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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
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签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案
% \( u# F1 b0 W5 R1.5请参照本章例题,编写一个C程序,输出以下信息:
0 A0 ]& G/ X- y8 v. h6 z: e$ jmain()) V! O' b! P; z i( [
{
: j& _9 E. _6 C/ h$ k" o0 Nprintf(" ************ \n");, {; M+ P. F) j) O/ K
printf("\n");
! X* `5 d$ f; T4 Uprintf(" Very Good! \n");
! n, o/ M' z/ ^" [0 ~$ J# V+ N' jprintf("\n");) \1 J+ O5 s3 m5 R
printf(" ************\n");
/ b& i9 }( _- c: ^; S}
5 N- [/ I5 r- A1.6编写一个程序,输入a b c三个值,输出其中最大者。$ U9 {* V! {: a* J9 n
解:main()
$ { v, L, ~4 ]$ r! i{int a,b,c,max;
+ i C6 \4 u; V5 u) kprintf("请输入三个数a,b,c:\n");2 {5 ]+ `7 J% P7 B% H" W
scanf("%d,%d,%d",&a,&b,&c);
" n+ _* H' n' W: l/ [' Mmax=a;
; C, U9 A- \% G/ eif(maxmax=b;; U& X$ v( Q/ V7 t }0 K: C6 y9 [
if(maxmax=c;
4 D( x6 C6 I+ i& G) J/ K/ lprintf("最大数为:%d",max);$ g9 x8 [( {# g+ x% V0 t
}. u( w# b( v/ B9 x
第三章6 u2 |6 ^5 B* F1 ?
3.3 请将下面各数用八进制数和十六进制数表示:7 M" N/ m& o' L3 t1 d9 k( U
(1)10 (2)32 (3)75 (4)-617' D! w' X, w* ?3 @6 b6 E! A4 C
(5)-111 (6)2483 (7)-28654 (8)21003
9 u8 L3 w* R, s: [, P+ n k/ X7 m* e解:十 八 十六
4 [* w+ @3 e$ _2 J3 J- d0 U$ m; V& S (10)=(12)=(a)
: Q* {: H/ e9 d. K (32)=(40)=20
s! S# t& b& L* {0 A (75)=(113)=4b
! L7 c$ P# H1 y) e/ |* H/ v; | (-617)=(176627)=fd97
C0 x5 \, i7 A8 d& y) y6 s -111=177621=ff91
; P2 _; T7 z6 u, y8 p 2483=4663=963
9 {& M; |3 ]9 _+ @! {2 ?6 R -28654=110022=9012. S, [2 T7 s1 U! s% Y
21003=51013=520b# G2 r, w7 J( ]$ Z: a6 x
3.5字符常量与字符串常量有什么区别?
- v) C2 ^) D" l; y- B1 S# q0 y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
3 I9 q/ p8 c6 r) Y& N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'.
& }$ G6 G# q7 Q7 g2 ~3.6写出以下程序的运行结果:( k4 E1 ?/ p6 B5 o% t* Q# g
#include8 a, Y& j. @. i
void main()& W- ~/ y% |7 H
{, \; @0 i% h5 [% n" b. d& C
char c1='a',c2='b',c3='c',c4='\101',c5='\116';
( ^2 u& a i" z; C+ S8 v0 j, k6 sprintf("a%c b%c\tc%c\tabc\n",c1,c2,c3);( K3 D0 ~* i( h* ]
printf("\t\b%c %c\n",c4,c5);
; S! e# I$ v% ?9 ~) F7 X8 L1 `解:程序的运行结果为:+ u! j% I, b" ~
aabb cc abc
5 Q6 n4 F& g! [5 q6 x. \, _/ G! ` A N
9 I4 y6 @) M- f) J3.7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,% h1 K8 C8 d6 e
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".7 M; s3 Y( a/ ~# B4 D
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
/ I8 x6 M" R% Y) j- V输出.% ?2 H2 s T1 ^* j( A( n' \
main()
7 L4 ~/ p" l2 R{char c1="C",c2="h",c3="i",c4='n',c5='a';
' C7 v4 ^/ K$ d/ Y; R, H' Nc1+=4;
* Q; {3 ?: \9 x% |' m1 ]c2+=4;$ Q# _" a4 s5 i+ [
c3+=4;
/ L( X, i8 O/ H$ k" T9 K$ Pc4+=4;
4 ^! V/ c$ f* wc5+=4;/ _2 A$ W D% L3 ]
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);
& x- z& }5 A l% N# V0 T+ H- U}3 W8 K1 B- r W- a2 S Z' ^) J) r& R
3.8例3.6能否改成如下:$ A0 |6 a- M4 @3 b& F8 j6 \
#include! Y" N+ x0 ^7 y6 f: K
void main()
: v3 [& }7 F4 n2 b7 q: [2 p, M{1 U* a& ?# L0 J) t; K6 p; o
int c1,c2;(原为 char c1,c2); a6 c, f* P) j; {
c1=97;
3 v' b- h2 C3 \* z7 F1 i0 yc2=98; }) V! x3 H7 S' C+ ]
printf("%c%c\n",c1,c2);, A7 N. k+ v$ E% d
printf("%d%d\n",c1,c2);' G, q) G. M9 J
}
* y& L$ S J2 x) Z7 `; j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.% e" ` p; V. f; n' k
3.9求下面算术表达式的值.1 w7 _9 x! R- i/ e% a) t% u2 t2 H
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
2 y& Y6 ]5 H9 q(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
. H/ K$ M: A: I: |3.10写出下面程序的运行结果:: m1 w; |/ f, {; k; }
#include
$ |; f( A! Z M0 ~' A& W% Pvoid main()
4 M4 t. f1 u% C R0 J{
- ~& L3 _4 r. M$ x! nint i,j,m,n;
$ @* e3 V8 p0 |' k" _) Ti=8;
- B \3 M1 I ^3 n- gj=10;( A' B( p' C& C6 |( m
m=++i;5 P8 p+ J# x" [
n=j++;8 o" S. [- Y! | F& v* {
printf("%d,%d,%d,%d\n",i,j,m,n);" {" @6 d$ G: Z( _' r
}! j6 O5 F- K @! P! m1 p! }+ k
解:结果: 9,11,9,10
2 j) F! s1 k# o8 G! T% F, p第4章
& \0 d! J# P+ n# u3 m4.4.a=3,b=4,c=5,x=1.2,y=2.4,z=-3.6,u=51274,n=128765,c1='a',c2='b'.想得 f6 t3 z5 \1 \* @/ J0 s9 X
到以下的输出格式和结果,请写出程序要求输出的结果如下:
; A8 ]% g% K5 k0 g* }! xa= 3 b= 4 c= 5& R6 H/ b: B' F2 l% d- W: D
x=1.200000,y=2.400000,z=-3.600000' j8 P* p8 W) K: M
x+y= 3.60 y+z=-1.20 z+x=-2.403 ]) i0 E" l. [0 d
u= 51274 n= 128765- |$ \8 Y4 q, Q& B) j* e
c1='a' or 97(ASCII)
, |3 r X0 [( G% @' oc2='B' or 98(ASCII)% |9 U5 J8 e4 U% O+ x4 L O
解:8 k/ \1 G4 j% j/ F
main()2 ~" i0 H4 g" M5 b; p
{7 S. u. f4 @$ L2 |( s& P& v
int a,b,c;
+ P0 p/ w0 b( D% ulong int u,n;) N6 y7 _" T& i! ~/ o& d/ }; B
float x,y,z;
3 `8 Z' Q* a9 ?char c1,c2;# [7 F& E' d0 j/ }0 Y2 O0 H2 @
a=3;b=4;c=5;4 d4 `4 e$ v/ o6 @( m
x=1.2;y=2.4;z=-3.6;
; r2 E+ X" \4 gu=51274;n=128765;
" ~4 R' w b0 fc1='a';c2='b';
, k6 Q2 [ F# p2 j8 x |& Dprintf("\n");
6 b$ V2 ]% ?* qprintf("a=%2d b=%2d c=%2d\n",a,b,c);
8 o+ A8 t R. t: i | p: Bprintf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
( @2 ]' S+ ?- |" o" }# P% l' sprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);
# u, ?1 i1 X+ D0 m( h+ p* Aprintf("u=%6ld n=%9ld\n",u,n);6 @$ r7 p ]) q( Q+ _0 Q6 E8 f
printf("c1='%c' or %d(ASCII)\n",c1,c2);
6 H5 \% Z6 z: ^2 n$ @printf("c2='%c' or %d(ASCII)\n",c2,c2);
9 v0 K+ |) X& \$ S8 | q}) h1 g/ I9 H% s0 n
4.5请写出下面程序的输出结果.
9 d1 C$ E1 C) G+ v结果:
1 B3 v7 R4 b4 E+ }2 ]2 i5 _, ~573 N+ r. T) d( Q, e. j$ j* C
5 7
1 S; v' y& b# i% |6 G67.856400,-789.123962
% ^8 m2 ~% l" Q3 n67.856400 ,-789.123962! l* e6 m5 u2 X$ m/ h' ~5 n
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962
8 J4 j, B( z" o0 ]8 h; z6.785640e+001,-7.89e+002/ o7 } N4 A p% O6 v
A,65,101,41# Z6 A7 N1 J I2 q$ C$ w
1234567,4553207,d687
6 n& ~/ P/ ~# |* G65535,17777,ffff,-1
- I( A/ Y% E7 o- p* N. j! Z) [" oCOMPUTER, COM
2 m0 s, X' x" @' D; T) _ A- }4.6用下面的scanf函数输入数据,使a=3,b=7,x=8.5,y=71.82,c1='A',c2='a',
1 S& R0 E+ C& s! {/ y2 ]. W问在键盘上如何输入?% J8 ?8 `' k" ?+ n! U% y5 i
main()
$ J, h! M1 n/ G) q: J{
# V$ p" B7 a( X( Fint a,b;
e. k2 H) B: f$ |: P/ \float x,y;+ g5 k7 c3 R2 x
char c1,c2;$ J: w+ Q H7 D' X
scanf("a=%d b=%d,&a,&b);
/ {7 N, a! v% v6 j" u$ h/ Lscanf(" x=%f y=%e",&x,&y);$ ?; {. K3 t( p |
scanf(" c1=%c c2=%c",&c1,&c2);" F' ?9 c. T5 i9 r
}
/ H5 R# h& a0 A" k: [解:可按如下方式在键盘上输入:
, W+ X2 Z- \2 b% I5 h/ B7 Ca=3 b=7
( K7 p5 y+ r/ U& j0 }x=8.5 y=71.82
$ l e- e. H2 m- U a1 \% N. ic1=A c2=a
0 z0 y6 Z% r& K, \, d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
1 W; k. R5 N6 e: ~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: T+ f+ s" c, Z, c6 H( {
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' g- u& k* I7 L6 K
初步概念即可,以后再进一步深入理解.9 x2 h( t* c7 L0 k* R
4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-# m; J+ o B. I( G2 }& x
3.75,z=57.8,请问& L& E O# N+ q2 W; h0 j
在键盘上如何输入数据?$ N' x$ p: k# B* A
scanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);( R I2 P% ~( h+ ]7 P9 S; G0 {
解:
. Q, J' h& K, t2 Omain()
; S1 U7 \1 N- X# ?2 c{
+ M: L+ s! W: dint a,b;
0 w8 f& w' o x$ {float x,y,z;
6 P( b! X: q" S( l* f8 N& ~char c1,c2;
$ f! J, E- N7 I+ ^1 Mscanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);; F$ X0 X2 G! i0 m
}8 ^( o7 s! F8 A: X% P8 e6 k
运行时输入:5 W& _+ Y+ v0 a0 ^! F; n
10 20Aa1.5 -3.75 +1.5,67.8) a% l# n, n8 @
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
/ R% v' k8 v( ?' f9 Q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( g9 o- y$ v8 r! K+ B
会赋给任何变量。
( q9 f. `/ v- l; n* ? r; C4.8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,
: }. q- ~4 ~8 O' K3 r3 O# G6 Z用scanf输入数据,输出计算结果,输出时要求有文字说明,取小数点后两位数字.请编
) P/ e: }# M. i! e' E! g- W程.( V4 q) J u. S p5 `1 }
解:main()
' Y" @& e% r9 }9 E5 t{, G% \( u, c- c5 P0 w7 D2 D$ {
float pi,h,r,l,s,sq,vq,vz;
/ ]$ h {- b1 C1 j7 h6 h S: Y3 Spi=3.1415926;
% K5 }$ f( Z& q1 P% I+ `# G2 m7 Uprintf("请输入圆半径r圆柱高h:\n");# |7 Q% T5 J7 `' p4 a# @5 J: g: m
scanf("%f,%f",&r,&h);
C# j" t6 B5 Gl=2*pi*r;) ^8 }4 d- e: _1 ^8 E
s=r*r*pi;
0 i9 M4 P" Y0 d% gsq=4*pi*r*r;
4 H& F7 i d5 T! Y" _vq=4.0/3.0*pi*r*r*r;
v0 E. J" O% }) Lvz=pi*r*r*h;) c/ m& O8 N" \. @, C9 w% j
printf("圆周长为: =%6.2f\n",l);
/ v# p7 M6 p5 g3 zprintf("圆面积为: =%6.2f\n",s);. ]1 E. t: d; o( ~
printf("圆球表面积为: =%6.2f\n",sq);7 R, ^4 u8 c' ]2 a
printf("圆球体积为: =%6.2f\n",vz);) t0 O+ c3 {' R V4 G) |
}
& d D' o; C2 }/ `0 O4.9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,
$ u# Q: L1 E0 ~5 c) @! k取两位小数.
$ }) P7 l s0 ^" @- v6 J解: main()1 O6 t" I* i3 P( [, i
{
* j+ u" r, m/ }2 r- yfloat c,f;
; ?" D9 M: w9 A# Rprintf("请输入一个华氏温度:\n");
! ]- `, G; t0 ` J( H, i4 B6 D& }% dscanf("%f",&f);
) {6 {( S; O% \+ c) kc=(5.0/9.0)*(f-32);
7 w, ]& e/ E6 S W! Bprintf("摄氏温度为:%5.2f\n",c);: f) E" Y" U) f& F+ L6 }$ y
}
- s5 v6 W) z3 k# z }; R/ g. ?( M第五章 逻辑运算和判断选取结构
% ^& r$ P; t: h+ N5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
+ z3 C: O6 j. |1 Q* qmain()
9 D; F; t, N, {" ~0 ?! [{; y0 t$ z) R* d7 B
int a,b,c;0 N0 {9 X; b1 G; i4 \4 Q
printf("请输入三个数:");
2 D% T* `! M! d2 E9 J+ Qscanf("%d,%d,%d",&a,&b,&c);
# M2 L9 x) ~# Q; Lif(a if(b printf("max=%d\n",c);6 V' F- Q9 |- Q/ t' r
else
, h) }5 u/ {2 @( |' X- c- b& A printf("max=%d\n",b);
2 c: Q W2 G9 }0 eelse if(a printf("max=%d\n",c);
: e5 p5 x# Y( v$ D( G, O* r else' i! C' M4 o& S5 P, g6 k/ @# d
printf("max-%d\n",a);% j3 a- F9 W& d2 X
}9 y0 w Z* A$ l$ N
方法2:使用条件表达式.+ o' l! e5 B% Q' W$ S# Z2 `6 W
main()
. o5 ~- Q6 m) _% V( O8 g5 w{int a,b,c,termp,max;
' p+ C: j3 p5 ~5 c& ^8 Z0 z: r+ L printf(" 请输入 A,B,C: ");
! Q% G" k; {' D, U7 p scanf("%d,%d,%d",&a,&b,&c);
: j+ Y$ h$ m! q: L printf("A=%d,B=%d,C=%d\n",a,b,c);
7 F# s& @" K! e- z8 F+ n temp=(a>b)?a:b;) k) @- h; u8 [! }
max=(temp>c)? temp:c;6 h2 l7 a, O5 h2 C8 ~ p
printf(" A,B,C中最大数是%d,",max);
7 h' t2 _1 ^) Z- Q- {6 Z( ] X5 Q}4 x! P/ ~' b: e) O! V; H
5.5 main()
( u( ]3 G6 H4 m2 P1 T( L{int x,y;
6 F d p- p) R6 C0 ~printf("输入x:");
5 C5 F3 J0 o- k; `# p) H; ^6 L4 w" ?scanf("%d",&x);: e+ K6 I" [/ M* G1 K# P6 A' h" }
if(x<1)6 ]9 b) z F5 E7 N( G* V8 o
{y=x;
$ `7 e) b; E4 b# i1 { printf("X-%d,Y=X=%d \n",x,y);+ T/ a) c( j8 S. h
}, H" q$ a; ]2 h
else if(x<10)
% x& R/ S2 N1 V& e _8 w. k! y {y=2*x-1;5 s) K( w8 G* l/ n z- N
printf(" X=%d, Y=2*X-1=%d\n",x,y);* I5 Y0 ?' p! m+ ^% \8 y
}
) A( T5 b0 C0 {* N: Q. ^% gelse' T- u/ R1 _' J' t# g2 _0 @& d/ Q
{y=3*x-11;6 `. I, u( s1 W( E4 b
printf("X=5d, Y=3*x-11=%d \n",x,y);
+ J9 K0 h# n) Y# K; o- v }5 S3 C$ S% g% L9 c" H
}3 p# B. M. J3 P8 G0 l1 U
(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! 5 S+ ?1 Y- f& M6 y) w
void main()/ y0 R+ N8 b9 v. D" s9 Z: d; B
{5 _$ E3 m: {6 L$ t
float s,i;
% ]$ ?8 l) m! O! g# S0 K3 I9 [char a;9 o8 x: F3 t6 j& v1 b# m
scanf("%f",&s);( h3 F' C4 L$ `5 N3 r: L
while(s>100||s<0)
8 R1 n$ c* i( k l+ K* z{. x; F# a/ G/ i1 ~) S2 p. G; c
printf("输入错误!error!");0 E. S: k. R2 U. `7 m8 u& f
scanf("%f",&s);
* Z/ W9 {/ \5 [& _1 F" t}
4 {0 S4 @0 d* ]: s; c' T( ?) ji=s/10;
& T- L4 c6 H. F1 Nswitch((int)i)
" ?) C7 s5 J# \{
' S. T9 n( D! Z* c& x9 l+ icase 10:& C4 B: @4 | w; k
case 9: a='A';break;
5 H3 [. C: y2 O& Xcase 8: a='B';break;" \7 w& Z) N/ K
case 7: a='C';break;
6 `3 Y( c* L0 f0 R5 rcase 6: a='D';break;0 d' w' m/ Z J5 ^' [& P2 G
case 5:
7 g% U, O. V' O6 D* I9 [case 4:* M# W O/ y3 c
case 2:
; @' ^& ^1 m! P: Q0 ncase 1:
' M2 p0 x) z! Icase 0: a='E';; v, f& h" G4 j' \% j, `6 `1 W
}- a: I. E; u# ~4 f2 M. [6 V
printf("%c",a);5 F- _: u: C" v5 W9 t5 Y2 ^& i
}6 I0 R- x: \' l; N: q* W& \ x
5.7给一个不多于5位的正整数,要求:1.求它是几位数2.分别打印出每一位数字3.
; S1 n+ m1 i2 P8 V$ n按逆序打印出各位数字.例如原数为321,应输出123.
