,other);7 k- [$ [# S ~8 V
}! Z, C% c$ x# }& g
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.' w3 }0 P" h, d" F/ A% P
解:
_* f) |3 r( z# O/ L2 d: dmain()3 F6 P( I) T- i$ Z [
{
4 O7 `. f( F1 _; Y2 xint a,n,count=1,sn=0,tn=0;5 l( ]% V8 C4 \1 t" R7 G/ A6 K# T
printf("请输入a和n的值:\n");5 ]0 r+ {( Y' S* e
scanf("%d,%d",&a,&n);" Q8 G. M6 f5 a- ]6 d4 ]9 _/ u
printf("a=%d n=%d \n",a,n);
$ O7 D5 A3 W. s* f% Y1 kwhile(count<=n), `/ U- _( m; L0 z1 l8 j
{
. Z; z2 m( x; v. z# P' gtn=tn+a;. |8 d3 n% U S0 A. t+ @6 p
sn=sn+tn;
; h& t# N- m9 u6 Z& T. r- f+ F( Qa=a*10;4 [" X1 O) f e; y7 a- k
++count;
; Y2 L: t5 ^# [9 M. _; u}8 f& n1 w+ x* e& h1 i
printf("a+aa+aaa+…=%d\n",sn);7 [ c3 B# {2 H& p
}
- q# X7 M) ~6 c6 _7 Q' U6.4 求1+2!+3!+4!+…+20!.+ x5 H" y1 c! z$ f0 b$ x! M
main()+ B6 G$ e$ y4 j) e) A# I' y( T
{
# c' O1 e7 F3 B6 S6 o' q. q& V, vfloat n,s=0,t=1;
' m" q' G. v) F+ |for(n=1;n<=20;n++)/ Q3 @: C! J( P" ]8 |
{0 k/ O0 z6 E' g
t=t*n;
' X3 m7 Z: g8 C8 ws=s+t;
) o) H* M& {+ P- }- v}
& _7 a; ?3 r. z1 T" `' v. xprintf("1!+2!+…+20!=%e\n",s);
9 Q9 H0 ]$ ~$ b" F}
7 e4 S3 V& A' e, h6.5 main()
n0 Y" X3 u$ U+ D{) S8 b9 V+ I: L
int N1=100,N2=50,N3=10;
+ |5 R1 e0 z1 X+ e4 _5 ?2 q; g' cfloat k; R3 o( O$ ] x3 R
float s1=0,s2=0,s3=0;
; b8 p- @& f. g$ H3 i- Ofor(k=1;k<=N1;k++)6 j7 Q. f& o* ^. a
{1 g! z2 N* }% d; L- L6 s
s1=s1+k;
' j: w; c% ^, g1 Y; |}
, }* T0 D: V& d3 {for(k=1;k<=N2;k++)
: c7 {. [: x7 m+ b* Y9 A{
. n1 g7 N6 `9 Q: Js2=s2+k*k;2 g/ @6 p3 s7 J$ i# k9 i* P" a
}
" H$ Q7 u/ e' d" Z5 r% Lfor(k=1;k<=N3;k++)5 v9 T! k0 k$ J- v
{
( ^$ ~* P4 V5 s% Q1 [s3=s3+1/k;
% H4 m/ U- @8 p1 E}2 f: p; q7 b8 B" }- C( o
printf("总和=%8.2f\n",s1+s2+s3);
' K/ n5 P9 Y) I# X5 B}
7 _0 r& x8 i1 U; T, D* F' F0 X6.6水仙开花
# Z5 J t- d8 B8 Fmain()
2 ^, P1 P" i" H+ I& f O* E) @{
: h: ?1 r7 h& W) D% D ~ v; fint i,j,k,n;) W9 L0 W& W3 x2 ]+ w; D; @% ^
printf(" '水仙花'数是:");6 S5 V5 |% n3 s2 t
for(n=100;n<1000;n++)
& N' Y9 Y, l, Z, C{- [; F! \4 b% ~& ?% ^9 ^4 \1 P: G
i=n/100;
) r9 s5 X0 c: l* s& z6 J% Oj=n/10-i*10;
9 ^- z1 t6 u+ n1 S `) b5 r) K) Sk=n%10;
# j* i+ |9 M9 g: o n) r Q" Jif(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
3 i8 ~4 @ C2 j{+ a' u. E; _7 e. Y2 s) p
printf("%d",n);
/ h) d; w3 d) g! P}1 J- Z$ N& L, a( h1 O8 X
}& J& |, w2 x/ c9 c l2 J8 k
printf("\n");
& D, Q) }7 F7 ]& P C}& g% [5 I4 h* o: T
6.7完数& e: i) F0 N! a/ i2 ^$ R/ _
main()4 M ^! @& Q Q i6 f
#include M 10007 F2 u. e! @( j
main()
) h3 X: S# F) A* J+ f+ ]$ l' b{
$ V: r( l4 ?$ H! V, D9 Yint k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;% S! I# s3 ^9 c- L
int i,j,n,s;& ^; [) L5 f0 T& n8 P2 N+ W
for(j=2;j<=M;j++)1 c$ B6 F, k+ I, [, C
{4 z- M$ _; I0 V6 M. v
n=0;2 n6 H! T/ d3 [6 W" B3 h( Q
s=j;2 F7 J5 f i& M7 [9 w% s
for(i=1;i {
8 |2 v( f" s4 C hif((j%i)==0)
@# o; t% v& w/ p W { P- z: w& S5 x* ^+ U
if((j%i)==0)
; {+ B1 T1 n$ B. ]1 B) N( Q4 i+ x) ` {" o9 j6 w, S7 \
n++;
# u' Q( s; X$ E; T# e9 j- L% x s=s-i;$ w1 c5 @7 V C8 _$ `# x
switch(n)& ]8 V& C! ?' J* G* F
{8 f* a/ r0 s9 n. Q& X7 v1 d/ ^
case 1:8 l; v4 v. ^$ Q
k0=i;
7 L# U/ ?% N u y4 V break;$ Z { j* \4 ]8 F8 g# w* ], u! ?
case 2:
p t( b7 o6 w$ O k1=i;
3 F" ^- l2 Z7 x; M6 n break;
4 Z5 W( U; m$ {, O& e0 Y case 3:' k! v2 ?/ B$ f; C
k2=i;
- R& J" c6 c; z$ L- ~1 x: _9 Y break;
) u* {8 V- w6 J, {# a* t- h* H- ?4 C case 4:
( R% h* h0 P+ l ^ k3=i;
+ j- \3 p9 O$ {5 R, q' ^, T break;
" {0 n2 E* L1 k' a' H case 5:8 V$ q& {* X i5 ~5 g! I- F" h
k4=i;
/ W* q$ G3 T# z" N break;2 Q2 T8 z8 Q# ]; u$ t" U% ^3 P
case 6:
: y+ m1 [$ N8 |1 _9 d k5=i;
+ B0 }7 D& k; U. A, b- a break;) I! f$ y4 @9 {5 m2 F9 Q/ |3 Z9 l
case 7:
7 B, _ |& h [! q O' ` k6=i;( n' E* o# |2 D8 K: d ?
break;
3 H2 ~2 v" T1 t- ^( R* m case 8:
u [( ^' V. i/ M) d: _8 s k7=i;
& m1 @8 H) [/ U- @: \% [3 x5 c break;
I4 A; t* O5 O case 9:
8 T3 B; Y o$ J. |3 g) a k8=i;" Y) u" W; `4 D$ }. T8 i! ]# |$ L; D
break;
+ y+ F& m7 b+ P" q- K- t case 10:" o4 E5 N: e5 |- c7 y, B7 N% E
k9=i;
" z( |, o- U4 Y: ?! M% j% ^ break;
7 }# \& T0 |; K' M; f# k }( Z& I9 o1 C6 W8 V
}
+ g+ g* J9 [# j$ _+ h4 r- t- N# _ }" C1 d$ ^( i1 {* {( Z8 P( M
if(s==0)$ v5 y4 t! n o/ f6 I% B5 x$ U
{6 Z! r5 @: C" Z- r X0 S/ h
printf("%d是一个‘完数’,它的因子是",j);
, f8 z7 b! G* V- K. U( Pif(n>1)' ^' I8 b- o* Z6 M t
printf("%d,%d",k0,k1);5 h4 i/ g! f/ e
if(n>2)
5 `3 X6 N* M: l5 m, M printf(",%d",k2);
% y. H$ k& ~: G# B) ~7 `if(n>3)! x8 g2 x# {0 C4 ^, D4 |
printf(",%d",k3);8 N1 [. t7 N9 F! c9 @
if(n>4)
) e; s( _; O m+ ^! x printf(",%d",k4);8 D+ P$ i* K& J3 L
if(n>5)
. k1 `" L% a5 g printf(",%d",k5);) ^5 N- x: _3 a% D
if(n>6)% o* k6 n( @' |( A/ R: V' ^
printf(",%d",k6);
' c# c. m/ D, {" g) @0 s; mif(n>7)
$ V. [+ e) |# J! T1 B printf(",%d",k7);
1 ]4 Y* K/ j J; o9 \0 J1 jif(n>8)& p! d5 y) U6 V
printf(",%d",k8);
; p1 S' a; ]" {" f8 n% [if(n>9)
, t8 h& |' ]( E6 h( G printf(",%d",k9);
; o, I3 Y, y0 j- p1 kprintf("\n");
; l$ L6 X( ]0 r* h- m5 @2 [. R }
+ {' A& J# h0 f! {}
z7 ^" T4 I% J/ V6 D; p方法二:此题用数组方法更为简单.
! d7 B4 H9 x; R9 {3 Pmain(): F; A4 }8 R" i r, ]
{, ~; \( E! S& v/ }4 c
static int k[10];4 b7 ^2 ^' a7 \
int i,j,n,s;! O U. c; m9 G9 X/ W& N
for(j=2;j<=1000;j++)
6 h. c: K- W# r{4 ?; Z% [. G& m' p, L
n=-1;7 v1 U& l2 x3 Z5 h+ P
s=j;/ `# y) J! s; }
for(i=1;i{
( t# E) W0 u) `- ^! Y t3 {2 Dif((j%i)==0) k# Y# d! n! M" m- D2 a" v
{
& s1 N% g- m( G( g4 ]+ e* Dn++;
* b: O* B2 n8 O6 q/ u- Ms=s-i;
; d8 t$ R) B0 z# a$ K4 ck[n]=i;
8 H* n. C& s8 ?) y& Q \6 Q* P+ K }
8 c+ O) w* C9 q9 A* b' t' p }( g* K: I( S1 [8 s" \
if(s==0)
) U; N+ F* H5 f3 a. \; b* I{& _" ~: y6 b' ]) E, e9 {
printf("%d是一个完数,它的因子是:",j);2 t" K% f: z# F& Z
for(i=0;iprintf("%d,",k[i]);" N, {) U; M* \3 h
printf("%d\n",k[n]);1 {4 {; h% m9 f C% j# H
}
8 D6 s+ r- Z: t; d}
h7 n# z$ K3 D, t# \2 f6.8 有一个分数序列:2/1,3/2,5/3,8/5……求出这个数列的前20项之和.
! t5 {# S7 b% s1 a3 s. S解: main()3 O5 Z% ^1 t0 J7 [
{
8 G8 }! d5 h) Y( eint n,t,number=20;$ C- M0 }2 a- ~' F8 x& O
float a=2,b=1,s=0;
2 n# o; j/ C ^& R! f0 I$ X! sfor(n=1;n<=number;n++)
, n( @( G" @. P. ~9 w3 `4 w{- `/ x( T7 t& `% O; G6 q. y( m
s=s+a/b;
, E* `( T1 d3 T- E; m# qt=a,a=a+b,b=t;
\) q* ?' ^; z* r8 H3 o}, ]/ P: m7 O: w- F3 c( J( K% F& O
printf("总和=%9.6f\n",s);
" ^# e2 }' W. `}
- A7 x3 [$ O5 w! j! G) z: h6.9球反弹问题& w( S$ a: u: J' J4 z# R
main()! r% h! t1 ^5 i
{: z/ x: O1 b% P0 v) G! p
float sn=100.0,hn=sn/2;" V! \6 a& ? p" J3 c/ P
int n;- Y5 Z/ P& b+ |) ?% `9 i6 z9 u; D
for(n=2;n<=10;n++)
2 P e) P2 h; ^1 r$ {{
. P9 v% N! L, G# A: w0 }sn=sn+2*hn;4 S2 e& S* |: I5 E2 b& ]7 q
hn=hn/2;' n5 h% ?" M- l* C) E2 z
}, u! b" A% X$ I1 \
printf("第10次落地时共经过%f米 \n",sn);9 p5 [0 h6 {9 ?: [0 A% q' f' R
printf("第10次反弹%f米.\n",hn);
: x! m' ^! ~2 f}
* e. D8 K) D! g6.10猴子吃桃
1 O; C1 d& R- r/ v) @main()$ Y9 @& a* J% e' O4 Y
{2 Y4 |5 \; v' D$ R6 x
int day,x1,x2;* R/ n0 J. z) X
day=9;: r' Y. [0 n5 d+ Q' H( J/ Q9 J
x2=1;
. S" R& A6 P" C! zwhile(day>0)
q7 ~) L# S% e{
: l1 h, i2 T7 v& B8 W Z7 sx1=(x2+1)*2;( O8 A( R# Z! z( I" x9 _
x2=x1;
* _: A' s+ P0 a" w5 z, w8 n# Fday--;6 [4 B. O( h) I1 N* t
}
5 o/ g: K0 O+ C& \" b8 \printf("桃子总数=%d\n",x1);2 |3 v. _) O4 T+ x0 `; L, e
}
) o2 S/ `' u; B" h5 x; s W& S* J% S( a V. C
6.12# j h9 x: [% s$ U Z
#include"math.h"
! }3 t: ^1 v9 j) q8 O& cmain()
" M* r" G5 o/ I' H2 _{float x,x0,f,f1;6 n( ?( `: Y8 F) s+ C
x=1.5;1 C7 C; t# _; L B
do: r; q/ H5 ^) J& y1 d* F5 W
{x0=x;7 f$ O4 e- X7 C
f=((2*x0-4)*x0+3)*x0-6;
" Y- p6 x( o5 r& O f1=(6*x0-8)*x0+3;
6 l. R/ }0 W! Z4 O% n3 Y, q/ u) i x=x0-f/f1;
5 W- i9 U% Q- r5 e }
/ D' ]! @# v! D* i) c2 Z* y9 z while(fabs(x-x0)>=1e-5);3 D2 i8 q: S, V4 N3 K, r
printf("x=%6.2f\n",x);/ v8 Y* F! e% Y# _2 H
}
7 X1 h* D6 m( \! P
3 B% m4 I$ H4 a4 W6.13: Z% M; I3 J: ^$ N
#include"math.h"
X3 N" q% s7 K2 o; Vmain()
3 Y) Q `4 h& H7 E{float x0,x1,x2,fx0,fx1,fx2;+ L2 I) _: W. n6 P) F
do
+ ?! b# U" y* H9 t$ u3 c5 H8 x) ^. n {scanf("%f,%f",&x1,&x2);, c3 |/ Q" D1 P
fx1=x1*((2*x1-4)*x1+3)-6;/ \5 O) y( @6 ^2 y, B
fx2=x2*((2*x2-4)*x2+3)-6;
" n+ d) G$ y+ ]. |+ r7 u8 t, W2 s% x }* r9 Z* s/ H" _2 u4 _# R
while(fx1*fx2>0);: A: i8 K2 ~6 Q
do( w, Z; w9 r! m3 y) v
{x0=(x1+x2)/2;
2 M) V) v$ ~2 F% `1 J fx0=x0*((2*x0-4)*x0+3)-6;5 L. l) u* W+ ^" S
if((fx0*fx1)<0) P9 _7 B3 w8 b" Y! K% Y3 z
{x2=x0;& L/ x' H/ C/ M+ I, V
fx2=fx0;
1 M7 _6 U ` R/ p8 C9 h, c }
& y; m- D% e. r else$ V) @# {# D, G. L( i( U. Q( w
{x1=x0;* r8 P! W& |& G2 N; ]# X+ V
fx1=fx0;
6 X- u. @" I- u+ a! ?0 S9 p1 ] }0 C$ K9 \: ^4 F1 c/ b' p- B$ l* Z
}
% _) F5 s$ A4 @6 ~/ d: Z. A3 J' [ while(fabs(fx0)>=1e-5);
5 Y# \: i' s/ K1 F& Z printf("x0=%6.2f\n",x0);
+ p0 U3 Y% i1 S! H, k* m Z}
' H+ T, d; d# G- D6 x+ T! Q6.14打印图案/ o; d, |) @$ [: F1 m' A
main()" j1 u# o; z: |. E V5 i
{int i,j,k;
2 S% w# U6 J6 W* a! o/ }& S- H for(i=0;i<=3;i++)
( s; M) e8 t& F) g {for(j=0;j<=2-i;j++): ?/ ~( k9 s+ I+ }5 @! f2 | [8 p
printf(" ");
8 M# E4 k' F6 U1 `! v! @ for(k=0;k<=2*i;k++)
: ?! z7 N! v. L' |* O printf("*");
" L* a; V! b' T" r2 u% M5 Q printf("\n");! z# l Z0 _' d' k4 I
}
- V* ?) T; [1 g for(i=0;i<=2;i++)) m. T+ e, R9 \6 A' j0 T+ H& V& s4 k
{for(j=0;j<=i;j++)
6 w' z5 l' b# _! n) P$ q6 ] printf(" ");
6 W1 }; D4 t, \0 |# t, Y* ] for(k=0;k<=4-2*i;k++)9 H2 Y- `/ z5 S# B _. |
printf("*");
6 }5 R) k3 h7 M printf("\n");
( { S" r8 Q' \# B }: m) v1 ], ` W: z) C( F/ O5 A' w- S
}& ^. N0 ^+ O* C" [. \3 p
6.15乒乓比赛" p" ~$ r' F6 a+ A: u$ i" q& g
main()& U- v; }! y5 T1 @4 C. O, ?7 S
{0 e( t9 [9 R# y9 b' D, w
char i,j,k;9 Q8 B/ |& o2 _2 b X
for(i='x';i<='z';i++)( ~4 _, o2 a) m# w
for(j='x';j<='z';j++)2 r0 J4 k9 ], M$ O
{6 Y# w' n6 e0 S$ C. K
if(i!=j)
$ v# j# Y l( ]) ]# {8 d }: Tfor(k='x';k<='z';k++)% z# A2 v( V0 D. `6 I4 v4 K
{
( ]2 P4 s d3 K: k- vif(i!=k&&j!=k)
& L% f: S/ ]% e$ x/ L) p {if(i!='x' && k!='x' && k! ='z')
) v* ~" J' ], ^7 B! n: ]2 F. p H4 w. eprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);
9 q4 d) e" F8 X }5 d% @# K6 v& W4 \8 ~5 |
} h8 J& ]+ f" s6 r l, i6 Q
}
+ H. |5 \5 S* s, B3 z}
$ G# K$ }% B* `5 U: o( `% @; { |' ?C语言设计谭浩强第三版的课后习题答案' R, ]) x6 a/ x* w7 l! @
7.1用筛选法求100之内的素数." |# ]/ L( l( G' X) }8 F
#include" i6 R5 u2 I0 `
#define N 101' [. h" S( z) ~: g6 h- B
main()
; ] i( ~" `; C# j{int i,j,line,a[N];5 n2 @+ t: f9 U. |
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
! X. |- y* g: _- T+ Y, t2 J if(a[j]%a[i]==0)
$ D* z$ s( W% O5 V a[j]=0;
0 J) `& c# |5 w9 w0 q9 _0 ^1 U- ?printf("\n");' g) L5 b0 J }6 `% K( ^( R; G
for(i=2,line=0;i{ if(a[i]!=0)
% j/ k7 D o6 V/ u m) Q2 Z {printf("%5d",a[i]);4 ?* Q, j; _: E# {/ D5 L
line++;
