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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/ l- l+ z+ Y# N) b3 Y! E
1.5请参照本章例题,编写一个C程序,输出以下信息:/ o2 |6 X$ n$ \: F4 H; L
main()
8 T, `$ K; S; X, l: H4 b* k' p- H0 M{! s' S) S& w. l- _8 p
printf(" ************ \n");
7 Y; i# w3 r$ g+ a+ nprintf("\n");
, c0 b. Z2 e1 a( Aprintf(" Very Good! \n");
@$ G1 c$ s. b4 Z. c; w1 F+ rprintf("\n");% N7 J" _! Z9 i
printf(" ************\n");% n6 x4 m3 p( w- C' F
}. Y: t1 x7 Q2 m/ E
1.6编写一个程序,输入a b c三个值,输出其中最大者。/ Z0 O8 Z# b( l- c" v* \5 \
解:main()8 p: x3 V* B) Y/ A! r: T
{int a,b,c,max;1 u% P* K) O% u) b7 d& W" F
printf("请输入三个数a,b,c:\n");
& e V& K4 M& K; M' ~scanf("%d,%d,%d",&a,&b,&c);
; N# P$ u- d* K2 l2 Omax=a;+ Q& U" V( a; p' x' ~2 }
if(maxmax=b;
/ K" N |% x6 X+ {' N6 u" W8 `if(maxmax=c;
5 ]5 N) j$ B$ C' x1 K) aprintf("最大数为:%d",max);
2 r" @% p& E7 N1 b8 ^( k}: O# S* b \: S0 U* z. g
第三章
& @0 K+ S, P! _# i6 J2 i) B0 y3.3 请将下面各数用八进制数和十六进制数表示:
8 b9 \* I) Q# K7 V6 A5 O' K(1)10 (2)32 (3)75 (4)-617& [; H& ~, K( B9 Q0 o; v0 F, e
(5)-111 (6)2483 (7)-28654 (8)21003, l: _+ q2 S z* Q; u: Z
解:十 八 十六
, t- j8 Z! K6 ^% I" r (10)=(12)=(a)
4 l! E, C; _: ^3 }2 \ (32)=(40)=20
1 r. l q6 z/ d$ E' q f (75)=(113)=4b" t0 w. O O. r; B
(-617)=(176627)=fd97) \ u7 ~% i5 m* E% u
-111=177621=ff91
2 t$ f" j+ [1 {& r 2483=4663=963
( D" ?$ A8 z, w1 R3 X -28654=110022=9012
6 {4 s( i; @% n8 V 21003=51013=520b( V6 |( q. A! K% i) L
3.5字符常量与字符串常量有什么区别?3 V5 c$ N2 q5 p8 B; q
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( |8 j+ ^+ o( s8 o+ s; H
而成,用双引号把它们括起来,存储时自动在字符串最后加一个结束符号'\0'.7 I7 L" B& @! C. V9 I; N, b
3.6写出以下程序的运行结果:
6 F+ y% @( ~0 f" M) Z4 D8 r#include( N6 i# K8 I1 a4 o' P
void main()8 e1 R' x/ `5 s3 K+ E
{; \& n) A* q: f0 }! c! T' {
char c1='a',c2='b',c3='c',c4='\101',c5='\116'; z2 l7 P) O; m% x& Q5 y6 K
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);- k' Z# s& j9 V# w$ d
printf("\t\b%c %c\n",c4,c5);; W1 E' F* d: ]( b& [% r
解:程序的运行结果为:8 u! F& _- j; F6 {9 {. a
aabb cc abc
( |1 m3 B8 @6 ^- R A N. W1 }7 j3 h: @& l$ f2 G Q
3.7将"China"译成密码.密码规律:用原来的字母后面第4个字母代替原来的字母,
1 A+ K6 | Z2 _& L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".
0 W. M. v. h! k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! U- u7 [/ Q6 Q. K, D* s" Q, i
输出.
n1 l7 e% R$ o ymain()6 Z. O0 K" l, L0 e! }4 `6 `
{char c1="C",c2="h",c3="i",c4='n',c5='a';
% ?3 O2 T" H$ J4 N& ^c1+=4;
. y0 n( m3 I5 x5 mc2+=4;" h( |, ^2 N1 ]/ N8 F) M5 W
c3+=4;
' B) I, R1 C# zc4+=4; p5 ^3 L) r/ s2 l. p
c5+=4;
! V4 W5 }5 ?. K, d& c9 v ?printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);8 p+ U# `: H( o( t. B# e5 y# G
}
4 T" E: ]9 k8 O2 k3.8例3.6能否改成如下:) S1 k* @7 j6 S& v6 o' Y0 _
#include5 E: z6 n# Y5 c
void main() ?4 I8 z9 y7 G. a9 T0 F, `
{
+ b# t! x: J% e k3 qint c1,c2;(原为 char c1,c2)
* D* p2 @4 T1 \8 g& y5 Ic1=97;
- f8 v0 E# |0 [c2=98;
7 G. B; C& o* e" Iprintf("%c%c\n",c1,c2);
9 R) c/ X" K$ x' L8 i5 Hprintf("%d%d\n",c1,c2);" x# C9 H# D, L( x
}
/ U& l" a0 j( {# p) ?2 w& E( f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.
* M/ }7 ^6 m% E* H% E" H3.9求下面算术表达式的值.
, z8 r& A7 \. N8 s j, A( G; o) c(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)1 |# L" a" y( k/ ?" s
(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
. d' o1 U' G n1 O R8 O3.10写出下面程序的运行结果:" z) n0 N. a; Q/ B5 R
#include- T y1 F. q2 K
void main()1 z8 c/ L* t( b0 F
{
# e& }1 h( m* F9 V4 ?0 Sint i,j,m,n;
" H0 B/ X9 h4 ?$ g `8 mi=8;1 y3 J- Q: G, z3 l* }
j=10;
! ]" D S1 n* p! W0 Km=++i;/ P+ _4 ~! N+ a8 p# D0 J
n=j++;
" |) p, e$ _4 F6 kprintf("%d,%d,%d,%d\n",i,j,m,n);
0 C2 l5 ?" T5 K4 _% C U}" B$ z/ K9 u {$ f# p
解:结果: 9,11,9,10 j. J3 O& i/ x8 r3 N0 u
第4章1 ]- e% J+ R# F
4.4.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
2 j8 C# s$ V; m0 [4 H$ Q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:# F! I% x6 c, u9 o4 n+ R3 t
a= 3 b= 4 c= 5: s1 _# h7 d D: N. W. K( Y6 _
x=1.200000,y=2.400000,z=-3.600000
3 X) f/ X' |' L8 M/ tx+y= 3.60 y+z=-1.20 z+x=-2.40& f9 J& \: Q0 A8 c
u= 51274 n= 128765
* J! ]9 V7 W. ^ m3 ac1='a' or 97(ASCII)
. M9 K) E# Q, ]4 fc2='B' or 98(ASCII)% |% b/ F8 P/ ?6 |0 b/ T
解:6 l+ N9 _% C$ V/ _; p% j
main()% r. h2 H2 V! M5 D3 D
{% Q0 `" e$ s4 n9 _0 w4 Q/ o
int a,b,c;9 S$ N U5 q0 B6 ^8 z& z& ^0 B( W
long int u,n;
7 E( p* \7 s @. rfloat x,y,z;
( n$ P" p) K# g2 bchar c1,c2;! h" w+ `' G1 m+ B3 P
a=3;b=4;c=5;$ W1 H" |- i0 g4 U- ?( P
x=1.2;y=2.4;z=-3.6;" p* q* n8 O5 s2 f$ _
u=51274;n=128765;
& b3 L* c' i" v9 _c1='a';c2='b';: ~/ E% y2 n- p% b7 i7 l+ P
printf("\n");$ d% N' i. r* F+ M5 _
printf("a=%2d b=%2d c=%2d\n",a,b,c);3 t5 Q- E! Z7 ~" g. ?+ _9 M
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
% A# s8 {; h; |4 g' Lprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);- i; _6 h8 G% _
printf("u=%6ld n=%9ld\n",u,n);5 p, b$ |, k0 T) A& V* N
printf("c1='%c' or %d(ASCII)\n",c1,c2);
/ n& s! \8 ~ \$ d, B* D/ M1 @0 Bprintf("c2='%c' or %d(ASCII)\n",c2,c2);
) o, i0 A( ~: W4 [1 ]- r' x}* D5 i; l' r9 |7 _
4.5请写出下面程序的输出结果.
3 g5 d4 x- _, _& R4 }2 f结果:) g! p- x& z4 o5 E% w0 r2 b) V
57
8 u: T6 r) _0 X! }" |0 G 5 73 u3 \3 O' a4 }9 }' A
67.856400,-789.123962; G/ L6 Y( z% v. N- k* n5 L* n
67.856400 ,-789.1239620 H5 A. D( v! x; V4 O- W/ o0 [
67.86,-789.12,67.856400,-789.123962,67.856400,-789.1239623 D2 @% F) ]7 Y8 Y0 ]
6.785640e+001,-7.89e+002
- A& H& W9 W0 ]( oA,65,101,41% _" [8 s1 {( q* k' n; B
1234567,4553207,d687
2 }- X+ d3 O- \/ d65535,17777,ffff,-1. [. q9 {" a i& @" a. t
COMPUTER, COM& u Y; u' S1 g- D
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',, s: i5 r) f8 U. S7 ?
问在键盘上如何输入?; w. Q1 F# |7 u) p+ h$ V9 O
main()
6 T' o8 d, N" }{
- P( F% r% h j& L/ ~$ y. Hint a,b;
$ F: l. W. b. T5 M+ Q( n% i" zfloat x,y;$ I5 T4 N* s" ]
char c1,c2;
/ K! x$ l: o0 G8 N' Fscanf("a=%d b=%d,&a,&b);) n! \1 M! v) x0 O
scanf(" x=%f y=%e",&x,&y);. l( H6 `5 n& z7 D6 C8 [
scanf(" c1=%c c2=%c",&c1,&c2);) X0 ~/ @9 S8 S* o
}- X! H$ o4 c* @& K v# @, s- i
解:可按如下方式在键盘上输入:
- a8 k# n1 F4 C+ ?. Ka=3 b=77 `$ m/ F' t) l# y0 U* l
x=8.5 y=71.822 ~5 q8 ?, I0 |3 S5 G' Q( Z' A* Q: O; C
c1=A c2=a$ D l" y c6 ?* h+ K: M- D8 R
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
- Z0 O: R1 }1 v- a3 j9 t' T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) l, s" |0 t0 {5 h! {% C( Q" j
入的"回车".如果没有这个空格,按上面输入数据会出错,读者目前对此只留有一
# E& y" N8 V. }# n" r初步概念即可,以后再进一步深入理解.
3 k+ J7 [; H- Y1 j) |' I8 g1 b# ^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=-
8 v; O: g2 F" E7 j% R; _1 p3.75,z=57.8,请问+ D& e8 V& C9 J
在键盘上如何输入数据?% e4 [* `- X& j8 k; i5 e
scanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);- @; J) G7 g h
解:
# W( n: z ?0 N( M: O& U8 }1 omain()9 o7 x' D, w" z: M0 j% q, P
{
; s4 e) v9 j0 b- Iint a,b;
# C" q1 x$ V; n5 e2 tfloat x,y,z;
1 t( d; ?" R8 J$ G, B$ G' Dchar c1,c2;* i9 w7 V# b- c! e
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
) h4 |* H( @: U4 R( p3 l5 y: C}5 L0 H: f; j) z
运行时输入:
8 Z" c, \' @9 B' v e! z10 20Aa1.5 -3.75 +1.5,67.8, }+ J. c! b& `
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& b+ u" T; [2 M( W
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3 U) U2 U; |& `0 p' C7 w4 u0 a
会赋给任何变量。' }2 ?9 v( f) q0 J2 }
4.8设圆半径r=1.5,圆柱高h=3,求圆周长,圆面积,圆球表面积,圆球体积,圆柱体积,- n3 [ ]7 \3 I& Y
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
J5 l9 @3 I! L2 } k' j2 `程.2 j+ B: @9 ^+ q: C
解:main()
e9 ^" q$ g8 {{" J+ X( n0 `2 f3 Y( q$ f
float pi,h,r,l,s,sq,vq,vz;( U' R9 L1 p+ L2 l" l
pi=3.1415926;6 N2 n! p5 t& R/ [
printf("请输入圆半径r圆柱高h:\n");
- @- P) v4 o7 a5 ]& H0 Hscanf("%f,%f",&r,&h);
, A- W! q. P& I" Z! ml=2*pi*r;% f9 T' L! T: n$ g! K
s=r*r*pi;
+ i4 `" S) c1 B, I3 }9 M$ V& Z$ Bsq=4*pi*r*r;
) H' d1 @; b& ~) c1 S0 y5 Jvq=4.0/3.0*pi*r*r*r;
6 E5 w2 f/ V( c" [# G( Tvz=pi*r*r*h;; a; {4 S' y& `; d$ k
printf("圆周长为: =%6.2f\n",l);
) m1 w9 r% e6 Hprintf("圆面积为: =%6.2f\n",s); y+ N! w7 V* V
printf("圆球表面积为: =%6.2f\n",sq);
& t: ~, u& g; _6 F( q& Gprintf("圆球体积为: =%6.2f\n",vz);$ a8 g% V1 `( } `3 m: a: l
}) u0 T1 n5 b5 _1 W5 w( m& i
4.9输入一个华氏温度,要求输出摄氏温度,公式为C=5/9(F-32),输出要有文字说明, d) t& n) z& I5 W0 E3 G
取两位小数.
3 c7 G; V# Y% g; B7 O; ?8 A4 ?0 v解: main()
3 [2 t- P( |# f& L0 F! z" u. `{
6 _' S0 q# j) G7 ?8 t5 v- wfloat c,f;
. q. Z8 i+ w8 i7 I; Eprintf("请输入一个华氏温度:\n");! W2 i$ i) z+ k. s
scanf("%f",&f);
* E& `! y- O. C3 ~0 v* Y! r) Ic=(5.0/9.0)*(f-32);
$ E/ |( a1 P# rprintf("摄氏温度为:%5.2f\n",c);9 X# L7 j& Y- d- Y9 J$ A
}
3 F; f8 F- J5 L, H4 G6 `第五章 逻辑运算和判断选取结构
0 J3 e& `1 N, P# {( W# A5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.% V* a3 V) P+ s- S4 g2 S3 K
main()5 s: x) N& {+ l) m/ d6 s
{$ n( Q/ s+ Z5 Z2 e* c: {
int a,b,c;6 B' C5 [8 R$ p& G4 Y, p$ S! o
printf("请输入三个数:");
; ]0 T# z- N7 D" y7 {scanf("%d,%d,%d",&a,&b,&c); q& T- d7 t, [, \" I- O! V
if(a if(b printf("max=%d\n",c);) t8 u ?1 q: |1 u) p2 e# x6 p
else
) r8 B, h" f5 @, H0 Z printf("max=%d\n",b);
- N ]% Y: e/ R: M" Delse if(a printf("max=%d\n",c);
. ^ c- T2 k1 x8 q' Q else# z) a; K- o7 y) x
printf("max-%d\n",a);& m7 L) V. B1 h4 x0 T0 `
}; _/ W, ?) v( k; w. a! d. X
方法2:使用条件表达式.
3 q2 _ }! O7 R6 |* S( |main()
- P. |8 Y6 q) M: i6 e/ m5 r- t) X* u* f{int a,b,c,termp,max;
2 N$ Y" c6 {9 G printf(" 请输入 A,B,C: ");
0 |9 ?6 ^ z( F- r scanf("%d,%d,%d",&a,&b,&c);0 ]0 q' c! _6 v1 i0 L4 D
printf("A=%d,B=%d,C=%d\n",a,b,c);* h' e' F- b" w1 D
temp=(a>b)?a:b;+ Y& `. ~2 {5 [8 \4 b
max=(temp>c)? temp:c;1 f. O9 {; m. h
printf(" A,B,C中最大数是%d,",max);5 i8 P5 @7 r5 _5 I! _( A
}
1 f/ w2 ?. |& M! i. ~7 k" o5.5 main()4 G/ q( X2 ^1 r2 m1 Z
{int x,y;
# _) n I; R, {" |printf("输入x:");
" r3 C" h* o3 l2 m% x5 I% Rscanf("%d",&x);& Z3 k' ?2 M" i! |. C w9 [2 n5 M0 g# z+ ^
if(x<1)" p1 i5 b/ R9 U" H7 K9 e8 R, ^
{y=x;
2 G* Z7 r1 l2 ?# J! } printf("X-%d,Y=X=%d \n",x,y);* {, H& a/ q6 R% o1 y
}
: j1 i3 @2 _, N9 Eelse if(x<10)% t; ?# X0 ]) C: B! [/ W
{y=2*x-1;- b, M% A9 V# h
printf(" X=%d, Y=2*X-1=%d\n",x,y);
! I4 P& y6 u0 t$ T9 U0 O }/ e' O0 b9 N. H, o) T# q2 D, l0 z# u
else
. W: @+ r6 i& s- n7 j. G {y=3*x-11;
" M; X3 e. r3 I4 q4 t" M' u% U printf("X=5d, Y=3*x-11=%d \n",x,y);1 c3 c2 Z% ?2 k0 R
}0 x" \+ @( G0 O. A
}$ M' \/ R+ r$ E ^4 B9 _5 f
(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!
( N, ~: V, Y; Y9 u( T5 {- ?void main()
9 e, h$ F8 q% \! R) F) z{
3 h7 x/ ]' Z$ y- f: Rfloat s,i;. p7 c8 X8 q. ^* ]9 g' b
char a;
( B- A' q3 l* E! E. ~- ^scanf("%f",&s);7 a0 ]* a& C, u
while(s>100||s<0)
; M! A8 @; l8 W8 P: M: L# G{
& v1 j, X6 W6 p2 P7 Qprintf("输入错误!error!");1 g# Y0 h5 a( B- f9 r
scanf("%f",&s);% h' Y6 T( h# }: O$ K1 l0 n
}
: P% n- V6 ?; ^i=s/10;
3 {7 V( {9 u1 qswitch((int)i)% N: w# b: T( M2 F' D
{! j' x4 K( b5 S5 M
case 10:% h7 Z: [) X1 P
case 9: a='A';break;" t1 N/ F; K. f
case 8: a='B';break;
) B4 ]" b# u1 Pcase 7: a='C';break;: G) {, \! _8 J. K( N, _7 \. K
case 6: a='D';break;4 m3 ]6 _/ z' M$ q P1 S0 W) ~
case 5:
! B# @; I. i; D3 gcase 4:
' f" _8 V, U6 Vcase 2:7 _# {) y1 a9 E
case 1:2 T2 l( @& ^: N2 q5 r
case 0: a='E';
7 H( t& @3 S8 t }3 l: [8 h: D) V, y}
4 ~7 F/ M, j# U# W" _printf("%c",a);
1 w& S! b I+ d2 E- c: G0 ^/ P}
4 w; v3 q( E# z6 }9 o2 l7 o5.7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.