S# B1 h0 |, ]2 }3 |- ?' x9 g0 @1 jmain()
d- l9 c- Z5 ~7 j/ ]. q {
! K; |- c4 j4 y7 D2 H2 f% \ long int num;) g! I1 q: j7 o* _4 B' ]
int indiv,ten,hundred,housand,tenthousand,place;: [/ @( G0 t) L/ C; r9 e5 D
printf("请输入一个整数(0-99999):");7 s; B) N- c1 F3 Z& X( s# f
scanf("%ld",&num);
" H3 u' z, p) F$ u if(num>9999)* h M3 @) T$ R$ K! D0 x' `$ f
place=5;
& ?. }1 l8 n( @else if(num>999)8 j! Q4 u, Q8 o' t4 s- H( e
place=4;
+ o d* C4 k+ w m4 c' l2 Velse if(num>99)+ q. X, h a) Q d3 E4 f
place=3;( T5 A P0 |" w4 | p3 l6 Y4 P9 W9 q
else if(num>9)' s* d' Z/ w; I# G8 X; c
place=2;
! d, C. _' U, ]else place=1;( v+ k1 E# I: U8 I
printf("place=%d\n",place);8 C* K3 I, f- [/ ]7 g
printf("每位数字为:");- }9 p1 ^+ z; i+ P. D+ C! N
ten_thousand=num/10000;: R' N$ k( I* _% q4 W& N$ I5 U5 A
thousand=(num-tenthousand*10000)/1000;
}% C8 z% R# @9 l: x Thundred=(num-tenthousand*10000-thousand*1000)/100;. c" v4 [( M/ Y7 ]5 f
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
* k9 K, y/ p2 ], s6 X* nindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;. f$ m+ h3 w; S( I/ D+ C
switch(place)
$ K4 v6 V6 }. c( J+ h. o9 H+ G{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
U2 F j0 a7 _; J7 _- O printf("\n反序数字为:");* h. n3 W; T- e5 f, Q; n
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand); U& `2 P' f6 @2 r4 c- s5 c
break;
. m" |3 K( M! u( G9 [& ^; Jcase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
$ V2 m+ C( Y% I$ `, f; P printf("\n反序数字为:");
5 ]/ \" P% U9 Z2 m+ M printf("%d%d%d%d\n",indiv,ten,hundred,thousand);, Q3 t3 I" v5 i8 C
break;& d" d3 u$ C& K8 I7 C
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);
( S4 p$ g' I: s6 ^ printf("\n反序数字为:");# k5 G6 J& w5 T1 J7 b$ }
printf("%d%d%d\n",indiv,ten,hundred);
1 _1 |6 e/ P: g% lcase 2:printf("%d,%d\n",ten,indiv);
8 f+ Y" R& V' r# `, f printf("\n反序数字为:");
5 R. V" x, N( T! K) }+ b printf("%d%d\n",indiv,ten);1 {. w ]$ v1 X8 a
case 1:printf("%d\n",indiv);/ a D( u- n- [8 Z
printf("\n反序数字为:"); u Q( @% a8 [: a8 O& A
printf("%d\n",indiv);/ _7 H7 Y( t& c# p3 T/ x
}' r, ]6 H) R! m6 z
}- j& K- x3 i+ h; q- e: {! x- |( x$ x9 |
5.86 N c0 ?( C. j5 ?- Y; e0 w
1.if语句0 }; @" [; x& ~8 P
main()
* J2 T3 i" K# Q' \0 _{long i;
8 N4 Q& i! @) g/ Z& W+ V) d$ N2 h! | float bonus,bon1,bon2,bon4,bon6,bon10;
' F: G: S# K' E bon1=100000*0.1;
4 Q5 K* y( _6 u0 S bon2=bon1+100000*0.075;2 d, R% q1 O! O- Q6 t1 q# v
bon4=bon2+200000*0.05;
" [' b! V$ w5 {, P7 ?* w0 ]0 {( o bon6=bon4+200000*0.03;' ^& f( w" T- I3 U: ]" H/ W
bon10=bon6+400000*0.015;
% p3 n H1 I+ L- z1 o: z scanf("%ld",&i);
7 d9 G, k0 A# k6 J9 A if(i<=1e5)bonus=i*0.1;2 m4 A8 d, r2 i$ i7 s
else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
% {. @" z" W4 J U+ d else if(i<=4e5)bonus=bon2+(i-200000)*0.05;5 O/ o0 e! g: ?
else if(i<=6e5)bonus=bon4+(i-400000)*0.03;3 T( ?' p2 q7 ]6 I* y
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
, g2 C4 ]: j! R: `9 H4 G else bonus=bon10+(i-1000000)*0.01;0 c6 N+ n1 F; k; O
printf("bonus=%10.2f",bonus);
- K3 ^5 }' u& k7 C}1 w/ B; [: V- ~; o P
用switch语句编程序
, z/ _5 W" ?* b Omain()
( ~2 W" I* U0 b" m, O7 ?7 Z{long i;/ ]7 H: h6 l$ W7 L4 f& Z! p, I
float bonus,bon1,bon2,bon4,bon6,bon10;7 Q4 X- v" Z. x3 E8 t
int branch;
3 q0 x/ t: P- o bon1=100000*0.1;
9 m' e6 Z9 h8 o& x& l, S$ e' l- V bon2=bon1+100000*0.075;
7 A0 R% u& v. X( h bon4=bon2+200000*0.05;
% Q2 x; M" L& k' Z bon6=bon4+200000*0.03;
7 l! g3 j# T7 o0 C, W) e bon10=bon6+400000*0.015;
3 j4 ~8 }0 ^7 d- o! u \ scanf("%ld",&i);& d! F0 Y3 N) `3 v
branch=i/100000;5 ]1 u O0 s$ ]' R8 B" M* Y0 K
if(branch>10)branch=10;, W; k6 M0 g+ N) [* \
switch(branch), A3 Q( b! B n
{case 0:bonus=i*0.1;break;! J- Z) B4 v0 c- m+ q, L
case 1:bonus=bon1+(i-100000)*0.075;break;
" ~6 a+ Y9 f) A2 \ case 2:! Z L' m' g8 [/ @
case 3:bonus=bon2+(i-200000)*0.05;break;
0 y3 i: w I6 U" R; N: q- R case 4:0 `/ ~- t& `, Q5 \
case 5:bonus=bon4+(i-400000)*0.03;break;, }$ q+ Y, B* V7 i
case 6:
# \: J! V4 Z- I( k) ?: q0 d/ u, M case 7/ f( o2 `& g" Y" h: j% W6 E
case 8:$ D- S: p' r9 r' e F1 f, h8 @
case 9:bonus=bon6+(i-600000)*0.015;break;3 K$ a5 l% X( L& C
case 10:bonus=bon10+(i-1000000)*0.01;& n# D. D6 j$ h3 D
}
. L( x: o/ l8 ?6 M/ f e$ g printf("bonus=%10.2f",bonus);
) y( n4 Z& e$ i3 x}
2 u- b9 a' T1 N: n" w! b) {6 M+ \5.9 输入四个整数,按大小顺序输出.5 m' M5 _9 C& n* X6 n
main()
7 ?3 O% j5 ?. \7 R {int t,a,b,c,d;5 q+ e1 n8 n8 E2 O2 {4 l9 j. S9 m
printf("请输入四个数:");
9 J! B+ i3 X) _8 S scanf("%d,%d,%d,%d",&a,&b,&c,&d);: V' ?* j* T r! @5 P
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);/ l& X% j+ d) f% ~% ] e
if(a>b)+ k- Y2 z8 c* C0 e& O7 }1 N2 d# }. l
{t=a;a=b;b=t;}
0 U% A; ]9 x3 p9 a7 N if(a>c)& j% c* _. G. s
{t=a;a=c;c=t;}. P* J- |; C8 |1 m3 A
if(a>d)" |/ H& H' S4 B* ^2 |1 K9 Q
{t=a;a=d;d=t;}
3 H9 g$ g" J% Z5 Z! \ if(b>c)0 E# Z) e2 I% y/ \; u
{t=b;b=c;c=t;}% p9 u% c6 H# ?( H2 t
if(b>d) |( R' \, r0 k7 A) W, f
{t=b;b=d;d=t;}
( V2 N" w4 ^7 @. r$ e+ {4 o if(c>d)4 Q% y; O+ _2 Z U9 @. m3 F
{t=c;c=d;d=t;}0 ?+ ^; k0 T1 b$ {
printf("\n 排序结果如下: \n");9 L, A Y) P! o0 `5 p) Z2 A
printf(" %d %d %d %d \n",a,b,c,d);
- h6 H! H/ ^% x/ g}
( T0 S. q9 K" @7 n1 r- q5.10塔( |# f5 h" @9 B% I# L3 `1 M
main()
( l) |5 |/ @7 D: E* i. y" I8 G3 R{
6 h, }2 X& X1 X1 k" y# W& P2 Aint h=10;* c8 d. w- T% U! p8 y" e. M/ V
float x,y,x0=2,y0=2,d1,d2,d3,d4;! f) d4 g% \0 Y1 V0 M+ e, t7 u
printf("请输入一个点(x,y):");. z$ g: M, k& m
scanf("%f,%f",&x,&y);
- L' \& u" b1 ?6 j3 O- td1=(x-x0)*(x-x0)+(y-y0)(y-y0);
8 F! U) w* q% p6 G. F- C% xd2=(x-x0)*(x-x0)+(y+y0)(y+y0);
) |: k4 ?+ O* `8 A! fd3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
* [ `; A+ L" }- w' \d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);3 F* x* ?% Y5 ?) e7 ?
if(d1>1 && d2>1 && d3>1 && d4>1), b2 w2 Q0 Q+ r1 y- h, I7 w
h=0;2 w, J: p2 [) I. v# \0 |
printf("该点高度为%d",h);
( w- y1 L! e0 S: n+ k! @ \; n}
, P. ^# c; H( g& y0 l ` X5 ~第六章 循环语句
6 g0 t7 |4 V" p2 \6.1输入两个正数,求最大公约数最小公倍数.
" k( @! q4 N% s- K4 @: ~main()$ m) Z4 k8 S0 v2 J4 I. N7 B
{
. W5 j: E& V% d/ jint a,b,num1,num2,temp;
2 d0 w; b2 R. \0 G( [/ Z8 S7 Nprintf("请输入两个正整数:\n");5 T9 h( ^4 h) a/ C% w) {7 N3 ]
scanf("%d,%d",&num1,&num2);, g/ P+ c! [( z/ |5 }
if(num1{% j8 j3 B! i8 k: A; x( \& d
temp=num1;! e# m; `' L/ }+ k, z
num1=num2;+ t4 E2 J: a" a8 G9 C# Y$ @) d
num2=temp;
+ B, G1 X4 F8 F; O4 [}5 f; C, e: A' ~8 a% }4 e W$ a0 }, C
a=num1,b=num2;# V7 J* j% |. V4 J/ A/ O) v% }
while(b!=0)
9 S$ | L8 o; R: ?$ ~& q2 T {' h+ ]8 a7 T7 U" g9 @
temp=a%b;5 ~1 Z" L* g2 g5 H) y: U
a=b;2 M. P; Z. O7 `+ ]" y z5 H- X: q
b=temp;$ W) b/ ^9 [# E) n, g7 U
}) |' ~+ l( j$ x, o$ b) F
printf("它们的最大公约数为:%d\n",a);
9 i' i" p4 f7 t& A. f- Tprintf("它们的最小公倍数为:%d\n",num1*num2/2);! I" q3 b: x0 ]$ c0 S/ G
}9 Z' X/ q) A! H. a
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.
/ o4 `5 V2 ]& m解:
( Z' Q) T. y$ J- Q; u- i1 o#include < >
& a5 W0 A7 o- [8 y' M6 ?, @7 t- d) mmain()1 h: w& N3 _ d
{
7 R; `1 ]7 a2 Q$ q4 xchar c;4 ?* O0 d% L5 ]6 }
int letters=0,space=0,degit=0,other=0;
1 k3 I& w( x' \) O6 hprintf("请输入一行字符:\n");
8 g% I8 Z' {+ J: Q3 lscanf("%c",&c);
, ]/ X% `' R$ Ewhile((c=getchar())!='\n')
$ g, r. w! U- k+ `{
$ ]% q# S5 y! N; C5 ]# c4 ~if(c>='a'&&c<='z'||c>'A'&&c<='Z')" G! W, B8 r! ~: a/ d
letters++;6 }' |* ~# R% W2 {* r, \& w
else if(c==' ')- C0 q' A/ N; |3 E# C
space++;, g7 l3 B% ^7 q( G" v, }+ x4 O
else if(c>='0'&&c<='9')
" e5 _- o4 s4 g3 v% ?( vdigit++;
' L& y( d- G/ z4 }) `# O r5 lelse( k5 z8 C/ }6 Q0 C" ~
other++;
! S+ |3 y) y9 s1 X6 w$ G}& p+ K8 m5 ^# c5 O
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
. O0 r+ `+ ], ?8 v8 p3 Rd\n",letters,space,
# H% x, x: t! Z/ x$ cdigit,other);
; Q- l* l' E+ A) t% Z4 j$ E}
2 H9 i5 H3 l0 x( d% _0 [6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.. j/ a: V) c4 _1 {$ f
解:
" e$ C& H& t; s+ [+ Dmain()9 d' F8 L# u8 F( f, t/ H
{
3 D; q5 V2 `3 F- o) R" y, \3 jint a,n,count=1,sn=0,tn=0;
1 \8 j7 p6 j3 ~4 {, X- `: b& fprintf("请输入a和n的值:\n");
, f/ W, S5 n; i% A- Y# k8 p2 mscanf("%d,%d",&a,&n);7 W/ w4 {, v; V4 d/ }- ?
printf("a=%d n=%d \n",a,n);
" \5 k# L2 n- Y! H3 Z+ Bwhile(count<=n)( o0 f! s# K/ g5 |) m) H/ ^! T" z: e$ ?
{9 `6 E, E" i9 r1 p1 A$ C- l
tn=tn+a;
+ v/ v3 |2 B+ Q% Z4 r/ h Jsn=sn+tn;
* I. W7 M! R+ I( F; w/ u& `a=a*10;
8 U; U% ~, Y7 P; l0 ?++count;
. @! Z8 T# ~8 A& i4 i7 e}& r, Z! L5 D% \- }8 ?
printf("a+aa+aaa+…=%d\n",sn);7 H" J1 m/ j3 J/ @; A+ H: Y
}
$ P d C% d) X+ B0 r6.4 求1+2!+3!+4!+…+20!.7 P+ M5 k, n, G2 x1 W5 Q* D
main()$ r: |9 x# s) g6 |0 {& S; p( h
{
. W- e9 s" [; D. V0 o0 w$ tfloat n,s=0,t=1;
1 f) ^/ t4 [3 X: efor(n=1;n<=20;n++): A8 ~. z1 W: \8 {* _" a2 o6 Z5 V6 E
{
" U% ]7 k( ~/ }" [t=t*n;/ T3 }) w1 w8 Y4 m( ^! _7 J0 i+ l; H
s=s+t;
+ w. Z/ s4 k1 o8 \! u0 ?7 L& Y}* X0 O3 {6 W( K9 r* A! g
printf("1!+2!+…+20!=%e\n",s);0 [3 E3 d3 K2 v+ C& C
}1 Z/ `( }: g8 D+ Q- F% s) o' y5 }
6.5 main()
% L' X; T2 o8 S+ ?9 Y0 i{' L6 _. I9 ?! N" q& p/ P( x' R( v2 J4 `7 z
int N1=100,N2=50,N3=10;
* s9 p% d- y R* pfloat k;; |. j! U+ `$ B) U' Z4 u
float s1=0,s2=0,s3=0;3 a" t. i9 S% \, h) H3 f" m
for(k=1;k<=N1;k++)
. s- I: k. s3 J7 X8 L) T{
6 p1 D9 \6 a0 H3 K& y; N" Ks1=s1+k;4 I" G: i# R- \: N. N
}
6 @0 h" E/ D/ @! O6 yfor(k=1;k<=N2;k++): X/ P( s- j. v, J
{ `9 J" V& ~+ ^# O0 a
s2=s2+k*k;7 m- z& m5 v9 {. g6 W( }
}2 T3 S3 R4 k3 U/ p7 `
for(k=1;k<=N3;k++)
0 ^8 r( |8 d, l$ t' l! L( b; h{& z- v( g" g+ V- F
s3=s3+1/k;
1 F8 R; Q9 R# m' O; t( y}
5 {( A/ d, c4 J6 n& }printf("总和=%8.2f\n",s1+s2+s3);4 w4 q, s- K' N4 f
}
" y6 l9 b# g% w* o4 l3 A0 |0 v7 z6 D; s6.6水仙开花: s$ h1 `# c z4 g$ P+ g
main()3 X9 V3 b/ ]! }5 p$ y
{+ |& D- G7 I5 [" {
int i,j,k,n;
' L9 k; o! A* v3 j( i6 n' q* t. fprintf(" '水仙花'数是:");
, m5 H$ p( s8 P( H4 r9 Lfor(n=100;n<1000;n++)' B: ]( D; o9 o/ P0 a) c6 j
{
/ R5 {& }: ?9 P( l1 \( `i=n/100;
# I' c2 Q( k2 O) \) j! Wj=n/10-i*10;
% I. h: y1 @4 m& gk=n%10;- O5 D, ]5 }' f1 F' u4 ~4 C& v
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
8 |9 {/ R7 R! W# }! {& i6 B* @8 _{( f5 i- ]! D8 K9 V7 t
printf("%d",n);3 r1 T/ C* f) w" K- @# O( z t
}# I& q: l' W4 Y* s
}
/ l! n6 u1 G' X' C0 l0 zprintf("\n");) {5 r- j0 X& `; w( ` L2 A- Z7 Z
}6 d' V0 d$ R/ J9 L; ~
6.7完数
& l, o6 a: Z" S5 Q( P- Mmain() F! y, ?6 A/ T* Y5 B# U$ Q
#include M 1000
1 A" t- p \$ K+ Fmain()/ L8 ~* |4 K2 j7 j' h! W; s. P
{' N8 Y2 E, W2 A$ {+ Z9 r
int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;
" K7 B8 J9 Z" I: C: |int i,j,n,s;
6 l4 y" q8 f) F, y& y% s( ofor(j=2;j<=M;j++)+ b' D2 b" x" h, t+ r2 w5 ~- `
{0 d3 ~# o2 C7 B! e6 P* d
n=0;0 t( _7 h% D( b
s=j;4 Y; @0 W5 V6 w1 a- A3 R( i. a8 U
for(i=1;i {7 y2 H/ a! l8 X/ x0 O R
if((j%i)==0)
$ ^( K3 g% W: u7 B$ k* ]6 O {2 m* A: k, \& y' [
if((j%i)==0)
! {5 z- c8 V* ]% P* ? {
) V+ _+ f2 B+ r# {$ H7 T n++;
2 ^) k- {+ O5 C s=s-i;
+ W/ V9 k2 J% n" z- _9 Y switch(n)
8 w, B) [- L* b; M {% w. O" O1 h8 @
case 1:7 V& X9 x/ D# D! i" O
k0=i;' P3 |3 B0 Z& p" e6 s
break;
3 o& t- p* H- v& c case 2:
: k5 H' t) A) l* J7 C# ~: m k1=i;5 J" m2 U$ F3 b1 \$ {$ O O* k
break;6 s: b5 d. c/ J: W
case 3: ~& J* j: I* O: a
k2=i;
" P: z. V& V% A break;
, Z9 r' H6 d# y8 J) ~ case 4:( O9 C$ T4 ? ?5 P/ G: Q; f3 }: p
k3=i;
" ?) u9 o6 z" o; J- u4 J break;) w9 h( u# \( a, a& C5 o, v% S3 c: S' @
case 5:6 X# W! C. [+ k8 @2 P! | h' w0 a& F
k4=i;$ \, R8 B0 p8 v4 j ~
break;6 M$ G& {& h+ S5 ~% g8 C6 B
case 6:
' Y7 h" [; ~+ {! O k5=i;! V4 ]' m' J4 Y
break;1 f. |! ^) l7 s1 H( F) @7 n5 P" p
case 7:7 L/ i! J, M2 `: d
k6=i;
# Q4 J; J1 N8 u' a) o+ x break;8 e" ]' V$ X6 z* l* t
case 8:
$ X, ]4 |- B! G4 {8 Y0 @5 p k7=i; Z- Q! K- h6 ~5 u1 D+ a- w( A% `
break; [' R0 }: h& C2 i
case 9:# p. q8 E8 s5 I; N8 U$ j. Y: W
k8=i;1 r) V4 D3 G# U0 K- S
break;
! `1 v: O. B. n- h! s5 x9 k case 10:
! _% R1 l1 {% d% E: R% Y1 A k9=i;
5 X' I4 t0 A1 B9 _ break;: J+ ]& u ~( H$ K, O, [: y
}
* X3 n& E$ j* a9 u, g! S }8 e+ ^- j* B) w( c$ b$ R1 z6 f
}
- L, Y* M" @% B# |if(s==0)/ ?" U* d* T) H; h
{
4 n% ^0 m5 z. ?3 {9 S; Vprintf("%d是一个‘完数’,它的因子是",j);6 v, D! k$ F7 I2 v4 x$ }$ y# Y: I' L6 D
if(n>1)9 `; J1 `" C4 Y8 E
printf("%d,%d",k0,k1);8 `+ M1 t( ^/ D! F7 D
if(n>2)$ v0 Y# j1 v% z9 \% b& s
printf(",%d",k2);, V) p. W* m3 D6 H& [% d
if(n>3)
9 e2 A$ K9 O- c- L( K printf(",%d",k3);
0 Q. L! \8 t" b* q+ s* jif(n>4)
3 f f3 s$ ]. y" s( D) ] printf(",%d",k4);
! K: s5 [2 Z$ {% R4 y yif(n>5)
" D& q2 O* w& u$ L printf(",%d",k5);
1 {+ m$ Q# B7 d* X3 |8 I7 V' Rif(n>6)
3 K/ g. ?( q7 e# i0 D& B$ `9 P printf(",%d",k6); Y$ H. v0 p2 `2 h9 H
if(n>7)
3 [7 x0 d2 h/ {4 d5 ]8 e printf(",%d",k7);) c9 f% E9 d& {# Y. E! y+ ~
if(n>8)$ K5 d) d& N! y) `
printf(",%d",k8);! _. e0 H; s6 c" a- b' x8 r
if(n>9)" H4 m6 i" x O& z- w; D
printf(",%d",k9);5 C& l$ O! z7 N
printf("\n");
& f( g( `; {/ {' h6 Z$ K9 Y8 ` }
7 @9 N! ?) K) K( X; N: c/ l' b}
" V s5 D, r5 P) a$ ^. q方法二:此题用数组方法更为简单.