3 b! F0 D1 {$ a3 a if(line==10)) g8 M ]. D: b0 o5 W$ P. i
{printf("\n");# {2 `) P* a( m3 J
line=0;}/ I% H7 k# l( R0 Y0 v5 i( }
}4 ^) Z0 x- Y+ j3 x
}
4 Y1 ?. z4 ^5 i- B# q, h7.2用选择法对10个数排序.
) Z; n" Z: \1 E#define N 10/ E3 M% h4 ~5 h. {
main()) E1 ?# U5 `. }0 R; B! z) g4 K% y
{ int i,j,min,temp,a[N];
" ~" L5 ?& A( R% mprintf("请输入十个数:\n");
8 ?& i+ B. J7 `: Lfor (i=0;i{ printf("a[%d]=",i);
: l8 Q/ }, }! s6 i# v scanf("%d",&a[i]);" g5 {4 }8 j# p; w+ u- M# V' M
}
" P3 ~3 S& @+ V% F7 a: k3 s0 zprintf("\n");! `* _1 K( g: W3 |7 b# s8 d
for(i=0;i printf("%5d",a[i]);9 w' y3 D# D4 L* X( y3 i
printf("\n");
* ^1 H5 K: }# x; Pfor (i=0;i{ min=i;1 k& J( h3 Q" C6 o/ O6 o, J, O2 T$ P
for(j=i+1;j if(a[min]>a[j]) min=j;
5 {* a# p$ Q( l temp=a[i];
' U) d! H0 k2 p/ a8 J a[i]=a[min];
k9 H: S) `' b# U" R. L' P% t4 q4 t a[min]=temp;
5 V; x9 p/ C7 j- u/ B6 W: _: ]}
5 l# e# o8 S1 U/ w+ A. G; @. vprintf("\n排序结果如下:\n");! ]/ f d; L. O1 `; J) }% S
for(i=0;iprintf("%5d",a[i]);
, p# ~- W; r6 l* X( D9 i) s w}% U. o6 `4 M2 H% I
7.3对角线和:
8 i/ \# K% I3 d8 |) Dmain()
; k' l+ d+ y0 M4 S6 A3 z# m{
" t) ?, P, t) |* G7 z8 ffloat a[3][3],sum=0;: ?, _/ o1 i/ R3 z' v
int i,j;1 C5 c3 R( D6 }5 I2 P
printf("请输入矩阵元素:\n");* [6 H8 q: d- @' p6 k
for(i=0;i<3;i++)/ V+ `* W' {8 t7 r6 S" @
for(j=0;j<3;j++)0 |2 L; y8 g, C, \" Z
scanf("%f",&a[i][j]);
) N9 M$ Z/ \5 M8 R7 L, i' b for(i=0;i<3;i++). x" D8 p7 o+ o- ~: p6 d
sum=sum+a[i][i];7 y5 \1 j* {$ T8 m" j- I9 o, A1 h$ R
printf("对角元素之和=6.2f",sum);. B3 f& ~, a$ g7 t
}4 ?# l0 V4 s% C3 t
7.4插入数据到数组
; ?6 q2 @+ T1 Y, D3 @9 _. {. ?+ y$ Hmain() ~- c8 D2 ` K4 A5 i7 m e9 o& O
{int a[11]={1,4,6,9,13,16,19,28,40,100};
0 t( j0 F' M; ]1 Zint temp1,temp2,number,end,i,j;
6 u& j3 j s' e/ ^3 ^+ o6 [7 Y6 Fprintf("初始数组如下:");
! N# h3 E/ B6 d2 ]/ Y9 @for (i=0;i<10;i++)
' k- {) R+ E( E6 @" |7 |1 Kprintf("%5d",a[i]);
, h7 D8 q- x) Y7 Zprintf("\n");5 e, e" ~- j! Q2 ^2 Q* Q
printf("输入插入数据:");8 u/ I: a# z% H7 e$ s8 \; n
scanf("%d",&number);
+ K# b8 m3 z/ n c! I+ gend=a[9];' M2 O& x7 l: b8 s1 _) M9 \
if(number>end)
3 y3 L8 ?; R3 G- |3 G* da[10]=number;2 I# m$ V& }% [. N6 l' N6 ?
else% r* b0 [2 O! H2 D
{for(i=0;i<10;i++)
6 L$ I# a( s! m: O# p { if(a[i]>number); b# u* w( N* X* I: `; q9 G
{temp1=a[i];2 Z1 x, C5 n4 Z1 p- g
a[i]=number;
9 j3 W$ W* Q) q- ^4 k for(j=i+1;j<11;j++)1 M& c0 t1 i9 k0 ]) e3 L; t2 K
{temp2=a[j];& z) U* ^6 x" M- y
a[j]=temp1;1 Z) ?8 d+ |; q" A* \
temp1=temp2;( q. \7 h, w F, w8 N' Q t
}
$ _' M( J+ j0 i! D: L) i3 v% Q break;. F5 D( D( `6 u9 N; {; e* f! x
}% ]/ \( i7 ]- g7 c; J0 R0 M
}
2 }9 J% {! R! S8 M0 s p }
4 L/ b9 J7 A' [ for(i=0;j<11;i++)) e8 k6 H. I% Q! [. q( b) S
printf("a%6d",a[i]);
" d) N( c( Q. k; N}
. }! |# D v9 E, I. b7.5将一个数组逆序存放。9 P4 O% Y1 u- y' p4 B0 Z
#define N 51 U9 b C) C: A: e, _4 z+ i, e
main()1 U' \: Q6 [$ n, @
{ int a[N]={8,6,5,4,1},i,temp;
6 z6 _4 r3 b; y. zprintf("\n 初始数组:\n");% T4 X" @+ g* ]1 }2 l: H' S( w
for(i=0;iprintf("%4d",a[i]);
) @7 w$ I; Z) d7 m% ]1 ffor(i=0;i{ temp=a[i];
# q. c9 k L, ? a[i]=a[N-i-1];$ q& f, J3 E- j" }0 {
a[N-i-1]=temp;
, B& k3 u; z0 @9 a( K9 d) Z/ Y}
# G# D1 O1 s1 W/ B# u5 m# Lprintf("\n 交换后的数组:\n");
3 c1 t3 ]( n& d! U5 A! d) t; Afor(i=0;i printf("%4d",a[i]);# L7 E0 @$ p. X6 x
}
1 G# o2 {5 `+ T- \+ ~7.6杨辉三角8 p" g6 S7 _: K* H! R
#define N 11
7 l3 T8 B- W; V+ |, }main()
* l# T# |) |5 T) L{ int i,j,a[N][N];
/ y" P5 } {1 p# r for(i=1;i {a[i][i]=1;
. i0 f$ b2 V% n a[i][1]=1;/ l5 M- c8 m% @
}6 Z* |8 {9 p( F' E
for(i=3;i for(j=2;j<=i-1;j++)
4 w# Z' ~- x# P6 y a[i][j]=a[i01][j-1]+a[i-1][j];
8 l O8 [8 P/ f# x* l9 t9 _ for(i=1;i { for(j=1;j<=i;j++)
* G2 I: e9 [0 ~4 d) I1 ] printf("%6d",a[i][j];
. X% X; k" e1 I% Z# R( K: m printf("\n");8 j$ r9 \0 u) q! \3 `% q \
}
& C& W4 N- O* a2 ^! |! p* y* \ printf("\n");
( u# s/ V) x2 P0 V, E! C& r}
6 \. x: F& f! @, Q( T7.8鞍点1 q2 \( n5 _- l' i( d
#define N 10! X! o- h0 ~, m5 l+ _+ f
#define M 101 a1 }% S) ]- s) T: c4 u
main()& l! ?" T: _$ O0 N
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;
! V: x( P0 f7 T$ q printf("\n输入行数n:");
) \/ B7 D, x5 {7 I* D# b" | scanf("%d",&n);
9 }3 t% D4 m6 e o printf("\n输入列数m:");
" P9 f; ]: }/ `7 Y8 N n* A+ k( }# z scanf("%d",&m);
* o' Q9 N5 H; K/ E$ d : e! U9 V# n9 x+ `* w8 O
for(i=0;i { printf("第%d行?\n",i);8 i( z3 k/ d& ?4 L3 G/ v
for(j=0;j scanf("%d",&a[i][j];
; N) M1 ?$ b6 ~ }
g. c6 N2 }/ j. c" K- V+ ] for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
' h+ j. M! i& {% m4 W* U pritf("\n");
* M# W, u) m1 ^4 h% w3 W4 T }$ g; Q5 r4 s' r9 C$ r+ U* n
flag2=0;
, ^9 w! i! e7 F% O. ^. J for(i=0;i { max=a[i][0];9 Y4 V7 F2 I9 A1 a a
for(j=0;j if(a[i][j]>max)) \9 q2 e7 r8 J! } u6 k' s( Z
{ max=a[i][j];
) Q; m5 f p' d! r6 r maxj=j;
' E. a% J9 z B7 v% { }7 N; w& K! w9 L
for (k=0,flag1=1;k if(max>a[k][max])
3 R/ ] s1 l' v( M9 w& z flag1=0;. A6 T$ V5 X' A' a f7 m
if(flag1)
0 {1 l. n! }, R* G! m4 O7 f { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);
! z' ]* i! [; X. v flag2=1;! l9 I- h8 r2 v+ k. {; u
}7 z! \2 D& @/ z4 z0 u4 U
}
8 v9 ~ O+ L: o! a2 f# q. u2 Kif(!flag2)
9 V% B4 f! G( y& R$ f7 O printf("\n 矩阵中无鞍点! \n");1 n5 n% q$ Q) K! {: G
}8 }8 ]2 g w! y2 D" o$ i- V
9 y0 B3 {3 H/ Q7 G" W8 K) {
7.9变量说明:top,bott:查找区间两端点的下标;loca:查找成功与否的开关变量.( h4 M0 N& q: q; v& m
#include! [8 {+ X4 u5 @* }- ~6 ~
#define N 15
8 f: L* w6 N( Qmain()
: _' M5 |: `7 M+ S2 c{ int i,j,number,top,bott,min,loca,a[N],flag;$ |% X$ ~0 G, |/ E
char c;
; P' Z! ?2 C0 y7 ~ printf("输入15个数(a[i]>[i-1])\n);
& j& u/ m: ?) W2 F4 i6 C scanf("%d",&a[0]);$ d& F7 ^5 j4 q& _4 L
i=1;
- b; ?/ h3 h g! {' s$ O8 I while(i { scanf("%d",&a[i]);
& l( O$ c5 X2 {& C if(a[i]>=a[i-1])
% M. D4 [6 B3 h, Y i++;
+ q( x, K) i8 }! @( J2 M) Q' O esle( n& V6 w$ B0 I9 `
{printf("请重输入a[i]");( e4 k8 N( e+ J1 o! `4 w) g- ~+ S
printf("必须大于%d\n",a[i-1]);
7 w- X6 c- k: @ }( t4 A3 d, J6 X! M& Y# {! \! t R
}" ]7 m& }8 i1 X% ?- Z
printf("\n");
# x+ g" |' G% b9 i% A for(i=0;i printf("%4d",a[i]);, x0 y- h/ @4 f/ o$ z1 Y9 G
printf("\n");
. j0 p$ G& a: w9 m: V
6 C& ^: z; W7 u& _# a5 H/ i flag=1;
% k+ g# E; s! X# P7 r1 w while(flag)
, U7 }' v2 ]( w/ F. q( N$ I- X {1 }% a" B H% X/ R9 B/ R+ o2 Q5 }
printf("请输入查找数据:");: R4 K" F1 E" M* g
scanf("%d",&number);
; [3 u$ b. |1 H+ d# e% C, U loca=0;
& d7 r K& r% ~& E2 Y* G top=0;
" [" B; J r2 t3 x7 P bott=N-1;
$ _; }5 A, C; J, i( v if((numbera[N-1]))
5 X) j6 f1 x+ Z1 m% Q: J loca=-1;
: V# _# ?' y* \ while((loca==0)&&(top<=bott))0 C$ p5 h: a# h: k& M6 m' s$ a
{ min=(bott+top)/2;
6 V. D0 j9 N( c, [ if(number==a[min])- i9 j" ]/ B: u
{ loca=min;. L& _/ X& r/ z# {+ n8 U
printf("%d位于表中第%d个数\n",number,loca+1);
$ t: p8 k8 W. Q- f n }
; G `: ?0 y$ x5 k: g6 M else if(number bott=min-1;
* Z* O9 @1 X! @ else
0 X. M- [. I6 g$ R5 o" A4 K6 [* Z' j top=min+1;4 E6 P% I" R9 D: Q
}
# J; E3 v/ @5 a" O' d9 f if(loca==0||loca==-1)
c- K) h& V7 ^$ V, i/ V3 m printf("%d不在表中\n",number);
, ^5 Z( A1 B0 E, ^ printf("是否继续查找?Y/N!\n");$ p: J* `5 E" k* `2 v7 @
c=getchar();1 C; \ V: T4 A2 X3 \
if(c=='N'||c=='n')
& B0 h* j( [# q% t. ~$ { flag=0;9 D5 {$ s% v2 q
}
8 L" K, _. v$ V9 X9 D& }}
8 f/ m( u4 s4 `3 U( G
j. h8 G/ m6 A5 i$ w0 ^ B$ z6 T( ~7.10
0 w( l; i8 ^% D& m$ Vmain()$ L& v. W8 O2 |% W
{ int i,j,uppn,lown,dign,span,othn;
3 [% n% x. X, w* m char text[3][80];
d3 d. N3 v3 r7 G* f5 u" V9 u) N uppn=lown=dign=span=othn=0;
5 d/ z6 C$ _' N1 H; a5 f- C for(i=0;i<3;i++)+ Q( T/ G: M. w q+ ^
{ printf("\n请输入第%d行:\n",i);
& D& i) I5 x( x7 s# B/ W+ e" q gets(text[i]);& A# T# F- t) X. \* w. M7 {
for(j=0;j<80 && text[i][j]!='\0';j++)" p. G% p/ Z+ Z8 ^/ p7 l
{if(text[i][j]>='A' && text[i][j]<='Z')
7 F3 |" t- B. P4 v3 n# l uppn+=1;
; {% \( |2 x: B% a3 N b9 ? else if(text[i][j]>='a' && text[i][j]<='z'); h5 j* M; {' K& ?: _
lown+=1;6 g* L, d4 m8 v/ V: L; E
else if(text[i][j]>='1' && text[i][j]<='9')6 Z! L0 S& ]4 U; j: ^# r4 ]
dign+=1;8 H" A: |, k3 A2 [+ ^3 B
else if(text[i][j]=' ')+ _8 L' {, L8 L! q: l% w$ a3 x
span+=1;1 o" I4 x; ~, B* w6 l
else
6 Z: @/ `! ]) z& l$ _7 f% ^. X othn+=1;
9 E* `- e2 F2 P* Z }: y b2 T6 q% l7 E/ c
}' R9 f) P7 V3 O4 x- ^7 F7 \' {. i
for(i=0;i<3;i++)/ n! @4 o1 @1 }
printf("%s=n",text[i]);
W- |* {. {& o3 O* D( b printf("大写字母数:%d\n",uppn);
`6 d0 r! I0 g5 X7 [% _; K; z) I | printf("小写字母数:%d\n",lown);, g! t3 B; M% A
printf("数字个数:%d\n",dign);
! S$ P+ G% I; }5 j; a6 u0 a printf("空格个数:%d\n",span);
0 }4 e0 J' p; q8 b. z \2 [! e6 L printf("其它字符:%d\n",othn);
: s" A5 `; o3 H3 F2 r2 ]+ T}
% ?' Z- v2 N2 C# K& W& x: f3 X% s# E2 ^# J/ M
/ U4 }$ N: g) h* R8 J
7.11
% Q! _ r9 Z7 _2 {9 q, Nmain()
# m: \6 O$ R2 Q' H {static char a[5]={'*','*','*','*','*'};
* t# s7 m! S4 s: ]/ x# J8 l9 K3 [ int i,j,k;) W: `) g- i3 J0 H# c& ~: ^
char space=' ';! v+ |' ~+ H0 V8 @8 _
for(i=0;i<=5;i++)! V' J- X" n! c1 F( S; }4 G) `- z3 X
{printf("\n");3 G. o/ v/ K8 H V
for(j=1;j<=3*i;j++)$ ]; f, f' }& y
printf("%lc",space);3 T. I# K, s. j3 C
for(k=0;k<=5;k++)
4 B% n9 N- M5 @& ]* o printf("%3c",a[k];
, L* C' l( D3 h7 o6 y& |4 q }, r! Z H: a0 o6 W, B- t6 d% K/ }
}% E. R+ D5 k+ l: n! w' X
7.12" \3 g1 P+ X6 V
#include2 ^* E, I! W& I" J% V6 ]. ~) k
main()( A6 X; H# m4 F* g
{int i,n;
) [: _- P6 N6 }! Q9 ^. g! H char ch[80],tran[80];& b' a" j2 D/ W( K# Z- k
printf("请输入字符:");; e9 t/ H( Z% b
gets(ch);$ m; C* o- \4 D- o7 }" B7 }
printf("\n密码是%c",ch);