9 ^$ U' C; Z# G7 _# K4 u8 u7 l按逆序打印出各位数字.例如原数为321,应输出123.9 Z1 l1 s$ t& l ]+ F& g
main()
9 o' ]+ ^. S3 i% t, K9 @( b; m {1 M# _/ e8 s! O4 a; [4 U
long int num;
3 j% I# Q& m6 q5 Y- q9 n( ? int indiv,ten,hundred,housand,tenthousand,place;
1 H' Z0 r5 V6 W- R# h printf("请输入一个整数(0-99999):");
, @0 m8 ?) V/ _" A: B" |+ B& i( r: V scanf("%ld",&num);* s$ @5 O' I- y" J4 g
if(num>9999)
# }9 c" L% K, J& y place=5;
) G/ k4 m2 y. h6 {: r, q! gelse if(num>999)/ E' a, M% h1 B x: J2 t. h
place=4;
$ @* [5 P$ x2 v1 s4 J! S9 ^/ c' velse if(num>99): {' k7 o+ Q9 }6 T5 Q5 \1 _' Z
place=3;5 A+ O% R/ M7 L
else if(num>9), w: s! o" m4 H+ t! Q
place=2;
M y+ S" m$ ^5 ~$ F$ `; m7 belse place=1;" ^" ]. f# N3 T' t; r; j& O. ~) s0 H
printf("place=%d\n",place);
: u/ B) @3 _" q) J+ X- H1 rprintf("每位数字为:");
% ]8 P! U, K/ P$ I5 b8 b0 Nten_thousand=num/10000;
6 k! a# \0 d6 P' Z) fthousand=(num-tenthousand*10000)/1000;* ~. H+ I2 i& C# a& h0 m* c4 R8 B2 `
hundred=(num-tenthousand*10000-thousand*1000)/100;
' M5 K! {; h4 X) f5 L7 @; |ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
) s' L- j; {) V. f( j, J) ]9 qindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;
' S1 _7 Z M: E5 ?1 Lswitch(place)7 Q+ y: \: }2 s" N! m
{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);. g" h$ h7 X; p+ m8 w5 @
printf("\n反序数字为:");3 t: e- B/ f* y+ I# [, p4 F' d
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);& b" ~1 u' v" l" V$ ]4 I6 @
break;
- k4 I8 a. |( T! ocase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);% {0 O: [9 c4 K. A9 Q
printf("\n反序数字为:");/ { e, ]7 y$ X5 L8 {
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);
9 j/ {; F1 g9 f$ L break;/ @! j( Q' s) R8 H3 x! o; \. P3 {, M
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);
/ Y! N$ U! F/ ?" m/ Y printf("\n反序数字为:");
# l h2 |& N7 N& p! f/ X printf("%d%d%d\n",indiv,ten,hundred);5 \2 C X( }( ^. T S
case 2:printf("%d,%d\n",ten,indiv);7 J4 C* n& j8 H% E5 h3 c) s
printf("\n反序数字为:");8 J/ b8 \3 q' Y2 B5 @" K
printf("%d%d\n",indiv,ten);, E' u0 w4 |, O- P% O2 G
case 1:printf("%d\n",indiv);
7 @' \# F9 ^" v) P2 R9 X printf("\n反序数字为:");+ Q) s* n. I* \! n4 _
printf("%d\n",indiv);; W" R; f2 \: d z! U: F
}6 F7 o( F0 j: B( h4 U' @( D5 a5 D g
}$ `. J) Y4 }. n! h) N
5.8
# z% @9 i9 w% e- I% X1.if语句" D4 d, y+ F5 X. D, W* T. o+ |- O0 }6 J
main()
3 \7 M6 r2 }8 A: L; R0 d& H7 [{long i;
7 X* c3 F7 G: u! d float bonus,bon1,bon2,bon4,bon6,bon10;
7 m# t& C- V8 y! n! {) Z bon1=100000*0.1;
% b+ b6 D* Y! _) d9 z) | bon2=bon1+100000*0.075;' e& j8 Y$ X3 f% a9 y" B9 Z
bon4=bon2+200000*0.05;9 f2 w7 D s. M- r* Q
bon6=bon4+200000*0.03;, l9 ]# l0 Z& ^/ b/ n6 I* P
bon10=bon6+400000*0.015;
, y2 _4 r' \6 l& m1 o, V+ W! q scanf("%ld",&i);6 b9 K* o L5 a' K' M( f; l2 P- J
if(i<=1e5)bonus=i*0.1;
4 ^% a' M& m' q1 J" Z) g else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
6 [. L" O" b9 A6 t7 x! A else if(i<=4e5)bonus=bon2+(i-200000)*0.05;; o- l2 |- H, v/ u! T9 k3 @
else if(i<=6e5)bonus=bon4+(i-400000)*0.03;
' G0 R: x& ?8 `9 ^ s else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
4 N3 W( g/ q3 h! X5 Y' d; h1 R0 Y else bonus=bon10+(i-1000000)*0.01;1 U$ a2 X8 }% Z
printf("bonus=%10.2f",bonus);( h; ^4 S! X7 u8 T3 r
}2 Z" n5 r: i% H+ ?
用switch语句编程序, @5 J. D* E& ]
main()0 o. b, R V; g T ]" t/ ]( z
{long i;6 X" H& n2 P, R$ t9 Z3 o
float bonus,bon1,bon2,bon4,bon6,bon10; _: R; `) c/ V
int branch;' t% V2 D- d5 W
bon1=100000*0.1;7 `& c" d8 K l. b
bon2=bon1+100000*0.075;# n5 r3 R+ v4 Z* t
bon4=bon2+200000*0.05;% d# c+ z. n/ ]# F
bon6=bon4+200000*0.03;
8 F4 a8 X; O+ N; |5 l- X8 e bon10=bon6+400000*0.015;
7 u% s+ @. B4 Q5 ^# |9 y' p scanf("%ld",&i);
$ z1 S& E' G2 B, }7 }2 X+ {, y& @ branch=i/100000;
x% Y2 z/ D' ]: q( A7 B6 h if(branch>10)branch=10;( V. b; e3 W% F, I
switch(branch)
+ r5 \ b6 J" e+ ^1 [/ w+ T' a {case 0:bonus=i*0.1;break;; u1 X5 I- W+ p. U1 u/ z( w$ A5 a
case 1:bonus=bon1+(i-100000)*0.075;break;$ B: p1 m8 d5 z' v% f* b1 K, n
case 2:
5 E$ G+ F* D# v- f case 3:bonus=bon2+(i-200000)*0.05;break;
* n$ e6 E% }4 K$ ?' y case 4:! o9 H9 T {/ `9 l8 K. B
case 5:bonus=bon4+(i-400000)*0.03;break;
" U( x7 i) q5 Y case 6:
( ]! W0 m1 D1 N w6 P case 7) ~! G/ g" H( b8 M
case 8:+ b7 b2 N) Q6 G
case 9:bonus=bon6+(i-600000)*0.015;break;( i8 c: B! d' [/ V3 v
case 10:bonus=bon10+(i-1000000)*0.01;
; d! f0 k9 T! y8 |6 W }
" |: ?* ]) ?/ U3 w2 ]8 n% [: {) _! N* i printf("bonus=%10.2f",bonus);
. G, l$ V4 G- u}
4 s; h& v: l* Y, E2 ?6 z# b! f5.9 输入四个整数,按大小顺序输出.
4 J: S# g7 d" \* A. Imain()/ w, A4 w+ ?; M0 T7 {
{int t,a,b,c,d;8 G7 `- g$ q9 \, J {6 h( [3 S* Y
printf("请输入四个数:");
$ g) \1 v# X9 W0 n, Y: B scanf("%d,%d,%d,%d",&a,&b,&c,&d);- ?5 N& l- l8 L6 T1 W1 ]
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
% s7 u# ]$ y5 `7 y# |# r if(a>b)
. B/ u# S8 Z# F* @3 A! v {t=a;a=b;b=t;}% A, ?% u& }3 _" ^
if(a>c)5 X9 C% U) C6 R! C. E; J1 C
{t=a;a=c;c=t;}
$ w3 F! t3 ~( z+ a1 _1 G if(a>d)
% y* R2 c! B4 P" n3 U7 t+ L {t=a;a=d;d=t;}! Q# b! h4 g( _ p0 H
if(b>c): n2 s# d9 x9 g. x; w0 b' p
{t=b;b=c;c=t;}1 F+ Z2 n1 l6 B2 d
if(b>d)
. h$ K+ O: S# ~+ o2 A) x. r( H$ m {t=b;b=d;d=t;}
1 p# F: g% r% s5 S3 ?8 ^ if(c>d), D k M- v/ [/ Q1 G
{t=c;c=d;d=t;}- E0 |; e* k3 O w
printf("\n 排序结果如下: \n");
7 M$ |% z. f/ b( xprintf(" %d %d %d %d \n",a,b,c,d);
7 N2 s1 l0 g# r/ f8 S4 e/ E}
; P. g, w* z, M- C: U% i5.10塔) e. o! F% h# [
main()
" `) M9 D1 L6 @$ \2 j6 e- g# Z: i; b{7 m0 _- Y( T+ v1 y# e8 W) C
int h=10;
2 Y( e6 O# L/ B' S `float x,y,x0=2,y0=2,d1,d2,d3,d4;
$ S- Z) s$ j4 ?5 C3 Z: L: tprintf("请输入一个点(x,y):");
/ a3 R" |" m! z7 C X( escanf("%f,%f",&x,&y);
4 c* k& M: ]/ X! H! W' ~! Sd1=(x-x0)*(x-x0)+(y-y0)(y-y0);
7 p0 G& D+ b4 e; I$ bd2=(x-x0)*(x-x0)+(y+y0)(y+y0);) ?" U( p$ @3 S. l. f8 w( }
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
" g4 X3 z7 l# ~7 Z1 _4 sd4=(x+x0)*(x+x0)+(y+y0)*(y+y0);
! u" v1 }) n# Fif(d1>1 && d2>1 && d3>1 && d4>1)8 ^" N0 C9 K, J& T5 {
h=0;; }( x- p& s, ]) Z
printf("该点高度为%d",h);
/ i/ _, }+ l6 s8 [}
% `) Z+ p$ _# n+ n3 ^/ t第六章 循环语句! F/ N# }" V# C& @7 I% t
6.1输入两个正数,求最大公约数最小公倍数.3 c8 z7 c- j4 Y3 i5 Z7 `6 S; z `
main()
$ g. }% p! i. P) w* b/ O) k. b{
3 [ R7 G+ _# P# p' ~int a,b,num1,num2,temp;" J. V1 C, p% }
printf("请输入两个正整数:\n");
- z: y! f3 J K/ o& r* m. rscanf("%d,%d",&num1,&num2);0 M5 N t! G8 z O. X
if(num1{% O! |9 r" O/ ]( N" x2 L3 u: e. ^
temp=num1;
( b8 D' L3 W1 ~3 D& onum1=num2;
. h& d' J0 q* C7 H# d% w% cnum2=temp;9 D' f7 b5 a( x4 c8 T- `5 h
}
" s3 P6 w5 D( N0 B# e) |a=num1,b=num2;* K" F& l v3 `$ ?1 X9 B+ y: X! P
while(b!=0)2 S+ z1 a1 X3 x9 g2 X
{
6 f/ E$ l: ]% i' \' X temp=a%b;6 T' Z2 W: J% o2 q* [
a=b;
( l! a' ]" ^5 L7 `( }( L1 G. g$ Z b=temp;0 ~& J* s7 V0 q8 \3 s, V: [4 `# E5 C
}8 ?) D& ^6 U; m0 L) C
printf("它们的最大公约数为:%d\n",a);
1 [- G' k0 m/ @2 rprintf("它们的最小公倍数为:%d\n",num1*num2/2);
/ y. J4 D; E$ J! m0 x: o% u}
4 g- j9 v9 s2 J) G& W: s2 q" E# ~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.# P( f, B8 K; U) Y N. g
解:3 i6 b9 r$ ^+ k# Q6 y1 O
#include < >
* P% _# r; J$ e& a2 H( q- Smain()
2 c6 |, ]! V1 K& }, p9 m, s{6 X& p3 _7 { a/ s/ g
char c;
, I c6 V B6 Y# O! q8 hint letters=0,space=0,degit=0,other=0;
* ^9 X3 j1 \( r) P* vprintf("请输入一行字符:\n");% |4 \7 G% M t8 \( R
scanf("%c",&c);6 J# I& r. o, y1 V6 C" j8 w1 `
while((c=getchar())!='\n')
* o2 n) S k% W; F0 o7 r{
}! Z2 z2 T+ T+ {7 M# F0 W1 ^1 mif(c>='a'&&c<='z'||c>'A'&&c<='Z')
+ L: w4 t6 p( }( s% n3 yletters++;
; D* C5 e M8 C9 t+ V! O, Ielse if(c==' ')2 B3 w% e% j. A+ B) R" f6 k4 W. t
space++;: B( c5 i" u) P$ ^/ i q
else if(c>='0'&&c<='9')* ^9 O. R+ R1 P3 S
digit++;
2 o0 a8 B- K0 d J$ p" A7 `* X7 yelse$ j: J. B; K- e9 K G
other++;
, f5 S8 x% i# y+ W1 [+ a}, F4 H9 ?$ r4 d7 Z
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
: u. g! W% d) H( Q" B' i5 Cd\n",letters,space,
' F1 }; W6 s+ F \5 w* ^+ a* x3 Edigit,other);
* L8 N, P+ |' H$ x3 s6 u}
! p* {* c0 a; i6 i1 Y6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.$ x* ~0 w `% a2 i2 T) Z; r
解:5 g$ o6 l; i( V# e
main()" c+ Y; O! A' M$ u0 K
{) W7 Q9 B( V1 E. {6 q+ e
int a,n,count=1,sn=0,tn=0;
, o: v$ X L$ R r u% f3 u( u dprintf("请输入a和n的值:\n");- j' a; p6 b7 T2 |# n V+ T
scanf("%d,%d",&a,&n);
( i& P! L# \" J( Z) c8 lprintf("a=%d n=%d \n",a,n);
- t6 a+ J# W% zwhile(count<=n)* s1 ?3 @: F3 v
{! m- _+ I+ w- w7 Y
tn=tn+a;# }: G7 V- c) i1 E5 K
sn=sn+tn;0 ?% }- I3 c) S3 F' m; Q8 @
a=a*10;9 i5 j* s) V/ O2 b- {) ~- b3 g3 ~
++count;$ R# M2 ^: Y) i% J: `6 s& ]
}
2 R/ F+ R! ~( d1 Xprintf("a+aa+aaa+…=%d\n",sn);
% \: a C$ m, O) T% p3 N: S}, | s1 @) h4 s. K x% t
6.4 求1+2!+3!+4!+…+20!.. y! i; Y- A8 V) {0 q% u( A$ t( H
main()
& p& E: f5 D; |; z{4 c. l2 o C2 N0 x1 e
float n,s=0,t=1;+ g8 N% v* b/ |& Q
for(n=1;n<=20;n++)
% J. [3 } v# n0 d5 K{1 J* u0 u* t' t6 ~
t=t*n;; c, R: V; a" n- S2 |* y
s=s+t;
8 x% p$ j6 X( L}
& I$ I5 ^' Z+ ~7 t' W! V# B' wprintf("1!+2!+…+20!=%e\n",s);
1 X" A- `% d0 K7 X' m}- J7 r# R+ Q2 r H+ `" V
6.5 main(); ~7 y, C0 U; x
{
3 m& }! X" m8 M" jint N1=100,N2=50,N3=10;
1 L! y1 g7 g* pfloat k;
3 W3 R8 I3 g1 h+ `* Z% \# nfloat s1=0,s2=0,s3=0;9 R) Q: ~" k _" |8 W7 \
for(k=1;k<=N1;k++)7 G/ R5 R. s* M3 o; b) B% [: y: a
{* u6 f- [6 l: W: {8 A5 E
s1=s1+k;- _6 q& M5 y9 s
}8 F, ^. O) ^4 s9 Z7 L& k' c2 C- R
for(k=1;k<=N2;k++)
& B! ^/ J) ]; x" Q v{5 b N3 l Q, b1 `2 P
s2=s2+k*k;
6 Y+ k$ b" J9 Y# X+ U9 Z# i& d7 I}
! h. P" Y Y7 V) v& e8 g; ufor(k=1;k<=N3;k++)6 e2 x7 Z! b* C7 [& S& z
{# w! e* Q- [ Z) R4 F7 d
s3=s3+1/k;7 G# i6 \7 D% ?
}
8 k9 ?: \5 E/ C* T+ @/ C, y; S9 `2 Lprintf("总和=%8.2f\n",s1+s2+s3);
% d# s2 U, g1 M6 X, ]2 A$ m}
2 B( K( O. s o3 O; e; }* R* i6.6水仙开花
l' B- g- j# ]$ {" Gmain()
0 U. r0 Z a c7 {% R, ]& r9 L{, p& t1 ]3 I( n1 I2 }& u8 z/ M
int i,j,k,n;6 t0 {2 u! N. n- k. F, K. ~
printf(" '水仙花'数是:");- f; w/ f: Z& b& b8 N5 ?$ V
for(n=100;n<1000;n++)
& Q: i: C: M; }: d{/ }! c8 ]' Q% V. @4 Q% [
i=n/100;+ i7 s; n7 U; i9 U; S8 p# X
j=n/10-i*10;
+ X ]" p: i3 q- [- ~0 dk=n%10;; g3 W8 O1 y+ W( C8 j
if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)! i$ N! g: D/ K- U: k1 R6 w m+ Q
{
" z" ?0 T# }& N" q- U. X% r" c: |printf("%d",n);$ P( T$ j4 d9 m! y5 u
}0 H, A# m, Y3 g7 s- X- r
}
' o; ^4 t0 V9 m2 c, w fprintf("\n");3 g6 k* Q2 d8 e3 C6 M) i5 x
}& f5 ^( ~' @# H$ y$ r1 @
6.7完数
, p6 ~ t2 o- @" t' {main()
V3 U1 H4 y* g0 N( r#include M 1000
, g" K3 b( H5 l" Z* r3 k; C e+ U" _main()
7 \. C: n3 |/ b{5 J) `7 c& d& O6 }+ Z
int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;
( s* K4 o7 {; }% Q6 K! S& @int i,j,n,s;: s, a, A U. x7 v0 y
for(j=2;j<=M;j++)$ v* @ \5 F \& x
{# c% D/ L# d2 p# i2 [3 x$ E& z
n=0;6 m- K a1 k1 V
s=j;
4 x: e8 {, t5 Z- Jfor(i=1;i {; \) _2 j2 j( a! A+ R2 y& l
if((j%i)==0)
; z: ^# C% r+ q4 J$ e% Z8 S {
7 `. W3 | W+ t" q9 E* s if((j%i)==0)
. W- E ]' ?& [# @2 e& d {
\8 t2 v( u. { n++;( V; E1 Z5 s! r0 j# F0 G: V& o- E6 q
s=s-i;
% i* @: L9 W7 ^3 d; r: j7 {' R- Z switch(n)
9 ^" ^* `2 ~4 \* x0 t; |$ N+ g {/ E: m4 m* N4 i! ]
case 1:# l1 j' j. c; ] n1 ]
k0=i;6 n! E, i0 l* r9 u
break;% r3 J" z/ D: M e' u8 @+ R0 g# Y
case 2:$ Z( \! e0 v' S D; `+ z3 L
k1=i;
, W N. U8 ]* R break;2 [: f* s5 I' S' e
case 3:# k" ]0 b& \5 w! r( r7 f; D
k2=i;! d. f- I: W' p8 B
break;
& }; x+ s( U" n3 C3 g9 \0 w$ j, B" B case 4:
! Z' X$ G6 _ C/ W) o k3=i;& ~3 Q+ s- m. I+ }
break;2 a, F! z. X0 @
case 5:3 y5 y, I$ G/ j) k7 I
k4=i;9 n5 b: s5 X0 H4 _; f& s
break;3 I Y8 H; ~0 [; f5 U
case 6:
4 @( H2 T6 x% ?# ^7 h& _; W k5=i;: p8 L* r8 s8 \; S- R* f$ s
break;
4 G- ^3 f7 a: c$ e( ^ case 7:8 d/ V5 P: Y; ^9 t9 s
k6=i;5 n" k( J; E! W- T2 z+ `% x ?
break;
8 s- x1 }1 ^! ^: C0 l! q. C* P case 8:1 X f( n9 k8 X: Z! w! h
k7=i;1 O3 {4 r: T* `
break;- f* C. a/ z' H8 U. w X( B
case 9: O* R# o4 h- U* G* H. I. k
k8=i;
( z& R5 \/ [# {+ }4 o1 Y# _ break;
5 E7 f, ~3 I1 p- c4 `& E case 10:- L) s G& q3 x+ S, i6 w7 y
k9=i;- [- t$ p Q: p' K+ R
break;
4 E& X. u! O- w }
) ?; u* \% j- _$ r, b } d& s" p* R$ j
}
# a- d. x. U6 v% I, H9 M8 eif(s==0)
& [2 q( ~- L: y' g: K {
0 h- ^3 r4 \8 o" Dprintf("%d是一个‘完数’,它的因子是",j);- `% ?! A8 f& C W6 A2 j7 Z: u
if(n>1)5 t2 e: k3 d! A+ l) r# ^
printf("%d,%d",k0,k1);
. b( v# Z9 h: l+ z* }if(n>2)* O1 ~# H. ^# N
printf(",%d",k2);8 }. p& |+ K, m$ T- i
if(n>3)! R3 [& {; V2 @" i% ?