: H/ d: F. a9 R4 M) ^4 gmain()
' i+ O* D& b$ ?$ t L* x{
& o3 `+ p* {+ { D, ~3 o5 v$ fstatic int k[10];
# @( _% g2 U$ r5 ^ B- rint i,j,n,s;% F1 L# [+ o6 A: A
for(j=2;j<=1000;j++)( F5 \9 T1 p, @' ~
{. I7 w1 x7 |3 T/ T' _! t, J
n=-1;/ A" N( A7 p! P# T0 K; O0 {3 a
s=j;8 Y8 w$ j4 {& ]& o, \2 G) E1 p0 P
for(i=1;i{
/ W# c* W! j2 r: @if((j%i)==0)
$ s9 H3 l+ {( G1 R{
% i/ n, D+ Z2 X' S: @n++;3 S- A% v4 z B! v0 W, p
s=s-i;
1 H: T+ O& Z6 R# v0 mk[n]=i;& M# l. R8 R1 q2 z0 P
}. u. O5 P: W$ c1 M" H
}5 O3 t$ k, e& ` W5 k+ F- @
if(s==0)
|" ^8 ?% T8 B$ n2 c+ u- Q{) l/ a @- w% {% b* S$ [
printf("%d是一个完数,它的因子是:",j);6 j( h* s& D0 T% J* X
for(i=0;iprintf("%d,",k[i]);
: }# x. w* v* Uprintf("%d\n",k[n]);- J; B( Q/ m( p8 Y
}
m' Q3 Q. W( P; ^& H/ H; C6 U+ E) t}
( e1 T0 e; B% ~7 Y7 q1 A/ }8 `6.8 有一个分数序列:2/1,3/2,5/3,8/5……求出这个数列的前20项之和.0 G2 \, t2 ]2 `0 _7 u
解: main()
1 J* z% y$ R$ n. P5 X4 E/ a: B{8 }- E9 B9 ]7 {2 {/ x
int n,t,number=20;) h8 D/ Q& V' X3 m* o" y$ G
float a=2,b=1,s=0;
$ M' l3 x J5 T+ m* a* Z7 R9 lfor(n=1;n<=number;n++)
' h9 m/ e' L! e% C! A{% L. k! [+ \( y+ n$ s$ f8 }- |
s=s+a/b;. L9 S8 g8 h/ {5 q
t=a,a=a+b,b=t;
) f) b& a; n0 [3 K}" V* B! F; D. I. M) j
printf("总和=%9.6f\n",s);1 n f" B8 P% m+ w! X: A
}
2 H L% Q8 u4 N6.9球反弹问题
7 U+ M- O0 c! `% A' G% N9 [* cmain()
+ g7 s/ c9 J/ `% K/ ~: `/ _{
7 B7 y/ r) ?* }, jfloat sn=100.0,hn=sn/2;
9 W1 ?9 b6 v7 C/ h1 I- Eint n;( ~9 w& z: G4 ^1 M# k+ ~6 D, i" R& C
for(n=2;n<=10;n++)
3 L( H) }# M. ?{$ r8 D; J# J2 {+ H4 H$ [' i5 g
sn=sn+2*hn;1 z' N; t. ]# T* b8 }
hn=hn/2;
V8 L2 P6 k$ T" }) K}
- \7 G( g3 H) Kprintf("第10次落地时共经过%f米 \n",sn);+ B9 t" x$ y6 \% F Z) U* O
printf("第10次反弹%f米.\n",hn);* d/ o* P/ x: k: M+ g4 o9 F
}" ?; ]3 N' p7 e ?; V
6.10猴子吃桃
4 {: b8 i# B" [+ q& K; pmain(); A7 @5 A. O* Q T% b
{
! U5 r x1 U# [ A7 c. J" Sint day,x1,x2;
! c1 C/ F! \6 f8 K2 ~day=9;
8 V3 U& G+ p; l% m/ S: s, r# }x2=1;
+ C4 J" Y" v" {0 I- a: ?while(day>0)
) h* ]% H% K7 I- B' y( x5 D: s{
! }# w; k5 ~+ {8 n: x p8 ox1=(x2+1)*2;
/ v6 @# {3 v4 `6 ^% Px2=x1;, X' m% H. m; y0 ?: v. C
day--;( g; F. t+ j5 r$ R$ t8 t
}% [9 F" X" x! Q( d' J
printf("桃子总数=%d\n",x1);/ C. U, n: i/ J% E4 k9 s% \
}1 I6 ]5 N& q E' F
+ P' A4 M, {4 L" w! d6.12
- {. d- [: G& l& k. u7 _% \$ y#include"math.h"
# N3 z* \1 r/ C6 z, O- Umain()1 U0 _5 C3 T( A. H* Y" @; w4 l
{float x,x0,f,f1;' J) Z4 R5 K9 Y
x=1.5;
: c8 X9 [# ]* a do9 } Z2 O9 N& Q- A
{x0=x;% M/ {* @4 I0 n, Y+ |: \; W
f=((2*x0-4)*x0+3)*x0-6;
v) p3 E1 ]4 n6 v9 g f1=(6*x0-8)*x0+3;
" D2 [% d4 b. g4 z5 z( x x=x0-f/f1;
' u: F7 s' B, ~4 [2 R2 X: m8 F }
. |0 s( k. g8 m, Z6 V while(fabs(x-x0)>=1e-5);
7 `/ |4 a* L S% {' I4 u& j printf("x=%6.2f\n",x);
# s3 `# X6 U& w2 @7 v/ B} H; i. c2 _' i0 P" V
1 |: N, t% L0 j6.13' {3 \; S9 ~6 e( m" f# c
#include"math.h"$ N' M3 }# \; U1 X8 m( X4 X6 e2 c# V1 Z
main()
! C6 I8 j/ _9 m; j& n{float x0,x1,x2,fx0,fx1,fx2;, E( B5 q9 @9 Q v: N+ ^
do
. s$ N+ ~0 \: u# Y) `3 a {scanf("%f,%f",&x1,&x2);
" [( \( c R& \# E0 ~ fx1=x1*((2*x1-4)*x1+3)-6;. X7 r, B$ h. u: U; o
fx2=x2*((2*x2-4)*x2+3)-6;' Z# T i1 o# L5 Q8 B
}
8 Y3 n# ~! C5 f while(fx1*fx2>0);
1 s% H4 |4 r: e; h do
/ G" D6 c5 w: w- |7 ^- E3 O {x0=(x1+x2)/2;
& U7 P+ `3 n3 Y fx0=x0*((2*x0-4)*x0+3)-6;5 K# v$ @9 a2 b
if((fx0*fx1)<0)
* @4 |/ F3 G: z8 t {x2=x0;# B7 j# _& y' Z
fx2=fx0;
& u' J0 z$ r" z+ Z( t }' N# U# d$ p4 G% \( s! F2 f3 H
else
" c; O2 Q1 T8 X; F& v {x1=x0;
2 @+ T2 S" p7 s fx1=fx0;
" D( g1 x6 d" x' q4 ^$ G: Z. V% H, `/ l }6 Y, ~" T1 N3 u5 q7 Q0 y
} Y$ n) u4 s' G- X4 Z% S8 C9 `( C$ K
while(fabs(fx0)>=1e-5);7 V# O; I3 ?/ T$ |* l. M
printf("x0=%6.2f\n",x0);
$ u" J# D3 G, |0 V7 l" F}2 x; m/ _) h$ k: u. K
6.14打印图案
$ N4 [- s: T2 _! G3 M$ s. m0 u( V1 Dmain(): E2 p, x( U# G1 y( I
{int i,j,k;4 b1 L% Q0 C6 Z1 Q C
for(i=0;i<=3;i++)
! n% @) q9 I. F/ d0 N+ B7 o {for(j=0;j<=2-i;j++)
& `; o6 {+ h9 G# s5 E% s printf(" ");9 V; h/ {+ `( a r. H
for(k=0;k<=2*i;k++)
! v7 M! Z* T4 y8 U" B" u printf("*");! z( ]9 M2 g. g. ^) t
printf("\n");; C/ h& Y7 R+ b) s
}5 `4 C! y1 `5 t
for(i=0;i<=2;i++)2 N: G7 B( V4 T- T) J" u6 G
{for(j=0;j<=i;j++)
0 Q0 {3 J! l2 V/ h9 S- Q, z printf(" ");4 W# h5 R5 ]5 d. h8 B9 ^
for(k=0;k<=4-2*i;k++)
2 @- e, X, h2 M* C9 F4 }9 ~ printf("*");
6 }) U+ K- P. _- ?5 Y printf("\n");
. W! l$ m7 u4 }# F( h }. G1 @! [5 W) Z
}; Y4 _- K4 B/ n& R
6.15乒乓比赛
) G& B7 Y: h0 e zmain()
, ?9 C. ?! L& B e% Z4 ]+ b{' Z0 w9 Q+ m ~9 {/ K5 b7 w
char i,j,k;9 j5 r/ q% U& Y$ Y
for(i='x';i<='z';i++)! }; z0 S6 R) Q- }, T1 ~
for(j='x';j<='z';j++)- w5 }3 `* V6 j: X8 e# _% W, i
{
1 E" V- T: `% wif(i!=j)' y2 J- V0 U" G/ C0 z0 l8 {: v+ ~( S
for(k='x';k<='z';k++): t9 _% f# f/ r( O. O
{; q! f- b* u9 s
if(i!=k&&j!=k)' t# w z* \+ P% H& [
{if(i!='x' && k!='x' && k! ='z')
( c( \7 c/ h( ~; A8 R0 e4 L) Qprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);) h% q6 ~2 q2 R5 Q L$ J
}1 a* `; G9 C0 A; j9 H
}/ P r9 G2 F8 @0 j" D0 }
}
& k# L( ~) ~3 ]- @8 \" _5 G}1 j, [7 \- h2 ?, c. h. f5 s
C语言设计谭浩强第三版的课后习题答案9 i. i4 Z- F4 o, [- |+ e! B# e
7.1用筛选法求100之内的素数.
( W. O) L" |5 U# Y( m1 a#include
: S6 f C9 `% j' K#define N 101; s7 C/ p( i, ~. a/ Z6 l; e
main()
& I- ~1 E( b$ `' [9 S; r( F9 e{int i,j,line,a[N];/ n# d3 o* r& T. t$ e+ ?
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
/ C! k2 t7 q7 Z6 k* L% O }3 ~ if(a[j]%a[i]==0)! F+ `+ _# l4 E8 Q% j
a[j]=0;
/ K$ ]# B7 n& K' {3 B. k! _printf("\n");# E/ G) b. {5 D- g
for(i=2,line=0;i{ if(a[i]!=0)
4 d0 R+ y& e5 ~5 R' K {printf("%5d",a[i]);
1 t, ]+ @4 }2 i line++;6 Y/ r. z V% p& U
if(line==10)
. F3 r8 I5 Y6 A4 h# S {printf("\n");
3 N. G1 i) G" E, N line=0;}6 @( l, w! [: F6 P; y7 \. r
}0 Q+ v% m3 ?3 W. ^; X4 a
}7 B' P( ?$ [! _6 B1 F" S$ z5 H
7.2用选择法对10个数排序.4 v. P5 e2 H% X
#define N 10
' A2 S8 q; N) e2 \main()
9 F6 P! s0 U' g{ int i,j,min,temp,a[N];; r* ?9 l j$ S" J) b2 b+ O. K; ~
printf("请输入十个数:\n");" t N: c8 h5 p2 U7 `
for (i=0;i{ printf("a[%d]=",i);: q/ V3 ~% E' O% ~. Z: K+ L
scanf("%d",&a[i]);( m I9 A1 J. N0 j3 p6 H( ^
}
' {/ w" I% W' H1 b7 e- l0 Eprintf("\n");
U# r- u: g, b7 [' ^for(i=0;i printf("%5d",a[i]);
: b' M. H" n2 s1 y/ Oprintf("\n");- J+ t5 i, Q9 ~5 H0 Y/ W c5 c
for (i=0;i{ min=i;9 T+ Y8 J7 Z4 K8 {3 H1 b
for(j=i+1;j if(a[min]>a[j]) min=j;5 T- ~; W. C5 B2 }
temp=a[i];
) }9 g! `& ~4 n+ I a[i]=a[min];) K1 \# ?2 P+ j0 v1 m9 S
a[min]=temp;' h# e* S! E) o' D4 w' L8 g
}
! ^" w J& r" Q/ g4 H1 Wprintf("\n排序结果如下:\n");4 h/ j. G8 w: Z4 o O
for(i=0;iprintf("%5d",a[i]);% j8 ?) r# ~7 B9 k' C0 {
}# ^! b$ @: [1 Q2 o: O
7.3对角线和:3 {3 i5 r* U% y/ y6 M$ O
main()* ?/ D. }$ z6 M8 U( a
{# O+ \- B1 Q8 E
float a[3][3],sum=0;
- |1 t$ r5 r+ Z! ^- a( \6 Iint i,j;
+ T( D1 e: M& x; B6 V7 K9 Y" hprintf("请输入矩阵元素:\n");; [) n; l5 `8 \. z6 W6 D2 Y; c/ L
for(i=0;i<3;i++)
# d, }5 i1 [% P for(j=0;j<3;j++)0 b. |$ b. t, W+ f
scanf("%f",&a[i][j]);# T7 h3 V, P- z6 z# S; t3 H: p
for(i=0;i<3;i++)" Q C: p- q: Q5 z
sum=sum+a[i][i];1 m+ V. G4 D) A. _# Q% u: b" a, [1 c
printf("对角元素之和=6.2f",sum);0 l8 G# z+ I m5 Y; Y) A/ d. T
}
9 Z/ H% y* c* j* V4 g7.4插入数据到数组
9 F0 x7 E/ q" U! rmain() Y, c) W0 V7 o* j, p. M
{int a[11]={1,4,6,9,13,16,19,28,40,100};; Q- u3 i5 ~6 Y; T. T; k
int temp1,temp2,number,end,i,j;. o" ~7 Y B2 h6 M
printf("初始数组如下:");8 g5 [6 d# ^/ l" w1 W
for (i=0;i<10;i++)
; X* p2 s: P! rprintf("%5d",a[i]);: X; c) c/ W5 Y) b- y2 w. W. o
printf("\n");" l! h8 z# ~$ Y) @0 c5 }
printf("输入插入数据:");5 Q9 K* G) P7 b" t
scanf("%d",&number);
y; s' x. ^% j$ Kend=a[9];
; e Q/ l( q7 e+ hif(number>end)9 n# B6 q6 ]7 \3 H+ L; w
a[10]=number;% Q. f, ~2 P1 H6 w
else
" {( P6 e: z8 {# M! ` {for(i=0;i<10;i++)
! b4 Z4 o5 A$ x6 E! |! }6 | { if(a[i]>number)$ i3 c' A3 I/ L# b: N8 I9 p. C
{temp1=a[i];9 W- _% U3 t M; v
a[i]=number;) w" {, W5 h7 P0 r4 Z
for(j=i+1;j<11;j++)( r0 e0 N( k) X- m
{temp2=a[j];5 d; J: ~7 v6 v# d' G
a[j]=temp1;
+ s( w; ?% H7 Z+ J temp1=temp2;. s g9 }# |! h! l* K+ R( a& ?
}) t! E: t" E7 x# W* l% V
break;. y! P* T& f' ?: s3 d
}
- a7 ^8 v8 P, M }; Z4 y( c$ A3 O- T$ _
}4 J9 o3 w6 c: _% x' ?* C$ a
for(i=0;j<11;i++); L4 w" h0 I" [3 X
printf("a%6d",a[i]);, K3 N# R" ?/ E3 S
}
6 @& f( l1 A% |7 n7.5将一个数组逆序存放。
# b3 T0 u* ~6 ~#define N 5
' ^1 X5 t/ y; N7 W* Fmain()
4 u7 z) D5 R, a# E{ int a[N]={8,6,5,4,1},i,temp;" p* y4 A; M/ F" c U
printf("\n 初始数组:\n");( W8 W! t9 N1 o7 ]( o8 I8 |
for(i=0;iprintf("%4d",a[i]);6 J6 A) F& h, t$ e Y) ~4 I7 T( Y
for(i=0;i{ temp=a[i];
) w; V! }* B( p5 w$ |" i% h a[i]=a[N-i-1];
- e. i; B- C. j( Z1 @7 K a[N-i-1]=temp;
$ l' m6 Y% m# _}) p) R+ Y' ]* f
printf("\n 交换后的数组:\n");2 y! t- u4 y0 q1 W% O D# o
for(i=0;i printf("%4d",a[i]);
, c# q4 c8 Z/ u, o}! l/ b0 O% }# {* R |
7.6杨辉三角
4 T6 p2 ]1 f" |, R8 Z#define N 11+ r, S7 ?, D0 X4 D4 p0 I
main()
4 j5 X* r8 S* r M ~" U6 e5 C{ int i,j,a[N][N];
: X6 v3 E4 A& o! X0 b$ w5 K for(i=1;i {a[i][i]=1;! s. Z9 T& [( W" W
a[i][1]=1;- ^+ [- K0 u5 J/ ^
}
+ R2 n- N* e9 V/ S! [* { for(i=3;i for(j=2;j<=i-1;j++)
" m* X, a' `! [9 a9 J' \. w1 ` a[i][j]=a[i01][j-1]+a[i-1][j];
/ L, F0 G, \- L2 }+ v for(i=1;i { for(j=1;j<=i;j++)) X# f( B0 D& \3 H% v& e' {9 R6 C
printf("%6d",a[i][j];
) I/ l6 `( t( u) [. i: v4 b printf("\n");
0 W% x5 e$ y! K4 T% t }
" g! N& \8 W/ u+ W2 a printf("\n");4 M/ E1 y' M4 H( G
}* F& a. z! Q! C/ {, Z. x# D
7.8鞍点
. ]$ F7 V( P. o' ^. m6 L0 z) J5 Z#define N 10
7 F \+ q2 X- `6 p+ p% W#define M 10
" P7 _ g; F$ A2 `/ _5 Cmain(). a- U8 T% R2 ^+ S# F5 O7 J0 U
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;
8 ~. [. C, n8 e printf("\n输入行数n:");# o) Y- _* s. g
scanf("%d",&n);
* S' v, `/ k5 f" {6 y1 t printf("\n输入列数m:");
' ]9 M' G. x5 a& A! y6 n5 o scanf("%d",&m);7 y4 L5 _1 h5 e) q
$ C3 n$ V1 L- l" O0 ~8 r& H for(i=0;i { printf("第%d行?\n",i);% z5 b2 H# m6 }4 W; C1 e2 |
for(j=0;j scanf("%d",&a[i][j];
) @2 _& ]9 t% T) q M: c6 T- r5 l }
$ N/ t2 ~8 k5 K8 J5 L for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
% q( C7 n! |; l3 e, s pritf("\n");
" r7 M j+ x! Y3 I( @0 X9 p. u9 A% M }
% s* F+ e K" k- ]- ? flag2=0;
: {" s" T' `; t9 w5 V W for(i=0;i { max=a[i][0];+ b& g6 P+ S+ H8 w5 a2 z0 A' D" F$ O i
for(j=0;j if(a[i][j]>max): R# ?" p& W# j! u: g5 R
{ max=a[i][j];& V# h$ h- B$ z7 N. H% h
maxj=j;+ o* P- h" a6 D* s
}7 O7 q0 q8 f* X7 v- w! A
for (k=0,flag1=1;k if(max>a[k][max])
2 v" d6 O% s' {6 T; g# w$ n flag1=0;7 z) k9 L' r) n9 e' _: B
if(flag1)
: L) N5 U- p B( X) v& U { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);
; R4 p7 Y) _: O. ~ flag2=1;* {/ o) Q/ S1 |. D* y! F
}
5 |1 n+ G! |% x( {, Q; w# n}! V4 H% a# j+ ]6 K. V$ p8 u
if(!flag2)
% d( D' D( n5 C' v" r/ V; O& ], O printf("\n 矩阵中无鞍点! \n");6 ?# F& Q6 u S' ?, j: w+ u4 p, C
}
& r3 O& T. L6 x3 h$ i# T2 l. M6 b4 S" e8 J t Z
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.% j7 I5 ~2 r0 d- P- ?