# g. P6 {+ X3 d3 B1 ti=0;
7 z; v3 J" W' V. m& h7 e( L* @while(ch[i]!='\0')- g2 B W. n, H M9 H
{if((ch[i]>='A')&&(ch[i]<='Z'))
7 U( R, I' p9 q( k tran[i]=26+64-ch[i]+1+64;
) V' T3 X" i0 A& w! [ telse if((ch[i]>='a')&&(ch[i]<='z'))) `3 a0 ?' G8 f8 T4 I5 D
tran[i]=26+96-ch[i]+1+96;
. A; o; c4 [# aelse
9 U8 I( D' R3 I9 M( Z) v tran[i]=ch[i];
/ V6 z# X6 I) h i++;- l1 t. J" q9 d! t
}% r, i. F4 u# v, B v( o9 @7 C* a
n=i;
$ m* O+ J" ~9 I: _printf("\n原文是:");# i/ `+ ?- R9 ]) N
for(i=0;iputchar(tran[i]);
5 F x& B D. K0 ^" u}
& N7 \& I7 Y# p& c% e0 a( F* _7.13# x% p; }, m" @
main()8 R6 m/ p$ k+ J+ n" P: C7 p/ I
{
- m- Q. d1 n% h+ Z; K char s1[80],s2[40];
8 ]( J% P$ ~% L int i=0,j=0;! O) D. m w8 L, o, \
printf("\n请输入字符串1:");! F% H, s) H( n; z
scanf("%s",s1);5 G' W( }, `/ p; d
printf("\n请输入字符串2:");; C0 |7 d S9 ]0 R
scanf("%s",s2);
5 b( [" ~4 |4 b1 \ while(s1[i]!='\0')
8 e+ X) H: T4 m( W i++;
- S* b9 R# c( X5 h q1 \$ Lwhile(s2[j]!='\0')
. J; R6 E; K1 c s1[i++]=s2[j++];
7 w2 z0 m( u4 ]9 T$ T% Hs1[i]='\0';
" j1 g0 X; k: }* N6 sprintf("\n连接后字符串为:%s",s1);9 \* R: B; K/ a2 A( p4 n2 R b' r
}9 E7 I4 R5 W4 P
- }3 _' n9 E( B, ^5 [# o
3 z G- u+ p) P$ u# N7 |
7.14
; a& U7 w3 _% d6 l0 \* g |( S#include
% n, F! ?- k9 z/ Umain()
/ r ^ ]& j+ Q9 V2 Z$ M# N& {/ c{int i,resu;4 Q& A4 w& K {* E4 a/ q8 y0 l& `
char s1[100],s2[100];% c. P; g3 L! F
printf("请输入字符串1:\n");2 x; g$ W2 M4 k8 L- u
gets(s1);
8 R( P( U) L% H9 M8 j printf("\n 请输入字符串2:\n");; n, _' f5 t2 w4 y3 ^4 D' `( Q; g/ X
gets(s2);3 I8 ~1 P5 t g8 g
i=0;6 l7 u1 `8 s& l
while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;4 f. T- ]. A% j5 ^4 p p$ V
if(s1[i]=='\0' && s2[i]=='\0')resu=0;, g' V7 U: e. _: m* H" w L
else3 W" R+ r4 D: [- ^" G% _' v$ P
resu=s1[i]-s2[i];( a" J) w5 ?1 ^% e9 O- q
printf(" %s与%s比较结果是%d",s1,s2,resu);" {" T* ]' n$ D4 d
}. n4 F ?0 {' A& }& ]3 B+ D
7.15
9 W8 v9 b+ P0 r4 [2 ?: ?& M9 [$ \#include
8 x2 j) B ]9 G0 ], a3 zmain(), T. T# v2 F. Z8 C& [$ O$ f6 i
{
3 c! A4 _4 j8 I5 m7 P char from[80],to[80];4 h, ~& \: r, n, d/ h ]" ^
int i;
2 P0 D$ R2 h8 u" s/ J7 { x printf("请输入字符串");
6 q, C0 V; l- }4 {) U: ~8 R scanf("%s",from);
, b8 R: t* ?' V+ H* E/ z for(i=0;i<=strlen(from);i++); h0 m# I1 \ Y+ J2 J" W. T" a
to[i]=from[i];: Q5 Q& ?) j; h" B* b7 a6 ~
printf("复制字符串为:%s\n",to);
1 M+ C9 ]' \0 M" z7 X4 T3 O; @ }: ]7 Y1 i% K' H5 @ E
- y: n) t) k/ i' Q9 I
) f" k$ T6 H6 o1 \
第八章 函数3 _! F# ?# m5 W- L) R
8.1(最小公倍数=u*v/最大公约数.)* c+ W, q3 P6 ] a% _2 y+ D
hcf(u,v)
* E! x% c7 {0 U* g! xint u,v;
, R: e0 r9 c+ k9 B' L9 n(int a,b,t,r;$ e# ?8 ^' E8 V+ J3 W
if(u>v)0 Z3 u& H2 |1 Y) E W; ~) q5 ?
{t=u;u=v;v=t;}, g1 c' W; Z0 k8 i- I/ q
a=u;b=v;5 c2 ]/ Q7 N* l/ a8 O3 c
while((r=b%a)!=0)6 t S- J) t+ s3 c0 e9 \1 b- w
{b=a;a=r;}# \# E% s# V( {
return(a);0 o" J; c2 r* |" m; @6 W& {6 j; X
}, s. i4 _9 e/ H; F2 T- D) X
lcd(u,v,h)5 Q7 [8 g( b a7 g' i8 H ]
int u,v,h;, F% E! `; J+ y: [( I$ M, B
{int u,v,h,l;
0 v/ B# x j5 ^1 O+ w1 i U! l scanf("%d,%d",&u,&v);
' y+ m9 G: z$ }9 w" E# \" I5 i h=hcf(u,v);
9 V9 i5 Z. H, D; `- d printf("H.C.F=%d\n",h);" }- j# B# _9 @7 _* Y
l=lcd(u,v,h);
, U/ |+ ?- M3 _) C printf("L.C.d=%d\n",l);: A3 l: D; K. N: j
}
; G$ k- e+ n& z8 n; N, S3 x {return(u*v/h);}
' o3 t$ \9 ]: n& L7 C b main(). a |. W+ A; C; O M# t: j5 l* i
{int u,v,h,l;" n* [& {" G1 h. X; N) k
scanf("%d,%d",&u,&v);
( K; q3 g* P. b. t2 A ] h=hcf(u,v);
! x" _7 y# u1 g printf("H.C.F=%d\n",h);
6 m6 A. g8 @6 t: Q! j l=lcd(u,v,h);
! Z" [* o- e; ] printf("L.C.D=%d\n",l);9 `: A; F$ q0 p5 Z2 }
}
" b$ X! X+ z9 z0 E9 k' a
9 @( N4 |- k0 w. D/ }; Z
5 }1 M1 V( m8 n1 B/ D2 { ~4 T# @, O; ?" Q2 a+ J/ F
8.2求方程根& o" ?$ k+ R# C* ^$ ]
#include2 n+ m: |) \+ I4 j8 y( A* r; ]5 D ^
float x1,x2,disc,p,q;
5 ?9 E1 q( e5 t" o Xgreater_than_zero(a,b)9 |0 E0 r+ [. }5 _3 K
float a,b;
+ t; i; n* w# f, ]+ x1 A{
8 B5 o! Z, |; rx1=(-b+sqrt(disc))/(2*a);* M! h& O5 `7 q/ a. f1 z2 w. c( C
x2=(-b-sqrt(disc))/(2*a);8 ]7 j5 X6 W S, i0 ^/ z
}% ?4 ?3 L4 J5 t, {% r
equal_to_zero(a,b)- M5 v7 o: @5 D8 |
float a,b;( U3 s2 O/ D: n
{x1=x2=(-b)/(2*a);}0 }6 \ o. s' N/ R$ C
smaller_than_zero(a,b)
1 D1 `1 j$ Y6 l3 n! Ffloat a,b;
( l8 J+ s0 L) H) i* k{p=-b/(2*a);3 D3 b) _. ]& z0 O$ J3 u W
q=sqrt(disc)/(2*a);
. [. Q2 H+ T9 E) v9 Y}! i9 O$ `. v s' A5 |4 D4 {6 A6 O5 B
main()0 p \( F$ [4 O' s* J
{
$ f/ E- V% {( q8 k0 r6 N6 K, W6 wfloat a,b,c;
/ ~% F$ l( Q' O S, @printf("\n输入方程的系数a,b,c:\n");
3 Y# |( p8 |. x) _& P2 Pscanf("%f,%f,%f",&a,&b,&c);
[/ X3 ]+ U+ z6 [$ Iprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);# [2 m$ T4 q; _' k
disc=b*b-4*a*c;. T. b/ m- D. w* W% N9 K
printf("方程的解是:\n");
7 O) a+ h) _& {, Q$ vif(disc>0). O& J X( _+ c3 _+ Y3 v( `
{great_than_zero(a,b);
* V. j7 f& o5 Q3 dprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
2 k' p. [0 y2 `& s; O+ b; l$ q}
8 C% a/ k* ]. R3 jelse if(disc==0)2 K, d( N, l$ f9 N; V* `0 }! J0 x9 {
{
* _: _- C* {' F* W: ezero(a,b);
# h) S' L2 C* R9 N" w6 G: P/ g( cprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
1 k$ H! k& F9 ^" }, [9 s }2 g! ]* ^' q/ c4 g3 ` A- f
else8 D. B5 @/ K9 ]# e5 h7 M+ Q
{
f9 [5 K9 H* q$ y small_than_zero(a,b,c);, _+ Z2 k. ]' c: L" v& A
printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);$ s% U4 C, h+ Y# u* X
}
# _- Q. l6 G& J1 H}# }. E$ o D9 z; c: f
8.3素数* _, w R5 V' Z: c+ p/ A! [
#include"math.h"5 c/ P9 [& x# B
main()
4 G* ]2 Y" S" U1 S3 Y" ^{int number;
^; F8 e( R( d* a: { scanf("%d",&number);' x, i' f8 [3 k6 l3 B) E* S
if(prime(number))5 @: l' V8 O( e; v- t; @
printf("yes");
; M1 Q4 ~$ `$ ^/ M/ F; O, N else1 B5 o. R. Q/ T8 j" f. {2 h/ w
printf("no");0 y! t: |) p+ M. t* ]
}# V4 p4 k3 c5 [# v+ e
int prime(number)2 \: _3 s' R+ h3 o0 ^$ o/ h4 O
int number;4 w8 o+ I) B' g! G* y
{int flag=1,n;
- m) H# v: h5 d- p" B0 N# V1 v for(n=2;n if(number%n==0)# B5 _5 N4 j0 O. z& N) z- j0 u
flag=0;) D! X" X* d* X
return(flag);
/ ]( r8 ?0 g Z* w( z) P$ v8 ~6 R}
% A' ~, I& R0 K' p; x* s
& \$ ?$ r4 r# p- b- p. ~
: v [6 W: O" Q# _" Q8 b3 [6 N, _; P' W
8.4
/ f, \! _ o/ T, W0 i/ q8 U#define N 30 c" P# z3 ^2 e8 ^
int array[N][N];: @5 Y, G7 D1 P1 ^2 S' j
convert(array)4 _1 G2 d! l5 ]3 V5 ]: W4 F
int array[3][3];
% N8 Q6 f5 A) v( T { int i,j,t;- B: q, \% Z- [
for(i=0;i for(j=i+1;j { t=array[i][j];, E& T) U6 @% E5 G- w3 f
array[i][j]=array[j][i];
8 m1 ]) D, P: A; Q0 J array[j][i]=t;6 r4 k8 h6 N* u; e; l% I( ^$ J
}" W; \( w& ^: J, k
}
! e( n# \2 c. ]% emain()
) k' X. A& i! I" S' a5 l5 S{
8 Y: `3 K: \; f* y* ?% ? int i,j;! Q: M% w z7 k& m4 E/ v
printf("输入数组元素:\n");. Z' ]$ K8 p% }; ?3 U _3 O
for(i=0;i for(j=0;j scanf("%d",&array[i][j];
" ~& l6 Y' Z! g, t1 \% \ printf("\n数组是:\n");! x$ B" ?5 ^( O+ {8 ~/ W6 Y. P+ p
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
) v+ S6 V1 A) ^+ ?* M1 x, l printf("\n");$ g, B* p4 q+ N* {% K# E" }1 [* v! m
}) P; V# Q& l6 T# Y
convert(array);$ W- g5 i2 a b, T! c6 k1 h+ r
printf("转置数组是:\n");
; a% ]& L8 o' u# Y8 _ for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
. Z2 b* n2 @1 x! C, s printf("\n");$ u6 E1 o' ~6 o: D; |
}
. Q# g1 G7 w& |8 Q}# c# @) _: E# S1 `4 d
, ]6 ?* B0 C) a/ c- Z& E! O
/ q o/ u% n4 } C" O T8 c3 y
, o- V, H. C& q0 C8 ~4 l* u8.5
) I2 N4 a/ z* f$ r; t# K8 b" lmain() d7 E/ D0 I9 a) P0 P& g) ?! w
{
: U" `8 X$ T( v1 Achar str[100];& ]: l( A7 k- n% j" W& s) N
printf("输入字符串:\n");/ H( H) X8 q8 X+ I( E! c/ o
scanf("%s",str);" G- }4 c. r( C9 t6 n _
inverse(str); m" X9 d$ @) o4 c; t
printf("转换后的字符串是: %s\n",str);
. }- X2 L( I- M* D3 q8 T1 Y. l. P}: j& I7 u/ w1 ]- H+ w7 S
inverse(str)
h# I$ Q+ @" J* \char str[];- _4 t4 `' h% X( r' F! B7 Z4 k# G
{
/ i- U4 P4 O# e char t;
3 {6 G e1 p2 T# v, Q( C4 I) n0 U! [ int i,j;
( d+ n, M8 H0 c g6 `$ K) ~ for(i=0,j=strlen(str);i {
$ ~. }1 z3 |$ q* B t=str[i];
2 m4 M0 h9 c3 ?0 I+ a str[i]=str[i-1];# |& w- p( M$ t! @
str[i-1]=t;
0 [/ \7 @( N( n, E/ }9 l4 R; d# v }2 f' n. H5 J, w: |+ Z% G0 e
}
8 O$ f1 H& Y6 i8 a! @
7 ]+ _4 q. y: {$ Y5 A* L
' d+ H! A1 L5 ?+ |( @9 T, R% T7 t/ k* X; B: f; Z
8.60 P6 d4 l' i) R
char concatenate(string1,string2,string);6 m' U: ^' B9 f0 f* q
char string1[],string2[],string[];# v/ @" O) N* e9 V
{$ ?" ?: s+ i t; |1 {
int i,j;1 p4 Z. j# s2 ^; }& r$ H+ @
for(i=0;string1[i]!='\0';i++)
; W; \' k* z, z8 u5 ?6 U string[i]=string1[i];7 p: N, N* G- ]) j' Y
for(j=0;string2[j]!='\0';j++)" t8 T% k# _8 Y* V
string[i+j]=string2[j];
" k5 k1 L' N" ~4 W. } string[i+j]='\0';
7 }7 l0 H1 q+ I7 }, b: o# W}
, s, U: c& e7 y9 T/ U) Gmain()
3 @! [: I( y! w* l{0 G4 ~8 w, N+ l4 `- e$ j. @9 y
char s1[100],s2[100],s[100];
- w! T1 ^5 U: u# d. y, D printf("\n输入字符串1:\n");6 h) \. n( ~( d
scanf("%s",s1);' r, u3 q' F7 e* y# Q9 q
printf("输入字符串2:\n");( H/ Z$ O2 x7 ]1 `9 b' Q
scanf("%s",s2);1 x; \3 V7 r/ p
concatenate(s1,s2,s);- P5 p5 ?& d x1 r
printf("连接后的字符串:%s\n",s);
: [9 D, m( @! \}
* x( A$ T1 o: v! Z$ D5 I! o, E$ Z" H6 \: z
C- y8 b' r6 |( S$ A
8.8- c/ M1 z$ u6 H. n6 }$ U
main()! M) t% e+ b- F
{% N8 s! y' r! k+ w
char str[80];
3 D3 e; k: f; t1 p printf("请输入含有四个数字的字符串:\n");
/ F/ p _7 U7 U4 C5 ]$ d: s( Z scanf("%s",str);
, L$ D+ r3 Q' j+ G, G0 H insert(str);0 g! E' S1 y! [' P6 |, B
}+ T1 r+ ]7 w' M
insert(str)5 j% L! D" q- h' W; `
char str[];1 j& d* P+ Y e t8 N, w* [' \
{; l4 v5 T" ], I" I# H8 e/ b. P3 C
int i;: c. D* c( V+ S. A$ X& Y
for(i=strlen(str);i>0;i--)
; w% f4 r) ]/ j" S2 Q- C: |! w { str[2*i]=str[i];( F$ f2 Q0 p' [3 E+ P
str[2*i-1]=' ';