printf(",%d",k3);
" ~' z3 u% k# r y2 l1 ?% {if(n>4)# |# p3 S* R: h5 {" Q
printf(",%d",k4); v+ ?* M/ \$ n6 Y. G s5 ~1 d
if(n>5)" D4 P7 H' p$ K! ]: b" K
printf(",%d",k5);! G8 \, P, E% h# c9 }- @
if(n>6)6 t+ q$ A2 M6 v, j+ {( c# n( F
printf(",%d",k6);
7 Y$ A4 J8 _7 P3 X1 |' g* X3 g" rif(n>7)
4 ]5 ~$ R: b& T6 ~ printf(",%d",k7);- y. b5 i+ F8 B
if(n>8)
3 ]# ^: H0 x+ w. N# _$ e3 U! B printf(",%d",k8);6 ?) t2 }3 T/ d4 U7 o
if(n>9)
- }) a1 j% b$ Z: ? printf(",%d",k9);- q, @6 F# J z( a5 ~7 P% C
printf("\n");
% ]1 f, j8 q9 P" B- S; u1 {3 n }2 G& ^$ i$ l- T6 |2 F
}
+ }/ H0 X w/ Z/ I6 s/ e8 X' l方法二:此题用数组方法更为简单.$ g9 N7 [, k3 R3 N. d5 ~" g9 n
main()) X* Y$ C7 R' o t
{
) S8 z4 @6 n$ h$ O! ~& t: } M, Ustatic int k[10];
8 M# x4 }, v% D, M9 }% Mint i,j,n,s;6 S2 }- I5 Y, ^
for(j=2;j<=1000;j++)7 l6 F2 ^: o' ^( w
{
0 v. r0 O/ l; S G+ }0 j+ [ ?n=-1;7 z3 v& l- @: Z1 L
s=j;+ `2 G5 u" n" n. j+ c+ F8 D! F
for(i=1;i{6 Q$ y7 p. D% `$ q8 Y- c6 y
if((j%i)==0)
. D% v9 \" a! C$ o( L. _( Q& T- z! q{: N- Z# ^% X" t0 X0 M* G
n++;
" s# Y# H- {& f9 {6 t- b4 | D0 a7 ts=s-i;- E: T# z* e! e' a3 `- x9 Q/ Z
k[n]=i;! D) S' E/ j' D7 K; z7 t
}
# F2 E$ n, b" k }2 S" q- r0 O, n- f x- V
if(s==0)! S$ i$ n9 L& B2 Y
{! J0 ?" r0 W2 I& m5 z* z2 w
printf("%d是一个完数,它的因子是:",j);; {1 l4 E- F# z! W! W: W7 I
for(i=0;iprintf("%d,",k[i]);3 l* o# w5 N% l8 T" J6 X# ~. W
printf("%d\n",k[n]);
n! @ h6 c$ R! R3 q. B* q}9 g5 [" m1 P- Q" ^9 V S; x* D: B
}
; o9 Q# O- e9 b; j: m6.8 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.: f$ M! v% T6 A3 ^' ]- k2 E, o
解: main()7 h7 w( L1 h7 z9 t7 S) n1 L. t
{6 m: S" m' z! x' `% M$ _
int n,t,number=20;
) ^+ L' n& |/ f6 m N/ Lfloat a=2,b=1,s=0;
1 K- D2 K. L, U \% ]/ |6 Nfor(n=1;n<=number;n++)
1 s+ ^6 R5 j2 H/ S4 g' v' I+ i{
{# v; q0 \( n Q8 qs=s+a/b; y- X l; U+ d( {* D
t=a,a=a+b,b=t;
: f" o2 o( k5 N" O3 M5 \4 O8 s% Y}
1 ?0 u7 ~) | n3 d' Yprintf("总和=%9.6f\n",s);
2 f& G1 P0 j h! h, b}
( i! f6 ~4 H; m6.9球反弹问题, T$ a& t/ R5 o5 V* `
main()
- L e6 W6 x% e8 M0 t \{. Z/ b7 g+ r2 d _* t( J$ L( X
float sn=100.0,hn=sn/2;; w( m+ h( Y0 U9 B- P8 v
int n;
& Z! s K9 {+ b, R6 u- Ifor(n=2;n<=10;n++)% c2 h& q- i% `' K( X
{
9 o! W$ i6 s2 msn=sn+2*hn;
' O0 y! k9 B+ t' m( ihn=hn/2;
b3 i8 [9 d3 D! v$ w}
+ `- x& {! ~ M6 ~printf("第10次落地时共经过%f米 \n",sn);- B' e4 y& e& I4 a! ~% c2 M
printf("第10次反弹%f米.\n",hn);
7 F" n0 `4 ]3 s* M; e$ S8 M}
$ a7 [7 J. _4 {6.10猴子吃桃
) F6 m" A! A& L& _" Smain()' Y6 Z2 Q, `# p
{
) r, K, Q1 c, R: u, C$ {1 d( F8 Sint day,x1,x2;7 Y6 m6 c& p) _0 |7 K U) R) b! y# J
day=9;
a" K- W" E7 h9 _x2=1;
8 }- E! d8 _# {' l4 Twhile(day>0)2 |2 N. Q0 C: {0 c' H$ J, H
{. [9 o$ P( T0 u& A& g$ P7 r; b
x1=(x2+1)*2;- Z9 b$ P# J+ n5 ^# E
x2=x1;% C4 U- f d) R3 W& y# ]& ^
day--;9 T/ \( ^( X. `, e/ P7 j0 l% A
}
0 W8 E0 H/ d j- O( D+ L+ @printf("桃子总数=%d\n",x1);6 G" ^; V' P9 ~9 Z* @
}
' o3 [# b( \1 p3 @. Q
5 ~' {8 X8 C! A: |3 i8 H: Z6.121 D# ]8 v6 K4 Y2 h! d6 \ S/ q
#include"math.h"! n" d$ s7 o4 j0 J# E
main()
1 N2 F: j* r, U! m# B+ y{float x,x0,f,f1;
7 L4 ~7 S6 `# u) J x=1.5;( n( N' C+ n3 D& D6 r( J2 w9 R# u
do
8 {2 m% m& `0 F3 ~/ p8 O {x0=x;7 q- u: t2 C& [5 u0 K2 s
f=((2*x0-4)*x0+3)*x0-6;
* j/ o7 N0 N+ g, b f1=(6*x0-8)*x0+3;( U& a. p0 @$ Z; k- f/ l& L
x=x0-f/f1;$ @, G. p# k: Z( {3 D3 H7 b
}
2 X. r d; E' x' s1 d; L2 X while(fabs(x-x0)>=1e-5);
. Y0 s. a( d! \; a/ b) | printf("x=%6.2f\n",x);- R7 I; a& r3 \) g# g
}+ t$ W+ D- T$ X" R5 ^) j# |: k9 \
; ?* V/ S2 ]+ d* W, W6.13& L' K& w! V+ |9 j! U" T
#include"math.h"0 ` b$ E# Y6 [1 \
main()8 ^* g5 z9 U7 W! ?9 A: I! ~+ N" W
{float x0,x1,x2,fx0,fx1,fx2;
2 _$ Y: L5 N5 Z7 |( V9 ?* q0 J6 G do
1 K" h6 ~: C4 V! d. W! | {scanf("%f,%f",&x1,&x2);
. t- ?9 F6 A9 ~0 f+ _3 G fx1=x1*((2*x1-4)*x1+3)-6;
0 t! c& Y) V( m! @! j5 t; r fx2=x2*((2*x2-4)*x2+3)-6;1 O/ f4 {8 s- z5 }
}
1 d8 r* f, _$ r* i5 @7 T( f- z while(fx1*fx2>0);
+ M0 V4 ?% w0 C do
4 p$ h" n7 d* i0 ^$ o {x0=(x1+x2)/2;
% `8 L; I {8 X6 R D* h fx0=x0*((2*x0-4)*x0+3)-6;
! ^7 N% Z) L* r0 C/ |( g if((fx0*fx1)<0)0 l) h e. D. z( J( L a* F
{x2=x0;7 b& ^0 H+ I0 Y8 Z# o1 e0 a
fx2=fx0;! b4 c+ Z% F" d$ v2 y) C: ^
}
; _# h* ^* {3 ?/ m( Y" J1 d9 O else
- c: t% P( R2 {7 q {x1=x0;
6 n9 O% Q( {( H' j! G& K fx1=fx0;
5 k6 M4 \# M+ g- c: w5 q }7 B& D+ F7 ?" `4 p: x2 w3 h
}( z1 \. D6 Z, t
while(fabs(fx0)>=1e-5);
; [) K1 _6 K* o printf("x0=%6.2f\n",x0);
p' q2 E4 c1 [) {}2 b3 x, v6 j2 H. S3 d
6.14打印图案
/ ]+ d! _ ]2 |, Cmain()9 I J% ~. D5 s" ]
{int i,j,k;8 l5 C* ?" r& Y0 f' X# x3 u! r
for(i=0;i<=3;i++)7 E% {, |0 j) S) r& m5 P( T
{for(j=0;j<=2-i;j++)
' h6 E1 ?; I% B8 t printf(" ");& O# G7 f( F: R3 A% A
for(k=0;k<=2*i;k++)
3 M2 f5 p; H/ l: q3 I7 M printf("*");5 s! N$ s c9 y3 Q% A# w: F0 w9 W3 \; Z
printf("\n");8 g P4 `& J0 l
}! S, S7 B& s; ]) i8 ^1 o2 Q) x) {, M
for(i=0;i<=2;i++)# W' N9 G3 s6 _1 A- `, h$ ^
{for(j=0;j<=i;j++)
8 p8 o& G/ v+ F' L/ } printf(" ");6 f5 S9 h5 `+ _7 @
for(k=0;k<=4-2*i;k++)$ K; Y' v# e! J, H
printf("*");& g! W4 M" X4 v; t- L M* z7 G
printf("\n");
. N3 m+ Z. S1 f, w4 K) M }
- o: I3 ~7 N2 `3 i" z/ k4 R' L# |}1 m: t) F* m7 P0 r
6.15乒乓比赛
8 m9 S, f# I! {$ Wmain()
9 f1 Y& l4 i4 T9 t* G{
3 C; y/ B" I+ _char i,j,k;
- {7 ^9 _' A0 ~, X9 \$ c' |for(i='x';i<='z';i++)8 p$ r( P4 J z Y; P- z+ X6 p+ G
for(j='x';j<='z';j++)
6 t3 }- |+ ?" n3 H8 S+ p {
F" x5 J/ k" Y) d% U& z5 dif(i!=j)
/ y' l0 Z: O5 j1 A$ nfor(k='x';k<='z';k++); b' Y6 F: S. e1 Y
{; x, ~: K: J. N s! O5 U
if(i!=k&&j!=k)
4 Z5 S& l' X: x2 f+ ~5 ?, z {if(i!='x' && k!='x' && k! ='z')
0 A( x, I" \) }* gprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);* Q8 m! }$ f& `- b; r
}4 C2 ?) h6 D7 b
}
5 n8 m/ N$ j( I9 ]2 g) y: x) P }
3 o$ c9 p1 Y% p9 |. o. i}# K- X$ j- H: M; F. `1 u: G& X
C语言设计谭浩强第三版的课后习题答案
/ N) Z' e8 A+ w& x7.1用筛选法求100之内的素数.0 o% q* J @( V3 n) Q
#include
9 ?" x! M( l! p1 p# T) p# N#define N 1011 W1 [0 [6 {; G: H( t
main()
2 c5 N/ \0 v4 ?2 p1 s; _: U{int i,j,line,a[N];6 H3 [) Z% R) u9 W' b! }
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)! ]& j) R6 A9 Y$ e% l. @9 ^
if(a[j]%a[i]==0)
3 E% H. B2 l' @4 C' a a[j]=0;$ e' \" G4 @8 z L: m/ l
printf("\n");3 h4 O3 B$ I9 k) |. W8 I5 n3 f/ V1 K" |
for(i=2,line=0;i{ if(a[i]!=0)# R; v! S' z8 W# w* |4 S4 t
{printf("%5d",a[i]);
& z, B7 L( V! L9 r8 B line++;; }7 c- u' \; V. I8 M
if(line==10)' N5 q6 ~, L* s: I4 b$ A/ \
{printf("\n");
2 x- |1 v* g: Z! X- |+ L. i/ _' q: V line=0;}( k, K2 ^8 B+ A9 n. S" K6 y
}
! u( O% l; q I: M- M}+ y$ F, y: {- V {
7.2用选择法对10个数排序.9 i7 {3 r8 H! _% l7 n& F
#define N 10
( A4 P+ q4 W/ r4 i% t( E( Umain()
: H+ @1 |" I( ?2 S{ int i,j,min,temp,a[N];
$ C9 V7 U8 s2 r/ X1 {" zprintf("请输入十个数:\n");
% d) }: G+ i, F) n! _for (i=0;i{ printf("a[%d]=",i);
0 O! q9 x: C2 I2 U3 E$ T. [ scanf("%d",&a[i]);
4 R8 T) }& n, _0 X}
1 y( m8 _9 b5 p0 f9 jprintf("\n");$ q! W4 t7 Z) W0 W) ^' L
for(i=0;i printf("%5d",a[i]);- j' ^) s0 N6 n! J4 n
printf("\n");
6 K! A5 I$ c' F' j3 Q, ]: K& Qfor (i=0;i{ min=i;9 b% X7 d# L4 |6 j0 U; P
for(j=i+1;j if(a[min]>a[j]) min=j;0 g# ]/ d2 o% r! `! [
temp=a[i];8 F0 F4 a h$ M3 F) J( o- N4 h
a[i]=a[min];8 M* Z" @2 K- ^6 ]3 K$ S' P
a[min]=temp; v9 ]% \- V9 e, Y8 Q2 M, \
}9 W* U5 u8 @; d$ x( ^
printf("\n排序结果如下:\n");; l. p2 ?% k. F" [' d" p8 D
for(i=0;iprintf("%5d",a[i]);) Y( [- v2 s$ u' C3 n6 K) j
}( u1 r2 ~$ V3 K. ]5 T
7.3对角线和:
" ^, m3 ~: E, \( Z5 {main(). u5 B0 B3 t5 N: b9 r5 b( Z3 a4 J$ ~
{
& E; @- q1 C, `% e0 a7 yfloat a[3][3],sum=0;
# c/ O7 i3 @2 J; a4 A" B: v( Cint i,j;
4 a5 a6 q4 s8 l0 f L- \printf("请输入矩阵元素:\n");
; `* ~2 w6 W n9 Efor(i=0;i<3;i++)% t6 e, a5 o5 i7 S
for(j=0;j<3;j++)% X! k( v [( B. e/ i! Q
scanf("%f",&a[i][j]);
- _+ n, f7 E; ^# r for(i=0;i<3;i++)
1 X: W( c* {; i4 ~8 d9 e8 ` sum=sum+a[i][i];8 }5 C$ J# P5 ?+ U4 S3 n" r
printf("对角元素之和=6.2f",sum);; D. [9 J# B' a4 q! J$ A6 ]
}
P& ?8 G3 ^8 B. d* V, b9 p7.4插入数据到数组
! Z, h) R' h, f3 J! b; q. umain()+ I5 H( h9 q H
{int a[11]={1,4,6,9,13,16,19,28,40,100};
5 K7 O! U% p/ J, `1 p1 }' J* Mint temp1,temp2,number,end,i,j;. r5 v% N: ]1 j$ _& Y. X, I0 N
printf("初始数组如下:");. O3 A' [8 O% O( y- Z
for (i=0;i<10;i++)5 k9 m* d7 B, h! @$ w
printf("%5d",a[i]);
7 S4 L- e8 s _$ E% w W4 fprintf("\n");
3 _& N/ D8 W" w; [printf("输入插入数据:");
. T9 d1 r4 _% g7 b3 Q' mscanf("%d",&number);. A1 S/ q- T2 y& W1 n. z
end=a[9];8 v; S+ ]) c" G8 t' v4 ]' [
if(number>end)
# ?$ O, V! q6 o- ]( t8 qa[10]=number;6 s. i7 ]$ w- E# Y5 u( d
else
) g4 w5 h% B; i; l: R {for(i=0;i<10;i++)7 ?: Q5 t/ X2 g5 J A0 j0 ]
{ if(a[i]>number)
* Y9 t q: c) w+ P) i {temp1=a[i];% Y/ u c* B X
a[i]=number;" `: D) Y* D; y3 v& k' q
for(j=i+1;j<11;j++)4 Q8 u, L2 V2 z6 C- A
{temp2=a[j];- u7 F; O2 T# j0 v9 ~
a[j]=temp1;
3 ^. [$ i9 E: L$ \) z5 T5 |- h Q temp1=temp2; P. s+ s, _1 s1 O
}
3 g9 ~4 s, }; L; k t( E break;
# z9 O: b3 a9 }4 i+ g } V" ^; x4 \6 Z3 Z: {
}
- }$ i+ d0 [& z0 N2 n }
8 ~3 A7 ^9 D; V- L8 u/ X for(i=0;j<11;i++)* o t) }# g2 l b
printf("a%6d",a[i]);6 P; \, d$ N' m6 r* K7 T$ Q
}
6 K" j0 x( T y+ |1 I7.5将一个数组逆序存放。
; c# s4 u4 {# {8 _6 i8 k! K#define N 5
5 h& z) |8 i6 e: E, O: fmain()
* ^) ]4 j) W. e6 \1 Y d8 G5 Q{ int a[N]={8,6,5,4,1},i,temp;
; u9 v3 R7 {% w* Pprintf("\n 初始数组:\n");
' ^( Q6 w, i; a6 w+ ^9 ^for(i=0;iprintf("%4d",a[i]);1 f" X& ~2 i& e1 E' j+ L
for(i=0;i{ temp=a[i];
9 w, i* n- c& H# k; o% r- Z a[i]=a[N-i-1];
" w4 Q9 ]7 L8 Q$ c6 O a[N-i-1]=temp;- m" m$ u' a v# w+ E9 [
}
. y+ n+ r5 I p( l6 f, y) Bprintf("\n 交换后的数组:\n");4 d, Z+ K! G2 k. L$ N
for(i=0;i printf("%4d",a[i]);
6 u, c. A2 g7 l5 ]% u5 _, }5 V' n}2 [# l# i/ d0 `! A! T* j
7.6杨辉三角0 T8 l o8 `- g# \
#define N 11
# g% |+ P$ ?7 s, ?+ u6 {main()
$ }2 b1 d0 L& Q1 d{ int i,j,a[N][N];
/ h5 J9 E4 S" I( W# T7 s* t for(i=1;i {a[i][i]=1;
; C! X( m$ z% A6 D) g3 Y a[i][1]=1;
$ v7 i/ w3 Q4 e }
3 f' p5 d5 s( e. N' E5 l; o9 q for(i=3;i for(j=2;j<=i-1;j++)
$ U+ u! b! l D. s! f# `2 O% Z% v a[i][j]=a[i01][j-1]+a[i-1][j];- \/ {+ N$ m7 {9 I9 o
for(i=1;i { for(j=1;j<=i;j++)
! w! K* E5 X' h printf("%6d",a[i][j];
% c9 ?; B% e) r9 T& z% B printf("\n");
1 E d4 M2 I. D! j+ J* E }* z& T- N g5 \& r' f6 X3 Z
printf("\n");
5 y4 G; Y: H$ P! I6 K1 `}
3 H% e7 L- a' X5 w, E9 \7.8鞍点7 q% z& k. a; `# z
#define N 10" A$ c" H' S. {- r O8 L) B% W
#define M 10
9 A# p3 \8 K& d! B- ?main()
7 k$ `$ }3 G( t{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;
, B' c3 K9 k! X printf("\n输入行数n:");
7 r3 v+ M9 b }( M) x scanf("%d",&n);
0 D5 B) b0 ?: s! i& G printf("\n输入列数m:");: m, ~* i6 K4 ^! b
scanf("%d",&m); W$ H" ~+ c9 ~. [
1 I7 N# L3 X+ j- t: y, F( H
for(i=0;i { printf("第%d行?\n",i); u$ \& c6 d) J
for(j=0;j scanf("%d",&a[i][j];/ q% f4 ], u+ n$ x
}
1 ?; T0 H) B1 v( r for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
4 l# y5 ^/ }9 x( O( G9 C4 T pritf("\n");' q/ x) { u2 J* W/ \7 v& m: g
}. v) M* a8 T2 ~9 I. M' D3 }
flag2=0;. U6 ?$ h, {: A* _6 w
for(i=0;i { max=a[i][0];6 \4 m5 a* y0 u/ p& j' n
for(j=0;j if(a[i][j]>max)1 P& _3 j4 r4 l% b
{ max=a[i][j];
* q! ?' d7 \$ D* y' C F( M maxj=j;
/ {8 P7 F7 i# R% ]/ v" q! h }
: }, {. g+ O& g for (k=0,flag1=1;k if(max>a[k][max])8 }( |0 t% B: T# E1 }# l
flag1=0;0 R( f' j2 M8 a
if(flag1)
! Y4 Z' J+ L0 ?) Q { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);# N* O# R T" f: G
flag2=1;
$ ^6 y/ ?7 a0 @! ] o( z }
# A4 L5 P4 J( |8 ]. I# Z- L}
1 r. s }# @" i6 e: ?, aif(!flag2). L+ Y) v% N( r: w5 }
printf("\n 矩阵中无鞍点! \n");+ ^+ f" l/ B& G k
}
- x6 \& f$ c+ u w, k6 N. J
0 J5 P5 c) R5 D+ J7.9变量说明:top,bott:查找区间两端点的下标;loca:查找成功与否的开关变量.
0 K& n D+ K$ l( Y#include8 W( `! G- Y I' L! P- ?1 Z9 m
#define N 15
- t) \* P9 s8 ?+ Q @3 M6 emain()7 `& g y! P, o7 ^/ |/ d2 r
{ int i,j,number,top,bott,min,loca,a[N],flag;
; \: y' |/ n9 `' I. l2 H char c;& r$ B: u, r j* v, ^2 G
printf("输入15个数(a[i]>[i-1])\n);
& f) u! U6 Y ^ scanf("%d",&a[0]);* a1 x d7 o/ n; C8 s
i=1;3 z8 S' L3 Y% {& j ^
while(i { scanf("%d",&a[i]);
' a" z" D" N) g" j" @ if(a[i]>=a[i-1])0 R: b$ R9 @& g; _% Y& h5 y7 ?3 {; T, s
i++;
0 z% q6 [1 |9 B; a4 u4 J5 k- L esle$ F# e' e( N1 v! V5 j1 J
{printf("请重输入a[i]");* ?, y P( o/ e% Q- {3 G* \
printf("必须大于%d\n",a[i-1]);8 I6 o' y" `, |+ w8 h) x: o: p
}
: k. M0 W- D% o5 N; C }
* J# y* P7 S. |* H6 H2 g @- d printf("\n");
- \* r- {$ e: V% x for(i=0;i printf("%4d",a[i]);% h5 m7 W$ L: t/ L) A9 Y8 C1 a/ \
printf("\n");6 Q$ W2 E8 `6 a& K7 V4 r
* ?9 R1 C2 I+ ]2 B0 Y/ N, U7 P4 p
flag=1;1 F8 u+ Y: m5 B1 X6 h6 m
while(flag)/ M# @ h E" ?