#include
) U5 u6 f! J+ b+ ~#define N 15
6 F. O, k5 L t) S' nmain()! k; F# \. J1 N; n5 j
{ int i,j,number,top,bott,min,loca,a[N],flag;1 h& o+ O; x+ |: Y7 E, j
char c;% D( k; ?/ f4 ]
printf("输入15个数(a[i]>[i-1])\n);
, f! Z3 Y! S j& F scanf("%d",&a[0]);
9 Y x# D0 x9 j3 E4 U0 W i=1;# l% \) E: f- [ \8 e1 o* R
while(i { scanf("%d",&a[i]);" u9 m. r) O: U) K% Y' Y# y
if(a[i]>=a[i-1])$ v$ {6 E0 d4 D k* J
i++;! S! j) c" J# B; F1 {' a5 y
esle& ?% [0 B$ S: `& W A8 [
{printf("请重输入a[i]");
! J! @: v' u, e- d! g printf("必须大于%d\n",a[i-1]);
0 C) Y+ |9 W, ~/ w' W+ \ }# J+ u" W, _! a; I7 _) z
}
4 J- U( I4 b5 O8 Q3 n/ @ printf("\n");
1 O3 G7 R3 T4 `- K( |- U4 N% p for(i=0;i printf("%4d",a[i]);
2 t" ]+ C# v4 Y5 u+ o" `* } printf("\n");! f# `; P' C# Q; o5 ^! |! e2 U( s
2 G* ~1 t& y' w1 L. ]5 k. f
flag=1;
1 \8 s4 N! l9 B* m( @ while(flag)
" }9 H5 L* [% k7 o1 X {
5 ]# ^& D) A; u8 d5 D, H8 j printf("请输入查找数据:");" i/ Q( }. u$ I9 R8 l( |! A
scanf("%d",&number);$ x G+ t" [+ W. ~5 {5 s6 ~
loca=0;' `6 ~" }- r" \1 t) D# W: L) X0 W/ Q
top=0;
" U. V3 s- K6 R% f* h- M5 |# J bott=N-1;
6 k9 \+ U" h. u if((numbera[N-1]))
- b) h; e5 I, H, `0 w N8 n loca=-1;
2 [3 I( L' i9 V" A" Q while((loca==0)&&(top<=bott))% K- e5 y. u5 w6 G' `4 `
{ min=(bott+top)/2;( V% e1 T- u& j( ~
if(number==a[min])
8 z- \/ e( @7 m { loca=min;
# Q0 S4 N# s; K5 | printf("%d位于表中第%d个数\n",number,loca+1);
/ d/ p( v! d: B: L }
0 H3 w4 ~3 u( q6 v9 k: b! I else if(number bott=min-1;1 s0 |: H" A3 r2 s+ o! }" N
else
1 s0 o0 j% W! c- n& ~ top=min+1;
2 X1 Y* C/ P& O, | }
7 f- h9 e; {+ V if(loca==0||loca==-1)
3 G0 S$ _, B6 {& i printf("%d不在表中\n",number);( e m4 i, C9 s4 p9 Y
printf("是否继续查找?Y/N!\n");
5 F+ F4 d+ c! N7 `& H" P9 p, D+ A# N( T# I c=getchar();+ l+ K' e: ^8 @4 A/ B7 i, e, l
if(c=='N'||c=='n')8 i9 O% X# J* Q) d
flag=0;
6 A& O0 e3 n* r1 { }
9 g, S1 C+ a! d! d}. s$ K: ]0 @0 q/ Q
4 q3 S4 D2 _6 C+ l9 c3 c
7.10) _9 s5 T! }) F$ n
main()
1 Q7 d) X+ \6 b1 ? O- P9 ^{ int i,j,uppn,lown,dign,span,othn;; B2 [; D6 |' |) C6 a
char text[3][80];. e" X6 I) p% }# i& k* y5 e
uppn=lown=dign=span=othn=0;
8 B# I; S8 k5 W2 D5 v9 m for(i=0;i<3;i++)
: H* o- ]- l. D. H. H7 f) W7 y { printf("\n请输入第%d行:\n",i);
5 ?4 _6 M2 h# h K3 v( w) ] gets(text[i]);7 e, O. u! D3 ]1 w# c
for(j=0;j<80 && text[i][j]!='\0';j++)
3 }1 A- A: L0 ~# m3 p+ L {if(text[i][j]>='A' && text[i][j]<='Z')
+ z: m0 {( d& Z uppn+=1;
3 h) u) w" _+ \! u+ X& ]( v/ X else if(text[i][j]>='a' && text[i][j]<='z')
4 ^ i; p! X+ k lown+=1;
) c) \, L1 R0 T; @$ N5 O else if(text[i][j]>='1' && text[i][j]<='9'); G: c! J/ n% L9 o% }1 `0 D$ X
dign+=1;
1 W. n7 L% R: g, P! P5 ^ else if(text[i][j]=' ')
% t0 R8 W$ D i span+=1;, R# e0 d( I+ T P
else ], o. _& l3 b& G U+ W
othn+=1;
- }4 Q% v) O6 m1 Y( i7 o# j }- w! }" \( g H' o
}. i# N4 m3 b/ k. S! ^& ~) s0 u
for(i=0;i<3;i++)! @ n' d3 I8 d% P+ N
printf("%s=n",text[i]);
3 }! n- v; W+ N4 R8 H printf("大写字母数:%d\n",uppn);
' \" L/ K5 {5 e g5 c% } printf("小写字母数:%d\n",lown);# I. c9 R) h2 ?$ I8 r, S
printf("数字个数:%d\n",dign);
9 p$ K. R" ], z! S% u printf("空格个数:%d\n",span);: Q( l$ W1 W& [# w. }4 P+ B
printf("其它字符:%d\n",othn);
+ `) w+ \' x( p7 j8 G$ G}
& X d7 P6 [2 F
; \% k( K+ Z ^1 e+ |
( l; k/ Z5 H5 F8 C7.11) }" i7 ], j2 l2 ]; M+ U& h
main()
- v# a! E0 n+ g0 |8 p {static char a[5]={'*','*','*','*','*'};4 k) q& m! @ [1 M' z
int i,j,k;$ m7 f5 t3 C9 Q; B( R+ ?
char space=' ';
" x$ ?6 e8 x+ b1 B- j9 P for(i=0;i<=5;i++)9 W& v: g$ C4 q: s/ C u J
{printf("\n");
: n3 Y& }3 m2 G2 u/ n: }4 j& ?$ o for(j=1;j<=3*i;j++)
2 H" h) N' _7 O, E2 R) r; q1 u# b printf("%lc",space);- K5 X: l5 l2 g; O0 M+ a
for(k=0;k<=5;k++)
J& l5 p& c2 b: p" {9 U. J, f printf("%3c",a[k];! u- v6 B, v$ j8 S
}! h: N c; a- o# v- B
}5 f5 w* `: {% c, A- j
7.124 B4 e+ H0 x4 Z9 _ ^
#include
# D; K+ J, J: {5 Fmain()8 T/ W/ I0 n S |
{int i,n;- Z5 y1 f* N" Q) q: N2 b/ j
char ch[80],tran[80];
1 o3 _& i" S4 m; K printf("请输入字符:");
' x; |" C9 W0 c$ K' ~) T6 f gets(ch);
6 C" J! U. G; E, b% v+ w8 wprintf("\n密码是%c",ch);( v2 Z2 M0 E( f4 s& T |
i=0;$ N0 P, J N/ R J% B8 A
while(ch[i]!='\0')( n, M" G2 j# v) d/ e, C3 U+ o
{if((ch[i]>='A')&&(ch[i]<='Z'))
+ ~, q6 p: Z: r r7 t tran[i]=26+64-ch[i]+1+64;
+ P6 R& X) ^( Z7 J: g3 q/ Melse if((ch[i]>='a')&&(ch[i]<='z'))/ t: k; k6 u; a" ^6 u
tran[i]=26+96-ch[i]+1+96;% o, D% ?# ?+ r3 z
else' o4 G" b' r% P* m
tran[i]=ch[i];, r, }% K. ~$ Z. B' p
i++;
2 y# B' g0 E- D}
6 N; d" o2 N% T. I$ H, ?1 {6 n; Hn=i;
) _! I0 u) v7 I$ M: nprintf("\n原文是:");
3 ~. {) @3 a; ffor(i=0;iputchar(tran[i]);
7 e, _4 v4 j G}
: M& U" G& U2 \* Y7 e6 ~' j: W7.13# Q# V0 _# Z! E- C0 O. n+ v1 R
main()% [7 a# a3 s' Z1 j4 X" e- n' x2 U( ^
{# U% n) r8 z" x
char s1[80],s2[40];4 ?* W9 A) E. Q A
int i=0,j=0; U, F- L; a$ N0 m5 ]1 |8 j0 S0 N3 Q
printf("\n请输入字符串1:");
, U' c6 X: ~1 Y1 S! A& T, G scanf("%s",s1);, Y5 r7 c1 l: \3 q, P% C
printf("\n请输入字符串2:");
, X! m+ @+ y6 T scanf("%s",s2);
0 Y1 s" q5 O \- K. y6 a, _' z while(s1[i]!='\0')
9 s; ~* R H5 U% G: J i++;
& } e( D1 Y# r1 a) }1 ]- S+ `while(s2[j]!='\0')
' u' [" C8 |& @- Q" R# N5 U s1[i++]=s2[j++];
' j3 P6 f5 F- ds1[i]='\0';
& ~) {5 I; V$ T4 U5 m$ wprintf("\n连接后字符串为:%s",s1);
[, n' Z3 f' r. r! i. i* m }) K2 j" K9 G/ n8 Z
: j; i1 \% V2 @
$ f! G4 v$ s/ z6 `3 {- \
7.14" g% }$ y! n0 O- K
#include
0 v4 V1 n1 o8 Mmain()
4 U- c7 w+ S7 J/ J U. R{int i,resu;
) P, Y. H8 M( j char s1[100],s2[100];
& p: z0 o' S1 U5 [5 T0 B5 j& T printf("请输入字符串1:\n");
w- `- @ o( x0 d& s H8 G6 n- z# r* e gets(s1);
- w6 ?5 A7 Q0 u! X6 k7 p printf("\n 请输入字符串2:\n");* S* D4 e9 {: C" Y8 Z
gets(s2); E! h; u2 S6 L. G: U8 D. t
i=0;
' d. P3 v' {& t/ L* B: D; \5 h! i3 U4 v while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;! |3 r: u7 _9 I! f" t$ _5 O
if(s1[i]=='\0' && s2[i]=='\0')resu=0;3 @; w. _7 H+ K: m) `
else% _ ]" y. t# z! S3 [1 R. ~
resu=s1[i]-s2[i];& _6 S3 P0 W* J: O1 u# n
printf(" %s与%s比较结果是%d",s1,s2,resu);
% }: Q8 W2 p5 y8 M$ ~- l& z}! G6 g$ k- g6 m4 c
7.15
, r- Z2 K$ ?1 R: z0 N#include. Z# j# [* b6 R7 q# K: W- H
main()- i: h7 `+ i& m! ^( \
{
$ i6 z' U8 A* H" R& O char from[80],to[80];
( [8 L% m& X/ T' R int i;
3 u2 K$ d0 w+ C7 M Z* C! b, i printf("请输入字符串");' ]$ [- {% y/ |" ~
scanf("%s",from);4 l' Y! v! D& k
for(i=0;i<=strlen(from);i++)- `' ~' h4 A) o* _- o
to[i]=from[i];! J: M+ G+ I. X* |6 y
printf("复制字符串为:%s\n",to);' J1 ?2 F- ~, f9 p. T
}" ^" R: G9 e' x; m; c0 y. }8 q
- p. h1 c0 i- f4 M: k. w* [4 z; ?
0 B5 r0 q5 ~( `1 ~, c第八章 函数
0 u& c, s% i! |5 g: u8.1(最小公倍数=u*v/最大公约数.)
! E. m! E" a# k' ghcf(u,v)* x% q; A7 X* g/ J/ \. n X2 X8 o2 W
int u,v;5 N8 f" K1 X5 w- s
(int a,b,t,r;
7 X4 t% d% s' }5 J if(u>v)% [5 X- ]7 e: {% }1 y
{t=u;u=v;v=t;}0 G) \( T/ H2 k' U0 n8 f2 D$ P6 [+ _
a=u;b=v;
, m+ c" X* |( p: }4 y while((r=b%a)!=0)/ Z4 c! i2 k5 n7 _" r4 R( ^3 g
{b=a;a=r;}. A2 F6 Z% |# u _4 G: z
return(a);: B6 r( A5 C! e" \6 n
}
4 @6 I" \% B9 f- g9 C$ @; z( f$ r lcd(u,v,h)
* g. E' ^* B$ p. U# d2 f( C int u,v,h;( q3 o+ u: O+ B: ]; G5 t
{int u,v,h,l;
) Z3 c B( t# M* ?0 j scanf("%d,%d",&u,&v);# G6 I$ K$ Q, o; h U
h=hcf(u,v);: k5 e' q/ n! }8 ~, w! G0 A7 l
printf("H.C.F=%d\n",h);
- b+ o$ C" A" k& Y7 G; ?! q l=lcd(u,v,h);
# e' l4 B2 |3 ` printf("L.C.d=%d\n",l);) x, z* I6 y; T) X X# z) }* q
}/ w7 N+ e# }9 q& S6 g' c
{return(u*v/h);}
1 P1 G0 U6 Z' Y9 Q, v2 x+ z main()0 c2 y7 U% V; p
{int u,v,h,l;
& j8 v/ N6 j2 H& @: C# f: [: A scanf("%d,%d",&u,&v);
+ u* U) `- g G h=hcf(u,v);+ U- ]- F: ]$ m, g0 ?* f
printf("H.C.F=%d\n",h);
8 G5 a, W; [9 l, H% E5 a0 I l=lcd(u,v,h);
( r4 q% K. q, ]4 a7 C% n% b9 X printf("L.C.D=%d\n",l);: u' W. a6 I; w/ J. z" R) ^8 Z0 {
}/ K/ c6 Y t4 j$ I
# S* R' E2 p1 X$ W
Y& v3 s5 ] z
, s; ?% V- ^$ K; T8.2求方程根) x& G/ `; q" y% `% x& [
#include
% k" w* a0 {1 e. s, |; hfloat x1,x2,disc,p,q;
6 a% N* M- y5 cgreater_than_zero(a,b)
+ r0 }5 j! e" `+ }' S# Ifloat a,b;8 T$ [- V' O5 L) o2 Y J
{
' o: C# q. V% D* ex1=(-b+sqrt(disc))/(2*a);
2 `7 v2 a. d# R. \) H/ ]# @x2=(-b-sqrt(disc))/(2*a);
6 T! ~- q% K: X& w: ]: Y: ?}
5 L3 d' W( A6 p: K8 K I. kequal_to_zero(a,b)' U$ K3 l8 V8 z# l
float a,b;, S1 A" M1 y) I; k; m+ Q' E
{x1=x2=(-b)/(2*a);}" ?: E8 N. ]1 {* R. N, H" X
smaller_than_zero(a,b)
! A" A+ ]; H1 K0 ofloat a,b;
- V+ ^* \9 T! R8 d! }{p=-b/(2*a);1 L, i2 V9 D" k. p$ }
q=sqrt(disc)/(2*a);7 ]/ V+ r$ t; k0 P- Z* e: T
}, H" ]* _$ a) w1 ], E s
main()
# \' G4 W) ~ q6 J0 _{
1 s7 @" R: Y" Z7 Y6 T1 n' ^7 O6 tfloat a,b,c;/ C* J. Q( y% l9 n4 y; h5 B
printf("\n输入方程的系数a,b,c:\n");3 z6 k$ g. L, ~1 y; h' Y
scanf("%f,%f,%f",&a,&b,&c);
8 t6 q9 a: R" Qprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);2 P! C* V2 n' w( S
disc=b*b-4*a*c;
' w1 d, [. Q3 |" O: p% cprintf("方程的解是:\n");* t( l+ M, D' U, O) Y1 s4 ]
if(disc>0)
. O$ ], q/ A0 z. I/ q{great_than_zero(a,b);
|) z) p+ ~3 p0 pprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
) E% c6 X( a0 V6 X& F3 w& e( x9 `}
0 l' k' p( S" x, L0 I% ^: W1 f$ l+ m% Helse if(disc==0)" q& L2 C {0 y; z6 h
{
; Z9 B0 e8 z& C* v/ k7 f/ czero(a,b);
# s! b+ [; o1 h1 [% p I! o% I( dprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
2 F4 g! d+ d: X, b' W# ?% K9 u }: p0 Q/ T9 ?6 z0 S) {4 s
else: C3 A4 R o6 r$ A% k
{$ f0 R" x' Y' I0 n0 L. V* \
small_than_zero(a,b,c);8 Z V% m w" }, |: H% Q% ~4 Q
printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);( X T' E; E1 t
}
6 t3 I% b( }0 |1 x7 F5 s3 b* }1 V8 T}
; P% K% o x0 f/ y- r4 z+ T/ g% I8.3素数
8 l1 n) x8 F" L! c, I; I3 u#include"math.h": V2 D! L( L, |; m$ O, s; I
main()! i4 @6 M6 M( U. l) ~$ d2 T
{int number;
! V: I: x3 M' K, k! @5 q' c scanf("%d",&number);
2 @" [6 ^. b& A% R5 N1 S5 w if(prime(number))
' F0 ~! C" g6 ^0 @9 J printf("yes");
7 M/ v! p* s, W4 x# j else
+ e& I3 f4 s! }& {0 a1 { printf("no");" x: w7 Y6 N$ f) q2 r
}4 `( A/ q& N0 D$ c
int prime(number)
4 A8 T* S9 U2 C" v" p3 N8 N N; K9 Pint number;
3 i) H& @! J& G( e{int flag=1,n;
2 E8 B* o2 }/ h; M$ \/ x for(n=2;n if(number%n==0)' e6 B O/ @" x- R$ } o9 }; o
flag=0;
6 n" i8 y5 _3 [ return(flag);! J8 E. H8 f' j6 R A5 {4 P
}
3 t: s7 k7 |3 ? @5 m9 g% r _" ]. K6 r& `8 T
5 e' P$ Q, ^$ E, y
; B. C6 @6 w) k2 D& r8.4: S& }; T) |9 n( R/ T* k6 V. D5 f
#define N 3/ k) `, o2 W* b5 Y6 @
int array[N][N];4 y8 @7 c0 O Z
convert(array)
1 }- `* a4 g- w6 ]: @+ f* ]int array[3][3];
. ~7 m4 Q) G, ?% a1 ] { int i,j,t;6 F, a3 s7 m( i V4 v
for(i=0;i for(j=i+1;j { t=array[i][j];
! Z* h4 E+ V% ^, d% K! l array[i][j]=array[j][i];$ q% _- W. [7 {' x' Y
array[j][i]=t;
2 f5 {4 W4 H( [/ B }
1 e+ ?: T( G) x, U }
" a _! B2 Q7 @& wmain()
) K4 K1 P X0 |7 w# C) x1 _{# M- ~3 K' @ Y: C
int i,j;9 b9 w, [8 e! H' q
printf("输入数组元素:\n");- {/ `4 E% p3 u: d }
for(i=0;i for(j=0;j scanf("%d",&array[i][j];3 ?$ ~ v# p1 F" a$ n' g: {- f
printf("\n数组是:\n");
. \+ l) h4 e1 Y2 K) ]+ w& w; D9 F/ v* @for(i=0;i { for(j=0;j printf("%5d",array[i][j]);( y v9 Z, G5 R w9 W! u/ O
printf("\n");- H! ]& A8 F+ A7 M
}