/ L# r+ B) X5 ~3 x }
8 h: Y* j$ \7 x, K# c) p/ l+ ^* } printf("\n 结果是:\n %s",str);# ?3 E% e9 u0 }3 z& |7 |0 I/ \
}
+ V, }$ e0 V- v# x7 U, h8 E: P. ]$ n+ f: Q% s8 b, m. V
5 y. p8 k; F- x3 L' @% @- i7 @+ y3 d% N# V- x
8.9% A! C8 s/ k& o" [ h% U
#include"math.h"& f) C6 k, T: ]4 ]
int alph,digit,space,others;
' w. L3 K P2 B& O6 [main()/ V1 P: K1 r1 I0 K8 m
{char text[80];3 q7 `7 M( e* H0 j, i
gets(text);
& z% ^/ Q$ f+ B7 f2 n: U2 T alph=0,digit=0,space=0,others=0;- w9 `( Y% f& x: h! K
count(text);
& s6 E) ?) i9 g `+ m$ ? printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);1 Z$ k' ?( }. x- R/ h5 A. h6 u c& g
}
7 r; E5 c d! ycount(str)
& i( M* k! [1 A. ichar str[];) j- m1 D0 I+ c5 v
{int i;5 {( E; V5 R7 k8 W& ?% j! y6 |/ j- j
for(i=0;str[i]!='\0';i++)7 Q7 d ]! `+ R, n* v6 y! E
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))# f3 t8 h* {6 W6 p5 G G
alph++;6 @7 A, i) J5 c4 n
else if(str[i]>='0'&&str[i]<='9') b* b+ f; l) E' S: A* l- _& a8 l
digit++;4 s: Z! T5 f8 a* O; f/ ]- k3 F
else if(strcmp(str[i],' ')==0)- a7 ]- D7 u% Z5 n# C3 T3 q
space++; H3 O8 v; L4 \( G! Z7 B1 y( P
else
2 {3 }- r0 _( h6 f8 m( _& |) I others++;4 V8 B) |/ j/ ]7 X/ t9 Z7 h8 q2 t
}9 C. i. H: l# ]/ e1 W, c
2 t! b* S$ E3 E- ` ^ c
: C, b/ \' c/ ]. E; M8.10
5 [4 n5 g" I0 ?int alphabetic(c);* H: x5 {. Z/ |) z( E
char c;
1 @6 {( r* v9 K! Y6 a$ r6 H5 C Q( e) j{3 x" d6 y+ F, j$ p8 L% ^9 T* N
if((c>='a' && c<='z'||(c>='A' && c<='Z'))
8 H0 {' x7 |; _+ c4 b$ I; {6 d return(1);! k' Y [% W+ S# M0 S0 \
else
0 W8 |5 k- K6 L3 x2 h, e return(0);. `0 ^/ C4 z/ x6 s+ J$ R4 L7 Z7 Y
}
. t: C) K1 b8 K. n* U R6 m! u4 g+ P' k
int longest (string)
$ e' X, l4 \, X+ zchar string[];
+ E0 i( _; F! W' d{: `' X- [4 Q3 Z
int len=0,i,length=0,flag=1,place,point;3 a+ D5 s+ ]' V0 i9 G1 t& b# _5 f. a
for(i=0;i<=strlen(string);i++)% {+ I6 v" l9 H% R9 f$ L
if(alphabctic(string[i]))
- ]! f5 C& b2 ]7 S/ n if(flag)
, l1 R) a: [9 x7 ~% ]% Y) h {
9 @. e' _ w* k4 p; O) o$ J% V point=i;
9 t. e& ^! _. _ flag=0;
5 N7 J7 i( b- ?" k Q! J }
) l% L& ?* h+ \3 `5 a9 k$ | else
0 r2 U% m" k, a/ f len++;6 }4 n L; O- _7 w: ~) d
else
3 }9 T5 [! \1 D/ e/ O2 O { flag=1; T# t3 E4 |' h" _, z
if len>length)
9 ?: i7 V$ P; R' M+ Q {length=len;9 D; i8 Y7 ~! X7 c3 X0 o
place=point;
% t5 k/ A. j; r0 z2 J) X len=0;( a0 Q+ }) |7 q7 m% _/ u
}- B' J% |& ~! t, P" r0 \
}- m% j3 d2 ^9 y) ~( b0 n
return(place);
9 q' t- @' j# P/ u4 A; v, X } ?. |$ @ L P% X6 N% j
main()
4 v" M+ ~9 B1 W+ t0 ^{
0 N8 f- n. u4 r$ s# T; Oint i;
( [% F/ P2 n% x1 G0 T1 x: Jchar line[100];0 k& \/ l2 {! P
printf("输入一行文本\n");4 `1 P. j; |+ C* ?: C) `
gets(line);
) F1 S d5 w+ A1 d4 n# n) |printf("\n最长的单词是:");2 S, W1 a; n. a' n1 J% }/ v) k
for(i=longest(line);alphabctic(line[i]);i++)
* ]9 F6 H) n, l printf("%c",line[i];6 ?- ~% ]8 q* i) L8 b0 w
printf("\n");
& r' R, @: k: o/ b. s}
! `, I4 O4 `# v) W& v' ]% ^0 I9 ]) I- C4 M: f; o3 s0 }
! ~: T# e& Q0 Q! X' V* q+ n
: h5 r% z- D+ ]! M
8.118 R* k& D9 L5 |& v6 d) V
#include( I5 A! R- I$ @' j
+ Y! Z0 ^& ^# N! r
#define N 10
" o5 i" J. I1 ~& f) O0 Q0 r4 Ychar str[N];
4 Q% [. }0 r% {( J! S) U' Ymain()
% O3 G6 ?, z; }+ L{
# J0 X- C) |& T* {3 z R: \ @int i,flag;+ y2 }) h+ s$ a5 V0 Q5 W0 W
for(flag=1;flag==1;)' L8 o& ^2 U1 m/ y7 G, d
{
0 C2 W0 _, s" P* M; K" E9 [ printf("\n输入字符串,长度为10:\n");
! K, V" I) f1 `& j/ F6 k8 @% w scanf("%s",&str);
) K8 U* d+ R- ?- c# J8 f if(strlen(str)>N)5 P% Z( [5 j0 n2 J: Q( g' Z
printf("超过长度,请重输!");
" Y8 u1 |3 [& z! U8 H else" a, A( y; `! u0 H! _
flag=0;
) c, ^' Q5 \& W, L! T% m}
$ z9 t b8 r( P% Qsort(str);* c; e9 ]# l) H0 Y0 N1 r/ T
printf("\n 排序结果:");& J* F! k) x: _& f- k
for(i=0;i printf("%c",str[i]);9 V0 z7 @, ^# V) U* e$ |: R
}1 y6 V! z8 F2 d. [ ]% S& w
sort(str)
" i' ~$ C' V' I, i* Y, Q5 n$ echar str[N];$ i+ F) [- G/ c! {$ e/ Y# s
{
: o' w2 {5 s1 u$ C, hint i,j;+ i' k8 B8 d* v% s( ?; i/ D
char t;
6 L3 H% {/ f1 Z8 U4 Nfor(j=1;j for(i=0;(i if(str[i]>str[i+1])# ~7 Z, M7 b& `/ K, ^$ z1 h
{ t=str[i];* k( a3 B7 J7 k) N z3 X
str[i]=str[i+1];
( E! d; j0 {% O N: x' F, z str[i+1]=t;
9 P- A/ A) n7 d( W( A# a }
( e+ z) D8 E% E" Z" e8 A}8 i6 J: Q5 u% A/ a- |
8.12
! j6 C! {8 ]6 j; u; M#include1 e [+ ^, A4 y# m! P; M% f
#include
+ D' t$ G5 Y5 i6 W8 bfloat solut(a,b,c,d)
# C$ M9 T- _" J' ]& N |) Ufloat a,b,c,d;
+ i) K3 ]2 A. g6 f# s, }: U{float x=1,x0,f,f1;
, s: _7 H' G0 z do) b4 U1 \ Y5 @+ {3 T3 y* X* x
{x0=x;- @% W) @4 I# |3 _. s M
f=((a*x0+b)*x0+c)*x0+d;
: `: m3 n# l8 O# C- l) d f1=(3*a*x0+2*b)*x0+c;
) }( u1 M3 a, q0 w$ [4 T x=x0-f/f1;
& @+ ?$ a/ i6 B% T- ? }
+ i; l( d! e% {( Y/ P4 ]/ B while(fabs(x-x0)>=1e-5);
/ d0 K& Q1 ~! r1 a* o$ ~ return(x);
3 G( Z6 G% g8 m}* }( T1 B- v7 @% A
main()+ L Q" f3 }! n; C
{float a,b,c,d;) O% P+ `4 Q" Z; {* T5 l9 V
scanf("%f,%f,%f,%f",&a,&b,&c,&d);
4 ]% r- E8 s, `0 q printf("x=%10.7f\n",solut(a,b,c,d));: s4 T& l! y2 E! c" E" }$ M: g
}
) e" T3 k- l6 z8.138 c9 M, w1 E6 y, D# }+ ~
#include
$ y% D( {, Q: f" M2 j- xmain()
* Q$ o+ h7 J' W1 D' S' r: d{int x,n;2 c# I/ F) u* @
float p();
# J/ R3 y) T2 S5 \. @: U2 k scanf("%d,%d",&n,&x);
8 x$ a7 L3 x6 P8 R5 h printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
; M+ U0 D9 e# `4 {# ~- F}& A/ f' F" B! g
float p(tn,tx)3 v7 U$ D( y/ u1 L$ R( E$ l4 s
int tn,tx;7 {, ~6 _5 I# n/ S. s5 X
{if(tn==0)
; M: V# c* @3 p9 T# G0 ^ return(1); n# l& {! o0 r* P5 C
else if(tn==1)5 z5 t; y; H( v+ j |5 [/ ]* T5 i
return(tx);2 a5 {$ v+ _" i* e3 j
else* Z# G4 Y- F# n( [7 C
return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
* J1 V6 E# i9 f+ |+ k) X}
0 _( B. K: l( H$ j8.14
8 f) o# w+ [4 g9 g2 E; C#include "stdio.h"1 H) ~: k' Z: v! l% w( }: j/ x- o
#define N 10
0 ~7 u3 {- L" [- a0 ~! Y- E7 z% R#define M 5
: K% \: T- x% D* r& v' I* Ffloat score[N][M];/ y4 Q. z* h% R1 t' W
float a_stu[N],a_cor[M];
' q6 ~8 B6 G8 v# r! h8 M2 Qmain()
7 x+ C4 u3 d3 A9 L/ k. {{int i,j,r,c;
$ k, ~' n4 [+ T- o- d+ n3 z float h;
. Q* T6 B2 B4 V6 ^ float s_diff();
0 a' a( e$ L0 A: Y, k/ F float highest();
& W7 d, S# [1 U3 B; E, R3 u, L r=0;: ]( J; w0 D' f
c=1;& @7 N0 Z5 g0 |/ {
input_stu();+ Y5 V, u+ D* D3 l. [
avr_stu();+ z* k! P, D& P A- o& C
avr_cor();
8 ?1 ]* {- g% a printf("\n number class 1 2 3 4 5 avr");
' E( P4 h# [% I! L2 b& ?/ [# @ for(i=0;i {printf("\nNO%2d",i+1);! t+ j. @7 Z# U
for(j=0;j printf("%8.2f",score[i][j]);7 \& O" ?' y1 y f# c+ k# P
printf("%8.2f",a_stu[i]);7 Q9 q: l4 h1 W) B
}, a, Y. C. x' W
printf("\nclassavr");
- r, v$ K1 A0 Q7 I) M% M! r for(j=0;j printf("%8.2f",a_cor[j]);
% \8 V2 u, Q7 W: ?) d' d h=highest(&r,&c);
# `: E Q1 g3 p2 p; `' `& x printf("\n\n%8.2f %d %d\n",h,r,c);; w8 y$ I. b! i& O" r3 |
printf("\n %8.2f\n",s_diff());7 d# Z' R4 `$ M# w& x
}
& x8 i* F: P; ~input_stu()4 A% v' J: l; G* ]
{int i,j;
& Y- I9 ]6 A3 X- U+ v( a8 { ?6 z float x;
) C+ ^* I" \2 o for(i=0;i {for(j=0;j {scanf("%f",&x);0 t3 e4 y6 I, H* E
score[i][j]=x;
1 \1 `! r U) b1 } X& g }4 f! N3 X, G8 V6 i5 _6 g7 q
}9 Z( |! U. K5 N
}
1 N- c7 p7 C: U' v( Savr_stu(): ^. S5 T I; |- f1 J
{int i,j;( v5 _$ J- G. a$ S- x5 W) t
float s;2 p4 c: e2 j8 q& @% ]9 A% {
for(i=0;i {for(j=0,s=0;j s+=score[i][j];6 ^1 Z4 F, v" {( ~- C% E
a_stu[i]=s/5.0;- C' q0 E' X) _- ]' x' `
}
9 ~0 S- w0 k8 d; X8 ]! F' a3 J8 g2 f5 e}
% n* |$ S. O h" B/ p# O! L- m! t2 qavr_cor()
3 w8 ^+ W3 `" w8 F) ~) R: ?9 }{int i,j;
8 u+ i0 D. `( o r j- ]7 W: _0 x float s;
4 P; `; E, B3 a) t for(j=0;j {for(i=0,s=0;i s+=score[i][j];4 {- E0 \2 }8 b7 D6 ], V3 m
a_cor[j]=s/(float)N; d8 M# {5 q$ D2 s" \
}, s* m$ l; m' q9 J4 w+ u
}
+ G# R1 m; k$ T8 Nfloat highest(r,c)! n6 w: C, Z/ ~" @- X/ E! x) q' z
int *r,*c;8 a9 X7 d+ b, _2 n) I
{float high;6 H2 v+ B/ n, z6 }
int i,j;3 \+ ~9 \4 e( L
high=score[0][0];
. H. v' |/ n; v) M% i5 j( g for(i=0;i for(j=0;j if(score[i][j]>high)& L$ g; ]2 U7 ?9 U3 m" ^* i
{high=score[i][j];
) y2 K" J' N/ ?4 |0 K9 `. @" W *r=i+1;
& k n1 d# u+ `4 W- b, l8 Q *c=j+1;
, L3 m, g& e: S& A2 q/ c* s. \' D9 a4 ? }0 J- k- G. F5 n3 z
return(high);$ X" f! R, a. A2 I2 m) B
}( P! f: j a% f7 w# j E! r
float s_diff()8 q' B. Q* D' o0 [; h P
{int i,j;0 q$ w( _: i b$ r
float sumx=0.0,sumxn=0.0;
$ h \+ c5 [% I9 U8 ~7 ] for(i=0;i {sumx+=a_stu[i]*a_stu[i];
+ S) F6 i) W6 V* {4 g, x sumxn+=a_stu[i];
* c# K6 M! l( W1 C! ~. g& X+ c }
1 Z& }3 E* w% S, w9 l: H return(sumx/N-(sumxn/N)*(sumxn/N));
9 z5 {$ ]6 m$ F' N4 |$ j) r}- q& q- D: i0 B9 D
8.15
4 C# e! O: p( o3 Q6 P& @#include
7 x4 h2 R( I. ]1 l/ y5 r' y/ O#define N 10( n, W8 V( |3 _9 n6 b$ V: K
void input_e(num,name)
) e- \1 z2 b( H& sint num[];
( O3 b5 U) Q/ F# b3 P" y& qchar name[N][8];
! h( h. |% e+ G/ Z$ h/ `( F{int i;. G3 C1 {8 o7 z, R* J
for(i=0;i {scanf("%d",&num[i]);2 L* C- }0 e0 L
gets(name[i]);
% }7 T3 f1 `' |/ u z1 y; g }
3 G+ Z9 @! a5 J2 S, k- T}
: @8 ]6 ] F7 K" \4 fvoid sort(num,name)
, Y5 g }* J# X' bint num[];
0 B1 j9 |8 A- j* @5 @, Hchar name[N][8];
" Y& }! P1 D/ ?{int i,j,min,temp1;
1 { \2 q5 S1 x5 L% D char temp2[8];
6 Q2 a0 l4 X( d) O) s4 F for(i=0;i {min=i;
( t6 G* _5 g( R9 A4 e for(j=i;j if(num[min]>num[j])min=j;
. I0 F7 R# \) |: {% j0 Z0 j temp1=num[i];
) Q P1 ~3 M0 j num[i]=num[min];
5 m7 _, S$ ?5 I- ~ num[min]=temp1;+ |( u$ f4 \* B" k; R
strcpy(temp2,name[i]);/ t' A; e- w. E1 [8 r3 y5 f
strcpy(name[i],name[min]);
9 a9 d# C& g: W6 k0 h; r# c# l/ g strcpy(name[min],temp2);. M3 b4 a# u8 e6 `) v9 e, ?