{0 t3 v! Q6 \; b: J" s
printf("请输入查找数据:");
1 ]" S* {8 }% K4 q5 X scanf("%d",&number);2 N/ L6 G/ K# m# u. x, d
loca=0;
% W' W/ G$ U6 n top=0;4 ^0 w! m" D0 P
bott=N-1;' g0 l" g, o* `8 k6 Z- @
if((numbera[N-1]))4 A/ X8 ~1 \. V1 W/ g" I; [/ ?
loca=-1;
1 r; |' E% c+ H6 h; D while((loca==0)&&(top<=bott))2 {# r% u% G, @
{ min=(bott+top)/2;: n7 \; p$ I( O" U) o& O
if(number==a[min])
3 I N$ M3 C0 ?5 Y* h6 Q3 r { loca=min;
2 I. v, p& ?6 A' p0 Z/ W$ [ printf("%d位于表中第%d个数\n",number,loca+1);6 N( K1 P/ ?: u. k% A
}4 B* E# k8 X1 `: D; F/ c: O* s
else if(number bott=min-1;, k* y* X& R& x
else8 Q" v8 j9 z% H
top=min+1;
7 `/ K, f" S* I; e( J }& r" D4 {! H2 ~2 H
if(loca==0||loca==-1)( ~1 \$ b( s8 ?; c% X6 F
printf("%d不在表中\n",number);
9 Z& {& Q9 f( ~3 E( ^ printf("是否继续查找?Y/N!\n");
, H) ^% R! P9 r; E c=getchar();& k/ r5 T* S; e, d0 g: c
if(c=='N'||c=='n')2 R6 u# P4 y3 m2 o; G! [3 P
flag=0;4 F3 O+ {% _. ]: u+ E$ Y, @
}
0 H4 G$ F- r+ A4 Q- B9 {- s& _}
: ?" o7 m, x- R8 C' H, | ' P8 }1 ]1 n$ z; | G
7.100 |4 M- k( z9 k' [3 J8 E c
main()( q A1 y4 P3 ~7 j
{ int i,j,uppn,lown,dign,span,othn;3 {' m. k( C! ^/ }, Y. v! |: }
char text[3][80];( v. P! L$ Q6 f
uppn=lown=dign=span=othn=0;8 h! g* r, i" ]& R; {
for(i=0;i<3;i++)( S) i: D1 A* ]/ v
{ printf("\n请输入第%d行:\n",i);+ l' V7 Q3 s$ F$ t- d
gets(text[i]);
c+ r, _( `# U# O% m) v' j for(j=0;j<80 && text[i][j]!='\0';j++)
9 W; \( P' D$ r; G9 S {if(text[i][j]>='A' && text[i][j]<='Z')
' f5 y+ @4 \5 R) [1 k+ w% m uppn+=1;! ~% [: W1 Q! D. B0 T
else if(text[i][j]>='a' && text[i][j]<='z')+ z7 `5 Q6 w5 B
lown+=1;
e9 r1 F m' X$ P S! o. T else if(text[i][j]>='1' && text[i][j]<='9')* }2 k* U9 k3 z; l/ u7 B2 F
dign+=1;% @* G$ r4 @# N4 N, c
else if(text[i][j]=' ')
2 J% t! e4 \( c b6 r& T span+=1;! |0 i' S' v' L' V7 i; g1 H; z
else
1 c6 v" T5 K+ L othn+=1;$ Q+ v" z g$ b0 M; I" \$ C
}
. X' H1 K& v& x4 x }
0 r8 K8 \% Y$ _% A' w! K0 H9 G for(i=0;i<3;i++)
' m& q7 `+ U; p$ @ printf("%s=n",text[i]);
* y3 J2 E3 Y' L7 x printf("大写字母数:%d\n",uppn);
7 t; F; l: ?+ @/ A4 x" Q7 H( i6 m printf("小写字母数:%d\n",lown);" L; E" f. j: r
printf("数字个数:%d\n",dign);
/ s O; L+ J, J# u9 @$ O6 R& E7 E printf("空格个数:%d\n",span);
& b0 e4 N6 A! N7 Q printf("其它字符:%d\n",othn);
; o5 N1 G/ s; s a! d+ d3 h& u}
9 `; s. n$ {0 a% B& z7 a
$ K9 `) g2 z4 A' j9 e! Q
/ L% [) P& w0 A. Q) K2 Y) G4 B7.11( q" R1 l1 _; P: h/ L+ e3 L
main()
; \; E1 h7 ~6 { {static char a[5]={'*','*','*','*','*'};
8 `- G" {$ T$ R int i,j,k;
% m6 T8 A( Z( M% w3 E% M char space=' ';( s: Z9 G. L" H3 B% k+ J" P8 u* {
for(i=0;i<=5;i++)0 A7 ?+ x6 a) O
{printf("\n");
e" b: A; A( C ?+ C for(j=1;j<=3*i;j++)
2 H& l+ R1 ~$ M- e1 t printf("%lc",space);
3 _4 y, {+ P0 @0 h$ I* G for(k=0;k<=5;k++)
n7 `0 Y* D( d: g% G! n7 f printf("%3c",a[k];
' d9 Q6 n [. N# O9 ^1 @# q6 A1 r$ n }1 I( c* @( A$ T# R9 G, g
}: t' A& Z1 I: \8 C0 Z ^1 t
7.123 [- Z. A7 T$ q; r0 z: h
#include C" r1 G$ i& S( V, A
main(), t# t8 @7 N' q8 I# m& o' U
{int i,n;% D4 I, R: x3 C" W
char ch[80],tran[80];* u0 R, w/ _+ {8 I2 K$ Z3 H
printf("请输入字符:");3 [) _8 E% y8 T1 O
gets(ch);
+ L, Q7 Z T6 Sprintf("\n密码是%c",ch);
, _ j- }" n: D& i5 x+ m; ?i=0;
0 P3 T u1 m3 J+ s3 Y+ Pwhile(ch[i]!='\0')
+ g Z5 k/ C& X0 h4 F, ]{if((ch[i]>='A')&&(ch[i]<='Z'))/ K% d q) U) e: v7 K. b
tran[i]=26+64-ch[i]+1+64;
$ {" x- c& W6 A9 T! g! kelse if((ch[i]>='a')&&(ch[i]<='z'))
5 |, A/ H! _6 v; Y6 a tran[i]=26+96-ch[i]+1+96;$ A; c9 p, ?" q; ~: K
else: f8 k; s3 U! v& W `* O% F
tran[i]=ch[i];
, _( c/ D) ~# b i++;
& {& I* D/ E- _1 Z- U9 R}- s1 ?8 {4 P( W
n=i;
) s. V/ r' {' G3 A! v$ Nprintf("\n原文是:");+ y$ A( q- ~$ N5 q6 @, S+ k: H
for(i=0;iputchar(tran[i]);
6 i; j, |& }5 b8 Y}2 I4 I+ u7 Q+ q I* p
7.13
& [/ I3 H) \& i3 Q* emain(). _+ N* L3 C! F' z) U* n4 ^2 d
{* H H, F/ j1 ^) S
char s1[80],s2[40];
( c! ]" d+ a' r% q! i$ p int i=0,j=0;4 M. {6 \4 g, i! U5 O
printf("\n请输入字符串1:");% [5 \4 z) C( f _3 B5 `
scanf("%s",s1);
' \7 H! ~! T* N printf("\n请输入字符串2:");5 b, Q; z% `( \5 w, ~7 O) L
scanf("%s",s2);: N: o( R3 B2 {0 c
while(s1[i]!='\0')
1 G5 U* ^/ U0 i i++; Q( J1 X, o# ?6 g$ U6 C/ Q4 z6 ]& T
while(s2[j]!='\0')6 c/ ]! N" B' m v0 W
s1[i++]=s2[j++];
+ _8 n% i+ T: d0 m5 I3 Hs1[i]='\0';
W0 U+ A6 W- l( p, B* R" Tprintf("\n连接后字符串为:%s",s1);* k* m; j' e" k: B& \+ o9 m. S
}
! Y% V4 ^0 M% s) Q+ a5 l! ~/ T. `9 z+ Y r4 G `/ E/ z
3 r# x# G. H& \% n7.14
. F2 n' m: m4 P2 W, E) N#include2 f' o# p+ J0 x# g* X
main()+ V: w B: e! f- X4 n
{int i,resu;
/ z" ^# |9 o6 Q& H, f0 B char s1[100],s2[100];8 }: t# w# j# t n6 t' y# T% B, Z
printf("请输入字符串1:\n");
# {6 W4 J' _0 O7 b- _ gets(s1);
( d& i U+ U- c printf("\n 请输入字符串2:\n");
I6 M0 E7 h9 F gets(s2);" S' o! R/ D S/ |% D
i=0;
# b$ ^4 G. F$ x' H" [. O while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;
8 o f: l7 c% t& y9 I1 N' c8 d if(s1[i]=='\0' && s2[i]=='\0')resu=0;
4 Q! b6 ]+ d* m! X! e- c; f) k else
: l) ]& _+ g: I! R" L9 r' _" @ resu=s1[i]-s2[i];$ C) @" G. e* Y G( y, V
printf(" %s与%s比较结果是%d",s1,s2,resu);
+ p- T1 i/ D" D# q% E) z% x7 `}
' g& s( j2 q0 O- ` h) Z7.151 ?: y2 A0 Y0 j0 C, q
#include% E m6 B4 m1 i: t, c
main()9 T' i$ n& F: A5 D0 E$ i3 K; Q
{; t# ?1 x% d5 F0 {* P( o* d
char from[80],to[80];& @1 _& O0 |3 ]- }$ E$ |& J5 n
int i;$ i( ~! z2 Z& T6 N0 X0 X/ \
printf("请输入字符串");
8 z* w1 T8 A% U$ F2 ` scanf("%s",from);
U6 Q8 ]) C0 n8 e" h q# \1 K for(i=0;i<=strlen(from);i++)
/ s' |% t& _& k4 i" c to[i]=from[i];3 B* e% y; v, @6 P
printf("复制字符串为:%s\n",to);5 j& n- \3 [" P$ d
}
" V: ], \! p& R% F9 {
7 A: b3 V8 s5 c; X
1 d7 J' I, Z, V6 O" e" Z7 R* i第八章 函数8 I" Z# b4 q! R& L( I
8.1(最小公倍数=u*v/最大公约数.)
. z9 \1 \& k6 H: }; phcf(u,v)
- U4 ?: d/ K) m! k0 _2 f R- [' Cint u,v;
$ L+ G# h) s' R* `% K1 m: u5 E9 Q(int a,b,t,r;& S( v5 f, s- O' B, s5 z
if(u>v)4 B+ z3 M' f8 L5 y" U" }$ j) _
{t=u;u=v;v=t;}
' l0 M7 \' E4 u2 p9 W6 V9 C. J! d a=u;b=v;
- ~' k: Z3 |$ u. Z8 ~ ~ while((r=b%a)!=0)
! C/ e$ h K7 e {b=a;a=r;}+ g! P$ s: \2 R' i/ M8 `
return(a);
, ^( _' q/ }8 w- S7 { v }1 t1 Z5 c5 T$ K% t8 f
lcd(u,v,h). J& K+ T+ ^5 [# W7 O5 l
int u,v,h;! o" e/ X( X" J2 N
{int u,v,h,l;0 N; D- h4 n2 r
scanf("%d,%d",&u,&v);
9 _2 O- ]% |- r4 O ` h=hcf(u,v);% t, A6 ~' U! G
printf("H.C.F=%d\n",h);
2 I+ X7 P. y" B l=lcd(u,v,h);
M0 O ] `( Q! [ printf("L.C.d=%d\n",l);5 h3 w+ k$ ]4 m# r. K8 u
}
4 G! N) z+ P: Y: ], [ B {return(u*v/h);}
) q8 W) h0 y- ?! G' }! K main()
. M7 Y0 c( `6 ]! s1 O) }7 B {int u,v,h,l;
7 q, B0 u3 k# S! r/ B scanf("%d,%d",&u,&v);
1 z. Y. d% }1 }3 x' ]2 J# M h=hcf(u,v);
2 d2 e7 O% l% }, L0 u! X" F' _ printf("H.C.F=%d\n",h);
W7 {. R, y! }) ?$ @$ ~0 t0 P l=lcd(u,v,h);7 C5 }; m2 t" S5 P2 ^$ G
printf("L.C.D=%d\n",l);4 O$ g4 ]' z: X+ s" t$ H% r5 }0 K
}* T: \* | D. p0 H6 Z9 n
0 v' W" i6 `$ t( _5 B) ^2 h/ \" E) V8 G$ B4 }
1 m9 [' u1 l. I g+ J" ^6 m$ S
8.2求方程根 N) C+ j0 D% `! F0 R7 r/ }
#include
" o- z- R7 G d- Q! Lfloat x1,x2,disc,p,q;, r) A( W( E) l, I$ p
greater_than_zero(a,b)
4 I0 d4 s T4 E# _2 V5 [" [; Dfloat a,b;% T3 Z' S( o& l& U6 M# Q
{
6 X; H$ n( z* h T, ^x1=(-b+sqrt(disc))/(2*a);
/ [- n. b: i* Ax2=(-b-sqrt(disc))/(2*a);% H$ N7 W( i9 q0 A4 ^. q
}
5 M) L7 J) O6 r# t# e. Mequal_to_zero(a,b)
5 h# v; i; W% E, k& sfloat a,b;
9 o, e9 E+ U0 F7 w( q h# C{x1=x2=(-b)/(2*a);}0 [( p3 o1 J7 {! h) a
smaller_than_zero(a,b)
, p: W `+ | G2 J! q' o* ofloat a,b;
1 Q$ n- F5 S9 k! w# z$ G; T{p=-b/(2*a);
$ Q' ^' e2 Z5 Uq=sqrt(disc)/(2*a);
4 V- I: g3 D8 Z0 P# g}
8 A( ^1 U$ i- g5 rmain()
$ f; H- J7 ^) o6 n{) ~' a% [5 r' l7 {& `4 _
float a,b,c;3 t6 Y/ o; S; _! ^% o2 o
printf("\n输入方程的系数a,b,c:\n");
, J+ R+ L7 T8 K6 e7 ascanf("%f,%f,%f",&a,&b,&c);, Q; \- ?& K( k, K
printf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);
7 i) p2 c! f! o0 |! }3 xdisc=b*b-4*a*c;9 }/ w: q/ f4 d; Z) B) R
printf("方程的解是:\n");
! B$ f6 v( @* p& w3 I. H' xif(disc>0)7 d8 L+ ]$ k+ G
{great_than_zero(a,b);
& C! n, s6 S' P. M% M. _printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
& w2 K2 X5 ?7 e" B8 O* ~* _}1 K3 X* F) O( R2 I5 t/ D
else if(disc==0)8 X4 T9 g, W. p5 D
{- A7 M" j( J/ \
zero(a,b);
8 Y6 x% R6 R, z. s! _, L2 jprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);% z/ y0 ?" t( T5 n, O8 C8 Q; Y
}
6 E' H; P0 m* H/ }% L) h3 U+ Aelse
. V4 r9 F- ~) @) k {( Z9 ]+ R6 B5 M; W$ r4 V
small_than_zero(a,b,c);* @+ \) |, \" @9 C9 I! T: E9 `
printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);; c( {* h4 T( i& k" k6 F' m% j3 @
}5 z5 [, G& I0 Z# O. |% H7 U
}2 \5 l3 ^( @4 j! y0 {
8.3素数
- I8 K. l2 E3 s9 K#include"math.h"( u1 f, u. E9 r) i2 j+ F' f( F( r
main()" M. h! z. `& S6 P& S
{int number;
* h3 y) f% g6 c0 a scanf("%d",&number);; V& s9 q! q5 D8 y: E
if(prime(number))
5 A' c: Y2 i* S- | printf("yes");& M1 f# U; E& M6 D. p: q5 V
else
; B' F0 v9 {# h: X+ V% f1 y printf("no");% x: I1 s6 _% w# i
}
& m" S5 q. ]4 n. s/ X8 I* hint prime(number)+ F$ E. ~, I6 K$ w9 o) ]7 r3 n( P
int number;8 e5 W U; x* g8 S$ X# p9 c y
{int flag=1,n;
! w9 M. O$ D+ C for(n=2;n if(number%n==0)
; e5 h- A) c- g n/ d7 K9 Z% O7 k% I flag=0;
# Q" B+ r% y: m5 e7 s: H return(flag);
6 L# }& K/ }( [ P) A}
( R1 Z/ M/ `. x" E) v
3 c L+ \5 W" C# k6 p5 s3 M! f- n8 [+ a5 M/ [* q0 o( g; L
! \( }/ j( ]) q* T7 }' ?8.4* l2 q, F# J! b+ T# z3 S3 B
#define N 3! |! c0 r+ F# r8 H% R
int array[N][N];
& `) |" W5 L X5 \/ C/ R: |/ iconvert(array)/ k: p8 l3 j q" B: }; T& i
int array[3][3];
* [' s/ n% l; _! v# t4 n* ^* b: P { int i,j,t;5 p0 e, u# y$ p
for(i=0;i for(j=i+1;j { t=array[i][j];
' P0 ?; v7 d0 ?/ o3 k array[i][j]=array[j][i];4 j1 Q x6 _. i4 _
array[j][i]=t;
+ D! _7 X A( ~; T0 x }
o: n. s" q5 I }
- }( `* i' c$ t3 \8 Wmain()
9 s9 S; L6 F3 P7 g: W9 L( w{) O! W R2 Q$ c L% i
int i,j;* V, g5 |2 E! h5 T" [- S
printf("输入数组元素:\n");
" r5 M5 E2 |' I" c. @/ ?1 H for(i=0;i for(j=0;j scanf("%d",&array[i][j];1 I( Q& H$ m1 Z4 O# k8 n% G1 \& \
printf("\n数组是:\n");: M% _- W! g" Z$ ?
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);5 l3 T: a* R5 ?/ x
printf("\n");+ r' R# v! L c$ j" T& L8 c- E6 g
}6 D i1 t+ d& L& s- K+ g- f0 M
convert(array);
' ~- c4 Y8 F, T- O# S5 ~ printf("转置数组是:\n");; [" l3 B8 {( ?1 {8 N- f
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);) I# X( }* m5 ^) ^0 B
printf("\n");1 N& C2 Z, J" d% b7 ^# Z
}
& y _: e$ i4 m}
8 ?) P0 m6 B0 _; |0 r/ |$ s) M6 ]! c! M. Z& h# X% B0 C* y
+ e; T' z0 L* Q
! g2 T* {7 v- o8 [/ p! W3 F! U
8.5
6 y- n/ F$ E1 K( z1 j. W9 ~* jmain()5 Z2 F x# r- e3 g
{0 U; Z! P0 e/ j$ L* C% m
char str[100];
3 Z/ ~; ~; C+ P0 F+ [ printf("输入字符串:\n");
: k3 i4 K: @- p" C+ U: N scanf("%s",str);
1 L; I+ D" i# e4 h7 W0 B0 u inverse(str);1 {$ I" @. q3 H) P6 s
printf("转换后的字符串是: %s\n",str);: }3 J- c6 G4 \0 h; T
}* I: z8 V1 t( x0 m
inverse(str)
$ z# B4 B& u' Z0 p# h6 Nchar str[];
; h) `) F6 C% e) y, q{/ O8 y3 O* E; P: Z }) ] A S
char t;
3 L2 `; A' _4 @3 s2 R$ F' P int i,j;1 n! D5 A& v- h" k0 z
for(i=0,j=strlen(str);i {
/ x8 {# F" a! d1 C/ N t=str[i];
( J# b) i. k* \" v4 ^, I3 J str[i]=str[i-1];
1 p! J) ^8 l9 d- v8 b str[i-1]=t;# l( _, R5 s8 a, H8 P0 w/ Q# Z
}
, k5 J' ~) i4 r' Y1 x( b3 {# F}
$ P' |6 I. ~" E$ t" R: z* {3 u4 \# C t- a2 h. e; ^
* x, Z W1 q( V1 V, o- |4 r& }3 f# U3 C* I: y5 C% `1 p
8.6
& M' l' I- v: r0 Y* Schar concatenate(string1,string2,string);
! C# V$ M4 `* mchar string1[],string2[],string[];
( O0 \4 S$ Y+ A- v* V# r4 M* `3 f @{
9 C! C: m$ u- Z7 {/ a2 Vint i,j;& D) O! H: C: I! s. d5 S0 w& w) m: E
for(i=0;string1[i]!='\0';i++)
: P# w5 e3 s0 ? string[i]=string1[i];
) }% i" B% c; k/ `5 K2 H+ S% Rfor(j=0;string2[j]!='\0';j++)- @2 V6 {5 b+ ?, Q, Q/ P" i# Y
string[i+j]=string2[j];& c5 L4 Q4 {+ j7 a" t9 R6 l" v
string[i+j]='\0';, w5 ^0 w; T! v7 ]
}
# [( P: I1 J% a) M% ]& J$ s2 Omain()
f& F, d+ N, H3 Z" d) |; W9 L{$ M1 _* d8 N% U: _# ]; e
char s1[100],s2[100],s[100];
2 M6 r1 F' j& I; z- k printf("\n输入字符串1:\n");; ^9 E2 u, l- t" a
scanf("%s",s1);; ]) K0 V. ?/ y: s0 F) y. S
printf("输入字符串2:\n");
% y$ f; N9 k8 Q6 L1 C$ J/ c/ Z' q scanf("%s",s2);
+ }* _' k- L0 ^0 `" `- C' t concatenate(s1,s2,s);
* S" L5 }5 | G8 U1 p, L+ c2 t printf("连接后的字符串:%s\n",s);
. `- _# S: F+ l+ A}# S) B# C( ~) _: U2 t) @
( x7 c# _$ K! J2 f! C3 w" C+ a
+ W4 e" m; l% M8.84 _( G6 t6 J6 v/ |5 e
main()- A F2 e J. X3 C
{7 @! q+ @, W* a" w! n5 D8 h
char str[80];
, Z. y- Z2 C+ M# Q6 K% a2 d, y printf("请输入含有四个数字的字符串:\n");
- e. Q( t/ g: L, _0 B: j scanf("%s",str);
. P/ K4 r3 [. h t" A insert(str);* ?0 e: u! L" S4 ?