& F1 q6 [1 _) M6 E2 E& J6 v convert(array);0 }6 S9 D% Z9 A) m
printf("转置数组是:\n");7 K8 l* K) }# L6 H
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
, M" j% R- p4 `$ H' h# w, c- o2 c printf("\n");
5 q6 H( ?9 p* p; |5 j. v0 V }4 @; t# R) C2 _ k
}% T! \" W. m) O9 f1 N
4 \. r) x$ I' K) f- p) X
. s4 l4 F& z3 L7 {4 i6 B- O
! Q% [; q( U: z/ Y s5 `8.5$ r2 T, W2 M3 c% H4 S1 L
main()
' g/ o4 g$ F( i/ y2 K, U1 i{% |( W; \- S' v; Y& P
char str[100];
, a0 b) ^# f2 L5 `* ~ printf("输入字符串:\n");
$ c0 @) D* Z& B! j& Y! N2 Y; ? scanf("%s",str);
0 q E# y# v: k' x) G3 q& Y+ Q a9 ^ inverse(str);! L5 p T+ W6 u1 q' E* e: E+ X
printf("转换后的字符串是: %s\n",str);
: P2 ?3 J! X$ z3 A}
1 t Y: t% z6 hinverse(str)& l( S5 M' H p
char str[];9 N$ f8 |, a9 F* x1 F" E) V
{
2 N9 U: _6 R5 @ char t;
4 Z9 a# ~* E" x0 R( \ int i,j;/ d+ ]' X' _& c0 J2 Z
for(i=0,j=strlen(str);i {
( K& ?5 q# l! H' j( F2 _" ~( V t=str[i];
' l! g4 O7 C1 y) M { str[i]=str[i-1];# Z/ ^3 ~- L1 H4 N
str[i-1]=t;
B8 m7 J3 K+ [+ w' Y/ w }$ L! s* Z7 [8 r& g: H0 l
}9 c, o. W7 o* B; J& U' ~* M" ]% e+ }
0 Y7 S, v9 s8 r5 z: }
' E: l9 s( e% F2 l+ ?5 r0 _% ^& }' j2 M
8.6
+ ]/ f {& u& ]0 Xchar concatenate(string1,string2,string); n+ b5 a8 j# Y3 Z, I6 P! s
char string1[],string2[],string[];: H$ K$ a; p/ X% Y/ H: O7 g
{
- _; ^3 T/ j3 x# s0 u. D% r* _int i,j;
6 j, q' g' L" L. v5 n" Z1 n: xfor(i=0;string1[i]!='\0';i++)) [* `* K1 q9 C& q7 }0 T5 h" U$ ]5 j
string[i]=string1[i];9 ?" c' Z% O$ r. T* b* l( H, R
for(j=0;string2[j]!='\0';j++)1 g2 c. a/ [, z) B }, y5 `: p
string[i+j]=string2[j];- s7 \1 _2 u' }, M6 w
string[i+j]='\0';
, \5 c0 |% `7 q& N) r9 Y, A* i+ A}
& I4 u. W& ~4 R% j- |! J7 \' J5 O ?main()
6 }: X+ Y; H+ B' w v6 o{4 q# S2 R/ J. k* X% w
char s1[100],s2[100],s[100];
+ [! S, Q7 D5 y: t% \ printf("\n输入字符串1:\n");) r2 b+ V: M0 y3 {
scanf("%s",s1);8 E+ K3 a3 A) ^" M2 `
printf("输入字符串2:\n");
- p1 A. W) V8 D! F3 I. Z scanf("%s",s2);- w, c, n \6 I0 w- r9 e$ ~9 S
concatenate(s1,s2,s);8 K. a* V, i) [0 _+ T
printf("连接后的字符串:%s\n",s);
6 L- o! p1 B$ @; h1 M8 x}3 b, P" \. w% y2 [" |* P; p4 l
+ ~2 Y- B& q1 [% d( D1 o) m+ O6 B* c; q4 a6 b, h( X
8.88 d/ @+ X0 M$ e/ I
main()8 X& J/ `7 g" T. j5 ]4 ]
{( K% H' q. ^& [* G& m( y5 J3 D) j
char str[80];& ~0 l3 ?# t- P B4 a
printf("请输入含有四个数字的字符串:\n");- o) L+ i9 Y6 i* v
scanf("%s",str);4 Q0 B+ l- |9 _& J* H/ c" J
insert(str);
% F* P& K6 x+ O- g V/ i- S3 d}
9 ~. D& u' W0 Qinsert(str)
4 d% s6 Z( N8 Y* r# `: j" A char str[];
& s( t' x. b& e. F6 F' U{
$ t$ b D; E+ Z& r int i;% Z' z3 H4 F5 F, k; f" L9 S5 E- A
for(i=strlen(str);i>0;i--)4 a4 W: r0 `4 H3 e8 X/ Q V0 K
{ str[2*i]=str[i];
0 m, N- b, Z, V! S: m& L$ N str[2*i-1]=' ';9 B( C; L: b" s' y" g( o) ~! Y
}
) A! Y* ?: O H/ N. Q6 ?7 ] printf("\n 结果是:\n %s",str);
5 P5 w) t( m- }7 { }0 }, h8 Q, x. d$ b
( K5 m. \' T/ E6 |. m; w
" O" K+ w% S/ M" a* q/ Z% K9 u) n& x9 R% z5 S
8.9
& ]- ^7 n; r. S: k" ^5 W#include"math.h"0 _; L! [; f$ y& n
int alph,digit,space,others;
3 P \5 o5 k0 Q2 I. B0 q2 smain() e+ @" P+ ^4 e" ~
{char text[80];
- s) u- S; r L, V+ O# V gets(text);) T/ i4 d& S7 @: {, p9 _
alph=0,digit=0,space=0,others=0;
- y" _+ V7 u9 B$ \0 j, x; \ count(text);5 @+ w* b d$ M# o6 q3 F
printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);7 l, u3 O, R- w8 ~
}8 B7 `4 e F4 X, u, v6 v; F/ t
count(str): Q, T& R. s- q4 O# g* G) M. f; s
char str[];
9 o3 ?6 B3 K% M/ ~8 l{int i;. T) i+ @3 t1 m
for(i=0;str[i]!='\0';i++)
( u+ e7 p) s. O: ] if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))0 d( }$ m# [' }" f% Q
alph++;, b9 F2 f: T- A3 K1 |( F
else if(str[i]>='0'&&str[i]<='9')
# p0 a/ K1 t5 M5 m) D- Q+ ] digit++;
) A2 p5 F" u8 z: { else if(strcmp(str[i],' ')==0)6 z% S9 |2 m8 l6 {
space++;' P# `% O1 P+ C
else) n$ ?( i( r! t G2 x5 V
others++;
5 D! T: O! ^) K" ~) k, K/ u}
! f/ x5 ^. R$ G3 P' N K, q5 p/ U& G6 m4 X' N
# q5 d5 x T. B5 d* ?" e8 E
8.10* r) J# o( b6 X. H2 Y2 j" Z
int alphabetic(c);
: B/ D& @/ ]; a& f* j8 x4 echar c;, D9 `, }- O+ H( x: c6 n, c
{
9 j [% O( G# b& k0 u; `0 m if((c>='a' && c<='z'||(c>='A' && c<='Z'))' d) {2 H& Q% o( j* n2 |
return(1);
1 N* }( w- ^. {5 U: p9 X$ Z else# K9 K, Q9 l5 `, c5 G, f- H3 g0 ^
return(0);; v( i& ]: ~1 B
}
8 c3 s R3 Q0 P2 G A9 n! B, O' W
int longest (string)
' n' I/ ?5 B1 V# N* Gchar string[];
5 J) Q; G( I0 n# O. w0 ]{
5 Z& \' A8 |2 M1 K int len=0,i,length=0,flag=1,place,point;# ]- v0 s. y: B; p0 H0 p
for(i=0;i<=strlen(string);i++)
( |* I2 i* }% Z9 _3 W( o2 I if(alphabctic(string[i]))) l P9 y- s: t- e
if(flag)0 b; d6 h6 j1 _& G3 _
{
% {9 {; m- ~" A" Y6 i5 j/ k' x point=i;9 M4 N' T$ o' K" r t: |
flag=0;
- d8 U p6 Q' k# q: ]) g7 w! i" k4 H }. Z2 w2 X& z1 ^, ]; \/ P: {5 F
else
0 v4 W( i7 c0 V* E0 [/ P len++;
' T* R2 s& @# Y2 [7 K else3 K% V4 l0 Q- A2 E+ d- |( U. S* b
{ flag=1;) O0 x& x1 P; d
if len>length)
% A; N& C% ?8 k+ l( a: y1 e {length=len;! C" W. K8 T) t B1 ?/ j, c1 u3 M
place=point;
8 K3 D, i5 V$ z% Q6 m5 N k len=0;( l! v; `" v( \5 y5 @
}
5 o0 J! {+ o+ S/ J7 ] o" d# l }' g: p0 l3 J) s: g
return(place);
O2 J G) Z$ k% t1 O }
k* J1 P3 F' A1 J l- m) Nmain(). Q! o c5 V1 _: X& l
{. f# q( l8 t/ N! y' I. I
int i;5 }( G: O+ h/ f# B/ R5 i
char line[100];
7 I y! R G% i: ^, _" E+ ~" Y; T' g/ xprintf("输入一行文本\n");2 L' ?: Q/ Y; F) u. _$ p
gets(line);6 l g" V" L1 D* w8 c
printf("\n最长的单词是:");( @9 g" I7 q; s/ i# N
for(i=longest(line);alphabctic(line[i]);i++)2 {; X3 @0 m5 o& F3 h- A
printf("%c",line[i];+ f6 W" J6 n, E* f" r( e
printf("\n");( `/ }6 w! K/ }& Y1 W+ h
}
& @* T. B: F5 @" A
" x: s( Q: A1 d+ G2 @7 B% I+ j& L; H% {' e! t
0 r; R. T/ Z. n# l! f& K8.11
4 s3 _5 u3 i/ N6 L#include
' Y0 s' w A5 A. h) N* N$ e4 R) Q$ {; h) l! ~8 c% v( l- b7 u
#define N 10
- I0 b5 m' c* {- Q) q0 {! `char str[N];
" T) J g5 U% D& X- Kmain()' _2 z0 X% }# P% G: g0 Y/ F T! T7 E
{& |- Z" Q2 Y( H- p
int i,flag;
5 {5 _3 s, }6 Xfor(flag=1;flag==1;)
) a+ d3 ?$ F% ~" Y{
9 }7 \6 E9 l/ c" y2 b7 U printf("\n输入字符串,长度为10:\n");! p" X! f* U( I+ }+ d3 Q
scanf("%s",&str);
' M% @* V5 C) s% ?1 O5 \ if(strlen(str)>N)" Z. ^( |" v9 ^3 C
printf("超过长度,请重输!");
* q4 W. M7 q# X7 [" d& T* a$ @7 x else" }2 x M1 t- v1 ~
flag=0;
8 \& }, ?' [7 L}
7 q f' S, }% [5 ~& o' wsort(str);& f3 W7 i' [/ l) D, G+ O: q+ k
printf("\n 排序结果:");
0 W/ x6 C. D' u% S+ }! yfor(i=0;i printf("%c",str[i]);
# k2 @2 {% B* a3 Y}
# P; L" F) Q& o8 Q0 R4 ?; D- Msort(str)) q/ @) B) n3 c. @4 N& D0 ]/ d
char str[N];( Y) j# N5 \% |# X; @) l
{4 x$ Y/ q% K# F$ T# s
int i,j;
; ]8 L6 U- M% Z# k4 d- echar t;
* w2 g, w$ `- Vfor(j=1;j for(i=0;(i if(str[i]>str[i+1])
( r3 [2 j9 y+ p5 ?: i: S( E$ [3 g { t=str[i];
* I; a" ]) f) ^2 ?* M2 r str[i]=str[i+1];0 {+ N1 I5 w+ t- }5 V5 @8 v
str[i+1]=t;
0 u# W! n* w+ M! X7 ^% B }
1 Y! a8 i1 u4 K1 A% ?}
6 y, T: Q# ]; d+ c) Y, S7 l8.120 T% j, L6 A( ^! U
#include \9 l' c' t/ R# ^+ y7 O
#include) ^; w4 b& s& c. L6 p; I) R' W: h
float solut(a,b,c,d)
+ {) x4 m7 h: {% K: efloat a,b,c,d;
4 I8 m' u4 p, E7 \0 B. Z4 |{float x=1,x0,f,f1;
. u/ ]: \8 f* w0 y G& w1 J do6 y( m) l; M7 z+ n/ F
{x0=x;/ _3 y8 L% q6 H
f=((a*x0+b)*x0+c)*x0+d;0 J6 w# H; P* G9 }& ~
f1=(3*a*x0+2*b)*x0+c;# b, X/ F6 D% i3 c
x=x0-f/f1;
) E, C1 V1 E* p! o }
# r; I& W% {9 f- l# A! l+ x7 T7 z while(fabs(x-x0)>=1e-5);# L6 x- X% }5 U# n7 c! k0 ]! j
return(x);0 S% m2 `& ^% k8 v2 U9 z, T$ p) d+ P
}3 z5 c' X' ]- R7 F: u; L
main()" v4 q0 L$ _' v. i) F: x A. g' R
{float a,b,c,d;
/ G3 p4 _& O H8 B c k scanf("%f,%f,%f,%f",&a,&b,&c,&d);- U- u/ t# x* J% |9 c$ P
printf("x=%10.7f\n",solut(a,b,c,d));
7 t K. h) P* t+ C% h}5 ?" T' A$ z' ?6 a1 t/ r
8.13
# I8 T4 ~5 ~5 b; e1 P v#include/ _, _2 _+ O' g+ M0 V2 p3 D
main()
$ C" `5 F+ l4 k# L3 l( o8 ~{int x,n;
9 I6 t% @! p* _/ Q float p();% ~8 t0 Y5 A. o: _5 i+ c- n
scanf("%d,%d",&n,&x);. ^1 ~3 x1 T* C, Y
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
6 |8 J" ]; N$ x3 a) S) _# i}, u% y8 H& U$ i& y
float p(tn,tx)+ p5 ~# F# P1 d* ?# q# q) S; `
int tn,tx;" J3 A! k# k2 z0 C7 U3 e r
{if(tn==0)2 x- F1 S* Y$ M
return(1);
( Q' B* W/ D1 G! X$ C! c3 Y else if(tn==1)/ ^! q" S0 y1 M: Z1 H2 `- S
return(tx);
# j) p" F8 F2 }. W$ |% t- K else: F4 t/ i3 |& w! @% r
return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
2 ]1 E: m/ C: X& f0 {+ v6 @/ ]8 v}
) i8 {) Q# f! x. u# D: O3 y% L8.14+ z8 A8 S* b0 k
#include "stdio.h"" C K0 M g t0 A& l) `
#define N 10
& E- e8 ~( p5 M# L9 D6 d n6 L#define M 5! I1 z8 u7 D& h- ~8 N- i
float score[N][M];
$ I5 V1 ^* ^, l# kfloat a_stu[N],a_cor[M];
: M$ s+ J9 ]4 _9 ~, R. q" Ymain()
' Q+ ^6 d: `5 `2 w4 G+ \( [{int i,j,r,c;
' D4 p3 c2 P: r. D, O7 E* A float h;/ F! @8 N7 R. u& ~1 c- K- S
float s_diff();
6 {4 ^( f" y3 C float highest();- c9 U( e0 ]# \) F5 v
r=0;
1 R, Y' S/ Z# C/ h" W& l* v c=1;
. S# k, [/ \$ n3 C! Q: F6 c1 U4 w input_stu();
& Z" ^3 B. |$ v avr_stu();( }. ^ q5 m; }/ y/ z- H# U
avr_cor();2 w0 r F5 `7 K9 v# k& P/ }
printf("\n number class 1 2 3 4 5 avr");
: D' m% I S! C( z5 }$ \! t for(i=0;i {printf("\nNO%2d",i+1);4 A9 X: f# H+ e2 h
for(j=0;j printf("%8.2f",score[i][j]);
% T3 b& j# J8 b/ U( i printf("%8.2f",a_stu[i]);
7 u( V. u7 U4 Y/ W/ z- X }
9 D5 U- ?/ ]1 L0 q' R8 t% j printf("\nclassavr");
. j8 a) T! t$ d) u for(j=0;j printf("%8.2f",a_cor[j]);) ~5 Z9 W5 ?$ c& H7 v! j: }
h=highest(&r,&c);" i, c! Z3 p- ?$ E) G: i( \
printf("\n\n%8.2f %d %d\n",h,r,c);
+ ^' m) |8 H* p/ o& _( R5 Z printf("\n %8.2f\n",s_diff());
* V( ~" |+ D7 B( v, D}1 m/ `5 s9 H4 e# c" W- j/ x
input_stu()
1 d6 G/ ^1 F' @/ P. i0 g8 k6 N* Z4 R{int i,j;) V. @2 H9 d& R* R" K. ?9 g7 M* m
float x;5 L4 x _8 o6 l, n! ^
for(i=0;i {for(j=0;j {scanf("%f",&x);
, i d% v$ q1 f% Y' M2 X, X score[i][j]=x;
% w. ]) \* M; Z( s2 y+ M; Y }( W+ v; U1 [7 `) e0 A
}
4 a# m9 ~1 w, h4 @; A}+ n/ J" `& [ S" ], B0 o
avr_stu()
" V% Y3 o7 `( g( z: c4 ~{int i,j;- J# P# v6 ?; B, J( W
float s;, }5 t7 J0 \. S+ ]5 j& W2 H# f
for(i=0;i {for(j=0,s=0;j s+=score[i][j];
; S( h& ]# n! L1 m% E, n a_stu[i]=s/5.0;3 K; t4 o/ L9 f2 w; O/ H. B
}& D! F7 {( z1 s$ g
}2 a9 ~: O7 V8 ]# f
avr_cor()) Y, p' o6 s0 y0 P' ~! o
{int i,j;& S/ p- G @/ d. Y6 ?
float s;
# O; H/ e! {' F7 o- ? for(j=0;j {for(i=0,s=0;i s+=score[i][j];$ C) w8 d$ o: O0 I: i
a_cor[j]=s/(float)N;
9 c" H& c! q- j3 h+ o4 Q }3 V; U+ n* R8 Z0 O* Z5 N& E. O
}
- a4 ~# [5 X0 h( I2 K* Sfloat highest(r,c)& B/ n8 o) Q( T+ c9 B
int *r,*c;
1 g1 ~1 f& p4 e3 x/ m{float high;8 b, L* M3 [8 _; M* K4 Q% s
int i,j;2 Z/ t0 o6 p( E/ K# q
high=score[0][0];
+ u9 ^; g9 ~6 A! T) O j1 I3 @ for(i=0;i for(j=0;j if(score[i][j]>high)
8 r& X1 u% A) p( V, i {high=score[i][j];) j6 H J2 w, [7 |% n
*r=i+1;
+ j/ A2 N* b% V *c=j+1;1 _3 G% i& ?6 |& W. {# D4 y
} |! J. s( K$ k5 a. f. L
return(high);
0 t6 h& r+ `* X5 H' W' l}
! q4 f1 W; S$ V1 N" Vfloat s_diff()
8 J. b. W* n* }8 R4 ^. V# j{int i,j;9 U! [; c {/ I$ s
float sumx=0.0,sumxn=0.0;
$ ]3 ]; [ q$ y2 P, F: E! \ for(i=0;i {sumx+=a_stu[i]*a_stu[i];" b, v6 D6 V) f; y3 ?