}6 U$ e9 c0 a" V8 b% P
for(i=0;i printf("\n%5d%10s",num[i],name[i]);
- }1 i6 w2 F _}
0 e( f! i5 n9 e \& m! Bvoid search(n,num,name)
; z( c _$ @; C. H! ?int n,num[];
% a! j! F& e1 k/ @- r' F5 C0 [char name[N][8];6 ]2 X( |# E% R0 H
{int top,bott,min,loca;" t- V) L! ]! D0 B, r& X
loca=0;
. ~' l4 F; J6 I$ O top=0;
) q+ H" r( v- f3 ~0 X bott=N-1;8 q) g4 J- b( N9 U3 Q$ I4 j" n, Y
if((nnum[N-1]))
4 O3 |4 x/ s' B# ]0 b loca=-1;" z8 _2 Z. i9 x+ r9 \( {$ N8 N
while((loca==0)&&(top<=bott))
% Q2 u- ^7 F: g4 j% \ n$ Z! B$ x {min=(bott+top)/2;
( R% V Y# |+ ]' { if(n==num[min])# n0 v; v# u* Y# K2 i. U! j
{loca=min;
' f; X4 S7 u; r" J5 p1 O) B printf("number=%d,name=%s\n",n,name[loca]);
/ t+ X I y* ^8 m }
5 u! F+ T/ N& o else if(n bott=min-1;
. e ?* d; u; Z2 W6 S4 ` else! I7 v) n( X' W
top=min+1;
# V* C+ D0 ?; g l. f }
5 u' n+ H; F" w if(loca==0||loca==-1)' c7 n, K. T4 {: N
printf("number=%d is not in table\n",n);2 d2 p3 w" |2 z: `
}
& U: _# d f# r' j3 ~9 h8 E: N8 J% ^main()) ^3 e4 X2 y+ r* N
{int num[N],number,flag,c,n;
# C6 F9 L# U' Q- a. W0 A$ k char name[N][8];6 `9 K7 ^: e E" X# B! h+ d
input_e(num,name);+ u8 @5 ~/ c% Y( U0 u C6 H
sort(num,name);
- A% `# C8 y7 h! b for(flag=1;flag;)
! T/ K% b! v2 E6 L4 N; I {scanf("%d",&number);! ?. G5 p: ?, q+ d; `, K
search(number,num,name);. R( A( a" A/ z" Y
printf("continue?Y/N!");
& }/ F4 k. y5 d. u- m' W8 l c=getchar();
b) w0 \9 I3 M9 Z, _# v+ y4 K if(c=='N'||c=='n')
! b' b" @2 \6 s1 c- ] flag=0;
, m: @2 O6 [% b2 `2 ` }
* H# M8 O: U& i$ W. g) L& V7 k& l}; @- ~: H4 U d9 q$ v
; o6 ]) t5 Q% X3 ?8 n8.167 G6 F) X# H9 e/ K$ h0 b; c
#include
4 F Q: T- G: T$ m* n#define MAX 1000
. ~" Z }% ], W8 }; Amain()
! G9 c( { V% |4 C' a Z( | _( W{ int c,i,flag,flag1;* l$ z1 s: f9 i# }5 f
char t[MAX];2 m- k/ o' R* `# c, h# s
i=0;
7 \' b- a0 _' L$ ]: z flag=0;
3 B/ \7 k7 M; m3 U) \/ z flag1=1;- G3 w: D* v1 J, x
printf("\n输入十六进制数:");
h8 ]* \: ?) t! |, q1 ] while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
$ \. h2 Y/ ~% ? {flag=1;7 o5 B+ H5 a. H" _1 w; R/ V w
t[i++]=c;
* K0 K+ f1 ]6 D3 t: K/ K* g5 v! f }# I" u9 P" h1 {4 X
else if(flag)
( t9 o4 M7 H0 \/ {- L {
`+ V6 Y4 r$ L- K3 B t[i]='\0';1 a. L4 @1 z O1 }% s7 p! h
printf("\n 十进制数%d\n",htoi(t));
' I: ^6 d" Q2 T9 o7 c2 `5 Q; ] printf("继续吗?");
x& K* t; x T3 s! _" ]/ H c=getchar();7 T# l3 ^/ n r9 i# w! P
if(c=='N'||c=='n')* [* N4 y2 e0 r @
flag1=0;" ~. v# k) ?' e3 D
else
6 W0 P7 D8 g/ o. E {flag=0;
" T/ p+ V0 s% @; U/ k- a1 s i=0;
7 ]3 W y- a/ ]5 |: o printf("\n 输入十六进制数:");
! ~5 U# z! m& q& M( D% o- g+ X0 C }; i! D3 X* Y3 G v+ W; l8 C) n, d
}" ?1 G' ^8 K. C
}4 S/ g9 J ^/ v* ?2 ~; ]4 l3 O
}& f. H' O4 l0 F2 V! k) L; g
htoi(s)( V9 L$ }9 p+ X0 V+ g
char s[];
% @) R9 B- n/ x! C, T{ int i,n;
9 R2 j9 j6 k7 S, }+ k$ a n=0;& g3 X4 k, {% l* F$ [2 S/ U
for(i=0;s[i]!='\0';i++)
1 r6 f0 J( o3 h {if(s[i]>='0'&&s[i]<='9')
& h1 k& T1 |+ k. i+ `; t3 e$ K n=n*16+s[i]-'0';
4 N: u1 r/ l2 E" R) y if(s[i]>='a'&&s[i]<='f')# Y% O& a2 e6 z) t7 }
n=n*16+s[i]-'a'+10;7 j6 `8 Q% m0 s9 P4 ~
if(s[i]>='A'&&s[i]<='F')
7 V) _" O" `7 P! e/ |" j0 W( v! W n=n*16+s[i]-'A'+10;0 M' S4 U O* w$ T8 g/ @
}$ W( C; @! ^5 o+ }2 Z6 `: Z+ n
return(n);
0 T9 ~" t- d2 A. i u7 i2 f* D) j}5 O4 Q U. e" n Y2 s7 b `7 ~
; X# o1 m( z7 e
3 `$ o/ ]# m" V9 o6 p+ X
5 t0 \. i' ~% u* X3 ~; Y8.17" D) t5 ]$ \; r2 {
#include
! ]8 Y" @0 ]% n D) [5 B) Dvoid counvert(n)/ j9 Q+ V+ v! |% J6 x! @/ H m
int n;) N1 K- J9 Y; L j: w4 |; l3 S/ e
{ int i;* L* ^( [, p; u4 j# ~* C- Y
if((i=n/10)!=0)
7 l/ {* ]" m$ z convert(i);
/ o4 M! S# A- b( {* g% M# K2 L+ ] putchar(n%10+'0');3 D+ ^, X$ s; R5 B& @
}
6 ?( O2 g5 k. R! ]main()
" D% C+ m E. a$ S: I{ int number;5 R8 E- H3 B4 [0 c* Z- F5 Z: k- n( w
printf("\n 输入整数:"); ]& @6 U! M4 Z+ d7 |. ~
scanf("%d",&number);" C( ]# Z& W3 u% X; h
printf("\n 输出是: ");
* W* u# D9 n( d2 `) Q if(number<0)7 c: P) g7 e: l0 ]/ Y
{ putchar('-');% x, b1 v- O2 H0 T
number=-number;
* x8 P l [6 h/ f3 |) w8 x2 S }
, M7 L& I! U% iconvert(number);
/ L; A2 }# t7 l0 Q}
1 `% c9 |) B6 J" e; |- e2 I$ s2 y. X: u8 @( E
! `1 ]1 e" c7 c* n
- T9 l! z. n# G) l, K7 |- G
8.18
( b# j' S' b' a* t" b9 hmain()7 y ^6 R# g2 d& `$ H. A
{
7 x( Q$ \: ]' S1 H int year,month,day;: K8 A$ r% H! S; [
int days;
3 o9 R$ z+ e7 w* I+ q printf("\n 请输入日期(年,月,日)\n");
$ u# f- W1 o7 g0 z/ w1 G7 a$ M scanf("%d,%d,%d",&year,&month,&day);
# m8 ]6 W; p7 T2 q printf("\n %d年%d月%d日",year,month,day);, x3 u, E4 ]3 y9 t1 W, Q
days=sum_day(month,day);% d' x: i, v5 O' `: l8 ~
if(leap(year)&&month>=3)5 C! }$ A( e6 Y4 q2 L& @: d; y0 b
days=days+1;9 C9 W2 ^$ T+ m& N
printf("是该年的%d天.\n",days);
: W+ x6 t& C8 o }6 B' d$ @ c# ~, Y _+ R o7 K( y
static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}4 x' J; M4 `2 k/ S2 ]0 B
int(sum_day(month,day)
4 c0 J) s6 |' f int month,day;1 K6 W* Z( @! j/ O6 M9 D
{4 D/ u+ W& x4 Z( \% o
int i;+ h* t, ?2 [3 M$ a: t+ q
for(i=1;i day+=day_tab[i];
1 |5 H% ^ {" d return(day);! W7 A4 p, t+ w7 Y4 l
}& u2 O0 {; k6 g8 H' f7 D
int leap(year)
' u! b4 ]" y3 p1 o% { G. O int year;- o+ y9 X0 ]! x' S. O: L/ ~
{. T) t7 h# Z V1 j
int leap;$ H [* M* q5 @+ s' R; l
leap=year%4==0&&year%100!=0||year%400==0;& Y) E2 Y. n+ }6 n: W9 v% S
return(leap);
) d! K4 _- Z" ?; D9 A" P }) R" r( k, x# f1 L* {
第九章 编译预处理
y% J3 O4 C6 w9 D' e/ k9.1
9 v# }* V3 O/ d. } T6 P+ L J#define SWAP(a,b) t=b;b=a;a=t
& @8 l/ Z0 y% m5 i9 Fmain()
# \+ f; N8 J: Q! S# a' b{) L+ D' Z1 d9 U1 n) j
int a,b,t;3 T1 A+ d4 j1 [2 C! _- T" T% p
printf("请输入两个整数 a,b:");
% D# n& E7 h# {scanf("%d,%d",&a,&b);
: L# \7 d5 g5 o4 O' CSWAP(a,b);8 M0 N& o4 C! W
printf("交换结果为:a=%d,b=%d\n",a,b);
* _1 W4 o$ i/ n}
9 l& v, K$ z! ?3 }
0 w5 Y2 W3 R. M( T( S- z. P
z: y) j/ U* b; g! B9.22 B! i" Z" b" M
#define SURPLUS(a,b) ((a)%(b))
5 N+ s) z0 ?+ a$ _) Umain()3 @. l* w$ Z. V, h4 y& z' o% A6 |( I
{7 a, k- h8 ~% P0 b" w
int a,b;: R8 c4 K, T8 `1 A
printf(" 请输入两个整数 a,b:");4 B0 u) {# d& t) @5 _3 K f* @
scanf("%d,%d",&a,&b); o, e8 }; _: X" y8 x1 Z
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));1 t6 F) J! w! z5 x7 k
}3 M8 g1 N% G( \" _
$ v# X5 x+ a& R9 t; Z) o/ X5 p5 X. Y
7 K3 Q% _$ p- [' n, U5 w
9.3
$ |% Q: g, v' K#include
- H" q. b, B( g; a+ t5 W4 T+ e#defin S(a,b,c) ((a+b+c)/2)
0 x1 L9 f: P1 S* F: L" l#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)- w- l- T& u0 \- I
c)))
3 S/ I9 v& E- J4 y: Gmain()& B8 Y" \3 i( P0 Z: Z7 d
{
7 B6 J& ]/ j( C0 X1 G" Z float a,b,c;! X: `9 l/ L D D2 {4 O# Y
printf("请输入三角形的三条边:");8 D9 v' k9 i0 f: A$ @- z% B: c" i
scanf("%f,%f,%f",&a,&b,&c);
& O8 B) ]1 B. t4 y* U5 c if(a+b>c && a+c>b && b+c>a)2 x/ [/ ]0 \% X
printf("其面积为:%8.2f.\n",AREA(a,b,c));: |% w! A/ G( {/ Y) s7 T9 |
else: p, z* y; e' n/ X0 L1 V( G1 t
printf("不能构成三角形!");6 G8 f" Z1 O5 P# {$ y
}" M7 [2 M' J9 l: r* h% G1 ~1 c
% _2 d# {4 I* D5 U! N3 A* t
: I( R: q5 E4 W$ H3 ~; n; a
3 |& X8 u$ t" C) Q. w+ R8 ~9.44 z" g. l8 ^4 s1 G
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)' }* V8 B* P. b! j7 C" p+ E3 n) U+ w
main()
" {. A! |9 {. c# |" f6 E" H P {% n# G T! ?" I" M" T8 m
int year;, n$ H3 h1 _) ^* y
printf("\n请输入某一年:");
( K) Q) T! @2 Q! m# t" T scanf("%d",&year);! j7 A& ^# ^$ p- X$ |
if(LEAP_YEAR(year))7 f# t s" D+ R3 K
printf("%d 是闰年.\n",year);* B2 ^6 Z1 Z7 \' Q2 e' O
else3 g+ d: L) ?+ M* T3 f
printf("%d 不是闰年.\n",year);
' y& b% P% v3 `8 j9 @ }/ g" F, P" ^' W$ A! M. b; q6 q
* U7 r2 M9 w4 W. ?/ I
2 X2 r8 G$ i3 ?; q2 E& b x7 L; f1 Y6 e H! ^
9.5解:展开后:
0 W4 f9 @/ e% x2 F6 ^( Cprintf("&#118alue=%format\t",x);& O+ j' w% s) k; I" D! K+ ~: ^# ~
printf("&#118alue=%format\t",x);putchar('\n');
. c5 `* t7 l/ l* L# F: X8 }9 Tprintf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');
1 `5 a# `+ k7 r输出结果:
7 G8 c0 @0 C. n. e&#118alue=5.000000ormat &#118alue=5.000000ormat2 Q- _$ `; w1 h- l7 L+ Q
&#118alue=3.000000ormat &#118alue=8.000000ormat, c9 E& Z+ Q$ ~) t) s
. w2 h5 |5 U* p; [7 J0 h8 z5 m c; N5 g
9.8
* Z, u5 n g5 g" L8 ]" e umain()! @3 k6 p" e9 w& v2 r. t& L' P* N
{5 K% E: ?8 H, B u: g! r& W0 T
int a,b,c;5 O7 r; ^3 F# D- g* f: V3 T
printf("请输入三个整数:");
6 N3 N- z/ |$ G scanf("%d,%d,%d",&a,&b,&c);4 m4 y u9 C. F) {, B
printf("三个之中最大值为:%d\n",max(a,b,c));- `. C" ^* p) x! I/ x
}
! t+ |" w( b9 r. @+ D+ _& _ max(x,y,z)
% ~+ G+ e2 c+ J, o% s p int x,y,z;& V) P7 y# _1 t
{
( X4 ^! ]8 n+ e% k, \ int t;
$ Z0 c) @' u" A9 [3 s& O( y t=(x>y? x:y);
. ?6 c) j& s2 \; x9 M7 I6 c return(t>z? t:z);5 ]4 _% x; r$ A# M! l% x, {
}! c# [3 P$ s" B4 s+ I
9 L& b' @3 Q E3 j/ M8 [3 m' K( e4 m; k1 e, |% }# h( P2 p
' T! B% D' w% B; l, k# [8 { X
9.10
. r o# O, m+ o#include
! M( P" ?% X0 D( L#define MAX 80* u; i) ]% ]2 T& h1 [
#define CHANGE 1
( G3 s, F; h5 D+ dmain() I1 G9 P2 D. l2 w) y: h% w; H
{, @7 B1 K% n$ S
char str[MAX];
& K1 }, F6 a! k int i;1 f V$ F$ u+ F' S$ f2 h& p
printf("请输入文本行:\n");: J. l0 s9 n7 ^9 C6 z/ \& ?. e6 C9 l
scanf("%s",str);, b) V1 J) A( s r3 ]
#if(CHANGE)9 s9 Q7 z+ F# t' h3 u
{7 w5 v0 \- q5 A& a& G0 M
for (i=0;i {
) q. q% b4 d( m2 O1 g( c if(str[i]!='\0'6 p/ @$ [4 f( M8 ~* {
if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
& L0 n# N: a8 f+ L3 n str[i]+=1;$ G, \+ m( z2 @5 j" l
else if(str[i]=='z' || str[i]=='Z')8 i' V; H3 F7 h/ I0 H* s
str[i]-=25;4 F0 v' F. [1 ~5 r
}
: k/ t) p' ?0 I6 g& e}& b: r) ]; T, }2 z1 Z
#endif
3 c, b- f2 N( p- Y0 k) ?printf("输出电码为:\n%s",str);3 S9 `5 T" {3 y" [' Q
}
7 |- K" }; v5 f/ o+ ]9 r* d第十章 指针
9 i3 \- [' r& w+ a2 B10.1
+ N5 v& d7 {# u6 X1 k: ^: d x4 Q- Lmain()
! J; f }( o9 z& P{int n1,n2,n3;
% E4 k, s* y3 ~8 M1 w int *p1,*p2,*p3;
" J% O7 e4 S/ m$ z3 G2 I1 g1 N2 U scanf("%d,%d,%d",&n1,&n2,&n3);2 f6 A8 N2 m6 b
p1=&n1;. r5 [, O7 Q3 r0 d
p2=&n2; {' i2 n- n; i z2 d
p3=&n3;6 V* V6 K p) m% w
if(n1>n2)swap(p1,p2);, m! U. h1 @+ w
if(n1>n3)swap(p1,p3);
4 u* ]. ^9 R7 b( Q if(n2>n3)swap(p2,p3);
$ Z6 Y; V! w5 W. {) Z printf("%d,%d,%d\n",n1,n2,n3);
: ]/ L4 o# \6 \) y& [ t- l9 H} E# |; ]" Y7 S+ e
swap(p1,p2)2 u; x! W& T; P! n1 S5 g0 P
int *p1,*p2;
; V# z4 D2 b4 g$ W{int p;. j, k' n) c/ c; v+ x
p=*p1;*p1=*p2;*p2=p;- C* `- \6 m$ j! O; d, E, m3 a
}
X' s" {+ j2 t5 B" q, M; {' K5 P2 c10.2
% h) ~! P0 t" u4 S0 imain()9 b- Q2 Q) B4 R( G4 f6 x- ^
{char *str1[20],*str2[20],*str3[20];% _, O. r" f% e% g& `
char swap();
* ]8 |5 F, V6 Y$ D1 x2 y; v scanf("%s",str1);. I9 m9 n9 ^1 U% }/ r4 a; x/ o2 p4 |
scanf("%s",str2);
# P6 X3 ?6 U2 e" Q scanf("%s",str3);# n3 L, r6 Z: j8 o5 a
if(strcmp(str1,str2)>0)swap(str1,str2);$ ~' z! n/ A7 y3 r" z1 m9 f
if(strcmp(str1,str3)>0)swap(str1,str3);% k% E! {* @5 |
if(strcmp(str2,str3)>0)swap(str2,str3);+ k, b# \2 s& T8 `$ x
printf("%s\n%s\n%s\n",str1,str2,str3);7 u [5 y$ U" V( i) K
}2 t7 f9 d" a$ P& X' `
char swap(p1,p2)9 y4 X# A" x+ |
char *p1,*p2;% R$ x) Q: H( R) L
{char *p[20];$ W* F) K6 _3 k. N2 N# a* x) w
strcpy(p,p1);
" q! y! ]9 _& L1 s8 I" p strcpy(p1,p2);
/ z0 Q3 i2 w# \" \4 I strcpy(p2,p);
6 \) m1 N% T! E}% v4 C, |# d) f! O
10.3
$ w4 R. K6 \) k4 l( M- Y9 jmain()
, `6 r2 ?$ l& h3 Y4 {{int number[10];# |, U! P( F7 |
input(number);2 @9 V" K2 g4 y. q( w4 c9 c4 k8 R
max_min_&#118alue(number);
; U! X: f+ k* T) c output(number);
8 o' [4 m) N( o, o$ q }
4 c8 w& Z/ r, Linput(number)8 `3 U6 j* Q9 v
int number[10];1 H. E% X6 u a9 z( X+ ^
{int i;" D( G6 p- y5 t3 D
for(i=0;i<10;i++)$ F9 s( ?8 }3 E0 ?