}
" X& V( D" p5 e# T. F% h5 ~insert(str). H( Z" [4 j1 B0 W8 O9 b# v& S
char str[];
( Z( } T' ]% F+ c{1 o' B6 z1 r4 ^/ w) W. M
int i;
# H9 h }- w7 x! N, f: E for(i=strlen(str);i>0;i--)! r. f) ]8 y1 E
{ str[2*i]=str[i];9 d5 [1 ]% i. _" B9 N
str[2*i-1]=' ';3 X+ S* ]6 s3 z1 T. f% {3 W
}
8 \; `* ]) Q1 {( O) d printf("\n 结果是:\n %s",str);
5 M) x9 T* Y& R% i; R/ P- j }5 I4 p; }- Q i& f0 A6 u2 [' L
8 z9 ]2 V5 ]2 O
( ?2 R$ A& p2 W$ u8 e: D# @% c) ?' i! _
8.9
2 p/ t2 @! }" p#include"math.h"
" z8 f" J4 { y( u0 Xint alph,digit,space,others;
/ p3 M0 ~ s8 Y5 T8 [main()) f% Z$ ^8 l% P! x. M4 D& y, E0 d* f
{char text[80];
* y, t2 E6 ^! f9 n ^ gets(text);
- r( K& [% I- G) W8 e alph=0,digit=0,space=0,others=0;5 [5 f- {- a+ |1 x5 m
count(text);% O X9 A2 K1 C! N/ T
printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);# v' T4 M, F* i, f8 W K
}
+ E. C. s7 H1 Jcount(str)
/ T% W" h7 |2 D2 ]char str[];# O, W! T" `, x- S1 [/ A
{int i;$ [$ h, P r* \) z, L. g5 ^7 c! D- T
for(i=0;str[i]!='\0';i++)
7 j9 ?4 d1 R/ ?/ Z$ l" F* w if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z')), p& I6 @. o# A) d1 M
alph++;
0 H3 _$ w& T( y else if(str[i]>='0'&&str[i]<='9')
9 T( x/ y! }/ Q* J digit++;2 A: k+ U/ O! y$ x% l A# r4 a
else if(strcmp(str[i],' ')==0)- T8 `3 n# q: `. f
space++;* M. _4 X; D) `) u5 N" t
else
: X% `1 X8 ?" p e others++;
) @' M6 H. w7 w' p3 [! d}# p& g7 s6 m4 \# v9 q
Y3 u5 \8 D/ G# v
, R* y4 [1 t2 e) S. B: n5 e$ O1 r+ \8.10% K* C, L* i+ b( [+ o: @
int alphabetic(c);
9 h8 ^8 f( }" O* D) Lchar c;
1 Y6 y& W( I- k3 T0 C3 q4 ~{
/ B5 J, {# u* ?7 C9 D; f q if((c>='a' && c<='z'||(c>='A' && c<='Z'))/ [; u+ p; b7 y L0 h! S
return(1);: y: E, `) f7 o; r7 A0 j
else+ ^, w' {8 [0 s) w' p( d# f" O
return(0);3 K; H3 o0 l3 h$ l0 X& M
}' p) q% ~& Z& a5 Z2 n3 J* G
% B( w' H; C. V" X9 Xint longest (string)
" a$ r1 D9 t2 C' N8 pchar string[];
$ j& E# O$ j9 h7 B7 T{" `; y0 b; C3 d% v" M
int len=0,i,length=0,flag=1,place,point;
& q$ y, Y! ^8 T, ?; K8 j for(i=0;i<=strlen(string);i++)
" d" l1 D8 b3 k' a) I* v, c; T if(alphabctic(string[i]))
" N( i7 F, ^9 l$ z) B- \" l if(flag)
0 M; }" z. [& t9 ^* { {
: w+ K8 d/ |& P) O point=i;
$ c% j+ `$ T, ~ flag=0;0 y+ F2 A; a4 g$ z, ?
}! Q; K: J( L6 z" X( }' M# U
else3 j/ V @; G; l3 H" x' X3 K
len++;
( s$ n2 c7 F* p+ ]6 _ else/ r H; C' J$ _9 E% {
{ flag=1;. c+ l) d/ V8 H- F: |# i
if len>length)
# Z [. z( _1 d+ S/ I& K {length=len;
; S! k3 m& T" [( }) ]9 g place=point;
6 N5 n, a5 n% y len=0;8 x& M/ K, V/ M
}
) S% |4 `' f. @6 f9 [* E0 G$ w( m }
& Y, z2 c2 _0 x: l# ?/ } return(place);- I% r0 a8 Y$ m5 [7 L/ f
}8 t! q5 Q5 I7 x5 P/ J
main()
# o( H+ U; f, |! z4 ^3 B" G/ u{
# O Y9 q. d9 ~8 b \* h* Yint i;7 V+ `* Q# {) G# g/ Z3 l
char line[100];
: ~) z- d! M2 v1 {) Cprintf("输入一行文本\n");
: c& ?% a/ p4 ?' C# N# L/ F' f' H& ^gets(line);; o1 `$ ~$ ?$ G8 d& p$ ~2 v
printf("\n最长的单词是:"); l( n/ ?2 t! `- n2 E8 Y* [
for(i=longest(line);alphabctic(line[i]);i++)
' K5 Z, [% m) h, {7 ^2 N" Q6 N printf("%c",line[i];
. P6 o3 x% z4 e. E, Aprintf("\n");
3 p! E' K, S- \. N1 a. ]! _# s: |6 X7 q}
$ I* } V2 Z: K) v1 j4 I4 \
. J0 @- r, O1 t% E# w3 T7 C9 M! Q7 T9 |. N8 x
' }7 I1 I: c2 q2 E9 y7 X8.11
* ?) b$ Y1 ?! U5 P9 ^% Y#include
5 J' H& `$ B1 n7 g# C# R
0 J( s# b, {4 Y2 T1 W& l#define N 10
' ~) K1 @* m; ?. dchar str[N];
& c- K1 |3 H" \+ [2 Q P, L2 Vmain()
7 L/ n5 h+ k& m) C' r2 ]& C, j{
8 p/ Z0 g4 y' `0 Y0 @, E7 iint i,flag;# z% t# W3 d/ O! A. N
for(flag=1;flag==1;)' F* h/ i7 v: V0 |% p5 ~% v3 I! r
{' e+ T$ i/ p) i" m% d5 t
printf("\n输入字符串,长度为10:\n");7 M* d- J( N, m8 m, s
scanf("%s",&str);
# L( s2 j3 E; M8 L5 \4 d" q if(strlen(str)>N)
8 c( `: l( ^/ g# V! x printf("超过长度,请重输!");8 U& w2 v& A. W7 d3 J- ]# u( ?
else
/ V" S& Y$ v" q5 e9 E6 ^ flag=0;
3 N- R0 S" V j}
0 m: J$ g0 y+ E$ p. P* wsort(str);4 j9 h! f/ F" c' v
printf("\n 排序结果:");
8 M9 f& d$ b5 H: q5 t3 M% Ifor(i=0;i printf("%c",str[i]);( |$ f, y' s1 _
}5 b) \8 X/ |- m' T9 O$ f) K
sort(str)6 R, U3 _' ?# A6 R3 g, K
char str[N];
! \: B; A+ u& j7 e8 i" ^' h{% z- L6 I( O" |- L' y3 N
int i,j;
" D# {+ W6 ~' c" n- x# Nchar t;
! k& z/ }' s+ s8 _6 A( c9 C$ Kfor(j=1;j for(i=0;(i if(str[i]>str[i+1])8 J& W1 f2 I$ D1 J6 t/ F) A
{ t=str[i];
. N( X s8 k' O( {$ [- [ str[i]=str[i+1];: {$ a- p1 A1 A z! G5 Y* |+ @9 C
str[i+1]=t;
' T( v4 n/ o, ?; N }5 Q5 g1 R; s1 S8 [
}
_" U% ~" T9 p1 U9 m8.12
5 B O( ?# Y& J% y; r4 M#include
* |6 p$ Y+ i5 f8 @$ b3 X& d#include$ y3 P( b3 ]- |
float solut(a,b,c,d)
) u' R3 N7 t! E1 dfloat a,b,c,d;' L7 F& ^ X- P4 s0 L! _3 w
{float x=1,x0,f,f1;
* r: {9 Z6 J) I. L1 o+ A/ m- [ do. t3 c' f7 f$ J |7 }% L' c
{x0=x;3 u5 \9 z3 r+ \) [; X2 p* @
f=((a*x0+b)*x0+c)*x0+d;
9 D. Y1 B9 L+ Y5 X( h6 g% [5 K9 n, }7 Q f1=(3*a*x0+2*b)*x0+c;! w( q: o6 X4 r
x=x0-f/f1;
- e% Y# w# |4 @" _( D0 n }
5 d! a! C$ j! E0 e" N. \ while(fabs(x-x0)>=1e-5);$ Y* J" e1 b# |7 S! J
return(x);! V j4 [6 W9 ]2 P; c
}
0 g8 L+ d! P4 ~4 emain()( J1 S) p) a$ c8 P9 k
{float a,b,c,d;
& k; c0 Y0 `- m+ f scanf("%f,%f,%f,%f",&a,&b,&c,&d);1 r# }5 l6 Y$ l# J+ _
printf("x=%10.7f\n",solut(a,b,c,d));
& U' H' O4 T( C: a. a}
! M f1 k' A2 |8.13
9 p) C- f- g- C* t7 e) `& ?#include7 h% X3 Y, K. w6 _
main()
% F4 v/ R9 r V* ?" {{int x,n;
" F5 ]2 \3 u: h% f+ a float p();* _. M- R3 S) H2 C
scanf("%d,%d",&n,&x);6 N {/ ?2 v& f5 M
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));# B. L9 X+ x0 Q" O
}
3 Z* x2 R% u+ d O0 A8 Y7 Hfloat p(tn,tx)
* \/ [& V; U0 H' M i; D7 Gint tn,tx;
6 k- F! \' y+ @1 _* E3 T{if(tn==0)
4 I9 i; w) A2 W( S3 V& s) h return(1);$ }3 Y: C' K% _3 }% u
else if(tn==1)
, {1 Y; J* B" M6 j return(tx);
# J9 h9 j* r/ z3 a: g7 d: ^0 S4 @ else
: c( C7 X" [* C# m- j' u return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
/ L& J y' b' A8 W S6 h3 X; y}
& y$ U- C& R7 f7 `8.14
3 s9 R: `3 D7 n1 b( s* \# s$ y#include "stdio.h"9 P T; }; j$ O) f0 E
#define N 102 D. k6 u& { |5 ?
#define M 5. W& X6 a4 t& i
float score[N][M];, E2 ^# }$ \9 Y" c" W
float a_stu[N],a_cor[M];' }4 k/ G' K4 L, W
main()8 l* @$ O; ~- k' ^
{int i,j,r,c;$ f7 ]8 p+ O! `
float h;
/ n6 _- P$ W( u6 r& [ float s_diff();3 B, n3 C; t$ `' u" E% }
float highest();, |8 J9 f* }4 w( k& a
r=0;% w* z: ~; Y/ ~. L' C! s9 c# }' ]3 l
c=1;
& M1 c9 t5 l" u1 q input_stu();
9 N) a$ Z6 t, ~! @ avr_stu();7 \& C6 @. C) a% H$ @0 y
avr_cor();, J7 P" d( T; {8 w
printf("\n number class 1 2 3 4 5 avr");, }) s9 M1 V" A/ M" G
for(i=0;i {printf("\nNO%2d",i+1);
* m8 E, D' X: C" Y8 x s0 v for(j=0;j printf("%8.2f",score[i][j]);6 r' O$ c4 p3 l0 ]6 A9 H" g4 e) p
printf("%8.2f",a_stu[i]);
1 q7 d \3 D9 P4 S& d }
$ p4 y' W5 O$ j5 `' O- s, p; ] printf("\nclassavr");
( j9 a$ j: L2 d/ u/ k" Z4 k& \ for(j=0;j printf("%8.2f",a_cor[j]);
% ~3 v5 }$ r/ g; T, ^ h=highest(&r,&c);
+ ?4 L' R$ _% q8 ? printf("\n\n%8.2f %d %d\n",h,r,c);$ F c3 w( E, B# C) ^
printf("\n %8.2f\n",s_diff());2 l% t4 \9 h1 e% l# ? @" w p
}" W+ ?$ }1 Q$ q# C F
input_stu()- @+ N6 F g% A* y
{int i,j;- ^- z! g9 `6 M g6 r5 b+ t3 a
float x;5 i% p" k" ~! h/ Y5 P9 e/ L
for(i=0;i {for(j=0;j {scanf("%f",&x);/ C& K3 N6 t6 a# H* p+ u+ c
score[i][j]=x;
; ^1 s2 ~% z9 Q }7 H) }2 f }
/ @3 l; a e- V) ^: L }
0 p! K# V# D% ]0 B4 }- K/ C}
, a' Z O' K9 Q% r. a; M% S+ Iavr_stu()
1 J% P% Z% t( B& I{int i,j;7 }5 i3 C4 ?* h, Q/ \& m2 c
float s;$ H; j! a! K- y6 U0 l/ z5 [
for(i=0;i {for(j=0,s=0;j s+=score[i][j];
3 x4 [7 P# p( b; a6 q$ f a_stu[i]=s/5.0;
; `( l$ ?( {2 T& r) m }
4 \, q4 p. \; {& q1 {; @6 Q$ ~}
% {0 L1 Q: Z7 K/ K0 ~* c3 Vavr_cor()9 u2 d* H9 x) }' e% U& ~
{int i,j;, |, \& b- L% l8 ?( `. x$ i
float s;
* d- J% A" M, m$ G for(j=0;j {for(i=0,s=0;i s+=score[i][j];
! Q$ A3 K( A: T' c# V) G Q a_cor[j]=s/(float)N;
- J, h5 w0 G0 j3 k1 q- x }; { p W! _$ H- w% e: e
}6 T* d& e$ P, A/ P7 @ _0 S: ]. i
float highest(r,c)
, |+ S+ j5 X% T" N; v7 o Qint *r,*c;
) G0 k1 V1 C/ r V. w! Q f{float high;
0 `' [' |" H( p+ m int i,j;
* J: |. ?2 n- r+ O7 G3 i6 G high=score[0][0];
# B. P4 S) C& I for(i=0;i for(j=0;j if(score[i][j]>high)
2 m+ _3 L, k8 W) ?# q {high=score[i][j];
5 p( d( I9 r8 q2 p# A( z *r=i+1;
0 @. i- F, e; d% P6 r8 } ` *c=j+1;
- b0 B+ A; ^8 ~: g0 G' G }' m5 b) ]4 _/ k; G! E# M* r
return(high);
) W1 N1 l* ]+ u% ?}9 j3 P! R* |% l5 S! G& S% V" K( ?% Y
float s_diff()
4 M2 m$ v3 ]2 F# w{int i,j;. x4 V" |; P W+ ?) k& J% \ b
float sumx=0.0,sumxn=0.0;
+ l9 \) j9 k6 D# C+ ?& e$ K2 X for(i=0;i {sumx+=a_stu[i]*a_stu[i];- ?; @% f( r6 F% H7 l3 m
sumxn+=a_stu[i];
+ z1 }9 @; k" K" @9 s3 d }
h3 T1 u- d7 _9 l return(sumx/N-(sumxn/N)*(sumxn/N));- d, y) n! U$ |3 ]
}
4 M9 ^; F2 s1 v" h& H, U0 a; P8.151 s* [) y% J- E
#include
3 W$ O1 \/ F; X; }#define N 10
" ]" G/ o K! N6 _$ vvoid input_e(num,name)
' i: x# D/ r9 iint num[];
* k; d5 j! I; m# [5 O8 s. u) @char name[N][8];! {1 _# C) ` Y0 H
{int i;
* C4 u7 e( y5 B for(i=0;i {scanf("%d",&num[i]);
0 L* ^7 k$ o; L0 y) Z gets(name[i]);
% N* w0 s2 a1 A. A- H' L1 N }) {4 y& a( w% p- r% ~- ~
}3 w6 w0 G" X/ d
void sort(num,name)
7 S% t1 n8 _$ e& [, {( uint num[];
# }; t) D, Y" @! X+ `4 vchar name[N][8];& P. j/ w% r# s. u; B6 }
{int i,j,min,temp1;4 v2 | w; D) x# n7 F* F
char temp2[8];
1 [( ~: {8 d% B$ k for(i=0;i {min=i;
' y8 U# F3 d% d: S$ @& A' U for(j=i;j if(num[min]>num[j])min=j;3 A* _6 |7 P9 [& Y1 S% ^0 A
temp1=num[i];2 o- D n* `7 m. J9 o
num[i]=num[min];' t: n; t8 o6 X) i. b4 Q& ]
num[min]=temp1; d6 V( {$ k6 D* f
strcpy(temp2,name[i]);
! v* y1 o, {+ |! n9 G strcpy(name[i],name[min]); X z' X8 i% I8 C0 H4 Y
strcpy(name[min],temp2);
9 i; v q1 A2 n4 [# k; ~3 ~ }
, B& a9 m6 b" Q5 l for(i=0;i printf("\n%5d%10s",num[i],name[i]);% n9 r# J& ]/ Y
}
: B1 E$ {0 s& b3 Zvoid search(n,num,name), | i- Y, W5 L, x4 E# t0 Z
int n,num[];
6 g; d9 ~0 Q; _2 D) Hchar name[N][8];
$ D5 w: Y/ ?" ^" P{int top,bott,min,loca;
* c* R- @3 s9 ]- ]9 T u loca=0;+ A2 G+ S2 z0 \0 ^# R
top=0;
, \4 B; Q( c0 U! v9 [: ] bott=N-1;
" E1 O# {" ~4 C1 m. D( v, u: G if((nnum[N-1]))
) X) w' l- f" y! L% r5 y$ S loca=-1;
, Y$ ^" \4 y' h& _ while((loca==0)&&(top<=bott))0 n( D# ^* F, r8 e9 o
{min=(bott+top)/2;
3 M: Q C, g/ I8 ~. n if(n==num[min])
6 v# o% N+ Q+ |6 r- H. s. T; ~ {loca=min;
. u" m, O4 `& T: j1 ^4 T) W) W printf("number=%d,name=%s\n",n,name[loca]);
) t2 {; Z. Q s! W/ E: c }
1 `/ ?: t1 o# ~3 R' G else if(n bott=min-1;
1 X! _. o0 u& \ f7 \6 X. n9 l else
, V$ L4 A2 z" o" c& o: E8 m4 `- l top=min+1;% G( C+ W: r* @2 g* S% Z
}& A3 m( n: z, f, n9 B
if(loca==0||loca==-1)
, k' T( ?3 M2 V printf("number=%d is not in table\n",n);/ Q& q4 `% c# Y! ]* o
}0 W/ G! V+ y' t( d0 Z. i7 F
main()
8 W0 P- N, V2 X9 r{int num[N],number,flag,c,n;+ {/ o+ ]2 h9 d5 C9 j
char name[N][8];8 L) v1 ]' s8 Q0 x1 n# F5 t0 @
input_e(num,name);6 w/ y c! ^/ X* d9 Q
sort(num,name);
1 [ L9 G, x0 a# X* O, E for(flag=1;flag;)( T. F7 {# a/ F- U
{scanf("%d",&number);
# V' Y6 x# u9 @- [' E search(number,num,name);& H! O# m3 _. u2 Q
printf("continue?Y/N!");
& z3 b5 y6 e# E5 ?/ n1 { c=getchar();# `- J# Y7 T: c3 E2 F7 i+ Z7 v
if(c=='N'||c=='n')
. V& A# V, f3 L- w3 b flag=0;
! j* M; \9 [& X% Z( W) n1 ] }! D3 S5 P$ J9 ^8 C
}
" B! L @# s- a( _3 i9 M
; q9 H$ C7 U" |% U$ _8 _8.16* t3 L7 z, j0 [1 R
#include
5 `9 t- { {) X8 W#define MAX 1000
! j+ n4 J0 M s& o: fmain()1 n( m3 ^3 h5 [/ ]# h
{ int c,i,flag,flag1;
& }" |* \; v0 m" f* N) m char t[MAX];% U9 y& Y1 ]: F3 ^' ?% l
i=0;( C# p# q) S; E
flag=0;
2 ]' `# ?/ z: O7 P$ d' C; g flag1=1;: f8 x& y8 H( y0 ]: P( M
printf("\n输入十六进制数:");
; N" U/ r/ V! ? while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
- x1 A5 D1 M2 _. y2 B6 _$ C {flag=1;
# `; @( P) Z; W2 `. Y# Z7 B t[i++]=c;( D. {$ A+ F- P: [+ |
}
* [% Z/ {! X9 z+ l5 i3 Q else if(flag)
( X) d1 c$ Q1 J4 _7 `$ D {- l8 v: z8 Q4 V' P
t[i]='\0';
9 Q/ [! h" A$ a$ }4 o; ]5 \& k printf("\n 十进制数%d\n",htoi(t));
: k3 \3 K4 d ?; b5 W7 m6 T/ Z printf("继续吗?");6 @6 o/ D0 @. x% ? S' R$ v
c=getchar();
, S% D2 D* I, \% u+ W. T0 k0 G if(c=='N'||c=='n')4 m1 U* U5 g) K
flag1=0;9 N3 S d1 l4 i9 J
else6 x! g" _0 f0 E- [3 X* a
{flag=0;