sumxn+=a_stu[i];' V8 j8 |( E: J( ]( r$ G
}
5 T% w/ G* H6 y8 l& q return(sumx/N-(sumxn/N)*(sumxn/N));
, u' f9 q* |3 [" B" S# }}
, Z( B5 M# D" x) f. L5 R# V1 k& D7 Y8.15* O; n. t1 b" ?% O; `, R
#include3 D2 q2 [6 {" n, a) t' O. ]
#define N 10
3 g5 }2 p+ O$ _/ \, uvoid input_e(num,name)% @# Q: r0 ?5 V7 p! U T
int num[];& [: h+ |0 R5 q: ~: H- E9 [
char name[N][8];
! b) e6 O' }+ u( _) Q9 o& E! ?8 Z{int i;( K; B$ E) u8 J: _
for(i=0;i {scanf("%d",&num[i]);
1 s+ Q$ v! s, |3 n gets(name[i]);; q$ ]4 Q( B; x" T
}
4 @4 K5 W9 d2 B" c. m}. Q. o0 n5 W$ x2 \: S
void sort(num,name)
9 M2 `# N3 d2 C9 F. J/ Cint num[]; E) c) r. m0 Y, F: Z5 F
char name[N][8];
0 W4 `! `& m& Y4 l1 d9 l _3 I+ Q2 @( {{int i,j,min,temp1;
% C$ X N; }: \/ N& i. S8 ?) Z, W char temp2[8];
7 w( o* e% h, m% ^9 D0 @ for(i=0;i {min=i;1 v/ g5 x7 w1 S$ A3 Q( ?* }
for(j=i;j if(num[min]>num[j])min=j;, N/ e ~ O, s {
temp1=num[i];6 L8 Z0 O& _. p, w6 t G
num[i]=num[min];
* f" E# L2 X9 y num[min]=temp1;+ T% s" e2 X g% `( v
strcpy(temp2,name[i]);, l9 a$ C0 g3 T! A* u
strcpy(name[i],name[min]);
/ C1 @, ?& S5 \$ P; M( s8 p: V, O strcpy(name[min],temp2);
1 O6 H- B, N* s1 y$ P" {$ i }
2 b; Y- W& R' C$ r! |; |' G L for(i=0;i printf("\n%5d%10s",num[i],name[i]);
; w( \( j6 u5 C}0 X$ ^' Z i2 L: m8 K/ R1 Q& o" l
void search(n,num,name)
) u% Q/ G( `+ A1 dint n,num[];
( F5 p9 \* i4 w2 [* i+ ~7 ~char name[N][8];* p8 f6 X" e* w; h
{int top,bott,min,loca;5 i5 _: q. R7 `, o
loca=0;8 T/ B# |' H+ b4 @. ?1 v7 F! F5 j, J
top=0;' u4 g6 a3 N' c1 x4 {! ^) w
bott=N-1;
/ W. h7 c% [9 N# R if((nnum[N-1]))
& q* x# x6 a; h* P8 m loca=-1;
& F6 u& X% I+ y6 Z/ w2 @ while((loca==0)&&(top<=bott))
$ E: `. K. F8 q# I! N+ k {min=(bott+top)/2;1 V9 l; h% T& F3 B
if(n==num[min])
& {1 `: A" O/ ~/ a6 C$ e7 V. Q7 N {loca=min;
' W1 p5 y& k! n! M4 Y' C* R printf("number=%d,name=%s\n",n,name[loca]);! ?7 G$ I: n0 A- I" r ~3 `, |1 `/ U. F
}
/ {; `0 F3 W7 M: w else if(n bott=min-1;
4 k: i( x& ?2 r, r else
% q- n8 |' a ^) N K$ L top=min+1;
! A7 i9 ]3 k- {% o: C$ r, }9 g( ]& B }
" c4 x$ J. W+ N: E! H6 z if(loca==0||loca==-1)
+ ~. u- }4 B u printf("number=%d is not in table\n",n);7 W6 V" f/ {$ Z
}
z1 j# Z! I V4 Q) kmain()1 c' u/ Y! z4 l4 Y1 `8 f0 ?: w8 v
{int num[N],number,flag,c,n;3 n* D& ?' P1 o# s8 @8 S( q
char name[N][8];
7 i [8 Q, A. m* C9 u. x& k% c( c6 m input_e(num,name);
" c. t b/ ?& G/ q* d sort(num,name);
* P1 P1 g: m, }8 ^+ I for(flag=1;flag;)
) l: M3 M4 {" N4 e u {scanf("%d",&number);
1 {! B2 t& r3 g, \8 X. g7 J search(number,num,name);& r! V5 K. \. y
printf("continue?Y/N!");1 t6 K4 a2 T( S. ?! `- P9 S+ Q! D
c=getchar();5 j; Z" L1 M, a- `
if(c=='N'||c=='n')+ r+ T! j" o; N" `0 a+ ]
flag=0;
# T3 t5 F8 l+ v4 R! r' o! D7 o6 b }9 f9 r' I$ v% O, k% E7 i
}( R" M' Z x2 I+ x& c9 N7 x$ t
" i3 B6 H) G# t
8.16: z% ^. a* u; s9 \" P1 K
#include
0 ~" {- N6 [! S: `#define MAX 10009 V" D2 @: v, d8 p, O- M8 z
main()
5 U3 y( C% W0 \3 G5 o5 e{ int c,i,flag,flag1;, f* ^" Q% ^/ v: O
char t[MAX];. A% S7 ^0 u$ P: n
i=0;2 |" O$ q& Q3 Q
flag=0;. C5 Q+ Q) `. p0 o8 J
flag1=1;
. \1 ]/ h4 ^# t. g printf("\n输入十六进制数:");
# d" F& ]: q$ ] while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')4 `% B+ X1 M3 [, T
{flag=1;- b9 H( p% x& ?0 ?9 x5 V
t[i++]=c;# q3 Q1 D( B# Z% Y5 l2 ~
}% N( V/ P$ ~: u( {
else if(flag)
: `; j0 @) t, l: m3 q5 v% R {
( E) t! x" N8 d1 E t[i]='\0';8 c Y7 v, Q) _
printf("\n 十进制数%d\n",htoi(t));, } K8 r1 o6 f$ ]3 P) m
printf("继续吗?");
. w' S, C& n. }' H4 ?* j c=getchar();! _% t# r0 [; c. u1 i4 S# E
if(c=='N'||c=='n')
: K6 @ X# `0 E flag1=0;
( {7 h9 c- A1 l* R3 k' ?# f1 e3 d else
7 p+ H! M1 ]% A( L) p {flag=0;' K; K4 {+ U0 j) W G; H
i=0;# J. s% D# b3 O8 @7 Z/ d
printf("\n 输入十六进制数:");: \* g9 y' `9 h; N* y/ Q- s% F
}
5 j. D5 r5 d# j6 a: j: D}
7 h: Q" e! z4 U0 B/ `6 ?}, r2 @( B: b3 ~
}
% x5 u, b& e4 c# V1 Ohtoi(s). W& M8 Q* j4 M& s
char s[];; ~) }+ D+ ^9 u# R: u; w
{ int i,n;
* |5 e) y( E* ^8 i' r# s, Z n=0;: X9 F# _5 z$ `
for(i=0;s[i]!='\0';i++)0 G3 E! ^) _' }5 O2 H1 L6 v( L. v
{if(s[i]>='0'&&s[i]<='9')
7 R: R" o2 b2 w( d n=n*16+s[i]-'0';: {9 E, | Y' j
if(s[i]>='a'&&s[i]<='f')4 _1 H# N3 `8 F/ N8 {; J
n=n*16+s[i]-'a'+10;5 S- I- ]; ~" [" ^
if(s[i]>='A'&&s[i]<='F')! q6 k) i5 Z, {$ i* B, H
n=n*16+s[i]-'A'+10;4 ?( R1 [" K! p) G3 e+ T
}
) N, @3 A L3 P return(n);6 z4 R# n7 b1 U
}( v* G W0 N8 N4 d' z
: U' }7 V* w4 y* F9 `, R
' r T7 w6 X' ~! e" ~( P; m+ F5 X7 w8 x
* [- H: g8 c. |, l
8.174 I+ t2 }" ^& X/ S8 J
#include1 h1 c: d- \. W4 q V* g
void counvert(n)
% u1 P7 j6 _1 F7 J4 T' Jint n;& j1 E/ u! ?/ W
{ int i;
% X6 V) f, |& _1 I1 a if((i=n/10)!=0)% N- y+ [0 ]* F1 _% {4 j
convert(i);
+ a1 D v) {0 k5 q2 P putchar(n%10+'0');0 _1 j" U: J2 f
}: x- X* @# _# @( e1 _ d- L8 P
main()
+ ~! A1 Y0 _# x: W9 g; t, a{ int number;% K! ?0 f4 N9 n7 H! B' o" c; s
printf("\n 输入整数:");; n/ I# y g' V) a, L
scanf("%d",&number);2 w4 h6 _5 c2 y4 k4 B# u" c4 Z. K
printf("\n 输出是: ");" t8 F7 A# {& m$ Q# s" _
if(number<0)
, U8 j' d; `% O; {: Q { putchar('-');3 J$ Z. g1 e% |
number=-number;: k5 a3 N! L9 F8 |1 u
}! L- f, A. T! I5 u5 |. C
convert(number);6 _& e' H7 c! s2 Q9 U
}' [# k4 R0 I# z2 y4 V6 z& }
0 K* H- Q8 A$ p% z9 H/ E
+ c$ x i* S! [9 I9 ~
; a7 i% k$ ^, p" R5 A8.18
7 U8 T$ Q5 P4 s4 C( Y, Q3 \main()" c( {1 q% V3 o& F
{
) Z+ ~) O4 q) P' G) C int year,month,day;
9 ?6 D$ [3 b2 d5 d4 C; W- f3 s int days;: n% c/ \- s% I7 C2 j$ F
printf("\n 请输入日期(年,月,日)\n");
3 _: d- o) M( S5 Q) g1 U" X9 K scanf("%d,%d,%d",&year,&month,&day);* i3 t u0 z- a9 f/ @
printf("\n %d年%d月%d日",year,month,day);
/ i% Y) y' l9 y4 l: f days=sum_day(month,day);* R: |4 h& f: l+ Q8 M
if(leap(year)&&month>=3)
# ^, ]$ r/ C3 U( s& J6 }% N$ S days=days+1;! k1 h( A' P# J& _( k. v
printf("是该年的%d天.\n",days);) I; H: o: P, {# x
}
* B; G1 c* T/ \- a+ {5 u, \ static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
& K' L# d; r2 K- f4 Y! d int(sum_day(month,day)
& j) e7 k. N; _7 D. b int month,day;* u ]6 C9 c8 I. I/ f
{
, l: C! x, L* k8 H* L int i;
" F- | u" G0 H5 f0 y/ c1 q for(i=1;i day+=day_tab[i];- `4 |/ @7 E/ ]4 m8 y
return(day);" K$ l& n1 z# }/ n% q, k6 S
}
' ^. ^& D9 f6 `6 G' _4 n3 ~ int leap(year)3 Q' D3 P1 p% x; |4 } j$ Y
int year;
) b) B: D G& ] U/ H) y" { {/ r! s& f- x* a2 {! z+ |
int leap;, d/ N) @) D$ ] o& L
leap=year%4==0&&year%100!=0||year%400==0;
+ t/ f3 R1 E, b9 z return(leap);
# d+ K- o& l6 |1 O4 P- L c }
/ | O4 K' j& e6 L! D+ }! J第九章 编译预处理. x8 t+ u$ H' g( J
9.1! T7 ?- Q; v& f8 f- C
#define SWAP(a,b) t=b;b=a;a=t
4 y, y, d5 Q6 L I- X. bmain()% _1 Z# u% r1 A- |) B# l" s4 G
{
4 D" j" k& l. |/ _3 l# |' P; q# wint a,b,t;3 B( ?2 ]" s7 z
printf("请输入两个整数 a,b:");
- L K F0 c* q1 rscanf("%d,%d",&a,&b);5 H% b) z# \7 [/ @4 @
SWAP(a,b);
$ ^& z. U- f6 O: r9 ~printf("交换结果为:a=%d,b=%d\n",a,b);, a5 M% Y; h0 I7 h3 C8 l
}
; u: [3 _/ A! z. f4 a3 W: ] , U% ?3 V" N1 b8 m# Z
% _" }9 R7 e3 @/ J
9.2; j0 ]- i6 w4 G, K) E- \. F$ H
#define SURPLUS(a,b) ((a)%(b))$ R/ c8 z9 \5 g! j( g% T
main()
2 `% [ o6 M8 m+ Y$ B8 y: z {
' E7 n j6 u3 D+ @5 F int a,b;0 H% r; y9 ^ U1 [
printf(" 请输入两个整数 a,b:");
6 s' i: N# w+ T& [0 y) b. e scanf("%d,%d",&a,&b);* u& f# \1 v. Y
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));8 E4 k$ H& a; `3 B
}
) @$ D8 w3 u0 c* u0 _1 h: X' S) e8 v$ w% m
( l p5 i; Z% H' {1 q# u9.35 _% o; o) C$ H2 c1 _2 a* r
#include6 Q- ~7 s Y* S
#defin S(a,b,c) ((a+b+c)/2)
/ w- y4 [- _9 E* K# `#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-
; D% {8 i% S! z# M- Q7 _c)))
) k) q) u7 u' D- smain()
) @, `3 f- h4 I$ K$ a {3 P. Q. h; g5 T! f: ]7 m3 w
float a,b,c;2 ?( X0 J+ L" K' y3 t& W. B
printf("请输入三角形的三条边:");
\; {: j4 F) \3 Y; ~ B6 j scanf("%f,%f,%f",&a,&b,&c);
0 |2 W' ~+ P) j# i5 R if(a+b>c && a+c>b && b+c>a)
$ T2 G2 g: f; P printf("其面积为:%8.2f.\n",AREA(a,b,c));$ M) v4 s2 s5 K; y4 r
else
- T8 k/ h8 k F+ G8 ?) y8 M printf("不能构成三角形!");, Y: K- I- K3 U& v3 c6 _) D& Z
}6 b5 K& g+ n1 N1 G
9 N) Q9 c' t% `
% k+ Z& _* V3 E/ h) H& R: C' Z. ]
0 c R5 w& L/ z' B9 Q9 v9.4; a' J* J! G+ \! D$ z& _
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)* I% o- ]" F' ]0 ?
main()
( F! h- ~. K2 j6 y1 L2 u+ ~: g {
, n3 [! w, N" W) W int year;
3 R7 ^- d" c R& }3 U printf("\n请输入某一年:");
# J* i; @3 F! Y scanf("%d",&year);
: s1 i# y; @& O; l* h B if(LEAP_YEAR(year))/ [- G: g) v* e- I
printf("%d 是闰年.\n",year);6 T; H/ a5 p C1 Y$ `# B
else* r2 L5 ~6 g$ y' F/ `# w
printf("%d 不是闰年.\n",year);
+ R3 ^( k" K. W) v, Y2 E; F+ v }7 d! G7 V1 h& e: ]' T3 Q# J% e
% ~9 G" v! } u9 |
- ~# t2 M0 _; t7 J# I5 k6 U6 N3 o7 `( X/ e: d- g
9.5解:展开后:3 m% z3 f" }; I- W
printf("&#118alue=%format\t",x);$ I# v, Y; G9 }' Z i/ o+ G. D
printf("&#118alue=%format\t",x);putchar('\n');* N1 b/ m: X9 A) n* B/ z9 n9 H
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');$ [4 i; n7 A: O- A2 K% \6 ?5 j8 I
输出结果:1 r- y( h. G4 ~& F" L
&#118alue=5.000000ormat &#118alue=5.000000ormat1 p9 H8 v$ Y9 w6 I5 n' v
&#118alue=3.000000ormat &#118alue=8.000000ormat5 u/ S" t+ H3 R& k" L2 N$ J
% G3 P1 Q- R: y* t; R- O9 l
" K* Q, A. ^+ g9.8' {7 E( Z6 h* a2 D8 b& O
main(); P6 c: w+ V, W, b2 B3 e
{, |& x* a ]( @& n6 ]- _% T/ g
int a,b,c;% ]5 F; b' j$ B
printf("请输入三个整数:");) d# E( J6 Z. {( _. s. X; p7 c
scanf("%d,%d,%d",&a,&b,&c);4 w! {2 T2 P$ ~. F3 |! [# W
printf("三个之中最大值为:%d\n",max(a,b,c));
! O( W: w% m$ P; M b+ L }. _( E4 k4 h+ h( x% s
max(x,y,z)4 _9 p: ]0 Y3 _+ x0 P
int x,y,z;) i0 O1 ~8 `! i! R1 @
{, L8 T& I# Y& s" j) C# D
int t;
0 W' t2 U4 y1 s: l+ x t=(x>y? x:y);, ?) n$ F. k) A) V* n
return(t>z? t:z);( z' k- ?# P% U! B0 ~7 o" J/ R( \
}
; R) s0 K* o: V }! u
3 o" P! q' s4 e
1 U' r2 f* C+ ~. K8 E+ r. [, y# R( U7 d I( E" C
9.10
" H/ n( s! V6 b- c/ B! w#include
6 S7 c9 o5 d& Y( ~9 c+ v#define MAX 80
% H2 E6 N2 S) t#define CHANGE 1) j) e! s, a$ C8 l% X7 k1 \
main(). R" E `6 W7 u2 }
{
3 i: a& R+ x* b" ]( y9 r- L, ^ char str[MAX];
+ ?3 t9 e5 x9 T6 O) o( E7 c int i;
0 I$ ^" E3 h1 T9 R+ e5 }+ o printf("请输入文本行:\n");
% P. s0 i% q& U1 C% h4 }0 w. f scanf("%s",str);+ E3 F f' W- Y4 O1 @
#if(CHANGE)
" j# M: ?/ `: T: z1 g) m {
8 @( W3 Q- y7 e2 W2 u; m) _+ W- s6 a for (i=0;i {
9 F& f, N4 L7 i, P! u$ v if(str[i]!='\0'
& o9 j9 ^% c0 ]# _+ h5 j3 e2 } if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
7 O1 [# ^. {, |$ I# R, y str[i]+=1;
( s( J7 A, H7 y1 U4 [5 S( K" y% S else if(str[i]=='z' || str[i]=='Z')
3 Z, e/ _; b: m& R! g str[i]-=25;* o: R( l$ `0 q
}
! R0 A* c4 }3 ^& c [7 V}0 S- g1 U+ b) S5 P6 ]$ A% k3 ]3 h
#endif
+ y% e9 P3 W) d$ T$ y, gprintf("输出电码为:\n%s",str);) @1 I7 k2 Z, v$ d& t
}
( Z% s. a2 a% T8 s3 x1 P第十章 指针
$ z# G( ^2 d/ A' p+ a) l9 }/ ~7 s10.1
* B1 d# s$ `! l% b& o+ u& A9 Rmain()
, Z4 G& w! y0 J; o! F{int n1,n2,n3;
5 h8 t; ~3 L: s! g- `0 h4 P3 \ int *p1,*p2,*p3;8 Q% R" Q% G" D( W: x
scanf("%d,%d,%d",&n1,&n2,&n3);9 w% D$ ]4 q2 F: Y
p1=&n1;
" e9 n3 `* G( _' ^% v# _ p2=&n2;
& U5 D. j+ i* _- c8 P, n+ J0 B p3=&n3;
. L P/ ?( a; w# ]8 g' Z if(n1>n2)swap(p1,p2);
/ i6 {# E9 k* {, `& I" i* a1 P if(n1>n3)swap(p1,p3);
; N' k+ t$ P# {5 X8 }3 @ if(n2>n3)swap(p2,p3);- b# T* S0 _( l! H* l$ h4 U
printf("%d,%d,%d\n",n1,n2,n3);
# S/ n" x% q! S}
+ K4 W8 R# i" sswap(p1,p2)6 Z& s- ], t4 L/ E; R
int *p1,*p2;
) h: B& a, F- S9 ], \% k C{int p;/ e/ N; L& z9 ^5 P/ y& w
p=*p1;*p1=*p2;*p2=p;0 y1 G% j# Q& M( L/ }- e
}3 L! k+ ?: u& G6 z `
10.2
& B6 F+ Q+ J9 s, zmain()- m3 Z$ o: [7 J5 G$ ?$ @
{char *str1[20],*str2[20],*str3[20];2 W6 s, |, Y. I; J2 B6 m4 ?7 _
char swap(); c! W! N0 Y6 p( ^, J. i. `
scanf("%s",str1);6 h# T/ C/ ?# B' t. [8 {2 J# {
scanf("%s",str2);
# q, N7 M$ `# h: b$ |8 E7 j scanf("%s",str3);
9 W, _1 ?6 p6 D3 ^ if(strcmp(str1,str2)>0)swap(str1,str2);9 Q0 b# I9 y* ^# {2 P) ]# j! g& i
if(strcmp(str1,str3)>0)swap(str1,str3);
) P: M" H! J b8 j$ q if(strcmp(str2,str3)>0)swap(str2,str3);* ?; {: l1 x9 e. R! h( G* z
printf("%s\n%s\n%s\n",str1,str2,str3);
5 Y, b4 a7 w& `* R}" y; Z. ?' K* Z( x' p; i
char swap(p1,p2)8 ?- i, g u4 C
char *p1,*p2;* v' g! I9 p, P
{char *p[20];
1 E; f* a/ h( p strcpy(p,p1);
" l w ~& x% C! F" K8 c1 d strcpy(p1,p2);
2 {. S; x3 j! X5 u/ Z+ d& l strcpy(p2,p);7 d2 t# G( _) u5 P4 l
}
0 ?/ S+ X: E) o6 f9 n2 j10.3
% ?! r1 B2 M ~, n/ H5 |1 Kmain() f% _+ f W, ]1 X. O
{int number[10];$ o& E" n5 n5 Z* S; l
input(number);
: X. v5 T/ X$ {5 _3 z* i max_min_&#118alue(number);
2 l, Q! V* J! J3 a3 a# b+ Z$ m# D output(number);
! L& P- c& p. n; Q" c* F. c }
7 j& v: w& f7 binput(number)8 `; D9 ^. M3 Z5 W
int number[10];) K$ I$ p8 | P
{int i;
( d8 v9 ]- [; }6 e for(i=0;i<10;i++)- z8 R+ J* S5 u5 G
scanf("%d",&number[i]);2 x, z0 U; [/ e4 [; H- Y9 V$ A
}7 f- R8 j0 K* M& X: n% n! F8 D
max_min_&#118alue(number)
" N' ^5 B+ r) U# s9 B# C7 q6 E {, Dint number[10];
) X6 g& n) j( M$ d- k{int *max,*min;# g/ h' v( U! L; u/ D J- b5 W
int *p,*end;% @ n( o1 P! L
end=number+10;; b0 s" M6 D j; `$ f" j
max=min=number;4 K& j- U9 \( X# }* W. i5 `8 k4 O
for(p=number+1;p if(*p>*max)max=p;0 y" ^1 v( R3 @8 ]" e' Q: l
else if(*p<*min)min=p;
+ _5 k$ v6 }8 e- s& k *p=number[0];
9 `- r3 t$ l( ^8 N$ h$ c/ V! Q number[0]=*min;
' N! u0 I) n; i. \! ?' T2 O' ? *min=*p;/ ~5 I$ M+ `5 X% D& Y" }& Q
*p=number[9];( g+ z+ g9 e' Q8 `2 B' k
number[9]=*max;
! n! P# U4 ]( R8 L) L: |' @ *max=*p;6 P. F; Z4 D& o# c! r- L# y
return;: Z t; H k t% Q
}
X( u% Y+ `7 B7 ?8 Joutput(number)
4 _3 n2 d' Z& q" g1 \1 aint number[10];
! A* ?8 E' v5 M$ i{int *p;
' Z8 L! C3 F2 p for(p=number;p printf("%d,",*p);
. Z- l" E9 }0 s+ Z; z6 o printf("%d\n",*p);& U* C2 u! G. w
} D0 ?) L+ v( s( v1 g# l
10.4
3 e/ f/ b8 N- a& Imain()! f) p4 {; O/ r" d8 X4 ?