scanf("%d",&number[i]);
' C7 u) }( J; x; `# }" y1 f}! }* [- k2 |! ^2 H4 F( k) d5 v z
max_min_&#118alue(number)
|1 t* P. K: r* @' g! j% Oint number[10];
7 }" f7 y$ K F0 N{int *max,*min;
# s5 Y1 r1 p& x int *p,*end;; t. O% U8 B5 A$ O0 N# e
end=number+10;7 p2 M' n5 m7 V( n5 e* P' G
max=min=number;
a( O! f6 W3 |: T- R9 S for(p=number+1;p if(*p>*max)max=p;4 U9 O. B% T8 S% a1 @& z
else if(*p<*min)min=p;
. r# m7 e1 x8 n9 Z9 z( V) L *p=number[0];
2 a) I2 c4 B$ y/ ]" d6 z number[0]=*min;
8 }/ L8 i- f$ t6 E V8 E *min=*p;1 ?: Q9 w$ C3 T' k2 j) E
*p=number[9];0 o6 H9 Z5 l- l
number[9]=*max;3 d& w6 D" i' g" H' b8 j8 ?% {0 E
*max=*p;2 a; g' l/ O6 g. L# z+ p5 i- e; ]
return;
4 ]# p+ o3 |& y$ }2 `8 Q e# K}8 w) p- }! u* Y' w* y4 Z
output(number)
, U+ A$ s [7 u, A/ }- Q. ^int number[10];
" ^! ^% D+ K) D- V- S4 @$ m{int *p;, S4 [4 m( g5 s: d0 R( J: y
for(p=number;p printf("%d,",*p);
E+ e+ {3 K( l) P$ D/ N: ]9 l printf("%d\n",*p);
; b& y* m, _' ~6 Z# D4 x$ {- U! d}
$ E" L. a0 P8 q. {' |3 p10.4
" L' u! x; `9 m6 d. K& Wmain()4 K* K u" }% [) i6 z; P
{int number[20],n,m,i;" a8 ?' k& l9 J+ T+ a6 k
scanf("%d",&n);
1 w* j) I. T3 Z- g& d3 ~) d scanf("%d",&m);
, L7 w; g, N$ Q# H for(i=0;i scanf("%d",&number[i]);
k2 h- m. z& a: [9 O/ P/ k move(number,n,m);/ k9 R5 k4 C1 X+ ?2 u, \; s
for(i=0;i printf("%8d",number[i]);3 O) Q3 v9 _; @" e# ?
}
$ }2 ~0 d9 S4 x% I$ V# wmove(array,n,m)* h" {5 _# }% L/ A& P
int array[20],n,m;
7 r4 B6 c# }; F/ O- x{int *p,end;% Q2 R" C4 f `
end=*(array+n-1);
; M# x4 o4 ~2 Q4 D for(p=array+n-1;p>array;p--)
- K: r) b: z! K. ^- u( m; p *p=*(p-1);4 `- _, J3 w1 Z# m3 f* b' E
*array=end;
/ L# @# \" \" H: G$ v6 |$ }9 U m--;8 X2 j( C3 Y' \8 W/ t1 S) a
if(m>0)move(array,n,m);6 g4 I3 k f5 M' m: D. r4 U
}" @( k( y: c- n9 T4 ?% p, E% E& D# @
10.5
' i+ g- w% w1 X% H3 l% t#define nmax 50
5 E9 ~/ s" O% ]' ~/ i! b6 Dmain()
! @+ c' A9 B( W{int i,k,m,n,num[nmax],*p;
/ y4 [7 E/ w# Y V8 C scanf("%d",&n);
+ Q- q" |" A' z% } p=num;- k" R! X+ q4 Q' i
for(i=0;i *(p+i)=i+1;
& \& x( G. V6 x$ v1 a. a i=k=m=0;# B7 ]8 `" @( B6 W! f
while(m {if(*(p+i)!=0)k++;; W* e- I- _5 Q# Q r
if(k==3)
2 q0 A9 r9 g5 E$ Z5 u. [ {*(p+i)=0;
8 g/ ?4 O% H: y% `5 ]9 J k=0;0 z7 e$ {& p6 R: S. ~1 Q
m++;1 R1 F% ~: j8 l J1 \
}9 G8 I+ Z. p3 ?6 x; T$ G a) I; v* [
i++;0 V* r k; C4 P7 o& Q. n
if(i==n)i=0;
/ @6 ]% P4 @# B) f }( D5 ?8 [% N! c& a. A( ~
while(*p==0)p++;
' ~* a2 W8 }. c' }7 V printf("%d",*p);
4 I& e9 t! e* D7 o) N$ I9 e/ m! N8 C}
& L' F; ^7 L2 t4 M$ S10.6% z4 _' n; |* N8 n9 y; p( v
main()
; N& T7 ?7 J6 a. } P& [8 T{int len;
5 `- R' t+ v/ n8 W# z# f$ h char *str[20];4 \8 V: A! P+ L0 n" M
scanf("%s",str);& B0 ]# y' S; p
len=length(str);
( ~( {& F8 K7 F, y( V printf("\nlen=%d\n",len);3 U( h& M: n# E8 y0 K$ B& Y
}
7 t; Q# t# `9 ?) N1 zlength(p)+ W; w4 p" Y) c `
char *p;) Z- E6 x0 O4 @5 c8 z. ?
{int n=0;
$ L9 W/ l0 R6 E while(*p!='\0')2 A! L+ v, A/ c
{n++;p++;}
# B0 g( ~, O1 [# Y( }! |' b return(n);
2 d9 }' L7 Z! z* v7 Q}% p% R5 Y. W# ~ u! @% _
10.72 X: T5 `* m' L! ^ }7 D
main()
+ L- X' `) J) t: S3 w{int m;. ]$ @, c2 s k7 h- I
char *str1[20],*str2[20];
; W$ e5 R8 X6 u( E# c scanf("%s",str1);% |9 n4 u6 d4 F/ ]& c5 u. `
scanf("%d",&m);* f) D* L1 J! |' Z
if(strlen(str1) printf("error");
. n6 t1 |4 x" w else/ Q' a7 C N' o6 ~
{copystr(str1,str2,m);
" H+ j6 Z7 A( W: A printf("%s",str2);6 p4 ?2 x: p, q8 A
}1 w$ p# x; Z6 e B2 G
}
; d7 I) b! }1 K( O, Ncopystr(p1,p2,m)
8 m$ o! Y) V% g' G4 {char *p1,*p2;- h) }" D5 \1 c9 O" G
int m;
+ R" _2 o( d9 ^. q. O$ j$ x; i{int n=0;! S% W7 U$ K4 }3 \0 ?7 e: t+ _
while(n {n++;p1++;}
% d d: p5 F# Z; ^4 T" C5 V4 k8 r: l while(*p1!='\0')
# K( \8 G H% C {*p2=*p1;
% A* `6 J7 f6 y+ A8 d p1++;' t0 [! K+ ], `0 s
p2++;+ Q; ]( Z* Z5 {& i9 R
}
: Y2 n& @+ G- Y% @6 O3 Y% J' j *p2='\0';
& m7 W4 C1 _) d( P# Q5 k# C- l}
! [% U4 h6 c* U3 B. _8 C1 s10.83 j {* Z, G$ z
#include"stdio.h"+ K- W' `& B" b3 ] J
main()
0 ^, p. p8 e+ O{int cle=0,sle=0,di=0,wsp=0,ot=0,i; B: k+ i1 U+ l% ?9 K2 B6 Y+ ^
char *p,s[20];
4 T! P2 P3 @* ] for(i=0;i<20;i++)s[i]=0;
: \8 l! Q( j4 ^8 ~7 c/ |/ Z i=0;
4 z% I0 h1 R" Y$ U while((s[i]=getchar())!='\n')i++;7 I( t$ g+ q" m& y
p=s;0 y: m; |) ~* a* p: e, i
while(*p!='\n')
5 J" w$ X# a* Z+ Q( m* m# m {if(*p>='a'&&*p<='z')
& q5 a P7 z8 S6 j ++sle;
' I# C k+ y$ q else if(*p>='A'&&*p<='Z')
j6 V. ^. C2 ]5 }/ E ++cle;
% z* u3 t$ c/ B else if(*p==' ')
$ ^5 f v! o# X2 I! A ++wsp;
+ u/ F5 B$ j2 X& }) G6 ?. q( Z* S- d; {; \% B else if(*p>='0'&&*p<='9')
4 {' r" G! U7 L1 o3 `# h ++di;
- n3 ]6 y5 E5 A4 H/ @' G9 f' w else
d4 m8 i6 T# |2 M0 } ++ot;
c9 I+ j3 c) l. K" L! E p++;4 I5 ]/ o, j u+ U, |5 ~
}, k6 Z! Q' L+ t# R0 m" z
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
- p# s5 A1 `" H# P- Q" }' e. R1 A* S}
6 h$ D, G1 c. I; G1 X. Z9 v; T10.9
% r+ j! a6 R: v$ _0 I, b9 M5 hmain()) S: `1 b3 V% e
{int a[3][3],*p,i;
" m# F4 u9 }* S: Q, o: d8 t) D for(i=0;i<3;i++)
2 _. |) V' ^6 |7 v' K5 X scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);
/ z2 Y0 O1 l z: ] p=a; @ w" v, Z U4 b9 d
move(p);
! {7 ]0 i9 v7 ^3 j+ H/ ~8 P for(i=0;i<3;i++)
) r- R* t9 E' V3 k& z) Y printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);
' N/ s5 Y' x& p, T}, ~7 P8 C* }3 I% C3 n
move(pointer)
: X E3 }! U5 R6 Cint *pointer;# p/ N5 N" _4 |2 s) \9 G
{int i,j,t;% V, v% N' L/ ]- x: L
for(i=0;i<2;i++)9 _( F u+ p3 {
for(j=i+1;j<3;j++)
: o& B ]' N5 {' O {t=*(pointer+3*i+j); y2 _# l; v) o/ f ]
*(pointer+3*i+j)=*(pointer+3*j+i);0 U) _* `8 g; ^9 @. S
*(pointer+3*j+i)=t;$ y: l+ T7 r$ e" v# p2 X" L4 R
}0 P8 }* f! \ _8 r! t
}& _6 h. z( U2 _/ ~( \2 k* G" c
10.10- u- r. `* f* ^3 }3 O
main()% b. P# K& o: W, l( q
{int a[5][5],*p,i,j;
% d" U6 _3 l2 w t' x0 p for(i=0;i<5;i++)
3 y/ {# N' n5 |+ h, ] for(j=0;j<5;j++)
1 z+ V' {6 F& Z! ? scanf("%d",&a[i][j]);' r; J/ F9 Q$ p1 Q$ z2 f
p=a;( B9 c; J0 s4 y3 d
change(p);0 @3 z1 u7 }, b% z" b3 M" u6 J
for(i=0;i<5;i++), W \3 l: c7 H6 i* {) H, d- F
{printf("\n");
' z' I. m6 v7 A- i for(j=0;j<5;j++)
- S/ j) e {9 @ printf("%8d",a[i][j]);2 ?: W9 \, F* `1 v0 B
}
2 B, [& B( {0 E; o3 A9 l}
" I+ e v* E* r9 s5 `! a7 Ochange(p)
& h' P0 ^' J J" O) b0 fint *p;7 V% P8 Y7 J% o$ X7 `
{int i,j,change;3 v# _% ~# [" r0 K: k
int *pmax,*pmin;6 |! e- d9 h$ ?' G. ? c8 U% x' @' x
pmax=p;, _) }9 E! q5 s( m6 L1 Y6 K2 p1 _
pmin=p;& Z7 p, v/ T3 {. O% k
for(i=0;i<5;i++)
) z8 j$ [. J9 f! ~# }; X for(j=0;j<5;j++)
4 m- _+ |) i/ }0 M3 r- {& R {if(*pmax<*(p+5*i+j))pmax=p+5*i+j;
& l$ j2 r2 X9 S9 M% m* ` if(*pmin>*(p+5*i+j))pmin=p+5*i+j;6 v2 d; f ] h! ~% ]- ?# W
}8 L# N+ W2 T+ O$ ^
change=*(p+12);( T+ P! B! R( l# C
*(p+12)=*pmax;
- s: m/ ~$ e) i* h *pmax=change;
6 ^& h: o ~/ G1 u( I) F$ ^' Q change=*p;
' h9 C3 s; ^8 [ *p=*pmin;
; P k ^' \9 V' f, } *pmin=change;
2 P0 c! @8 X9 ^0 [' ] pmin=p+1;
: |; ?0 i/ t" U5 t5 o$ | for(i=0;i<5;i++)
! j% F+ w+ r% U for(j=0;j<5;j++)
a( C# z, t% Z: [ if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;. {, Y( h' e& l9 w# A
change=*(p+4);6 `7 W$ w7 v9 l& x/ U: C
*(p+4)=*pmin;" P! F7 }1 M" ]) t Q
*pmin=change;) d( U' v0 x4 O
pmin=p+1;8 N# o P( V3 W) J/ K3 v
for(i=0;i<5;i++)
% F3 F- J: ~9 a: u* Q8 a for(j=0;j<5;j++)
* A+ Z# Q b# t K+ \) O. K- W" M+ @ if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
1 y) {% x% A% i4 I# q/ G6 X pmin=p+5*i+j;+ C7 H: _! y' x8 ^& e
change=*(p+20);" p: f3 N) O' x2 J
*(p+20)=*pmin;' [9 ]+ Q& ^5 A- l5 [3 ~
*pmin=change;2 }& y3 }, N: b+ {
pmin=p+1;
; z) K* |1 S7 P; |; e+ } for(i=0;i<5;i++)& n. ]! H/ n" ^ t# ]6 |
for(j=0;j<5;j++)
8 \- A/ o: ?* ~: f, J& ~" k) Q if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
& c, ]! @5 g; H8 M# y H &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
: `1 Q+ n. z- B5 F, D5 g, H8 ` change=*(p+24);
- p v9 c5 x: t/ l2 Z. p *(p+24)=*pmin;
4 m$ {# f( ]; j& l9 p+ U *pmin=change;
4 g* ~: z3 k" h6 p}
* ^! D1 c, n6 W0 W* E10.11$ N# F6 G' @& C+ f! ~- O* m
main()
' E2 }8 L# b7 }+ R) @{int i;+ ~% G. A3 c5 ?% I
char *p,str[10][10];
6 Z5 Y% D( D( m* Q! x- { for(i=0;i<10;i++)' D! z; i+ h: [8 }) v) {4 Z1 q$ q
scanf("%s",str[i]);: u, p* {0 p) B) r5 M1 s
p=str;7 ?- Z- ]1 z3 w
sort(p);2 x- S1 B3 u" G
for(i=0;i<10;i++)' {! V/ _6 U, U4 R; z0 k. R4 W; R
printf("%s\n",str[i]);. p2 I& d2 H. h( d1 f* r! S K! E
}' T# E @) _2 h8 r1 N K8 G
sort(p)7 h$ X ^( \- f; I W- h& }5 s
char *p;
2 M: I, T8 q! B }& T{int i,j;* I: G- J; r: G. A/ e- j, w
char s[10],*smax,*smin;/ B3 G/ h% L3 ?