) v+ }3 f3 x: p! d* N6 d i=0;
) q2 b2 Z/ l0 I, H: @2 e printf("\n 输入十六进制数:");& K; \) ?0 M q
}+ {, }- Z: W$ z3 p8 F
}- l2 ?; }4 l! J# f2 C" O0 L
}
& m: z5 n6 g/ }3 a8 `( q9 r}
& q& t( x8 a# y9 z0 S( y% L! X) bhtoi(s): G. s8 ?) V; R7 a$ r) A2 [1 U" v4 ~
char s[];; s9 U1 J$ f$ i) | e
{ int i,n;
7 _1 i8 |% |2 m/ W n=0;
+ f1 M, S) r" f) v0 Q* H for(i=0;s[i]!='\0';i++)- e* Q* ? E! o1 S; O7 N
{if(s[i]>='0'&&s[i]<='9')
, m# n8 [3 D2 L n=n*16+s[i]-'0';
- \7 f( u3 s: Q" E/ {8 a' Y( J if(s[i]>='a'&&s[i]<='f')
& S# T( N+ D- e* U n=n*16+s[i]-'a'+10;8 P0 W& u& b! y4 ?+ j7 }
if(s[i]>='A'&&s[i]<='F')) C, B0 ?9 I* [
n=n*16+s[i]-'A'+10;
2 m+ W! Y9 w5 F) o6 [! ] }
* x- O3 s* |8 A6 g: @- t& M return(n);% E2 a! O& z1 d
}
1 d; f- G9 }7 I x+ \
7 N% ~0 L( t; b
# {6 x7 W( K0 S( f3 o: _* J6 K3 Y* C; z, t! L
8.17
5 o9 J7 X( h# n1 q#include
+ W, L2 O" e: p, p" Ovoid counvert(n): ~1 o3 u8 S1 T6 H9 o/ h9 w; v
int n; q! M, O4 x2 t3 M( T4 f% T
{ int i;
. F! c6 p9 f/ f1 {% Z8 e# q if((i=n/10)!=0)
& d `$ e* v4 w, a7 i+ ^ convert(i);
+ h% e4 j9 E# G5 R N- X7 u8 V putchar(n%10+'0');: F8 q% K! c) z# l$ M. S. h8 J
}
9 v: N/ L, @9 U+ h" imain()1 F% a( D) C. t1 q3 ~# a; f. T
{ int number;
?) H0 v7 u' c printf("\n 输入整数:");0 J: |8 H2 b1 ]/ ~ P9 ~$ r
scanf("%d",&number);: n) @( V0 q/ b4 t+ m/ l/ a
printf("\n 输出是: ");5 w' c) r9 p4 l3 [/ j7 C% w6 Q+ ]# |
if(number<0)8 I4 T+ a7 i) M0 P. B/ M
{ putchar('-');6 |" |+ l' B/ o0 ]( N! P4 m
number=-number;
) O3 S3 J O) e2 u }6 K* f$ Z8 W5 w- E3 H
convert(number);2 N2 S9 z) C) Z/ O2 }. U
}5 p. T$ ]" D: e ^, s7 l' O9 [
; J4 E0 k7 F1 u: p, ~7 f) i$ O# l& E" y8 I
1 n% c3 w5 |# F1 O; l1 R. z4 @7 a8.18
7 n; g$ F4 n3 \9 ^3 Ymain()
# t9 u) o. h- T$ D7 W* T6 P& y4 e{2 a% {. u; w' Q o! A! Y' Q( G
int year,month,day;8 C( [$ e$ Q2 c: N
int days;6 a6 b y3 ^; Q" Q8 o* e0 o* f
printf("\n 请输入日期(年,月,日)\n");7 D3 _7 ~/ B4 M7 a9 c
scanf("%d,%d,%d",&year,&month,&day);% i6 J$ u* f0 E( B" g
printf("\n %d年%d月%d日",year,month,day);
! C( G3 C( |, I' \( M0 I; T days=sum_day(month,day);
3 @- h7 l# Z: [+ A* ?. Z9 } if(leap(year)&&month>=3)" q7 S) M4 G. o3 m
days=days+1;
' {- N0 r: d% [* S) a printf("是该年的%d天.\n",days);
7 `: `' }6 V- }! x: Q8 z4 b$ @ }
1 C+ P0 ~* P) _2 `( n7 K; |4 B static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; C# h# W7 i/ k1 x" ^% E
int(sum_day(month,day)0 y8 A5 o! J0 C2 w9 }# V0 N J3 \: \
int month,day;
) d" y. c) U$ D2 v5 L* J {. y+ x! ^; [7 U3 o
int i;
# n0 l* l$ V3 v6 Z for(i=1;i day+=day_tab[i];
- \. {, `9 z6 q; d& A return(day);
( s: x7 _- z5 t }
- \: W; M7 A9 O int leap(year)
k' Y$ V6 u& F' }+ s int year;
# N$ d Q* v, g0 g3 R8 o2 a1 b {
2 s3 m) g3 s4 j- ]: z$ i7 K! M int leap;
7 q G$ [5 \. a& m/ n leap=year%4==0&&year%100!=0||year%400==0;
6 m" t% u7 @$ L& ]/ ] return(leap);
/ }7 G( P. ^" V$ ]. }' o }# s: R4 E. r+ {0 e5 ?9 D
第九章 编译预处理
1 Z8 V' W0 }: n; D9.1
4 f! }6 a! Q; ]5 F9 o' X#define SWAP(a,b) t=b;b=a;a=t
* O: Z) S1 n {6 E2 r" K1 Ymain()5 ?& b K" g. b0 |
{
: ]4 c) r) L/ r; k2 P' [int a,b,t;* p1 G% e' I$ a* z0 @ t
printf("请输入两个整数 a,b:");- b/ Y' x, G* e8 o
scanf("%d,%d",&a,&b);5 R9 f0 o! \" |; d0 ?9 B+ Q
SWAP(a,b);2 v# _2 F4 ^0 X7 ]1 ]
printf("交换结果为:a=%d,b=%d\n",a,b);
, \! a2 X+ g. q% v' W6 U7 w6 B} ; {0 h" I% O/ I7 q3 e% ^
) q7 V# H+ o) R' S2 v- u- c6 x
4 S4 j) |1 P k4 k* ]5 E/ [
9.2* N& A& l' x# o8 P4 l7 ?
#define SURPLUS(a,b) ((a)%(b))
: p7 r1 U6 A. Pmain()
5 P, L, I7 c1 A- a) S9 y, \# O {% @' v. |2 `( u8 N# R9 E
int a,b;- Q- M6 F3 O" @7 O
printf(" 请输入两个整数 a,b:");; h8 \- _/ A- g9 m; x/ m! b$ J O
scanf("%d,%d",&a,&b);) E; Z, a& j* W v# j
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));
c+ [; U- o9 k }
, C6 ^) S# z( t. m9 G" @% Q C" ?5 C1 {8 Z# E/ a$ E0 B5 i
1 L6 E ~ A8 \( E0 Z
9.3
7 Y5 {3 R7 R Y5 y, A o# m" B#include% n3 h" X3 y- b
#defin S(a,b,c) ((a+b+c)/2)
2 l" ^7 G% D6 E+ E: D9 }#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-% j. k/ J# y( C
c)))
5 {- I; n8 C+ [5 o5 {main()
8 ]/ D* T9 g) j2 Q0 `! h9 Q {
- `0 Y. g( O% Q: d! Q/ S float a,b,c;
3 K$ l/ }' i3 J, H5 Z0 H printf("请输入三角形的三条边:");
1 L' \! d8 g. L( l scanf("%f,%f,%f",&a,&b,&c);
& ^! T! L- N9 `6 O8 S if(a+b>c && a+c>b && b+c>a); }0 I: G2 } v
printf("其面积为:%8.2f.\n",AREA(a,b,c));# w7 |% @5 D0 R. a
else! q2 f+ \4 P+ B& n
printf("不能构成三角形!");
- b C8 I# M: H7 O: w9 a5 x }+ {! i! F# n1 O: ^& _, ?
7 h. Q A. Y$ _$ p: z7 ^. _! Y1 u4 P9 d6 u
/ Y! \! ]& v5 B+ ?! h. i% | l+ X# P- M$ l$ d4 o8 @
9.4
# P- { |, ?- Z, z$ e+ M9 s" [#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)
+ _/ ^$ [! x8 T% Nmain()6 c# D4 ?9 _3 f2 t
{& Y5 z$ ]# A9 P0 z! e' ?4 [
int year;
4 H4 t$ G8 a# \9 d1 s, R printf("\n请输入某一年:");
& E" `2 ]7 w9 i7 s9 R! L3 ]% ` scanf("%d",&year);
# x2 ^. f- @( B% X if(LEAP_YEAR(year))+ D2 ?0 {7 b8 g- L5 q7 H$ Q+ I
printf("%d 是闰年.\n",year);
R* ?: d L- R6 R6 Q5 X else) d- f8 y* a4 [1 [5 D
printf("%d 不是闰年.\n",year);+ k F. z6 A% S/ Q9 V
}
' j! S% k: J$ P7 P6 [
+ U) h6 r' K) p, S* P+ t" I$ `! k0 K9 T3 \9 f
/ t- z& O( M9 o# F4 K: u1 h: j
9.5解:展开后:! _5 ^5 V, _* r3 ~6 ], n
printf("&#118alue=%format\t",x);) K8 Y2 J3 z% u8 \, q7 {; X
printf("&#118alue=%format\t",x);putchar('\n');5 G9 z8 ?% W" }, U
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');
2 [' w' B+ w; z$ H- w0 x输出结果:& F3 \' o. P$ F$ l7 W
&#118alue=5.000000ormat &#118alue=5.000000ormat1 `6 E! q2 `: b# N4 m) j( y
&#118alue=3.000000ormat &#118alue=8.000000ormat
5 o, q) f; l$ J: c! }# c5 a" ~5 A# A9 U9 q2 r7 H' H
7 B8 T D/ w( \- R9.8; h) q: t7 g* n0 F
main()
M7 w+ Y. r# X# _: D4 Q. T- B {
/ C" Y, y* j$ b. l1 c6 p int a,b,c;0 z9 a% J) U, A3 t3 Y- E7 J- P' Y
printf("请输入三个整数:");" j% V: V0 B" {/ n1 ?- T
scanf("%d,%d,%d",&a,&b,&c);
. ?# g8 E" x8 V4 d- D) e& V/ [ printf("三个之中最大值为:%d\n",max(a,b,c));
, `5 X5 ?# l6 U/ X+ L) H }; A1 e9 q5 }& z% f* i% I: M! R
max(x,y,z)4 {; ^" G3 e6 R% U
int x,y,z;, [6 i9 l* \0 K: O& n
{
8 {% W2 l9 d7 s: h' A" o int t;, I0 }0 ]( I- D: n
t=(x>y? x:y);3 n8 K, [& X# [! {+ y
return(t>z? t:z);: t* b4 s5 P: |% s8 X6 Z- W, D) u
}9 @( G# L! n0 k0 j( D2 ?
, @ ?+ m7 t" m" k/ a) |
' M1 `3 x3 [/ F u3 a/ ~( t! b
" j. H. a9 ^' y4 S7 [( l# n$ M9.10
) L. T/ K" I% F4 ]3 e v4 N#include1 \" ]7 ^% l0 A0 J
#define MAX 80+ o: ?( B* |. B. }* g
#define CHANGE 1# {3 Y2 A9 y2 Q
main()
/ I# _& e1 Y4 `0 L* ?3 K{
) t, z, y5 I7 q- } char str[MAX];3 e/ e+ z" B7 a1 U) ~; b. ^
int i;
7 v( a u' O- { printf("请输入文本行:\n");8 v1 Q8 g/ k5 m9 B+ l+ E$ f
scanf("%s",str);
' @7 [- V; S. F9 I6 b #if(CHANGE), y6 s6 a' J5 U# @
{
+ L) c/ B$ d: Q( m for (i=0;i {
l- _* i2 z/ i4 j* T& f3 v if(str[i]!='\0'
0 q1 d# Y3 @4 }. p$ w+ D if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')$ |1 Z+ s7 y0 y& f& j
str[i]+=1;
0 T3 L/ R2 p& G P else if(str[i]=='z' || str[i]=='Z')
- w( N! V+ r, w* |3 C str[i]-=25;: P) G f6 A( h3 C
}
! p1 A! H V* Z( U}/ Z& H3 [1 K* t! }1 P9 T
#endif m) _. U5 o6 B' I4 o: G9 R
printf("输出电码为:\n%s",str);" b7 M9 h& i+ I2 t1 ~
}' s0 Y9 [& h! [: K. |
第十章 指针
3 C1 G8 X" x: C9 ?10.15 b' {6 l3 I) l9 h. O# u
main()
+ _5 \' d1 D- j; s+ d' C{int n1,n2,n3; V1 `3 e( {" ?7 i& w! U& |
int *p1,*p2,*p3;
$ b( c4 P/ j0 P! M8 L7 z0 L scanf("%d,%d,%d",&n1,&n2,&n3);2 d* b5 E9 r; [2 }/ [3 D& s
p1=&n1;' s! E7 B, s0 U+ \7 d x6 m
p2=&n2;4 d3 P, m' Q" |4 Y/ j
p3=&n3;
; ^. W! P: A, N3 P" x0 Z; Y if(n1>n2)swap(p1,p2);
1 [# C+ |* x" S0 F( M3 f if(n1>n3)swap(p1,p3);, c4 {2 _" {. H9 m2 S% g, }
if(n2>n3)swap(p2,p3);& g$ P0 |, t# o* i/ T% ?
printf("%d,%d,%d\n",n1,n2,n3);* G( Y# B3 ]/ {$ y: ]- f0 r
}$ G7 G* h! ^3 O+ r, k/ K
swap(p1,p2), {( J5 m& p7 z$ n9 w
int *p1,*p2;! t; J- i& D9 L' S" y5 x4 ]* h+ P
{int p;
! V; Q$ G" O2 S5 y" Q6 q p=*p1;*p1=*p2;*p2=p;
3 K; i6 t' a5 k/ \( @" w0 [4 q N. w}
: h. _ C4 |5 u L! J# ^- K10.2
( A% i9 I! f0 o" C. L; C1 Umain()" Y! X1 v8 b: X* g$ j3 B8 Q
{char *str1[20],*str2[20],*str3[20];
3 A' f( S, b4 c) l, A char swap();, V9 J9 i3 S! K( O. {7 ^* B
scanf("%s",str1);
" |) n' [% f: Y7 x3 W scanf("%s",str2);
; q7 j8 Y2 K! _: W$ n" E- J/ k scanf("%s",str3);
( {0 S8 n2 L. l) B* R+ E H( f if(strcmp(str1,str2)>0)swap(str1,str2);
% O- d7 ]1 n0 U) A* l if(strcmp(str1,str3)>0)swap(str1,str3);! g4 P5 }. a7 b( j% W$ f
if(strcmp(str2,str3)>0)swap(str2,str3);
% H, i0 t: g0 y' L printf("%s\n%s\n%s\n",str1,str2,str3);' ]. L' Y/ |; h
}
9 O/ S# K! u* j' G9 r4 Dchar swap(p1,p2)
# m+ M' F* Q" @4 t3 kchar *p1,*p2;5 M) h' E# k9 C9 M' i+ \
{char *p[20];+ _5 t% B, H6 x
strcpy(p,p1);2 m" G$ B3 d3 d
strcpy(p1,p2);, B# e/ r) E2 N( T: x$ c* D6 I( r& F
strcpy(p2,p);. e4 R. ~" K6 u$ {. g
}2 \4 g1 K& C9 o6 A! z- u( p5 u
10.38 Z, ?4 w8 D0 E7 b8 r Q2 \3 A* M" K
main()
& Y1 Z* z1 B1 s z% J& D{int number[10];
1 n8 E/ f2 a+ _: `; }3 B2 c input(number);
: ]7 R" b# o1 R. t max_min_&#118alue(number);
: [0 R- d3 E2 Z" a- U0 [+ `0 J output(number);
5 q3 e6 g/ _, r& H* f7 M }/ u+ [8 b# y U9 v0 J) Z1 E
input(number)
1 G3 V y$ y5 \6 `4 R, u2 {! U2 D* Yint number[10];9 I2 p, n1 C2 k/ [
{int i;+ y- E, F5 |, L" W- B2 C0 ~
for(i=0;i<10;i++)
( V( Z* N+ [0 L( q$ _) U scanf("%d",&number[i]);
" s+ O E o( ^1 U9 g2 h' k}
8 j' f- `( _! b0 Y& \max_min_&#118alue(number)* b$ Y9 c+ ~, T
int number[10];
5 n0 f- u9 }! A6 n; E4 D' A7 A{int *max,*min;
! Q* w' w0 Z- o: c% z% O int *p,*end;
; |8 f( s/ s& g4 M9 g7 b& _+ | end=number+10;1 \1 H3 Y$ j4 J, b6 N
max=min=number;
+ C/ x* z: A! } for(p=number+1;p if(*p>*max)max=p;
w3 d8 X9 X l$ Q) K& ] else if(*p<*min)min=p;
* O3 J) h9 U) y$ ]& N *p=number[0];
; w4 o/ n3 i6 P) F5 p/ y/ k6 R7 E8 ?+ U number[0]=*min;
; G+ I+ H% k, ^2 i *min=*p;0 e, D3 x. q; u1 j* G# X
*p=number[9];
! m4 o0 b2 w: O; s number[9]=*max;
/ L6 X& a3 f' n+ @ ` *max=*p;1 r: M/ Q9 L4 M4 l2 i" |) ^
return;- z# K2 _* l V: `( g/ \
}5 n3 y# F4 z2 y0 N
output(number)
; Q3 ]! q$ Q* Cint number[10];
9 f/ l0 ?( f1 `9 I{int *p;
3 B; ^( Q& V5 H) t: L/ o" Q for(p=number;p printf("%d,",*p);7 R: O5 Y2 E% o
printf("%d\n",*p);
% o0 b* r" G4 f0 J! L; o5 N}4 }4 c% D2 ]8 d8 u$ v
10.4
# }. G4 }' k( emain()6 s! S1 e; ^9 F3 X( c
{int number[20],n,m,i;1 K' Z' ]) c: V9 q0 U
scanf("%d",&n);
- g$ s9 m) s3 `1 C2 e) l( E( R scanf("%d",&m);: j, c ?& L( C& U" d
for(i=0;i scanf("%d",&number[i]);' L& P: `; S4 F+ W) Z
move(number,n,m);7 t( I c+ i) @7 @; E
for(i=0;i printf("%8d",number[i]);
m; a) X; X" o) v: e}
8 X/ m9 l7 W2 @2 {move(array,n,m)8 o2 T) a8 q3 F+ S9 z
int array[20],n,m;
1 O* E, J* ~$ V/ _{int *p,end;! X/ g4 s# p4 Z7 n. q! G1 ?
end=*(array+n-1);! k* p* j2 e2 R' H8 ^3 x; A
for(p=array+n-1;p>array;p--)
# \; z" n% j0 [% e2 T: k *p=*(p-1);' ]( ~" H0 s9 g
*array=end;5 ~* U/ I0 a" C2 x Z$ K B( [
m--;
/ H" f# T6 u0 h8 O+ D6 M if(m>0)move(array,n,m);
3 c% j7 w) E4 w4 z: x8 j}
: C; `* E4 A; E) j10.5$ Y9 o1 Z! a& r- \2 K8 ]
#define nmax 50
1 j8 X& o3 P! U; n/ e* k2 cmain()5 v, W% D1 A% L! U6 u0 t
{int i,k,m,n,num[nmax],*p;
; @; m: r; z8 L+ x3 w3 ] scanf("%d",&n);
3 w- l/ `% Z; V' Z# u p=num;
/ B H p$ U) ^: I+ J4 W( | for(i=0;i *(p+i)=i+1;
% b5 ~, M3 [3 l% Q( [! `/ a i=k=m=0;% V2 ]) k& y% F
while(m {if(*(p+i)!=0)k++;" v+ N* C o3 T1 a7 \
if(k==3)
9 n! O- T& `% N! h& p" ^ {*(p+i)=0;
" d* n0 U0 ? d6 z8 F+ J k=0;* c* x) W% j8 E! [# S
m++;
7 F. g0 O! v2 u* T3 Z }4 l3 H5 o9 y. K- ^+ J4 i
i++;) w! [4 a# t# d
if(i==n)i=0;
/ P, T& ?2 S$ G: ?9 \4 a# k }6 R& u0 u' }3 s0 ~( |0 U
while(*p==0)p++;
5 ~( ]. W; v7 H) i2 f printf("%d",*p);
" S8 T; } z6 l& T}
. @3 n0 n# F& y10.63 F% }" E7 o4 J6 h; e
main(). m: S# c% D. m* ?- s, ]! R
{int len;# U% E+ h$ U1 h" a8 e5 p; S8 n
char *str[20];" e2 u( c0 r- C V5 n
scanf("%s",str);
. ?, `3 f; K- q! f+ `" g: \ len=length(str);
9 H- ]9 a# q; S- v; A9 o/ E i" I printf("\nlen=%d\n",len);
4 B, `- _9 j9 {* {6 \+ S# e7 ^6 H4 R. V}
# I+ `, J6 e- s4 Vlength(p)
8 {& S/ @3 F3 vchar *p;
3 t9 D' A3 ]) L4 p! f8 i1 T! N{int n=0;5 g0 E) O0 Z3 V7 A
while(*p!='\0')
4 }! y3 T$ o, a) s {n++;p++;}
5 o. R: z6 ], q8 f7 r8 ^ return(n);
9 m0 x- v+ F5 b- ]9 t}
* f! W% i4 {% p4 \1 a10.7% ~# z' e5 s- O ~
main()7 j3 {$ h1 l7 h& y
{int m;4 d* }8 p" ?7 T8 p) o) ?