{int number[20],n,m,i;
6 u" g6 w, i0 O8 _( N3 A scanf("%d",&n);
F# k3 B) g/ m" m$ J1 ` scanf("%d",&m);
) h* k8 A* B- U0 S9 _ for(i=0;i scanf("%d",&number[i]);/ @ o f' I( M- `
move(number,n,m);
8 i! M' H( o) X2 d for(i=0;i printf("%8d",number[i]);! ?9 x; e2 }- W+ P
}
0 |$ t6 a; S/ V# `move(array,n,m)5 b5 G$ w4 W( }. _/ g V9 j( k' D4 G
int array[20],n,m;
* q( p3 |1 }, A/ ]/ @% W; k/ X{int *p,end;8 }/ u0 l, G' o% Q7 l. b2 ^
end=*(array+n-1);, }1 S+ Q0 z( d& q) H
for(p=array+n-1;p>array;p--)
" a3 ?) K9 ?4 _8 w' W& L; N/ D% S *p=*(p-1);% G6 [' i6 e4 ~0 L7 c
*array=end;# [3 v) k4 H9 u( `5 z
m--;
) j+ }7 \) ?, ]; a$ |+ g if(m>0)move(array,n,m);% x* l+ K- ]& ~. \" m! S+ W
}
1 x$ y! c7 K2 J! e& B10.5- V0 ]5 x; r3 N8 f
#define nmax 50
7 a- u5 u8 A. j4 Omain()
# b6 C2 l5 z4 G6 ^! s9 P+ G{int i,k,m,n,num[nmax],*p;
3 d2 B3 ^" U( h* j" g+ ]& O scanf("%d",&n);6 W+ r* g- `3 H: Y/ Z, l
p=num;+ }( h! c! F" Q- Z. G; ~1 ~% E+ o& v
for(i=0;i *(p+i)=i+1;
6 u( A8 I+ }: j7 G: Z. i% L; w i=k=m=0;
: _' G8 b7 a" b& k while(m {if(*(p+i)!=0)k++;
& T/ b( z0 E4 a+ o7 }& z if(k==3)
: {* f! N' U n0 z8 Z {*(p+i)=0;
7 ~" h- {6 u$ \- _4 p" g k=0;" c+ a! `5 U1 D0 E# G
m++;3 b5 @; K$ S" W
}
" ]5 K! z3 j: e& L4 c i++;, U! u& p/ q q: I
if(i==n)i=0;* o [- e+ |! a8 I
}7 ?9 V; | s( B3 T! r
while(*p==0)p++;7 z# Y4 H$ S0 a" @* A ` W) ^
printf("%d",*p);
' f# p" Z: w s& T}0 X, e4 l. U. b5 m( i6 S8 |
10.6- c: a. t9 W4 {$ E D. s0 T
main()/ m$ V" d, [4 _% ^: F, s8 u
{int len;
* F- Y3 o3 n/ ]- h char *str[20];
. K9 }4 p% g) k* O- o1 e5 g scanf("%s",str);
/ A' ~" Y; B) ~ len=length(str);* F( }4 y& ? H$ Z
printf("\nlen=%d\n",len);4 j* _6 s1 M; S" A3 z% S
}: Y0 F. r. a0 z7 m, O/ m4 t* R
length(p)1 [0 Q+ b, s9 Z
char *p;/ ~0 ~) a: k; b4 d+ J- W
{int n=0;& q8 a# V3 ?) J4 S' O# t
while(*p!='\0')
% Q6 M" ^" P9 x Q8 H1 f7 o {n++;p++;}7 Z2 M) R2 z; {- h) r) D
return(n);
1 Z6 U+ U$ ?3 a+ {* B i}
4 v$ C7 G7 t" J! d10.7
+ ?, v) E) e- {# l0 p+ ^main()
+ K# Y* q! ?. O' K" B{int m;
3 o. I; ?! P- ~0 A4 @& X char *str1[20],*str2[20];- N+ j) R% W5 m* P D9 k3 o3 @ Z
scanf("%s",str1);
2 P4 s- H( c: _$ }1 g; | scanf("%d",&m);
; o/ U* |# Q5 K$ S* N. P( B if(strlen(str1) printf("error");
0 i9 ~4 X; P% b1 ]- P; E- J else
3 M, `: ?' T% j {copystr(str1,str2,m);0 l. G: O8 _* y' X
printf("%s",str2);# D8 d% h. y5 g9 }: `/ B3 q1 D2 i
}
' G" i4 h/ j+ u+ A, |3 V}, X/ E* D/ O7 {
copystr(p1,p2,m)5 S& q2 F( x4 q- @, o$ Q
char *p1,*p2;
8 j p+ Q: d4 f( j! Q; W# [int m;
, V! ]2 G$ K/ B/ P1 |( I9 m2 C- O{int n=0;! x: M7 o- I( V0 l' s0 j
while(n {n++;p1++;}, n6 z# ?: f" h
while(*p1!='\0')5 l7 ]4 [& L; k& j, {! L8 J
{*p2=*p1;& ?' `6 \+ K2 ^3 K4 p
p1++;7 g+ F; A2 x+ F, |# k
p2++;
4 L" _+ q- \' F$ V1 e. L6 a h S. e }
0 ]2 h1 \2 d5 ~, b- \ *p2='\0';
q6 ^8 U0 |3 y+ X' I1 c" l- {}
1 ]7 k" [9 E g: {! g10.8
3 ^4 O( ?/ [& Y#include"stdio.h"/ p7 b" r3 A0 |) g
main()6 J8 I- O! U7 g& f% k* f" _# D
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;, B- C% _% N4 R1 Y+ b) b* N
char *p,s[20];$ n/ o& M1 ?, Q' g, t
for(i=0;i<20;i++)s[i]=0;" j9 P2 n' L y8 p* q0 M6 r, O0 j
i=0;
( T4 e2 r8 k" ` while((s[i]=getchar())!='\n')i++;9 L4 E1 T: ~, {
p=s;
7 p, U M2 ?0 I9 I while(*p!='\n')$ F N$ w# q. t
{if(*p>='a'&&*p<='z')
, H! x9 E" v# h; v ++sle;5 j3 ~+ t; ?: C6 Y2 c; q# c' R
else if(*p>='A'&&*p<='Z')2 [7 H) {6 ~* q D9 q0 ^
++cle;
# |0 I1 T3 y' S else if(*p==' ')( v( l' s2 l9 H+ ~/ {8 @3 y
++wsp;
* i) k) X; d' b$ N, @ else if(*p>='0'&&*p<='9')
+ R4 t1 N2 Q& z( g ++di;
! j0 r; q) G$ ?/ a else
- x) Y. `+ S2 { ++ot;7 B' ], c- V j6 d1 L
p++;
$ d9 R' @( o+ h8 Q# K" h }8 Z* H- H3 ^) E3 I
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);+ P& a7 H3 M) L9 A. h; n/ G
}
6 q( S% r. F, L& E6 c1 o10.99 J* e E* P( [. X, ^
main()5 u n- E9 E1 ?/ [0 n( L/ c
{int a[3][3],*p,i; C3 Q4 l" T. \8 w8 S
for(i=0;i<3;i++)
5 g' J E/ S B i5 J6 z scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);0 c; e2 t/ j6 V1 @ H4 m+ ]8 P% A, M0 j
p=a;, y; }* ^7 E2 i% k1 @, E9 ^
move(p);
% u! j7 I T% h) G& \( n- \ for(i=0;i<3;i++)
+ d/ q" M$ R& a printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);% m" r. }0 Q+ C) P# Y9 I$ P; ?: h R
}! r5 I1 D$ w1 E# k/ V7 v
move(pointer)
8 S4 m4 V# F$ R5 s; Nint *pointer;
0 ^* _, z2 D+ @: C! W{int i,j,t;) Z& [9 R9 q) @& m9 D" I6 C
for(i=0;i<2;i++)' v6 X5 X4 d- F" M f
for(j=i+1;j<3;j++)
0 D2 e: |$ O. e# r- L" B {t=*(pointer+3*i+j);; x3 P& \5 }) A, C3 a8 {+ J! N6 h
*(pointer+3*i+j)=*(pointer+3*j+i);
0 a+ }% b9 i. t7 z3 r/ C% p8 N *(pointer+3*j+i)=t;. B' z. A: y/ c' J" A3 A/ g
}
7 K1 M" [4 J0 v4 W8 P' H' ~$ H}
: d2 X+ a0 s) \ ]: O10.10
1 H4 M: W2 `' y# E! v, C* cmain()/ Z% @; J. |# Y- J
{int a[5][5],*p,i,j;; c0 m8 S3 r* `. l/ Z
for(i=0;i<5;i++)+ \ l4 F4 d% u- ~8 M
for(j=0;j<5;j++)) u2 w5 a( ]5 d( n8 C* _- @% |
scanf("%d",&a[i][j]);7 z1 m9 T0 M7 S
p=a;$ v, s4 ?* N9 w$ c+ L+ }! E
change(p);; u: a1 M7 F* o2 D* c
for(i=0;i<5;i++); V5 m; k+ X W! x3 G9 ?" a
{printf("\n");8 y$ R8 p2 u( G% y# O# W6 K
for(j=0;j<5;j++)% J" f, B. I' C) e$ Z) g
printf("%8d",a[i][j]);
, r# Z# ?# }/ p0 c0 X5 k }+ I v: Z* e8 z& j! O
}9 J" ]( k0 u7 b9 g& P9 f
change(p)2 Q8 @4 M- ~2 Y9 u- |( m% ~, a
int *p;
5 u) f4 h4 S( i3 `9 ~/ M{int i,j,change;! x0 E9 d+ C0 {6 S
int *pmax,*pmin;
9 N; ^& M8 z2 n# ]0 T pmax=p;
3 P) C7 S0 A- [6 E pmin=p;- D [3 U7 h1 i8 c+ H. @
for(i=0;i<5;i++)2 c ]" x: J* H, }/ M, v
for(j=0;j<5;j++)
9 a" i# l9 N# M! a; |0 J {if(*pmax<*(p+5*i+j))pmax=p+5*i+j;/ Z2 l9 T) Z2 z* b6 F5 B" A( M
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;) Q1 u; ]6 v! U# s9 T! p
}
|7 _( h1 W5 M; n( O change=*(p+12);
2 ?: M' f2 {9 \+ k+ f4 {6 N *(p+12)=*pmax;
4 G4 m+ y7 C2 L *pmax=change;
+ f3 G; H% c1 Y change=*p;7 l2 L1 O! ^# I Y6 C
*p=*pmin;* u5 o- T$ X5 ?1 t
*pmin=change;
+ b7 {) u' h+ h& U# h/ s# ? pmin=p+1;5 _! Y$ R! B9 r% ^8 f: m
for(i=0;i<5;i++)
6 r2 N3 |; K% ~! [; g for(j=0;j<5;j++)' i7 G' J _# [! R7 R; p \/ a7 x
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;% ?5 O9 F4 f. `, ?* h
change=*(p+4);
& p4 C* y% Z; U# C m1 c9 P *(p+4)=*pmin;
1 J- r# @0 K& p$ P *pmin=change;
; s) f/ c- N9 W' n pmin=p+1;7 Y2 g( @( `! ^5 g* _& ^
for(i=0;i<5;i++)
/ U% X$ Q" b1 I0 f! z for(j=0;j<5;j++)
: G. M1 F7 C4 \ if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
' S% w; D. F1 W pmin=p+5*i+j;
5 D# _( q! @, }! k! Q- ~& n change=*(p+20);
8 {# y' f- e% K5 c4 d4 w *(p+20)=*pmin;
2 F+ E% j# f$ l6 T d1 z7 z9 {8 ^ *pmin=change;) \0 F0 i% f9 M1 i
pmin=p+1;2 s6 ^/ U. |" S) F2 H( b9 c
for(i=0;i<5;i++)- ^1 b7 E/ m5 y6 ]' h( @: @+ \
for(j=0;j<5;j++)7 D9 Y8 b I$ O7 d) w) N8 ^
if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
) T/ v4 t( I# X% R &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;$ j0 V0 Z- Q" _/ m+ \& E
change=*(p+24);+ I5 c4 a- F+ _- C
*(p+24)=*pmin;( j7 ^6 K$ a+ m/ c* B9 I
*pmin=change;
+ C1 U+ s" I7 U5 J}
! z1 u' {0 `9 d, H b+ n" A& q10.11
' G) Y7 i) o9 i1 q/ F5 f1 `; z& [main()7 l, P) X4 N' i1 l8 _. d% J& I
{int i;" z. _6 r! v7 i" d4 B
char *p,str[10][10];
& b7 f" ^& V7 Z' N for(i=0;i<10;i++)
( C. h) d& U3 T& ?$ t9 R6 \ scanf("%s",str[i]);! \. \8 @4 X$ F- M: F2 l
p=str;1 z' n1 y: t; Y7 k3 e; I
sort(p);
. h: W1 P' X& z. w for(i=0;i<10;i++). ^# e2 d, h5 |0 |1 g, d" P6 ?
printf("%s\n",str[i]);+ S; v, ^6 b% J8 }" T! m
}' ^! i+ I; E( b: Z t
sort(p)0 l# X. }* u" ]4 a( C
char *p;1 _# \# j+ r+ [8 T8 P/ y
{int i,j;$ z* [: ~, F+ J4 o
char s[10],*smax,*smin;
* Z8 f) N" f _8 A! m0 n for(i=0;i<10;i++), l2 y2 x- C0 E9 k( Z
{smax=p+10*i;) W5 J- M" f1 W$ d
for(j=i+1;j<10;j++)
5 r v# m' Y- e7 E* a2 O {smin=p+10*j;
$ [3 I( I: k) u/ e4 E/ c if(strcmp(smax,smin)>0)
1 L2 `$ Z) c, m3 d8 J: v/ L5 t {strcpy(s,smin);; j8 k5 x$ q! {6 y+ P: S- R+ h3 b* Z
strcpy(smin,smax);4 c! N3 v- w- `, Q J- g4 G
strcpy(smax,s);2 M* |- O3 ?2 m# H3 N A
}! ?. }: e8 H* c4 v) `, ^
}$ E' f) Q' L, D4 @; M" G
}
7 |3 t0 U9 K! M p6 f z/ W9 W+ @- h" {}
. x( ^% n1 a( z; J: B; d1 v0 p10.12
5 Y' _* f4 b7 Y e#define MAX 20! j4 @, D' W' _$ S% Q: F0 Q
main()
L7 C9 s, }* H9 p{int i;1 e; N: b$ H' E7 h6 |% O8 @
char *pstr[10],str[10][MAX];
5 ^1 N6 V. q: s" b2 r for(i=0;i<10;i++)
$ x# [5 y6 u: S) l8 L pstr[i]=str[i];' p3 }$ W& x6 ^* o
for(i=0;i<10;i++)
3 g, H0 h7 p# z scanf("%s",pstr[i]);1 Q0 G/ s, }6 ^5 U
sort(pstr);7 a8 V9 q' ^0 J4 N
for(i=0;i<10;i++)
. j3 l! T# F& ~. L printf("%s\n",pstr[i]);
9 _. [' k; W. y}
; W P9 {1 G) k' b( wsort(pstr)6 U L( T: v& v
char *pstr[10];& _ I6 W. Y1 J: H8 q7 S
{int i,j;
$ _# I2 E9 ?% z! [) x char *p;
* Z" O' j: {. ?3 c for(i=0;i<10;i++)
4 k4 R+ _, c$ u9 u {for(j=i+1;j<10;j++)" `4 x2 ^8 Z# S- g6 A& F' |
{if(strcmp(*(pstr+i),*(pstr+j))>0)
- c7 G9 X/ I! z# g {p=*(pstr+i);
: X8 S' _/ y7 V! t *(pstr+i)=*(pstr+j);
, z5 D6 w, x0 g# p *(pstr+j)=p;7 K& d2 M7 `0 O
}
, u+ ~* i1 X! J; p* E+ A }
( U% v6 O, B) A# W$ q" b0 ?/ _" E }
, [% j# o2 H7 O3 l6 X}7 k( r2 f% R, f( K2 ~
10.131 L0 U6 ?; }# T, c5 r
#include"math.h"
) }0 t+ \7 Z5 j( f" Q* o1 Q) lmain()
- o) o. L. @( L{int n=20;
3 U. I: M, V5 F- D# j2 y% l float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
& A& V* C& k I. ^2 ^5 ?' ^ scanf("%f,%f",&a,&b);
* ]+ J6 g" E( R6 r scanf("%f,%f",&a1,&b1);2 N2 N& R7 M j* g2 ] x5 I$ o
scanf("%f,%f",&a2,&b2);% m$ { R- u+ @$ }0 {
p=sin;# @' d6 F$ }1 V& |) Z3 U
c=jiff(a,b,n,p);! o& U1 b7 z1 E( r; P# l
printf("sin=%f\n",c);8 \4 K0 C( T9 e8 l. c' G8 h p
p=cos;
5 t- X4 A; |2 m- [ c=jiff(a1,b1,n,p);+ J; L' L- k* h% W! o
printf("cos=%f\n",c);4 w7 C/ c- T" y; j" |0 {
p=exp;$ @9 H: r" [! U2 R* C# N
c=jiff(a2,b2,n,p);
( n0 r1 i9 b6 h: I% t printf("exp=%f\n",c);/ V; v( U- t6 N5 v0 u' g8 Y# l- U
}
! B# u$ O, Z$ i& F" v" O9 B. Zfloat jiff(a,b,n,p)
; V% u5 A- j$ N U1 y- |4 }# G; H+ }float a,b,(*p)();$ w; |% O2 v4 A- f( q4 H
int n;; p: b L$ v a9 K& i
{int i;) k- K, G" Z: |% k9 ]# R$ q/ i, ~: i
float x,f,h,area;
: D' ~* S' f+ b" N1 \ h=(b-a)/n;
9 j& J6 [2 y# P3 F2 N3 O0 V3 e% z% Z x=a;6 i7 n' c) Z3 z- b, m1 q) C
area=0;
3 G6 w; p" Q: }7 o! m: F for(i=1;i<=n;i++)8 [! M1 j+ c, r: v/ m2 S
{x=x+h;
3 ]6 s) y6 ~' ~% u6 L area=area+(*p)(x)*h;
2 ^3 f+ g9 ]* Z( s }9 x; u0 V+ G+ s# G; I2 w
return(area);7 |+ a1 O0 W9 `* x
}
9 q) ~4 Q; R" @3 }10.148 A; e- S _8 @+ W3 _7 \
main()1 f8 I' Q1 ~0 v! _5 c& ^& i
{int i,n,num[20];; r4 x8 n5 }+ B
char *p;' ]& P9 B" q9 I0 O! G" t: B
scanf("%d",&n);
% H, E9 w& N/ `9 ^0 a/ A, X for(i=0;i scanf("%d",&num[i]);; Y: J$ S! c# [5 D. V n
p=num;
: ^' F3 j, C- b- R sort(p,n);
$ o( r6 [" v4 K for(i=0;i printf("%8d",num[i]);
2 n9 f" K9 M* i4 P% z! W}
( D3 t# n6 Q6 s \sort(p,m); e r/ X9 A1 A7 L& v& N4 ~; g
char *p;- s! E2 i: t2 T- N1 {" I! r
int m;
- n$ _! B, Z8 Z1 Q{int i;
2 p+ G: T7 ?# H2 B7 N* v" [ char change,*p1,*p2;
. Z- V$ R L/ G: M# O0 C# f for(i=0;i {p1=p+i;
4 e0 q- m$ Q3 `; {2 ?% Y p2=p+(m-1-i);# J, `; F& p' d* U" ?1 t
change=*p1;* n ]3 a# S; m- h6 t
*p1=*p2;$ q F/ [2 O8 c: c, n- U
*p2=change;
- j/ y& @" f! P3 R3 ?- k$ H; V }
' ^" D( \' d% K}
1 f4 R8 B0 f3 f9 r# E+ c10.15& ?* P4 t/ v. p5 C
main()
! M1 B9 \* O) M7 d B{int i,j,*pnum,num[4];* s, m! s: n+ F1 t6 l; g) F
float score[4][5],aver[4],*psco,*pave;
/ @& E8 k& O' b char course[5][10],*pcou;0 A G$ J7 E4 ~
pcou=course[0];
) I: ~2 e5 b$ G5 m: ?5 P; t; P2 U' l; U for(i=0;i<5;i++)# h7 E+ N; e' V+ f
scanf("%s",pcou+10*i);0 p: b [; H% }$ v' r6 V, y
printf("number");0 U4 `' @% w4 K% f! h$ `
for(i=0;i<5;i++)4 N. g3 p: b3 w7 T
printf(",%s",pcou+10*i);
, K: b3 @/ e; a9 c/ j. c" u* \+ V printf("\n");
' [# K- d4 D0 O. k- a; b) M psco=score;3 O/ z/ ` R6 i1 ^- E) v' J
pnum=num;
& Z9 ^, p+ q1 D+ [+ t7 w for(i=0;i<4;i++)
, B5 p2 c; c# |$ ?) S2 y* D; { {scanf("%d",pnum+i);
* ~' G8 U7 f) i: n- H2 e+ \# ^& n7 j for(j=0;j<5;j++)0 }8 i, `, k: b" X8 Z9 A: N
scanf(",%f",psco+5*i+j);
7 Z8 o% T$ l! Z }
" B9 n6 A1 u/ |2 F& @* o pave=aver;4 Z. \' h3 K# n7 U/ ]3 X; k& x) U: P
printf("\n");
# O9 C6 o* q- r1 d! U avsco(psco,pave);
1 P0 h0 B- T! P& o0 h7 G6 K9 g avcour1(pcou,psco);3 \1 f3 G1 ~6 X! R; h
printf("\n");; f) C% W L' S
fali2(pcou,pnum,psco,pave);! U7 z( b! _( S, N8 i$ d
printf("\n");
$ n0 A& A) Q1 M good(pcou,pnum,psco,pave);; w, H$ v9 I9 g) V% r& L3 _
}
~% u$ `0 o8 J; ~. T* e! x3 Zavsco(psco,pave)
- g' w8 T4 y$ @* Zfloat *psco,*pave;
+ a8 K; ]3 K3 D- s3 a% R9 V) q/ d$ ]{int i,j;
7 x2 p# z7 T0 h4 P0 l3 P float sum,average;6 ~# e4 o' {, r: A# T7 Z
for(i=0;i<4;i++)
) K2 b, ?1 J# m2 o {sum=0;
* Y6 l+ m3 V# M. o2 g( r* n" K for(j=0;j<5;j+)0 K, s/ j6 O% y& v; ]( N
sum+=(*(psco+5*i+j));
2 E. H4 M; x8 A( x% ]6 ^9 Y: k average=sum/5;
, v4 h% a7 y! `$ ~- I$ Y( b! c( { *(pave+i)=average;5 A7 p4 X+ D: }& ~- n, R
}% O9 s/ X6 l9 x5 D8 p1 _' g. p
}
+ s* D* T! C0 B$ V. m& w) wavcour1(pcou,psco)
- C0 l0 \, a9 |! n: f' dchar *pcou;+ t) h" v( V$ @/ ^1 O9 c
float *psco;9 \1 O8 f$ {# C4 n. g3 V& `1 l. s
{int i;
% a# k& G4 J5 f8 I9 | float sum,average1;
+ r: [: v: w- p, a( ~+ `# y sum=0;8 q. I) l9 e& Z- \
for(i=0;i<4;i++)
. c! p! X8 t! C$ d; h! R& F/ x sum+=(*(psco+5*i))5 [8 c9 g8 p+ M
average1=sum/4;2 L" N' s- k# D) U- c4 _
printf("%s %5.2f\n",pcou,average1);0 n9 [/ B& Z! }3 T, j9 k/ B7 u0 e$ T
}
* _. V* b2 @6 r, d8 V: v0 pfali2(pcou,pnum,psco,pave)
' J6 I9 K. w/ Uchar *pcou;0 S& `' h D1 K% ^) h, @4 Q# Z
int *pnum;: v# _+ `. _5 ^& ?7 Z
float *psco,*pave;6 w/ o/ K/ h: T3 o" J
{int i,j,k,label;! K t% p4 V# \4 x4 C" v
printf("\nnumber\n");$ @% _5 v& p% _2 C
for(i=0;i<5;i++)- c; K! K+ K2 g3 }
printf("%-8s",pcou+10*i);
+ q o2 L% A& G printf("\naverage\n");) V9 E8 B/ q; l. d0 k+ C+ l
for(i=0;i<4;i++)
- y9 \" R7 M9 q+ w8 a% O0 P {label=0;+ c2 @' V2 h4 R9 i9 u
for(j=0;j<5;j++)1 u" b% z' ^3 j9 a+ u% d/ _; J) \& ?