for(i=0;i<10;i++)
/ @' Z' E8 f0 f0 C2 ~! O* f: P {smax=p+10*i;, {; ?0 j5 Y* G. d0 i
for(j=i+1;j<10;j++)
' k; r1 x6 Q2 { {smin=p+10*j;. c% `0 j1 @* |' s2 [6 @
if(strcmp(smax,smin)>0)( s. \# K( |7 ^; z
{strcpy(s,smin);
1 _" [ Q/ m0 z+ [9 h( y+ b strcpy(smin,smax);# S; F# ` I: e G6 ^/ X$ x9 Q
strcpy(smax,s);$ C! z; C/ G1 E- c0 | ^
}
& `3 V y$ g( y {5 Z }
* e7 v) T6 A" V; a3 H$ d8 T; v }
" W" j- K2 a6 u. |}; L: m5 x& r4 D @
10.12
$ h7 q. S( S- G# r2 F( [/ E#define MAX 20
! b% [3 l+ w6 o7 s$ b0 Q l/ C) `main()& _6 h- E# Y# U
{int i;5 S$ E3 F, K8 B2 a
char *pstr[10],str[10][MAX];4 t5 c' r+ B+ `4 S P
for(i=0;i<10;i++)' g: |, C' G: c# B3 e
pstr[i]=str[i];
, w: @5 H. Y1 T$ \ for(i=0;i<10;i++)4 q4 r* k9 i; B# |% d' u
scanf("%s",pstr[i]);- E* ^$ M6 w: J0 ~! {" w& j# W
sort(pstr);
- y: L; k9 w+ w- l) z for(i=0;i<10;i++)
: |6 |$ g$ e8 m, G" K1 Y printf("%s\n",pstr[i]);
- y: e8 f" N5 A}7 z; \6 o, n& }# j" A; z h
sort(pstr)" V2 M. y( s; T0 W( {
char *pstr[10];
' ^! h" L! j w) P" N7 r( L5 Y{int i,j;0 _+ Q4 W3 z7 h- J1 A! C; w, }* ?
char *p;
' f6 ]6 D$ `, l5 h for(i=0;i<10;i++)5 `; [2 {$ j& m2 O% C* j
{for(j=i+1;j<10;j++)
$ @# V: u6 d- e+ D! q* } {if(strcmp(*(pstr+i),*(pstr+j))>0)% d5 r! X1 P+ }2 Z5 Z1 @. o; @, X4 ^' a
{p=*(pstr+i);
, ^- i' R( z/ E *(pstr+i)=*(pstr+j);
- F" Q. H8 s0 O$ S: }' I; R v V *(pstr+j)=p;- A8 c3 F) l) Y4 _1 r& C% N# D* F' ~
}; n, C1 j" G T/ X
}
) x& T7 C3 `* |2 f- ? }
. l0 G& S3 G: Y}
+ T6 T3 z m# y, l: X10.13
& t. C1 d3 ~% I" I#include"math.h") n/ V. q/ z4 ?" W E8 F; K
main()
1 v7 b% _" V6 l$ h8 c: d2 j{int n=20;
# H3 }' L2 C6 M, C- a8 a h( V float a,b,a1,b1,a2,b2,c,(*p)(),jiff();! d0 U5 y/ \* T" U) \7 q
scanf("%f,%f",&a,&b);8 s) O/ X; ]. \, R7 b! F( x) ?
scanf("%f,%f",&a1,&b1);* B4 y+ ], k4 @
scanf("%f,%f",&a2,&b2);
% @$ Q2 m4 O9 E- z- Z1 H p=sin;9 f* [" R* K7 Z% d! _
c=jiff(a,b,n,p);( ~6 j) a9 [, x* R4 u# X( M* I
printf("sin=%f\n",c); Z+ ~" G8 L9 O2 @
p=cos;
6 k, W; Q1 E( n# P6 ~2 R+ k c=jiff(a1,b1,n,p);9 ?1 g6 l3 v5 w; s2 s* w
printf("cos=%f\n",c);
, e3 Q+ a9 [: L/ K6 ? p=exp;
& h2 e3 n8 n1 ]- _' t/ S$ x7 [ c=jiff(a2,b2,n,p);
1 k" p; o# u7 q* L2 p! a printf("exp=%f\n",c);7 V/ f' S1 R7 j8 i/ X
}
6 u8 _0 E. t7 Z8 q) A$ Ffloat jiff(a,b,n,p)0 O. t- C/ ?; a2 I" U- t; ?
float a,b,(*p)();+ {, V0 t: m$ i
int n;4 j! }. c+ A8 l
{int i;# W% [. _& v- v" C: ~
float x,f,h,area;
# I" {! i8 [4 M) A& D. S) t h=(b-a)/n;
" [9 G. c3 {& f8 K( x9 \$ A x=a;) K/ P' U. q- `- K7 \
area=0;
b" T1 R- t* q" [. o4 o for(i=1;i<=n;i++)" B& y% A0 _, s. u% H
{x=x+h;* G! r+ c0 v0 W* L9 _
area=area+(*p)(x)*h;, e/ G/ N% F" [' L/ Q
}" L7 E& b" g7 y9 z
return(area);
' f( L8 E& `, [; S: G M9 t}% R" a% v) U6 e2 y8 s4 U9 M( d$ o
10.14
7 P! |0 ^! C$ T& u7 smain()
, X) B1 y( q3 \/ ]. F8 g{int i,n,num[20];
! b2 t+ U [! O& _$ d char *p;
$ K3 K V' J+ S1 y scanf("%d",&n);6 y0 O ^" q! B+ e3 E/ r& P
for(i=0;i scanf("%d",&num[i]);
' q3 J+ w; c5 S: r p=num;! z$ N! Z, T- H( R
sort(p,n);$ l! @' T- {/ l/ R6 ? n- I5 Q# p
for(i=0;i printf("%8d",num[i]);7 R0 l. z8 f8 \0 q# c( P0 l8 D) K" x3 [
}
Y" w! F% _& H6 lsort(p,m); X2 ]# z6 t8 W* P* E+ h
char *p;* Z; B/ `9 b; J7 t+ x
int m;
9 H$ G; j, M! D+ I/ d{int i;& P* U# n% i8 \7 H
char change,*p1,*p2;
5 N7 z* ?6 B: Y z3 T# o for(i=0;i {p1=p+i;4 ^' |4 V) U9 N" q1 j K
p2=p+(m-1-i);
$ q3 D% e4 i! Y9 }! S0 ?* D change=*p1;" [# m8 J# k, w
*p1=*p2;+ M, V! w; |5 p- }+ y" U/ P% l
*p2=change;
- Q; [" i. n6 Z$ O1 x% Q }1 L9 f7 C* F6 h2 y
}; q" W- u$ n+ D
10.151 U; A% {2 T4 k2 Y
main()
) t2 B2 ?7 }0 Q) r c5 p6 k+ Y{int i,j,*pnum,num[4];
+ P' o! I7 m/ l: b) H3 U2 ] float score[4][5],aver[4],*psco,*pave;
* c& c% _0 W6 X char course[5][10],*pcou;6 p F3 F( H s+ n
pcou=course[0];+ o: w. M% a! {5 p" o) s
for(i=0;i<5;i++)4 W- j# a& u, u9 L/ {3 E$ ^+ n
scanf("%s",pcou+10*i);" }4 i* N' m; H
printf("number");, P# N7 e$ Z3 \% I
for(i=0;i<5;i++)9 {- g( L" S# E3 u/ M6 q4 `
printf(",%s",pcou+10*i);; y' y4 k6 p/ n5 j/ g0 v5 b
printf("\n");
! I+ t& W" c( K0 O$ K5 } psco=score;6 U% [9 \; g. |' A4 i2 Y
pnum=num;
9 s! q6 s* Y* }6 \ for(i=0;i<4;i++)( I0 J: z. v' [4 n x. h
{scanf("%d",pnum+i);) H" ~1 L9 O) V; J
for(j=0;j<5;j++)
4 ]6 H" V' @* z# h- o9 H C scanf(",%f",psco+5*i+j); h# u7 L3 U9 N$ T7 U4 m
}6 b5 ?; O) \( w8 w
pave=aver;* J, R6 K: c7 d+ z1 ]7 `
printf("\n");
& ^; x0 L9 c, t5 f1 |& k avsco(psco,pave);+ M B/ s Z. l: t* D7 b
avcour1(pcou,psco);3 ~5 E0 {5 _) X; a- M( p9 G
printf("\n");
6 a/ _: V0 E7 k2 R* X fali2(pcou,pnum,psco,pave);# }" [' g8 E& m; Q# q0 q9 Y% O
printf("\n");2 B3 b" H& I- S5 Q8 k; ]* t
good(pcou,pnum,psco,pave); p4 Q9 ]6 O6 h1 ?0 N
}
9 ], `+ b, Q4 g7 R+ r/ T' ?& tavsco(psco,pave)2 y/ k% u' D/ \: S
float *psco,*pave;
; p* l# f2 p% V9 L0 i{int i,j;: e$ c4 B7 E3 t) `4 Q+ [' M
float sum,average;
+ B& e$ T# y% }$ j/ f) v, j& V4 T for(i=0;i<4;i++)
6 l8 \ s" K! I7 u1 |( F8 Z {sum=0;. e1 S" w. N; t0 v3 y( v- I' L$ y( N
for(j=0;j<5;j+)
+ {. v+ V: M- y" Q. j! [4 c6 s9 q sum+=(*(psco+5*i+j));
2 P R: {1 v, o _ average=sum/5;1 \; r1 V+ m+ y( U& V4 b" m; \
*(pave+i)=average;
0 k# E' D0 D* W: a0 h9 u9 Z' ` }, E- c! h9 w: C- W& B; V3 V' |
}
' _( i& ^8 S$ ]! B9 j% X8 B A" q$ ]avcour1(pcou,psco)+ T0 Z& D% G( Z) Y) ~
char *pcou;; M2 N' b& L* B1 A5 s
float *psco;3 k4 R4 m/ P8 M- {
{int i;
: K* D. B/ X8 x* x: ] float sum,average1;
8 P2 S8 G! Q( S/ c6 l+ C sum=0;+ m7 y v9 j- X' H
for(i=0;i<4;i++); d* b" E4 q' N+ B- j
sum+=(*(psco+5*i))7 o. F& j% W5 f% {5 |
average1=sum/4;
) }1 u( d6 S8 H3 w printf("%s %5.2f\n",pcou,average1);7 y8 i$ N/ E1 q) Q; d% w1 |0 ^ T
}
* p# H& ?( X+ ` X: v* Zfali2(pcou,pnum,psco,pave)
& t% h& |1 |/ y' w* i6 tchar *pcou;, g* x! Z( u3 l7 d6 q' T4 J1 ^
int *pnum;4 s4 E7 K9 E+ {0 b# U( \
float *psco,*pave;+ T: Y! l: o! A, t+ l3 d6 o
{int i,j,k,label;
, l* a4 A3 T9 z2 S6 o* ~ printf("\nnumber\n");! g' e; f8 G" d( r/ T9 ^
for(i=0;i<5;i++)0 M" u0 h8 \: A7 {
printf("%-8s",pcou+10*i);# l) O" D6 Q4 E/ U$ H
printf("\naverage\n");
7 M# G/ M6 j9 T5 W for(i=0;i<4;i++)8 h! H- j' K8 N" M7 H( a! d( l! \$ j
{label=0;
5 z7 |: x* \0 B9 N3 Q# _ for(j=0;j<5;j++)
# i# V- L' q# _% r, [- Q7 A if(*(psco+5*i+j)<60.0)label++;
% B/ l3 @2 G T5 W& s: m if(label>=2)
2 r7 Q/ s: @0 ] {printf("%-8d",*(pnum+i));+ H7 i3 b% Y; a; d" t4 k
for(k=0;k<5;k++)" E; m: O3 s, n4 H$ {4 v
printf("%-8.2f",*(psco+5*i+k));
! C- ^. `1 P8 H/ ?2 c. p printf("%-8.2f",*(pave+i));
9 w( B$ X% q8 @' E" g5 Z" C } W4 M; x) u, P; G
}
8 g3 x9 w! U6 \ {# B* G) ^6 q}
+ j" x: F5 y7 D* ogood(pcou,pnum,psco,pave)
6 K* Y y4 I* k& s9 Echar *pcou;6 q9 x# ]1 _& y( I/ }8 _4 W
int *pnum;
8 T& R3 L( t3 i# |8 s. ]. E3 Z* U0 xfloat *psco,*pave;
4 E( B6 Y+ D: [9 v; H* Q e2 B{int i,j,k,label;) T+ {$ R+ g2 j) L2 j( |
printf("number");# \9 [4 M, D& ?& g1 O/ Z
for(i=0;i<5;i++)) B7 E: s- j: H$ i2 M3 o. H: p
printf("%-8s",pcou+10*i);5 `* ?8 w7 O0 d# e
printf("average");
, [$ L+ P0 L# c q2 a, ? for(i=0;i<4;i++); e& Z0 t7 C+ H& j) [ R
{label=0;" d) M0 K' Y* k# r3 z
for(j=0;j<5;j++)
2 H# V! g! C3 z; o a6 P1 P7 \% | if(*(psco+5*i+j)>=85.0)label++;' v% z: p$ r( M/ S* a1 O$ C7 }
if((label>=5)||(*(pave+i)>=90))
2 _# F4 j6 E" y- F8 r( E {printf("%-8d",*(pnum+i));* l- S. _& d% [0 } k, o
for(k=0;k<5;k++)$ V% O% @' d( C0 ^) ^; Y
printf("%-8.2f",*(psco+5*i+k));+ E4 I9 |6 r4 E: J7 E' O
printf("%-8.2f",*(pave+i));
( F7 L% w9 I7 |5 T }
* ?% J9 C$ R+ U T3 G }
a7 a: N) r0 H+ Q. U6 ^! d" r4 c' T}
. l; H( ^$ k+ ]3 t6 h' W10.16
5 {; H; d8 c4 }! L#include"stdio.h"7 u' F8 Y; l w% s
main()
; M% i9 B9 `2 \4 E{char str[50],*pstr;% i! e* U2 n: r# E- t5 S. u
int i,j,k,m,e10,digit,ndigit,a[10],*pa;, f$ L* p! P. E; Q
gets(str);3 A+ e% r0 x" J, H! B, I% h# E
pstr=str;
! r. t) p4 u, f( v ~9 ] pa=a;
- ?# U6 x o4 U% C! e! f( U ndigit=0;
# b; v9 U$ M6 z) I, K i=j=0;- B, Q0 ]4 V/ d" ~7 j; E& |
while(*(pstr+i)!='\0')) V7 V9 M Q9 h+ D
{if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))2 `" x+ n4 b' T1 X
j++;
5 s) a( P# Y5 ^& b: C) b else
, `6 q6 A( A; ~6 _; Q+ ` {if(j>0)
) C. ^. ]% d/ e* z& J7 w4 ^ {digit=*(pstr+i-1)-48;
8 z1 ]* B7 ]/ C# J( [/ f" q; } k=1;5 q+ J K9 N# A$ O8 i" U3 R$ ?