char *str1[20],*str2[20];
! y$ ` H- N% h6 A' C1 G scanf("%s",str1);1 r" Q5 _9 V; B$ M
scanf("%d",&m);
: v) i: R- M' Y% T if(strlen(str1) printf("error");, S+ }: U: @( @$ {
else
: e+ S# I: e9 p5 n% W$ k" O {copystr(str1,str2,m);3 W6 H. ~; `+ J' B n
printf("%s",str2);; Y- J# |7 o# j& H, s
}
. w, R& {% f; J7 p$ K9 K" \}
, x1 ]$ K$ `: \# D% |" `7 I1 h& Bcopystr(p1,p2,m)9 m9 [$ E' `) d! U) i& q
char *p1,*p2;
* a* L* a$ X, m ?0 Dint m;
9 h# f, a8 w$ w7 c3 @{int n=0;. r6 R' a' }8 z. [- V0 _
while(n {n++;p1++;}9 I. \* {! m2 D; ~5 Q, T5 e
while(*p1!='\0')
% K- e, V& g m r {*p2=*p1;
$ C# E0 F* k% T6 ~ p1++;
# k8 N* d1 z r+ x$ S, I, ^( u* m p2++;# q; Z8 W# V$ F- D3 s! k0 U7 G
}. w6 `. u" Y, r: z9 A
*p2='\0';
/ L: U/ N4 C- t* ?' L3 b}
9 T, z" E" d, B1 y) X4 V10.8
2 ^3 N- _" j* c- h, v9 j1 ]4 R) E3 E. p9 f#include"stdio.h"
& D2 p: p$ g! h8 c* l2 K. bmain(). W" A9 p- E* [, I, S( r
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
' N5 `) D' h7 F/ |+ V- { char *p,s[20];
. I3 g9 n/ @! B for(i=0;i<20;i++)s[i]=0;
/ J( S6 g' D% L8 m9 D$ {: { i=0;
" p. v0 x( S9 }' d( Y while((s[i]=getchar())!='\n')i++;- q; I& o/ w% y' e4 d. q
p=s;
: S# T. z+ Q- \ while(*p!='\n')# a) v6 B6 s# Z/ Y4 N
{if(*p>='a'&&*p<='z')2 A$ S' k3 S4 B. ]6 H
++sle;! k9 d+ ]3 [9 ~0 {: ]$ u) j
else if(*p>='A'&&*p<='Z')
: K! Q+ ^: u9 D# k3 e ++cle;
2 P1 H I4 z; \# j% N else if(*p==' ') g* l& f/ i; `( }
++wsp;1 N6 B _$ B! C7 Y9 ]
else if(*p>='0'&&*p<='9')1 m- ^# A9 z t# X( f1 E* s
++di;
, t: l0 x/ o, n" F% V: c% V4 B else
9 [3 T3 q$ ~5 _9 ? ++ot;& M5 o* H9 {7 D0 g. F y+ p
p++;) P! ~% ~) H2 e7 h' e% s
}
4 d* J+ a( s. P* g+ |7 j printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);3 C V7 F8 ^+ S0 l! P: D% @) q5 r
}
1 K- L5 h: l$ v" I' P. J9 X) S- R10.9
- y1 W8 _( q5 U4 Z, \main()8 O+ w- U! H; P
{int a[3][3],*p,i;6 W7 _& ^) O9 s8 t& S6 p0 `# u/ p
for(i=0;i<3;i++)
8 _3 Y4 p! i5 ]3 ]& } scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);% z; I0 ~+ N* j8 e. i& J
p=a;
' W! d( C& [, H7 d- \- t8 M: Q+ Y move(p);
8 f0 j, }) H4 ~: R for(i=0;i<3;i++)1 Q/ _/ q% C* E4 f+ m
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);
! ~% e/ Y' y5 f, q+ x6 n$ a$ A}$ Z. a- c7 s6 E- _' V4 \; F1 q5 @9 O( K
move(pointer)
) \) {% X3 ^3 D3 kint *pointer;
% t6 p9 x) B) p" N" o{int i,j,t;
5 b+ z, U1 m7 |$ j0 R. w9 `) `. } for(i=0;i<2;i++)( y' J2 ?( f: d" ?# n
for(j=i+1;j<3;j++)
: o0 ~$ j; F; p4 H: a8 ~ {t=*(pointer+3*i+j);
/ i0 N% W. E& m& v7 w @; [ *(pointer+3*i+j)=*(pointer+3*j+i);
5 S: W! H5 x8 a2 |! p& b4 E *(pointer+3*j+i)=t;
' {& K" D9 p+ E3 h$ w" }# I D }8 S4 t$ T8 z$ i6 _) j
}$ ? J! h: a; u% W7 b; `/ Q) }; V
10.10
" A( F7 C& W* |main()
$ G7 g7 p& Q) q1 U. D% h{int a[5][5],*p,i,j;
5 Z$ A8 y( i% Y/ y for(i=0;i<5;i++)
t/ m3 o' y$ u' u for(j=0;j<5;j++)
, @. P# c! V) `: b+ l scanf("%d",&a[i][j]);
/ j, e+ q# L# P3 Z, x$ [ p=a;
. t q5 b f( y& ` change(p);. ~" ?. b. c; _
for(i=0;i<5;i++)
1 r' e t( t" W/ c) M3 k {printf("\n");
2 S# Q3 t" ]" m4 F' ? for(j=0;j<5;j++); K4 A: Z* J/ _1 i- t
printf("%8d",a[i][j]);
# _! |# @2 ^7 \+ u }
+ n( y, e- k: J3 E}' X, l7 s3 _; z b6 l; _
change(p)% \0 E0 ~; U- ~
int *p;
3 o/ `4 Y ]) W{int i,j,change;- a, f$ B( E! ?) _, ~
int *pmax,*pmin;3 H. q9 K+ s* P, R3 P) K$ e6 Q
pmax=p; B4 q8 i2 q+ W) H/ g
pmin=p;
' t& b; T8 y; ` I for(i=0;i<5;i++)/ f, j, a7 b! |- s
for(j=0;j<5;j++)- O4 E% T# b& n# n
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;
9 U x4 L1 i( w* G, T if(*pmin>*(p+5*i+j))pmin=p+5*i+j;, T5 [" C" m" E/ `. @; ]
}6 A4 e& I* [. G' P" C
change=*(p+12);
" I' Y2 X7 O5 T' e l *(p+12)=*pmax;' O& g. U3 ~( r& G
*pmax=change;" S; s, L# J8 Q {9 @6 \2 F
change=*p;
& s; b8 w/ C* s' P2 ^ *p=*pmin;
5 e3 g3 V* j4 }7 i% s *pmin=change;
e/ J9 J3 t% R" R& h pmin=p+1;
0 q' a9 E+ p0 V/ {! Q/ J9 R% R for(i=0;i<5;i++)4 g3 a% u$ ]4 m1 U! t" m# S9 H
for(j=0;j<5;j++) S3 h2 p9 r9 u8 A5 M5 M
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j; k0 u2 |* j! W3 l
change=*(p+4);
# _3 L% E0 n n' |: M, j9 t *(p+4)=*pmin;
- X# C3 K5 h3 o/ P3 N6 ] *pmin=change;0 R6 \; F' V' G2 Z( \4 e& t7 V
pmin=p+1;% m3 }; t7 k# x
for(i=0;i<5;i++)
6 l) u+ h/ [6 T o for(j=0;j<5;j++)
% o2 r7 A$ x7 X5 S if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
4 H# T* Y* N L* {/ ^ pmin=p+5*i+j;
% I$ a$ S* G* Z3 R6 g change=*(p+20);9 B1 v0 M1 t: f
*(p+20)=*pmin;
* W3 o! s$ z! i; y *pmin=change;6 M8 ]& @/ [- c( @, h
pmin=p+1;& x2 z! t5 v( d) S# t" {; _
for(i=0;i<5;i++)
3 {+ A8 ~' x1 q, L/ R for(j=0;j<5;j++)
/ A6 g; F: O/ H. k if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
7 y6 z! E2 p" I% s( O9 s &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;- t9 a3 ]1 B& o7 O7 s5 c
change=*(p+24);
# }6 L/ {6 G6 h( q *(p+24)=*pmin;. R! i% H! x& n! P. c" a9 V) o
*pmin=change;. {6 q& f# y5 x" G' r& N
} _ H: A1 C$ B" ^+ L1 w
10.11
! m( z! A& b. H5 _- u7 }5 U6 I4 `main()1 p! k9 P2 c: @4 K% P8 a; p: y+ L
{int i;" }1 x! f; {% n3 J/ w
char *p,str[10][10];1 `4 d" g: @; ?* a
for(i=0;i<10;i++)
* h' ]0 h! c9 q1 I C% M3 j$ C scanf("%s",str[i]);
, a w, h; M2 G: j p=str;
; i, o9 v1 w& d( U _4 i7 k7 D sort(p);( g! n, D" D2 N/ }
for(i=0;i<10;i++) O3 ?7 ^- `1 Q2 _
printf("%s\n",str[i]);, X1 H( H" o6 L8 E
}' L' L* e' ^5 M6 s
sort(p), s9 Z; n$ H' i; n: e3 ^% L7 U6 v
char *p;
/ h ~' j9 e2 ^' y" v6 i2 O{int i,j;# V$ F5 S% @! y, G- L! C' k
char s[10],*smax,*smin;. o9 h, r% i9 J& b, H
for(i=0;i<10;i++)- h. a* N8 z; d2 }; {, M6 ~
{smax=p+10*i;
1 X% p N7 |( n- k5 h9 Z* E" F for(j=i+1;j<10;j++)
5 n. M3 e7 [: P {smin=p+10*j;
. w1 X7 N. g: z% U- j if(strcmp(smax,smin)>0)! P" I9 {: {, V, v# N
{strcpy(s,smin);" H7 d# Q+ o) ~3 D& J3 u& v9 @2 g
strcpy(smin,smax);
( R4 X$ q6 W) g1 l3 B* N2 d4 [7 _% A strcpy(smax,s);/ Y( @ W! g3 ^4 p
}
$ _: U: A( H: Y8 H% W: J }3 q8 `1 V* }5 n
}+ d8 L9 E. U/ G; q% C' h1 S# K
}% q* t3 ]3 u* V6 W, Q
10.128 t# h# L3 s. ]! T: {8 `7 h
#define MAX 206 f+ {, `6 M3 N* J
main()
1 ~2 X# \2 \5 g$ y% ]! K' p{int i;
+ W( \/ w5 @2 V7 a char *pstr[10],str[10][MAX];$ j: q5 t4 g$ M A; {
for(i=0;i<10;i++)! ~/ x& w, q6 ~# l( n
pstr[i]=str[i];& }# m: r2 `2 r* L8 l
for(i=0;i<10;i++)- S5 p( X5 t; ?1 J, T( h# p
scanf("%s",pstr[i]);
z# [% Z) |* f! x sort(pstr);) k6 L3 A) u1 W6 w: ]
for(i=0;i<10;i++)
8 a/ b! d; e7 ?5 l printf("%s\n",pstr[i]);( |; Z! u, [' r' K
}
! C- O) ]* o" bsort(pstr)7 O/ G! e; T& X& Q. B& O
char *pstr[10];
# F& v0 I$ l: M$ ^! N{int i,j;
; J* {, y* A& Z) h, }& b* e6 N3 h' q char *p;
8 ^5 Y h. m4 w6 x for(i=0;i<10;i++)6 P! c x# [ G2 U* m0 G$ j* p
{for(j=i+1;j<10;j++)% o5 }- F1 K: F6 k0 q
{if(strcmp(*(pstr+i),*(pstr+j))>0)& z; F: r# R) ]% N5 `
{p=*(pstr+i);+ k. g6 o- B/ B8 T
*(pstr+i)=*(pstr+j);8 i$ @- L) D' y
*(pstr+j)=p;
" }; J% E; N* E: D& c& { }) \/ l; p8 h: G9 ^
}
) @7 J: D* s \4 e/ ~ }3 r: b5 y& }# `! [ @$ d1 y
}) P, S9 W( {: ~1 k6 ?
10.13) v5 a9 X2 Z- Y) {, w! C7 t
#include"math.h"
8 R. K* P c5 r1 g6 ymain()
# i% S0 o/ ]3 u- ^) u3 O. R{int n=20;
- y' A3 ` j: T% b float a,b,a1,b1,a2,b2,c,(*p)(),jiff();' W' A. ?; j* y! R2 z) l
scanf("%f,%f",&a,&b);
* x/ Q/ ~/ v/ n s4 u scanf("%f,%f",&a1,&b1);
2 G# \' v& L% I/ U, S( t scanf("%f,%f",&a2,&b2);
/ N' I% e: q' v- H; u p=sin;
0 n" {6 A& L. p) ^5 Z9 _3 x9 q c=jiff(a,b,n,p);
# l7 h" r5 v. g2 l# q* S! A printf("sin=%f\n",c);
+ b6 D7 Q S- C, A3 c+ }4 I7 J p=cos;7 v+ r; y N5 r: D) z
c=jiff(a1,b1,n,p);
% \+ E2 t! U7 U* n% N/ i- T* X) p printf("cos=%f\n",c);4 C5 I) w7 n# m: a3 Z
p=exp;
/ w- m7 i: F( T7 d I/ ~ c=jiff(a2,b2,n,p);% e9 g2 }5 ]+ A+ i! q& n
printf("exp=%f\n",c);
0 V7 A* A I; G; o$ Y8 M; s}
* j( R2 k! S1 A$ Y; Ffloat jiff(a,b,n,p)$ s9 X. J' L( [! X- T9 G
float a,b,(*p)();
/ M J( I$ a) y, P/ Nint n;1 `& r' N6 r/ f3 V3 k
{int i;5 V# t2 t% K+ a6 ~
float x,f,h,area;2 |4 \6 ? p3 n7 A
h=(b-a)/n;
! i$ V0 {) D/ L x=a;1 g- {! G3 { d3 y7 U
area=0;" @2 H& G7 e \7 R: X" u2 V3 M
for(i=1;i<=n;i++)
7 h+ ?4 m7 J4 }, r {x=x+h;
2 C& ]7 f( |$ X9 ]! u5 u1 Y/ P area=area+(*p)(x)*h;8 S; w5 X E% @+ Q
}5 c' Q6 ?7 t# `9 i$ k
return(area);! k$ g% U; k3 V& q: E
}
- V# N5 ^4 Z3 }1 z7 D; ^" v10.145 G6 U( j, g% c% ]$ L3 ]$ |2 {' L
main()
$ `/ o; S3 l6 m; I) r$ j& U{int i,n,num[20];' v; c0 ?. R/ s% F. X$ W
char *p;
( i! |0 f2 T5 q: ~: b scanf("%d",&n);
) C" q6 ?& v$ x/ M s$ v for(i=0;i scanf("%d",&num[i]);: E4 } Y2 q5 D
p=num;5 [* v( j/ b) D2 _7 d9 o7 p
sort(p,n);
: V9 @& W! r! B; F" z7 h) Z for(i=0;i printf("%8d",num[i]);
% d# r2 _: d1 i}3 q2 h9 V6 b) E: `
sort(p,m)3 `# l3 U+ y9 W+ C% R( C
char *p;
' m6 u! f& n7 P& Z4 l, l$ @- q, `, G8 Cint m;
* W4 f& n- v' W O* `6 u{int i;& r( L" {( E4 b
char change,*p1,*p2;( z* G" k- u6 M" f7 s; L& S
for(i=0;i {p1=p+i;" e6 A2 H; q# n2 m9 V
p2=p+(m-1-i);
- W' n* ?+ Y- S% a* k0 _ change=*p1;/ T9 C" W/ b4 K. z8 B' n
*p1=*p2;
2 E2 N$ N4 A+ e* K *p2=change;% b! n. t- v/ m* r$ e8 F
}' S5 }/ A; G: I1 P: H
}
/ _- [, t) z% ?8 Y' `3 H10.15
6 R" x! i5 O0 X3 y/ l/ @" y2 t- emain()
* ^$ X$ `* J" O$ X' {$ b9 l& c0 f. S{int i,j,*pnum,num[4];
* k8 n/ Q. ?0 x1 {) D' i* @ float score[4][5],aver[4],*psco,*pave;$ W s- G* T8 M
char course[5][10],*pcou;2 f/ Y% t4 @- I# i. x Q+ j. ^ `* r
pcou=course[0];
8 Z' L0 T) x0 n. m/ F for(i=0;i<5;i++)+ Y7 t3 O( V) h3 q; i7 }8 N
scanf("%s",pcou+10*i);) e& ?! E [. J8 ]/ I( L- I& j8 w
printf("number");3 x" \6 E/ |% T3 w2 [
for(i=0;i<5;i++)3 T; ^. l& N& s7 e2 K8 E' f6 l N
printf(",%s",pcou+10*i);
6 ^, r7 r: @' K, m" d m printf("\n");: g! @) w3 G6 {& z O8 @5 J, _1 K2 o
psco=score;& j, q% _% J# [, ~4 _4 \3 i
pnum=num;8 ^+ P7 w; [; O( J' F6 X
for(i=0;i<4;i++)
M; C% f* I9 V. U3 G& J {scanf("%d",pnum+i);
& G% z3 P( g# w+ A# u for(j=0;j<5;j++)
% v3 G3 t6 T6 `& t- }/ d) E8 T scanf(",%f",psco+5*i+j);* n* {: X& w1 M! G
}7 x( V( L- u; q) W8 Y4 K( u
pave=aver;2 ^8 V, E$ p5 ?# @4 j9 p. c
printf("\n");
; Y/ W2 M2 C3 C I* @ N avsco(psco,pave); q' F; O" b. F7 ~8 a, p
avcour1(pcou,psco);
; ^7 |' o; M `1 ~% [) l/ x7 X printf("\n");
6 t l( j8 \( b& p4 W fali2(pcou,pnum,psco,pave);
6 X' ~- p2 ?* x Z printf("\n");
V5 ~7 V& d: ]- M% P- O good(pcou,pnum,psco,pave);
8 a0 X, j. h7 u p}0 K i/ a; S; z6 i8 x, e) t
avsco(psco,pave)# U6 M+ H4 N) r1 [% F7 y
float *psco,*pave;7 P/ P7 t+ p. `8 Y9 A+ O
{int i,j;/ a4 c9 d8 g# i) }# _( f+ o K% C/ n+ u9 U
float sum,average;8 m- Z1 \( S% ?/ q
for(i=0;i<4;i++)
- }5 G$ I) @/ D {sum=0;
" V/ Z6 G3 n; ?+ S. Z for(j=0;j<5;j+)) C( p- i+ [& f; E/ @4 e( k
sum+=(*(psco+5*i+j));& V" P! I! B7 B8 c* s7 e' y5 ]
average=sum/5;+ q9 P! H; e$ x% k9 s1 M( U
*(pave+i)=average;
7 s; b U* E( c! ~ }
' \# K+ }9 f9 j+ d# r) t6 J! [}
7 r' l! @: |7 e' K3 j9 }' eavcour1(pcou,psco)
3 |* S, {8 Y# E1 E* r6 Bchar *pcou;0 E1 `5 N0 d7 `+ m _1 i8 |
float *psco;
1 Y' m0 n2 J! C{int i;! F/ B$ m* N; D2 n1 L: `, C$ ^* \
float sum,average1;; o; ~) n/ v: Q3 u& D
sum=0;
! j5 q4 r7 {& M; x9 V4 u for(i=0;i<4;i++)
# f& B/ F# i/ p, L' U; E9 d" x: i sum+=(*(psco+5*i))+ D2 H& S6 U4 x/ F6 J" n* ]
average1=sum/4;# v9 O) r% F; c: R1 b
printf("%s %5.2f\n",pcou,average1);* \$ b+ v5 {0 f" [ y( i5 S1 q
}5 s" G$ K- ?: \2 `" ~5 b& K1 Q
fali2(pcou,pnum,psco,pave)
' U& q& a6 d' w+ W6 h4 echar *pcou;) U* c9 E" C) R' l1 d5 A
int *pnum;: N, x8 Y# @* F) F* j/ y2 Z( g
float *psco,*pave; ?* m* Q, d8 I' E. M9 B
{int i,j,k,label;
! p4 M9 V# T$ I0 T2 Q1 f$ q printf("\nnumber\n");
9 y9 Z. _( f6 f' \ for(i=0;i<5;i++)
0 A, } i& }2 _# a printf("%-8s",pcou+10*i);
1 p H7 e, I' a) P8 c" C' I printf("\naverage\n");$ A& r3 o8 s6 ]2 U
for(i=0;i<4;i++)
4 @% J+ u7 T3 L7 R$ U; z {label=0;
$ k* m! [ S" H$ S# T for(j=0;j<5;j++)# R8 G5 ?6 p$ C, K
if(*(psco+5*i+j)<60.0)label++;