if(*(psco+5*i+j)<60.0)label++;( p! ?6 Q1 X$ ]5 v. ^
if(label>=2)- e; l+ v0 o; z! }0 {# A. S
{printf("%-8d",*(pnum+i));& m) y! j1 d. f. D& ?
for(k=0;k<5;k++)
- P8 X# r9 v5 q9 H6 T0 p printf("%-8.2f",*(psco+5*i+k));% I! H: t$ h1 e8 b; D
printf("%-8.2f",*(pave+i));0 M0 c9 o% V. e
}
( p/ \; c. }& V+ P8 i }1 Q$ U& g/ f6 Z0 h% R
}
- D! U8 ]+ V, U$ Pgood(pcou,pnum,psco,pave)0 Z* d# u; {4 f& Z$ `
char *pcou;0 k" C2 f9 {% h+ t8 z2 B& K
int *pnum;' B6 \+ f* L" p! R* z6 ^( H
float *psco,*pave;
- I! A" K/ G: |& S# F8 M# F{int i,j,k,label;
; D) B' v" X; ?' a1 w" P4 j, p printf("number");" E7 O6 r. W7 b8 e1 ^+ z
for(i=0;i<5;i++)
5 [3 ?" v! f( |9 L) Z4 C( y printf("%-8s",pcou+10*i);+ ^/ x- M! N; s: Z2 c
printf("average");
5 c5 q1 T" Z; P d$ e$ r: Z: H for(i=0;i<4;i++)' @6 ~) h4 }4 g/ @' m; J+ D( ~
{label=0;
4 Z( P2 v4 m4 \1 @. |7 e" d for(j=0;j<5;j++)4 N+ }3 _8 ^1 E2 J- d/ \
if(*(psco+5*i+j)>=85.0)label++;
0 K# f5 n8 T6 ]" p. { if((label>=5)||(*(pave+i)>=90))
/ M0 O# c8 d* s. f8 D, G {printf("%-8d",*(pnum+i));
+ P( g9 {6 ]; V$ U0 W' a. `1 n7 u for(k=0;k<5;k++)' D) L1 c' ?* X) d2 \/ o
printf("%-8.2f",*(psco+5*i+k));
6 M5 C+ t7 L) A' D$ u printf("%-8.2f",*(pave+i));
: O$ G* P# F- x/ l% e }0 u/ C, f# U- q4 t! w" C
}
: X1 x5 L+ k. i/ c9 s}
4 ~& U6 y' ^, a7 S/ z10.16& X9 ~. s8 u+ y3 x6 G k$ J% T
#include"stdio.h"/ P# _/ G) T! c% D$ k" }
main()( s/ N' g1 P7 ~# }, e( |8 R- \9 s) v
{char str[50],*pstr;
$ y+ H4 G s# V/ ~, F* d/ d" T int i,j,k,m,e10,digit,ndigit,a[10],*pa;
! _+ k2 y! P1 t' g1 C/ m gets(str);* {8 T" D$ h, N. k# A2 c
pstr=str;
- k& h$ R6 O; v$ K pa=a;
0 P7 R& H1 V8 G ndigit=0;
3 A3 R- \( ]' {0 ^7 x5 ?! r6 o i=j=0;
/ E) z1 L" d7 Z9 L' X3 \ while(*(pstr+i)!='\0')
: u3 D4 s& k. W7 d# D {if((*(pstr+i)>='0')&&(*(pstr+i)<='9')). o5 B* C+ J" o1 E: n
j++;
5 J8 a9 _/ u% B$ |% b else6 d) }0 p+ q7 t7 M
{if(j>0)
8 P9 f5 ?! V% u5 T0 r' K& } {digit=*(pstr+i-1)-48;5 S6 @1 z. H0 [
k=1;5 `& x5 W- |1 C
while(k {e10=1;) B/ R( j/ F! o/ [, Y$ K
for(m=1;m<=k;m++)
5 M) n6 w1 R A1 F+ M4 a5 k j+ v e10=e10*10;; i p, u/ e7 P9 N2 d7 H) y
digit+=(*(pstr+i-1-k)-48)*e10;
. `2 s* Z1 x% w4 o$ O( M' I( { k++;
6 \" F$ [" L5 {0 [7 d }
/ B m# B4 `9 F *pa=digit;
( A& M( {! q+ A( [4 R ndigit++;3 X. \5 U4 m5 b5 b
pa++;( J9 k2 b9 z0 u- M, F! l5 u
j=0;
' X- E9 `+ M% \" K: _0 F7 A* W2 Y }. b3 r0 ~$ ~7 F5 R
}7 \% H$ E; ^, [" Z+ W2 D, k' R" O, o
i++;
# \, F6 W" E, D" o }
1 [5 p j0 [ {7 r if(j>0)' f, Z# E6 [$ y9 `) K' `
{digit=*(pstr+i-1)-48;
! p1 t; L# O3 f- ^3 V k=1;) e* N2 m1 |) y/ r6 e
while(k {e10=1;2 |& M; X$ ^! {9 G' f2 j
for(m=1;m<=k;m++)- v" e) M. v" |4 ], c
e10=e10*10;
+ W+ l' L+ @$ p1 t3 B4 w3 a; x, y, Q digit+=(*(pstr+i-1-k)-48)*e10;
1 g7 F' ^1 m- T6 m w$ @ k++;3 m& t% ]( [( s+ w6 W' l. m
}8 i$ d% h2 c3 h H0 F) Y
*pa=digit;& |3 j/ d6 Y7 A( s% I# E
ndigit++;; J, w" y1 F5 O L
j=0;1 V* _# h: i& `0 n4 g- X4 C
} 0 M- ^: ^, G! l7 d( O/ | v
printf("ndigit=%d\n",ndigit);
! }" {8 N; c s) A) ? j=0;/ E2 x: f w0 Q0 ]- s
pa=a;
, g2 [5 T7 V. F& M, Z4 y for(j=0;j printf("%d",*(pa+j));
. N, z2 ~$ x3 Q5 a8 I}% S: A. _4 I- j4 J$ G$ \
10.17
4 S& Z7 s( e; ]( d/ Y: Tmain()# z- r* `8 D H3 `5 u$ Y9 ` p& r2 S
{int m;+ V: ?$ z% O( ^% L- X( c( [
char str1[20],str2[20],*p1,*p2;
2 ]% |. M+ y- T: N" L7 |- C! u$ F scanf("%s",str1);
7 H/ p2 D# \: P! a. F( ^ scanf("%s",str2);
$ x" Y$ U9 \' Z* o" s0 W/ o p1=str1;
) r8 O+ Z% d8 l. D7 a p2=str2;. s* P- T% d9 M! B( ~% z* v
m=strcmp(p1,p2);
8 o6 e" U* a5 t! e printf("%d\n",m);3 Z# K) @3 Z5 \' m0 D0 Q0 g4 X
}
o. A1 [9 P( D! Z6 mstrcmp(p1,p2)1 |7 S2 O. U3 j5 W3 U
char *p1,*p2;
2 Y: W) S# ~' ^% s0 A{int i=0;
& \, T, |) {) p8 w: @ while(*(p1+i)==*(p2+i))& x- |' Y1 {: J& g" t1 ?' w& o2 d9 `9 N
if(*(p+i++)=='\0')return(0);
* U9 k# j+ W8 [ return(*(p1+i)-*(p2+i));5 a' {( L' w" b
}% c1 }+ V" z# C0 f# E8 a, T
10.18* q7 F. i: N- o
main()" c |: P1 d* M7 S @: `9 {) u. Y
{static char *mname[13]={"illeagl","January","February","March",7 x. \' \/ q) c. l# j" T
"April","May","June","July","August","September","October",
9 \' M W9 D8 W) M: H7 Q& L# G "November","December"};
^7 t" v/ z4 Q2 B+ {) }8 U int n;+ {) h2 Z" P! X/ V1 a% ]& n4 Z1 d
scanf("%d",&n);) e u3 Z t+ [# I* Y' F
if((n>=1)&&(n<=12))
# p* }. U, b1 J2 W# U0 T printf("%s\n",*(mname+n));7 D4 G, J6 q2 S4 S2 p- e
else
1 n9 [+ H. }1 s9 |7 x printf("error");8 Q4 P* w* v0 U+ D3 K. F! T
}
6 T' a& M, {8 u5 n" T10.20
. g1 Q' b1 z/ B! M) _, rmain()2 ?7 f6 L: G- a9 P3 I. t# @
{int i;
0 I* _' [9 u4 W& u4 A- V1 \: i% S char **p,*pstr[5],str[5][10];
$ c) N) ?' m; t) o( k9 n! i for(i=0;i<5;i++)6 b9 ]- C# E" p5 J: H
pstr[i]=str[i];9 g: a; ?, |! d) p
for(i=0;i<5;i++)
9 W+ w5 N. }5 T+ [7 J& @) j scanf("%s",pstr[i]);
% H8 z8 w7 u* I p=pstr;5 g) J) o% \" [1 D* G
sort(p);
. F# O0 O8 _+ h+ _' m3 e for(i=0;i<5;i++)1 I) W& `% s1 [9 |! [; |" ?' |
printf("%s\n",pstr[i]);* }3 S: O, k, s, }7 t' C
}
2 f$ k2 F& Z V! B' `/ Asort(p)
1 Z( x& E! W( G6 e E3 fchar **P;/ p& V _# u1 }+ I$ a6 k
{int i,j;8 q C& D$ [9 I/ p) S: u4 X0 z
char *pchange;; p9 {# v9 ]7 P6 |' t
for(i=0;i<5;i++), H4 i& w2 G- C% @# q. X
{for(j=i+1;j<5;j++)
$ ]( p! f4 ?* ]# g6 O$ k {if(strcmp(*(p+i),*(p+j))>0)
* }. q7 W9 X% L% H0 w {pchange=*(p+i);
. F% o+ c) ?! W2 a *(p+i)=*(p+j);
1 \. {2 e) F$ D* ~+ |! D0 t. `' o *(p+j)=pchange; o2 Z# w% e( w9 s- a7 I# f
}
9 }: F. J$ h3 ?$ ] }$ C7 }+ P7 H5 `. ]% A. x
}6 S# O4 ~7 N! _) r! S9 G
}
' r5 d) a: r: T8 t2 q10.21
/ r6 X. k1 }, j: V2 Z2 Hmain()2 w3 U, q8 L5 h" R1 @+ x7 j
{int i,n,digit[20],**p,*pstr[20];, |0 o/ O* `) f# y
scanf("%d",&n);
; g4 E! I4 U7 ~* `3 O3 e' S/ D for(i=0;i pstr[i]=&digit[i];
( c1 t* e! z- D; V5 |+ P. C9 i _% U for(i=0;i scanf("%d",pstr[i]);! u" g5 u/ J) H9 @- ~- t
p=pstr;" v! i; K2 z9 @% L/ W! F# `
sort(p,n);
2 V1 y! K, h( A1 G for(i=0;i printf("%d ",*pstr[i]);6 w% s2 S3 {! c1 t
}
) l) C3 P. J+ ^0 i, u k& S! nsort(p,n)
: \/ ]' A- l, I4 c% D9 nint **p,n;$ b" L4 s4 Q$ M+ p
{int i,j,*pchange;
C5 o& K2 ~2 l for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
4 L9 O7 ~) H' i0 ~0 X. b3 x* t8 D; \ {pchange=*(p+i);
; s+ L/ w+ v$ {# k6 W! |$ N *(p+i)=*(p+j);
6 C& L) X6 x: M4 l5 a, G! J *(p+j)=pchange;$ |7 H1 D1 `" A. Z3 g* I
} P6 Z5 F: O; ^7 b
}4 d4 @9 T9 V- \' A$ J
}
7 p/ a% P0 |- z! ]}, z& M7 {; v% k' G. k" O
第十一章 结构体与共用体
* Y( }: U, n) A' {11.1$ I# C& W! h: u3 T8 m
struct
6 u9 E% J4 M: m1 _/ `- Y {int year;
% `5 e9 x$ |2 w9 K int month;
! K1 J3 z; l* t; p) p int day;
1 t U) [- w. { }date;8 k- d- [4 ^7 J1 D% }8 t
main(), w5 E! L: g# n( G
{int days;
8 S9 _) }2 W3 G. ]& \2 o3 }+ D scanf("%d,%d,%d",&date.year,&date.month,&date.day);
* P5 z* \1 \1 v: z A: L switch(date.month)$ m' _ h [9 g- O6 I$ o9 q
{case 1:days=date.day;break;" I7 a! ^( Y5 H
case 2:days=date.day+31;break;
6 n9 F ?1 h* s* o6 e4 O case 3:days=date.day+59;break;
; o# Y+ J. Y. n* q7 M5 X2 |9 W case 4:days=date.day+90;break;$ p1 W9 A$ }. w* L* `
case 5:days=date.day+120;break;
6 [/ ^: T" V9 _$ c$ | case 6:days=date.day+151;break;. W6 c' R+ w& O0 J
case 7:days=date.day+181;break;
5 E; g& x* C; W, J5 O7 A8 E! v- c7 q7 P case 8:days=date.day+212;break;
9 e. T9 J) }9 O4 H/ b$ y case 9:days=date.day+243;break;. N& c4 a' M! }4 B4 m. B
case 10:days=date.day+273;break;
# B7 _. E) t% U9 J2 c; I p case 11:days=date.day+304;break;$ O. F/ _9 g, K9 B4 b
case 12:days=date.day+334;break;
6 N8 X4 v; o+ l, n$ W' Q. W7 }# k }
- `9 m L4 `, y9 a/ I7 G5 r if((date.year%4==0&&date.year%100!=0||date.year%400==0)
4 }3 f" p$ U' ^; w% g" g &&date.month>=3)! i' Z- \) Y3 P
days+=1;" q+ Q. z2 O8 p1 g
printf("days=%d\n",days);
0 y( z2 ]$ A9 O+ {, E}; K9 R# q' W2 H7 ]; v1 b+ E
11.2
. o: U5 p/ T* b2 Cstruct dt
- a% [- Z6 H0 \! W {int year;
h3 D, j/ O5 R+ A$ q' T; z, _$ E int month;
8 R& u7 E2 F: u" d0 g int day;, A& \( V6 C* @% y
}date;
# L- P# I$ L7 Vmain()% @! ], T. i1 ` W8 y5 M
{
) Q: U; P0 c5 ] K6 d6 R" ] scanf("%d,%d,%d",&date.year,&date.month,&date.day);
3 F: @: d9 u0 i printf("\n%d\n",days(date.year,date.month,date.day));
4 ]$ v" R9 f! K% `) R}
0 A8 W4 ~/ r+ l. N1 b+ Mdays(year,month,day)
/ A" {' @! q4 k* l& J6 r" x6 Rint year,month,day;
, A, @$ T+ E1 ]{int daysum=0,i;- a% f3 V& b2 _& ]0 j' E- L% B9 r
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
' W i, @6 v& P& I' h: e for(i=1;i daysum+=daytab[i];
! I+ g; k8 } d daysum+=day;
1 _% T" e5 H) x+ t; Y2 F if((year%4==0&&year%100!=0||year%400==0)&&month>=3)$ s1 V+ `6 Z/ q% s
daysum+=1; U7 t/ d9 ~) m1 ^/ j; c
return(daysum);, a" ^2 ^: A1 a) R3 j% k
}
4 ~1 p" l" W/ R% N, j1 a$ y+ b; L11.33 n& s9 C/ I# t
11.4
& x' [0 Y" y1 y3 Y#define N 5
: ]: s3 Y3 u$ D0 ^struct student
) H2 Z8 v# P& v {char num[6];* K/ p8 |4 s( W% M8 n
char name[8];* w" s2 Q3 }8 }8 X2 \1 o
int score[4];9 ~7 u3 Z- ~# W7 `
}stu[N];. [2 W; o' s) A9 Z
main()! r# y. N. D7 Y4 k/ B' }- v
{
0 v% |) Q$ J5 g5 }6 i input(stu);* j8 J% w3 P* Z2 j" G; S
print(stu);4 u; b' T. t) ]5 N N7 C, M- W
}
+ }5 ~) r* U( B" u1 p2 D$ Rinput(stu)3 p- r( M& k6 d6 z( S% m8 i$ ~
struct student stu[];
$ b( G/ s j; Q4 J3 h{int i,j;. [: C; A- N6 z
for(i=0;i {printf("number");
8 Y+ w: R8 L; ~5 n scanf("%s",stu[i].num);
& ]( ^, L5 d& {' A0 ]) r1 j+ B printf("name");
1 }, p( i; T& w scanf("%s",stu[i].name);( G6 v* Y6 l, W' s
for(j=0;j<3;j++)# ?% ?8 q3 q+ s2 }1 l1 l$ v
{printf("\nscore\n");. C5 B4 @$ U0 x/ n+ d, |
scanf("%d",&stu[i].score[j]);
7 M! t! L8 y. S }
, M) U2 Q r7 |9 p) V$ m0 z printf("\n");
$ B9 N3 g2 n4 j) o+ J9 e5 ]5 m }
. A6 v% A H8 I" ^' x4 H9 R} Z) H& A7 {- x& i' d. P* w
print(stu)
" q! L$ x) p- O; e2 _struct student stu[];$ q3 c. c3 t1 q1 ~4 Q
{int i,j;8 @! r* L- Q2 H0 V3 F$ ?
printf("\nnumber name score1 score2 score3 \n");
x7 u" H7 f/ ?- _9 p for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
, R/ ]/ W2 Y# l+ I for(j=0;j<3;j++)
6 U. I# }3 m! \0 U. K printf("%7d",stu[i].score[j]);
6 M+ u! b5 V! `2 j5 G: q+ L printf("\n");4 U& O* e g- o2 B K
}
4 w8 z' T* B) O% o' s}, m8 L; `& q+ J4 a- r, c/ m
11.5
7 A- V! a5 P: o$ p0 p2 ~' ustruct student
. T# @+ G' [) L& y; T: Y {char num[6];
\( w4 H- b. _1 m4 d" j char name[8];3 e# B' M2 d+ M3 q
int score[4];) Q; M% {4 P, g; \' B3 L
float avr;4 j$ H3 I1 G/ t) l7 `
}stu[5];" `* m2 n5 d# z' ~" z2 U
main()
6 R7 ?' e' I! W% x' J }( N{int i,j,max,maxi,sum; z* B2 X% P0 f- q2 A) f3 O0 N, k
float average;0 P F3 y0 b' n) Q# f) n
for(i=0;i<5;i++)' H0 T' y6 h2 G/ J/ b
{printf("number");
: l# @2 b! Q5 }& K/ b scanf("%s",stu[i].num);5 Z. p0 z" N% I) q1 Y' j& o
printf("name");9 ~; B6 V; V8 G- s# N
scanf("%s",stu[i].name);7 ^& D7 t9 o9 i
for(j=0;j<3;j++)
' w" Y0 j3 Y0 A C, K4 z4 k- n1 i {printf("\nscore\n");
+ i* W& y* X) w, h/ @( N4 Y scanf("%d",&stu[i].score[j]);
7 Y, ]0 `! o: L& o+ p3 [ }8 o2 _+ [% a j9 l
}6 W! l# d; z9 q8 k: L
average=0;
! V" K% T3 N. f2 p- ]; t% U max=0;/ ^ p" g9 W8 t* E3 T6 S
maxi=0;
& D! _7 n+ c; ~& H+ } for(i=0;i<5;i++) v J% n+ ^1 S3 p2 Q' W8 M! @8 H
{sum=0;2 \0 u. U5 I. c9 O8 l8 ?. P7 v
for(j=0;j<3;j++)" R z7 R! l0 r( y' {6 w
sum+=stu[i].score[j];
* N3 p7 b, i% y9 I& h: c stu[i].avr=sum/3.0;
4 Q, K9 b& j1 E average+=stu[i].avr;( ^2 C% t7 K2 A5 L1 u
if(sum>max)
7 w2 q) c1 v0 p* L3 { {max=sum;) n1 R: m( m; z/ q7 H$ G
maxi=i;
: H: T" f" z: i! R# u }
/ H6 _% X# v$ K: C3 }% J }6 J, X. ^, I, l
average/=5;7 g$ k8 S7 |4 J% h
printf("number name score1 score2 score3 average\n");
4 s; X3 J$ L3 L; ~ S0 R for(i=0;i<5;i++)
% B6 U2 r" m2 d. q {printf("%8s%10s",stu[i].num,stu[i].name);8 } D* m% K3 u4 |- ?
for(j=0;j<3;j++)! ?( f- f1 O5 c$ \& l( s
printf("%7d",stu[i].score[j]);8 B& e0 f/ ^, }9 a
printf("%6.2f\n",stu[i].avr);: Y5 `. z, E& v3 S) k
}. r/ _' N( ?4 i0 g* e
printf("average=%5.2f\n",average);2 r y6 R( E {! |8 h4 q5 N
printf("The best student is %s,sum=%d\n",stu[maxi].name,max);6 N) Y* ]! K @
}
* y9 ~* ?" Y2 d1 D1 D5 s
" [5 M5 S: o' i0 R |
zan
|