while(k {e10=1;
1 H# z" F0 c3 _$ q) v9 D for(m=1;m<=k;m++)
~# s y* u& _7 q" O- K' O, ] e10=e10*10;+ D' }+ r; e2 b3 c
digit+=(*(pstr+i-1-k)-48)*e10;9 t0 \$ a! Y& z4 d9 L6 D1 C' ]5 n" ]
k++;& T; j3 [% L5 \
}
" }0 b, n7 A1 C8 b( L) s8 h! [) Y *pa=digit;/ F) ?* l$ K, _2 ~' d- q
ndigit++;
" N& `0 g7 b: X' } pa++;
/ |! ?4 U; U1 R0 ?# m3 |( x j=0;( v9 ^: T7 v: c2 F
}
, p9 @1 s9 |, s* N. ]) n }$ {8 T! V( @1 p' L7 F9 F/ y
i++;
3 t0 o# B- {& k, Z8 V7 o, Z( e1 z+ r }, }1 `# e7 G0 X) c' c- D; _
if(j>0)( a8 L/ Z5 Q9 d" N4 O
{digit=*(pstr+i-1)-48;
2 b0 h0 z8 d$ F& ^7 Y! c8 o k=1;" d4 j6 Y* l/ U+ P
while(k {e10=1;* n$ ~" s$ B" H& |* t# t
for(m=1;m<=k;m++)
, z( L8 n1 c0 m: C9 N e10=e10*10;
) _9 r. K y! y" @) @) O% { digit+=(*(pstr+i-1-k)-48)*e10;' n0 E' x6 _( Z a/ t; J
k++;! D+ i' g* O1 ~& U, l% Y: ~- v
}
$ `2 u5 X2 n7 w* U& v *pa=digit;) U# d# `6 e' L: a5 ?) I
ndigit++;
2 T, o, X$ J- V7 v! Y T; i# | j=0;4 e. M$ o# F' Q& ]: k4 \ K1 n8 l- W
}
5 f0 S! J4 u- ~4 `' H6 C, Q* X5 E printf("ndigit=%d\n",ndigit);
6 F" z- X* m1 Z) H( u; B2 T j=0;
+ i* T/ c. }2 J pa=a;8 w! d7 v) i& E, \4 O3 K) N9 @
for(j=0;j printf("%d",*(pa+j));( E, `" d1 @* y+ o4 Y
}
! j* ?6 C8 g: |* N- W2 _10.17
- Z2 ?+ B A1 L; emain() l/ p( J6 D; T2 Y
{int m;
# q. h2 z% C6 G& K char str1[20],str2[20],*p1,*p2;% X' C4 ~% a5 u+ ]; a4 `& R
scanf("%s",str1);2 [2 A, w# i$ G4 h/ I, ~
scanf("%s",str2);
7 q; n6 t$ ?4 y p1=str1;1 |* O9 U! H/ D9 U: G5 U2 T
p2=str2;
5 f2 y: J$ G* i0 C S) B m=strcmp(p1,p2);- K2 c! r0 |* v
printf("%d\n",m);; }+ f3 N% E+ y* d- `2 ?3 x. T4 y
}5 V0 I3 v6 ^/ H3 W5 u# ^, z
strcmp(p1,p2)
* u# R5 I' h) U4 }5 a# [char *p1,*p2;% X4 ~/ z* j6 Q2 q; Y7 G
{int i=0;
3 X1 C& I Y4 G* t while(*(p1+i)==*(p2+i))1 d+ {0 h8 O4 b3 }* _! A
if(*(p+i++)=='\0')return(0); C6 D( q& ~( s- J6 ?& i5 g3 I
return(*(p1+i)-*(p2+i));
9 o: E- h" H) ?) |: w}
1 A8 ]% k0 p: e1 _7 t0 Y: ]10.18" f ?- b5 a3 ?; v6 D1 Y) X6 ^9 {
main()! f" F& n% q' b+ ]
{static char *mname[13]={"illeagl","January","February","March",) d. M. | l$ T
"April","May","June","July","August","September","October",
! @3 w' U" G* [6 @9 |$ m9 v- } "November","December"};* O8 r) [4 @7 x5 O" J
int n;
* c6 g1 p2 x. P( |5 B scanf("%d",&n);
; j) l! }9 |7 D8 H if((n>=1)&&(n<=12))" w. S7 y6 w1 {$ L5 K# \
printf("%s\n",*(mname+n));
/ u* Y) y, P! H1 h5 e2 | else
1 }1 w( U0 n# {3 H n, ?( X& ` printf("error");7 I T; i c d$ v/ A% O/ [0 L3 D
}
8 }# s4 U4 Y/ S h/ h10.20
! f( c# a( H1 ?/ \ s* Qmain()% R5 w+ }/ F) j9 T+ p6 E/ M
{int i;4 R2 z( `( _; G! @0 P4 r) W( M$ A
char **p,*pstr[5],str[5][10];
2 d0 F8 M/ _4 N% S5 Z for(i=0;i<5;i++). U8 V, x2 u* k6 J* @& k! O
pstr[i]=str[i];
7 q |3 Q* m' i _4 _$ ` for(i=0;i<5;i++)
; F* C* M/ a8 o; d* g( `# D scanf("%s",pstr[i]);8 D9 f: n8 }8 t. ~1 D* L! S
p=pstr;% }; p% v) D" S3 i% ]( u
sort(p);
3 q8 W9 r1 K% J \* g8 v for(i=0;i<5;i++)
7 ]/ F+ h5 Z+ s4 i/ f1 T6 L printf("%s\n",pstr[i]);0 v% G& B$ d& a5 r5 M& {7 H* C
}
1 l, W0 z& G5 B$ F$ D; D+ a( N Fsort(p) g' g& W# F+ [" y; d4 a% x% ?
char **P;
4 n' S: [4 G# \6 v4 G- f* p{int i,j;& d3 [/ b) I! C5 J3 H: j
char *pchange;
& |# L: G% V& r4 E" {$ a for(i=0;i<5;i++)
9 o3 D- t0 D! p, D: L( k6 P {for(j=i+1;j<5;j++)" g8 h. T7 W J m( }6 D# U5 ~
{if(strcmp(*(p+i),*(p+j))>0)
% G1 d/ B7 [8 U! h {pchange=*(p+i);
" `. g# g$ D$ k *(p+i)=*(p+j);
* C, u: ?/ Z; x( I *(p+j)=pchange;: q) ?2 o' c* g P. G% w
}" {' y) s( @4 ]- G& k1 M
}
, J0 R$ L+ a8 x! @3 Y( n1 l }6 H) |* [9 ?) c1 p' e/ Y0 a% c# c
}- j5 Z+ L) D) ~: h, a; X9 O# K4 R f
10.21
9 n4 X' `$ g/ h$ c( Imain()
/ Y1 p6 ~) V$ A' A3 t1 N{int i,n,digit[20],**p,*pstr[20];
& K/ P# M4 f2 a5 k2 j& I: u$ P6 L scanf("%d",&n);9 o# ~% O/ F: C8 _# O/ G
for(i=0;i pstr[i]=&digit[i];/ K8 h) u) J5 j7 U, W3 Y; ?
for(i=0;i scanf("%d",pstr[i]);
# o+ u( Y, @+ \1 A$ F: v+ d/ v p=pstr;- \ C* B% e9 I8 w( @ Q" T
sort(p,n);$ f$ L+ }8 Y/ L) B( y/ y5 Y
for(i=0;i printf("%d ",*pstr[i]);! Z# ]2 M7 {6 ]
}
4 L( e3 r" F! R* q" t* ~sort(p,n)# g- B; R7 q* I) c
int **p,n;* Y4 F+ J! @* ?5 i- l( ~, i; E, N- |
{int i,j,*pchange;+ U6 ?/ t: k) E: S: T
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
# x4 y( a. Q! O1 q$ k, K# V {pchange=*(p+i);( a' I. u# g% [
*(p+i)=*(p+j); R$ x4 _' i( d
*(p+j)=pchange;
6 X8 C2 o& a8 T# `7 z r; e }
0 T7 D8 `, j* } }
, q0 G% {1 V4 U: R) \5 O }0 m7 B7 U1 f7 z& r
}! H$ q/ l9 I Z1 l; E% n8 k# ]: D
第十一章 结构体与共用体2 h3 {! l0 ~) F! k
11.17 Z0 ^9 x1 K" k' I4 ]! W
struct4 K6 ]$ ]2 B" i8 [7 }
{int year;# u1 r( p; z" {6 H+ m: {
int month;
7 g3 j+ A0 ^1 c: _ int day;
' U% b) `0 D5 W- y" P }date;
! s; W5 s/ D: b1 O6 \1 B) b# ^ ^main()
5 N6 c8 g- s4 W3 e: B" B! \{int days;9 Q: T1 d1 `) |6 Q7 o0 w
scanf("%d,%d,%d",&date.year,&date.month,&date.day); }- x( U: \, S# J
switch(date.month)0 F- ~! F2 } J3 g2 ~ l! z
{case 1:days=date.day;break;' j7 g1 B h( y( m3 K4 z* {" ?
case 2:days=date.day+31;break;
3 [4 V9 R" j9 n2 {4 P/ I9 }0 @. w1 ^ case 3:days=date.day+59;break;
# L6 C, p: } B case 4:days=date.day+90;break;
4 ?- t. U8 q5 B- { case 5:days=date.day+120;break;
9 P3 Q7 A' t. ~" z8 O/ Y7 s case 6:days=date.day+151;break;
; f9 w( R3 h8 n" R4 I0 T case 7:days=date.day+181;break;
& ^- n" _1 }' ^( i case 8:days=date.day+212;break;4 p/ U# J" A$ z3 w
case 9:days=date.day+243;break;
4 z# i6 U, c& }9 Y! f case 10:days=date.day+273;break;
& Z/ c( C3 G% s- M7 C1 \ case 11:days=date.day+304;break;
: w) o0 l( c& E3 ]" d case 12:days=date.day+334;break;; r) N, M8 ^3 q" O' f8 n) N
}# A$ i, |: f8 u ]6 x! ?2 [
if((date.year%4==0&&date.year%100!=0||date.year%400==0)
` s( @$ U1 N; b ]" u% C# H &&date.month>=3)
# n ]* a- C7 q0 M8 q6 g2 i3 n days+=1;
# F6 ^2 D& a2 p$ E" K4 I printf("days=%d\n",days);. K( l3 F `! b
}9 w0 o; z( p. |% K. v/ F1 P
11.2' `8 d5 ^: G/ v+ m; n3 [( g
struct dt
. X* T& }, J/ i6 k& O {int year;
6 W: ?+ S; q6 H2 x3 } int month;
7 {* V9 t, B( V* q8 K int day;- o, m3 \# \, O5 `
}date;' p9 T6 C6 Q6 g9 u" }4 q& g$ o
main()
8 p1 |; d& `9 e; h5 H$ W/ O9 G{
/ t6 j1 k8 U" I! I- k$ O# g J scanf("%d,%d,%d",&date.year,&date.month,&date.day);
( z4 k; F! g: i( g1 ^ b; M' j printf("\n%d\n",days(date.year,date.month,date.day));' `, y! S% F" Y% ?% E0 K% n% f
}
. H" f4 }$ n5 a, E0 H1 j- q5 }# Gdays(year,month,day)7 Y2 G. m- j \7 s
int year,month,day;
: h# F& z9 u& v2 t; a{int daysum=0,i;; I. s' f8 H4 G0 D
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
! ^* A1 m8 j0 h5 F' T4 q1 H8 k% E for(i=1;i daysum+=daytab[i];+ S- b Z+ v+ W. R- t
daysum+=day;
0 }! M& N. h |8 ` if((year%4==0&&year%100!=0||year%400==0)&&month>=3)
l4 u P& u5 h, N& r4 D8 C1 { daysum+=1;
& b# {: @4 c0 f5 u- S% F0 h return(daysum);
: \" G2 `* K/ ]5 s}- y- |9 V& r1 l$ h' w y
11.36 C# A7 m7 s/ v ]8 P
11.48 b( t% |% @5 {% b: p: ~
#define N 5
& ~% E v6 T3 S( A9 M, Dstruct student
' ~6 e1 O) H+ Q7 _1 G {char num[6];
! Z4 ]# T0 e/ ^: C# |) C# T char name[8];
1 _( S1 |' }) p( T int score[4];
5 ]( z& m6 k; d! n7 n }stu[N];
2 a3 b/ q) L* m/ y. Umain()
0 C# ] F6 T5 x- D& R{7 q0 t: b( S4 q# n9 X7 m' g, ~
input(stu);) G7 {4 Q& x5 T0 W ]
print(stu);
3 ]% c( F: k( W+ N+ Q}; Z+ K% B/ y- g1 a9 I
input(stu). x; H2 a- Z$ f
struct student stu[];
& \6 O* a* }" R{int i,j;0 @& Y+ L* n& N/ Q
for(i=0;i {printf("number");5 T4 I0 q. r: R6 y: t
scanf("%s",stu[i].num);
+ \, T; V9 Y- J) D$ g& g) Q printf("name");
# o- @, C. m9 I+ X" K scanf("%s",stu[i].name);* N I: l6 [1 V, d y0 \, N
for(j=0;j<3;j++)( \! i$ Y8 v, e1 G& } {
{printf("\nscore\n");' [6 Q5 Q! a h6 ?( n6 K
scanf("%d",&stu[i].score[j]);
3 @5 P6 Z- ]1 [/ [( C8 L }
' a4 D% i! ^* U" D7 C7 G$ m printf("\n");
$ C; p8 O" e* l' D }
$ h/ B! R3 Y) N. I}
) L& ~, w# ^: I t4 Y9 dprint(stu). O2 ^: h& `) }
struct student stu[];( f& E- K+ x% f, n* w% F! I4 m; z
{int i,j;
& m+ H' P* t4 H; U" j/ {* P& p printf("\nnumber name score1 score2 score3 \n");
3 [2 J8 S, W% g! c; W- R for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);1 A8 m3 M$ Y: n9 s
for(j=0;j<3;j++)! B5 z& z; ]+ S' |$ S, N
printf("%7d",stu[i].score[j]);, h0 P7 [. E/ Z/ q0 n0 B
printf("\n");9 ~) X6 B2 `0 H- u
}
P8 x0 V- E! N/ l4 w; _}. r' m# H# S5 t" M8 ]! Z; Z
11.5& O5 Q' e, R6 Z% k
struct student
( r( G/ l( A" ~% D$ q0 @$ h {char num[6];
5 K" K9 A% ~, y' Y5 `/ m- S( \) P char name[8];
, c6 a# k9 f% E, E4 [) U, \ int score[4];
: p/ F5 E3 U6 z8 j4 L float avr;' {4 Q; Y& P) r9 v6 j6 b& d% M
}stu[5];
0 i/ T* ]2 m, L( O! A0 C0 Kmain()4 A# s" `1 a% j- w
{int i,j,max,maxi,sum;
% {4 A) }0 o/ S& c) B& h0 B9 N float average;+ [; Y6 k, G( \8 w( F( Z' C8 T& a Y
for(i=0;i<5;i++). `9 F1 B+ Z# T7 l$ L: X2 A
{printf("number");% [/ V! x( k+ V
scanf("%s",stu[i].num);
; c$ y5 ~" E) `/ L* I printf("name");
; v2 m7 e0 V# f I; C scanf("%s",stu[i].name);: ?7 C8 A. X8 c8 Z0 S X
for(j=0;j<3;j++)& K# g- k% x% l: l+ @1 z2 m6 U
{printf("\nscore\n");2 \. C& k. b4 ^1 w0 ~5 |! c# S
scanf("%d",&stu[i].score[j]);1 P! a4 P9 v* T0 e
}
* }9 H; K, S' X }$ I# h4 E( E; v# c) O3 s
average=0; W: Y. _1 G6 i; ^2 u3 V
max=0;
- S$ }/ l& H1 E6 e: z0 h, ]1 }& E maxi=0;1 L' Y# d( s+ b
for(i=0;i<5;i++)
8 T8 r0 r4 }5 U" z9 \* ^ {sum=0;
: c, O B4 U/ U3 d) w for(j=0;j<3;j++)
, ?: l/ R$ x* p9 _6 y; \ sum+=stu[i].score[j];% I* |0 Z! \ W; D3 C
stu[i].avr=sum/3.0;
% D) Y- V$ b9 l# E1 [; h average+=stu[i].avr;
( h0 a0 w* Z* c1 _" w7 E if(sum>max)
3 x+ J5 Z U! L( Q2 I& K2 j5 T {max=sum;
4 s1 r: X/ @, e# E% b maxi=i;4 I* o! E/ s2 ]% ]/ \. f
}, f v0 Y8 @' n
} B3 C8 b6 D% J+ O* h5 w
average/=5;
% A0 H6 U5 K+ c) z+ z printf("number name score1 score2 score3 average\n");, P2 y9 n2 k5 l
for(i=0;i<5;i++)
" w1 ^* P, S5 q* Y0 u& ~' B {printf("%8s%10s",stu[i].num,stu[i].name);
( ^0 w- ~6 k. C& |$ `- H- G for(j=0;j<3;j++)6 n5 [& R% p) t: Q0 ^& l! ~0 j* k
printf("%7d",stu[i].score[j]);
9 Q7 B0 `- t+ T, E4 ~ printf("%6.2f\n",stu[i].avr);
( e. U% V" f6 H" J: Y4 G }4 k; B% l! l" m% r; ?: l! @/ |! Y
printf("average=%5.2f\n",average);* _# {5 `/ }- |& ]7 s, w
printf("The best student is %s,sum=%d\n",stu[maxi].name,max);
! D+ h- P* s! R}7 W1 S0 T$ h' P* n
! f4 k4 a, c( b* ~0 C# [+ H
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