# x3 j( m. B& r1 d& W6 B6 x if(label>=2)
# z2 ~. Y" O( ]! | {printf("%-8d",*(pnum+i));0 o6 \% H# H8 y1 w, C! ]0 ?
for(k=0;k<5;k++)
/ q8 r& f n7 J printf("%-8.2f",*(psco+5*i+k));7 P9 n F. J8 \3 P- o4 x9 j
printf("%-8.2f",*(pave+i));
% U8 a2 o1 _$ [( |! M1 B }+ m; w4 T) \( `
}" V. _# X4 k/ w1 X; Q
}' d; c: |! u- {7 t5 w1 `
good(pcou,pnum,psco,pave): Z, d' U2 b+ ^& q, e, \" `
char *pcou;
. i# m) x5 { T# F( _- l8 Oint *pnum;0 ]& U3 O) ~! P; B% ^; Q$ c
float *psco,*pave;
0 a4 A! `( ~* A, \{int i,j,k,label;) V, x% ^# j. R% b; q4 K: @! Z. x9 D
printf("number");
, ?+ D" c. B9 k, p) B, D for(i=0;i<5;i++)/ r7 k. ~3 e# c9 B& M I- e
printf("%-8s",pcou+10*i);
' e2 b5 x' ^* ~ l) l5 j: ^ printf("average");( j. x: K- Q: g( S! y i/ f7 i, }- o
for(i=0;i<4;i++): U! q; [) [6 }& N/ K
{label=0;
$ B- J6 i6 ~! y$ c& h: b. H/ w for(j=0;j<5;j++)& B! @5 ~6 `' n9 `; z/ M4 X+ h
if(*(psco+5*i+j)>=85.0)label++;4 q, }, F0 q; Z; A7 i2 H; [. M
if((label>=5)||(*(pave+i)>=90))! M- T, @2 E! H, ~( q
{printf("%-8d",*(pnum+i));
1 _4 S2 C' D* D& d: Y4 F7 P for(k=0;k<5;k++)# l& F$ x) _% a
printf("%-8.2f",*(psco+5*i+k));
- t1 J( }+ C6 M/ I printf("%-8.2f",*(pave+i));7 m1 M: W5 w' G' |
}
/ c- ]% f0 ?, Z' C3 T. s }
3 h# a2 A$ i; ?7 u}
C Q; v* y4 D/ b! Q, C- p4 K10.16/ N5 y/ \* L, A8 j8 B
#include"stdio.h"
9 v% g" @% Z+ imain()
7 I5 P# ]4 @7 \, B{char str[50],*pstr;! q4 }% S' K1 b0 s1 Y" n' S2 p: u2 u
int i,j,k,m,e10,digit,ndigit,a[10],*pa;
+ Z' Q ?5 j7 w6 \- \, p0 H gets(str);
1 @# V1 y" v# |" n. t6 @ pstr=str;. d+ S/ Z: n# {; ^
pa=a;
1 ^8 P- A0 F6 o3 k$ H ndigit=0;
) a; u: y' x; A8 w4 _1 F i=j=0;
6 p, x* }( M1 c& I1 O while(*(pstr+i)!='\0')
1 q6 J- k( g+ ]) t9 q {if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))! n+ M* R. `( [2 c! G- X, l
j++;
' h) w1 D4 ^, U else
! \% K" `5 y2 C6 H; @% y0 P+ @: [3 X9 J {if(j>0)8 u" z% Q7 e u
{digit=*(pstr+i-1)-48;% x4 G$ p- A8 }7 U- m
k=1;0 _1 V4 ~6 z$ O7 K
while(k {e10=1;
" h' _- s1 l' J! S# H/ y# q% { for(m=1;m<=k;m++)
( u0 F2 Y' S' O) t" T3 A e10=e10*10;$ l! w3 ~" [% a4 T' q) Y7 n
digit+=(*(pstr+i-1-k)-48)*e10;* E8 T) d% e2 W9 x4 o
k++;$ i9 B3 N0 U# n( N: q& D
}
^. X2 e: z2 h! A4 a *pa=digit;+ }$ ^& w: Q1 v. i$ p% C/ S
ndigit++;
* i% D, ~: P$ }5 P2 b3 N6 P pa++;2 V, _( E2 Q% f1 a" L+ A1 ?0 F
j=0;3 L) C5 [: z+ Q9 s6 N5 K2 U
}9 B) Q' V/ m2 J" z
}2 q* N8 x2 d0 Z3 p% U! P4 B
i++;
1 F" j/ h1 n% ]. t! ^. y6 Q% a }$ T) v6 t) n/ R" X4 R: G$ w
if(j>0)) V. L. A- {0 A" h# G p
{digit=*(pstr+i-1)-48;
$ i2 r4 s! Z& P+ S" E k=1;* }% f3 d G( w" F' D* U+ a M G
while(k {e10=1;- ]7 \7 U) M: Y1 G/ ?" e
for(m=1;m<=k;m++)5 p7 r& p# @0 f" }
e10=e10*10;
# F5 f# B2 ^9 I n/ P: W digit+=(*(pstr+i-1-k)-48)*e10;. E; Z6 J5 B N& `0 m
k++;
7 l; z# T9 S# ]6 ^# k$ @2 c }
& V' x) {; w3 m( B( } *pa=digit;; P8 u0 `4 u& ]5 x$ N
ndigit++;
0 y" s o/ l2 z. f1 X5 Y1 ~ j=0;& s2 B- K" L+ P8 M& Y! O0 k
} ; ~1 H0 p1 T+ P V" E9 N( k5 Y" u
printf("ndigit=%d\n",ndigit);
2 i7 g, ?8 l2 }8 l) P j=0;8 M K* t- |, C, Q' w3 F2 r
pa=a;
: t5 i6 g* I; k8 b6 G: A! O. N for(j=0;j printf("%d",*(pa+j));
6 s( r% N1 M O3 T}* C* z+ U P% X+ F* c5 c
10.17
. {* S. R* L- N/ Q8 omain()
& ^4 K0 g* _" P4 M) u- F: a- S2 g{int m;, T% w0 m- q( E/ s9 b5 z: b9 D( n
char str1[20],str2[20],*p1,*p2;
0 k1 H# @7 {3 P$ s3 m' }* F5 f; w scanf("%s",str1);. r6 z. ]; `( u, T" G* t4 ~
scanf("%s",str2);
! t' L+ a, ~. z- H L# g: d: p p1=str1;& M5 y0 v1 ^. \8 i' e! W# f
p2=str2;0 q& e% k x) N2 B% e
m=strcmp(p1,p2);
' M9 k0 R% _# y) p, r% q" Z printf("%d\n",m);- r3 Z8 m2 Y4 V5 M
}
9 h, T4 u: U$ X8 q- }* Y# z& fstrcmp(p1,p2)! x1 E2 k) }) n: {
char *p1,*p2;
- u6 U; K3 ^7 s. N{int i=0;
3 K" x/ l* X, e! |7 C8 j- a' W$ | while(*(p1+i)==*(p2+i))
\1 a. m5 l d9 N if(*(p+i++)=='\0')return(0);
' P! b% S% u5 S return(*(p1+i)-*(p2+i));) u3 M1 I' |" I1 U2 G1 c; e
}$ ~- `/ s% C8 c4 {& y8 T
10.188 X9 d: u @, o3 }" P
main(); k, r$ _: ]+ Y
{static char *mname[13]={"illeagl","January","February","March",' l3 O- A; W' a( P: l! `# m
"April","May","June","July","August","September","October",$ j" l* _, @% T# _; d
"November","December"};
) I$ X7 G" ?. S& t0 C z# N int n;" \) G# O7 F' T( Y- N4 o
scanf("%d",&n);3 Y d Y8 L& Y+ c0 q5 x
if((n>=1)&&(n<=12)); m, o( Z7 A' s+ K/ _. b2 Y
printf("%s\n",*(mname+n));; }2 |* T5 }! G! i: P
else
' ^9 @; y$ e6 n3 ] printf("error");( W7 _( T- U; a' l
}+ C6 O& B. f; ?' }4 Z
10.20" N8 X' T/ g! Q! Y' U# X1 h
main()' _; e& D: s: ]/ Q+ ]) l
{int i;# j( y r0 G" T$ E8 Q" L
char **p,*pstr[5],str[5][10];& I2 k" w3 x4 i) I* s8 B$ `
for(i=0;i<5;i++)
8 S8 S; M) V ?; s) L pstr[i]=str[i];
/ `! o( I+ C6 D. ~& c+ L& a for(i=0;i<5;i++); x4 I8 p. a- s- U. e; P
scanf("%s",pstr[i]);
. e( p3 ^1 }# O7 I( Q' e. Z" @ p=pstr;. |; u* F7 P. t! V# n, [4 a, g1 U
sort(p);
9 C9 T" ?- T6 b! ~4 l1 ` for(i=0;i<5;i++)! F% W# K3 X4 ?
printf("%s\n",pstr[i]);
$ H# G9 Q; e3 s) H}) {/ i! r' a+ H* R
sort(p). `; W0 Y s! a0 X0 Y" V5 x; I
char **P;
/ v3 M1 j) G$ J! g5 ^" l{int i,j;4 e2 q6 r& E- T
char *pchange;
, A9 P* q) U8 v- b6 a+ W* f for(i=0;i<5;i++)! @ D1 K: K* A: W
{for(j=i+1;j<5;j++). i; ~; T2 O0 \3 U0 s6 U
{if(strcmp(*(p+i),*(p+j))>0)
3 n6 R/ Z7 U C. H% A {pchange=*(p+i);) H3 F! b7 z3 Y( c9 }4 T( E- b
*(p+i)=*(p+j);
; ~/ I" x2 ^3 ]. {( H- `3 G$ a1 ~* L *(p+j)=pchange;
% @) D b4 H/ I: A1 y }
8 i( {5 g7 G6 U }$ o8 F+ H5 |+ K) l0 L. N( c7 \
}
/ a# X( ~" h& u7 E}
$ T% Q, M2 g3 r8 t7 O. w, |10.219 v; B9 x+ M3 C B. D
main()7 W( P( B$ ^. X& I# K8 ]( A
{int i,n,digit[20],**p,*pstr[20];$ s, y/ I4 S3 a$ h4 c
scanf("%d",&n);
, m0 o0 P" d, A9 |; o- W. L3 a for(i=0;i pstr[i]=&digit[i];
- y( s$ ?3 m4 D6 \ for(i=0;i scanf("%d",pstr[i]);
8 N1 `) v; V, x p=pstr;
( W* V0 t+ o& Z! e; T( V' k1 y2 X sort(p,n);4 ]# Z7 j/ Q% ]/ O, c
for(i=0;i printf("%d ",*pstr[i]);! q' a3 ~# |$ j6 E2 e
}
+ P% U6 Y" u* T; j% |sort(p,n)
4 ~# {' I; U2 F- ]+ q( K" t: _int **p,n;/ k& x" g- ^- r5 T
{int i,j,*pchange;+ C+ t+ @* l2 w1 I |0 I6 G2 n
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
# M3 v( C+ i; g" k6 I {pchange=*(p+i);. |5 a# A3 X: f. O+ {
*(p+i)=*(p+j);- ?6 z7 P9 `, u
*(p+j)=pchange;3 g, }9 {& ?9 _6 Z8 Z. m6 v
}
' S3 @, t" j5 ?6 y, b4 E }
5 W, X0 L+ C8 b( ?/ n$ a }
) O+ z& t/ j/ Q7 v- |, n}
5 l' A! p3 L) [8 z+ b6 `4 W' Y! f第十一章 结构体与共用体
1 Q2 x: P+ W" J11.15 D% n, z$ `4 s m
struct% Z+ C; Y# v" _+ K1 X/ \
{int year;+ X* l4 y& W7 d
int month;
3 n: `5 D" q8 j' Q3 {5 s, H( L7 a int day;
* `9 o" @1 K. M9 o7 B4 P }date;
9 Z, _2 m% m' W' N) Cmain()
; p- f7 h- i x! h( C+ l{int days;; @4 x1 y- b% u6 d1 @# J! n
scanf("%d,%d,%d",&date.year,&date.month,&date.day);
' x! s: {! H8 a4 ` switch(date.month)
, J9 ~0 E4 h1 R {case 1:days=date.day;break;8 |4 a; w1 l$ o% S6 K
case 2:days=date.day+31;break;
) u% F( [1 q6 N! x2 a) Q case 3:days=date.day+59;break;. [- W$ n7 A( x1 \5 A f
case 4:days=date.day+90;break;1 r% M! ^1 \7 K/ e: O! d& _
case 5:days=date.day+120;break;* j$ E: S( q) x& p+ ^9 U6 O3 c
case 6:days=date.day+151;break;% o- R" R, u4 n+ E9 w& U" b+ @) S
case 7:days=date.day+181;break;5 O* b$ x! O! }) @4 W9 L
case 8:days=date.day+212;break;
+ J1 m5 V. t' f) B0 ~- d) F case 9:days=date.day+243;break;
# N4 o4 |8 b2 P case 10:days=date.day+273;break;' v3 b% U& k6 R; o# g3 j
case 11:days=date.day+304;break;8 S3 O$ r" t }1 t/ B/ P
case 12:days=date.day+334;break;
' u5 s9 f9 V* K1 K4 r. g) c }
, U2 W: H' f6 B4 S7 q: l2 F" j if((date.year%4==0&&date.year%100!=0||date.year%400==0)
: g* J& _, r. _ ]% r' j- U &&date.month>=3)8 h, ]' z* v% F1 T3 s, }) w+ m& W
days+=1;: A. d2 s& s- Q* |
printf("days=%d\n",days);. N) ~* R8 l7 M
} w1 M7 Y" S: n+ [* D+ A! W
11.2 n& d- ^* m g$ `4 Y
struct dt
" ]! _, A0 W9 n# a# c* Y) K* Z {int year;
9 l( `0 |5 x: E) L1 V! z9 e int month; I6 K u4 V6 h5 f
int day;
8 C/ k6 l2 \3 L$ C6 m0 ^/ y }date;
/ E/ p2 ^" `# C/ M: H; f) v8 c4 @4 Smain()
. H9 T! p* h/ P7 r( k0 S{
" e. r7 g8 b# t scanf("%d,%d,%d",&date.year,&date.month,&date.day);7 R9 G9 [, d8 a
printf("\n%d\n",days(date.year,date.month,date.day));# _9 c/ k( ^$ k! }1 c
}
$ Q- [& t7 s! ^- t, Y7 Gdays(year,month,day)' m( o& h* c. Z
int year,month,day;4 ^* ~4 h$ q: _9 U# q4 b* e1 M
{int daysum=0,i;/ A( P4 {5 r0 i1 m) O
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}* Y: f$ g9 `; ]2 }3 @6 b3 K
for(i=1;i daysum+=daytab[i];
- C+ F) Q# o9 s1 P% e daysum+=day;. ^6 _4 u) P$ P
if((year%4==0&&year%100!=0||year%400==0)&&month>=3)" I/ d2 J- R; j) q; M
daysum+=1;
0 j& `* d: [. x' S7 Z return(daysum);
8 O5 u: Y" a T) s$ @% y4 N) Q}6 `1 \6 Z3 Y, X) d0 Z1 u1 K2 }
11.3
! Y+ O9 p9 Q( e: I* j11.4+ M) H" }) k% T W$ [$ y
#define N 5 M, W# l5 [" n7 U
struct student0 m- i) j! I3 i6 \+ P
{char num[6];: Z! Y" u" d# L. @0 ?8 i
char name[8];
6 S; Z, W8 ]# i: h int score[4];
+ m& k% o# ^: n" |3 z; R. T* d }stu[N];
0 I+ i$ [! a; ?* Y* Zmain()! p6 [5 O8 L* p" [
{
( y& O' ]! {" u0 ] input(stu);
0 X/ f9 c& \# q& K print(stu);
7 j" V5 j3 [ m}
/ _' V5 N( r) rinput(stu)
2 Z) H2 e4 U L, @2 w0 _/ Wstruct student stu[];( }1 C4 B7 i" A& z2 G
{int i,j;
* v- ^- K: c# z% T for(i=0;i {printf("number");0 ^/ Y0 \2 R5 c
scanf("%s",stu[i].num);2 O) a" ?- ^7 ~- t5 J7 t* ^! T
printf("name");
- e3 Q' }! p5 g! i7 a7 h6 ?( \ scanf("%s",stu[i].name);1 J8 q1 O6 A& A1 L- m! M3 d3 q
for(j=0;j<3;j++) r- E4 O1 F& e0 \
{printf("\nscore\n");
# j1 R9 n$ }" A0 m! [$ b scanf("%d",&stu[i].score[j]);! A+ U& A% {7 k) z+ I1 q
}
+ \7 C6 K& C, ^0 m7 f# u1 w- f5 s printf("\n");. V" Q5 w1 }/ W# p# {6 k9 Q( o& i
}5 V/ W' B4 N1 G6 c. w! K
}- v D" S; {8 e2 r: K; u9 s
print(stu)) E3 `( }6 a0 y) c. a, Z1 z
struct student stu[];3 R7 v) M# P/ E2 ^8 `) p* e
{int i,j;
9 U& r- K4 ?& S- n printf("\nnumber name score1 score2 score3 \n");8 m. P0 V) v" @( z; w. p+ q
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
: i+ z7 I7 ?; f for(j=0;j<3;j++). A+ \5 ~! u& c6 S# D. m- v1 f
printf("%7d",stu[i].score[j]);
; e0 _% r! M ]6 P5 W* _ printf("\n");3 T: s, A9 M. b1 c! {$ i7 t& a) a
}
! H; I7 d* |& J; v; Y/ K}' l& D% U: e3 V h
11.53 ^5 U9 Y* G( `7 F5 B9 ]4 W
struct student
' p& d: Z3 R; F {char num[6];
; m. y% ~1 I4 z5 m Q' s% z char name[8];
' ~, Z% a$ @* o/ \7 J4 t( o int score[4];
- ?1 q$ e" }. m. r float avr;
; }0 Q* s9 A6 q: r" ? }stu[5];
+ h( t5 u1 u) g; `. b% Xmain()
9 I9 Z+ W8 |0 Y- @{int i,j,max,maxi,sum;1 D4 N, Q3 D9 \- I% {/ H3 o
float average;- ^9 E4 I( t% Z1 q8 P$ ]
for(i=0;i<5;i++)
4 q7 c# i. _! X7 }9 _/ F! G {printf("number");6 _" K4 ^6 R5 ~: h7 J2 y$ I2 U
scanf("%s",stu[i].num);
: a5 h. t/ e+ V, S; y printf("name"); \- q& X1 H% x' K8 S+ ~6 ~
scanf("%s",stu[i].name);, j8 T" y' Q% u
for(j=0;j<3;j++)
% B; A# q1 \) P, g5 Q {printf("\nscore\n");4 ]4 K3 l s; f
scanf("%d",&stu[i].score[j]);
7 c9 a8 \: F- H& B }: R3 z3 ]0 _/ a$ I0 f4 E
}
' `, z [ R8 ^. W average=0;
) o9 ?8 |- r: A6 l0 ]) o4 ? max=0;0 _" f, {" z0 b4 ~$ l
maxi=0;2 S/ [" Q0 d# ]! z3 k. m% K
for(i=0;i<5;i++)
0 w: [* \7 ]/ g, b8 J! g {sum=0;4 e- F( I. D* K7 S% s2 Z, A
for(j=0;j<3;j++). f1 y6 M# ~+ [6 A
sum+=stu[i].score[j];0 V" Q6 Y. H! T1 w" ^
stu[i].avr=sum/3.0;9 ?9 a' U: Q3 L0 t, K
average+=stu[i].avr;5 O- H4 x' U n' Q [
if(sum>max)
, H7 n: y% Z8 ~, O# M9 u {max=sum;
4 ~1 N, n& J) R2 A2 o; { maxi=i;
- v! i) \' q$ @8 R4 H+ ~' i2 }$ T* K }8 v- n$ N8 E: S! m7 Z
}1 i( V' U5 m/ s8 r
average/=5;
& {" y" W# V3 _7 Y printf("number name score1 score2 score3 average\n");
' E0 u; W9 g% o2 p8 e( _ for(i=0;i<5;i++)
2 p' v. v8 D/ @- A; y+ R, }4 ^ {printf("%8s%10s",stu[i].num,stu[i].name);; \9 v5 b8 T2 Y! ~
for(j=0;j<3;j++)
0 C; `) w5 K7 H U% A/ A# X+ | printf("%7d",stu[i].score[j]);) J% D2 L( s/ y& v
printf("%6.2f\n",stu[i].avr);( E* L1 d6 q$ [1 j
}
/ K7 r" l2 [7 N printf("average=%5.2f\n",average);
}& q' L" \0 e8 [# H+ m printf("The best student is %s,sum=%d\n",stu[maxi].name,max);
2 O5 d3 i7 i M}
$ p; g' k. b. C* T, t! i, Y4 A7 J* W) [/ B6 @: |2 F, h( R
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