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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
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签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案
. u% B3 t6 N5 p4 E% |1.5请参照本章例题,编写一个C程序,输出以下信息:
3 g% V J3 a; y7 y7 t: D7 Emain(); l c( ]0 ~- }' G2 ]" h* f9 V
{% Q4 w" x9 M/ o) r! ?
printf(" ************ \n");
% [9 E. \* y* S# p" d: jprintf("\n");' M7 }' N* S/ q# _5 l* _7 F+ }
printf(" Very Good! \n");
. U( h- Q! V+ Vprintf("\n");
6 C3 O5 i7 j$ e' g& a" vprintf(" ************\n");
& G& t8 J2 I! q L. z}
: O# S, Y7 N3 T8 R* V# r1.6编写一个程序,输入a b c三个值,输出其中最大者。
8 U5 I4 ~9 d3 s2 @: t; o E, f解:main(), l4 F" ], @7 U) R# B
{int a,b,c,max;& d& A) h- C: t0 ~& [3 V
printf("请输入三个数a,b,c:\n");- B' ^+ U9 q5 a; \! ~! Y, m9 ^
scanf("%d,%d,%d",&a,&b,&c);6 @# X+ N* }" z, G
max=a;
3 z( b" z! A: Z- U" X. |if(maxmax=b;0 O* _: t& u, L, q* {/ a- C* w
if(maxmax=c;" e5 x6 P6 D$ c' e, G/ ^
printf("最大数为:%d",max);; a; q$ i3 {0 O. | b8 w5 i
}
% A) g! \7 S, h, A, N; |第三章
1 I9 u4 T( R% d2 p0 A* K3.3 请将下面各数用八进制数和十六进制数表示:" b" e6 w- V$ H/ Q) g
(1)10 (2)32 (3)75 (4)-617
/ C) x, G, \' U) a. H1 L! F(5)-111 (6)2483 (7)-28654 (8)21003
0 u9 r N! k* I9 J解:十 八 十六
" z `+ Q: ~, L( j3 R/ T7 T( p2 m (10)=(12)=(a)" P/ h# G# X4 Z
(32)=(40)=20
+ R% _4 n. q! [ g0 `9 s+ W (75)=(113)=4b! A/ e; E7 a1 c/ s7 O! G( b, x
(-617)=(176627)=fd97
4 h9 T' T1 E4 W- m M1 U -111=177621=ff91+ p2 ?& s9 v; {( X. z, g0 w3 d1 n- y
2483=4663=963
( V# @6 `* l. L# M# \ -28654=110022=9012
8 o1 R: h# c+ F# x 21003=51013=520b
c' z* [0 h, C0 [3.5字符常量与字符串常量有什么区别?
9 \0 o3 W% ~) v" a9 X0 G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
0 T9 v4 k; Q) _0 v6 a3 _) s) w0 C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'.
$ s% s# i! x$ P3 m5 E {- f3.6写出以下程序的运行结果:
$ a8 {# F2 U ^& b4 ~#include
1 J5 f& y+ i p6 i; m Evoid main()0 f- [! E& h: K
{
3 {6 u" [$ ]* x2 wchar c1='a',c2='b',c3='c',c4='\101',c5='\116';5 X+ P$ x- @& V: f5 Y* G% Y
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);
0 p" `) d6 w0 n6 ?printf("\t\b%c %c\n",c4,c5);" M, r, y. A8 `0 h' Z1 P- v
解:程序的运行结果为:- C' E7 J: q. {2 E+ ]
aabb cc abc8 y3 B( {& ~3 ]
A N
3 m, z6 F( E, N2 u& G2 y0 o3.7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,% `$ j4 H' i5 {0 _
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"." X8 X% k% G9 B9 O# G$ w4 H
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6 B- h6 P H) }
输出.
8 Y: m- A- H1 l7 ?& M5 k! ?. Gmain()
% R# R/ z; o8 |, m3 w6 \# Q3 M{char c1="C",c2="h",c3="i",c4='n',c5='a';% l* C5 [0 H3 i# Z" c5 A5 {
c1+=4;3 b8 s8 E; T! J) v, w
c2+=4;
% V+ V; r, V. N; a! k3 m. Ec3+=4;- V1 N8 d( n5 @" x2 \3 B$ W; ~# O
c4+=4;
i" P, J% |% u5 }0 q& w' j& Ac5+=4;: w! }. p% o& F$ J- ]+ I# z1 g( @
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);. i z) |) ? t. }0 A5 e, o
}
! o* w8 @2 C/ x3 X- G3.8例3.6能否改成如下:2 u4 c; y. H5 ~. G7 {8 [1 c( g1 h
#include
" l. l$ q& [! n" Y# \, gvoid main()
/ q* w1 c& \4 m2 V+ ]1 p/ u/ ]6 N{ ?. a T5 m; K# u y9 B! w
int c1,c2;(原为 char c1,c2); L5 S+ P! I( d* A- a; S+ D
c1=97;* h* G3 W1 ^# G/ U: {3 L
c2=98;: m' ~7 v4 h$ \/ F( _0 G; g+ u" ]
printf("%c%c\n",c1,c2);! A8 J: n" f, T! R+ a1 f
printf("%d%d\n",c1,c2);) M4 h* x+ r0 o# @; m& }. p
}
, E, I1 k m. y+ @: p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.- s' q# d2 k! ?$ j6 L- b6 @
3.9求下面算术表达式的值.+ G6 p! `' @$ e( w' Q4 N
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
+ V# A1 ?7 j7 U1 @8 y1 P9 V# U- R! c(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)
9 K. z: b! R) E7 D& i1 t. q3.10写出下面程序的运行结果:0 H, _3 C8 g* p7 n: c* [# n# ~- I d
#include0 Y4 t# w0 {2 |6 q0 i9 E, c
void main()
9 {/ I6 |* W: w: c' }{
9 ]; q, i8 k, r+ m9 e7 H4 Uint i,j,m,n;
( a" R+ z, U( ki=8;
% e9 f3 w v" h8 e& B3 {3 [6 w+ nj=10;) O! G" N7 A6 F
m=++i;3 B# z: j2 N. j7 F! x6 `/ F* R7 L
n=j++;& ?: U, i6 v, s' Z3 U" [
printf("%d,%d,%d,%d\n",i,j,m,n);
% c6 ?! |) x" {3 Q. X J6 l6 m' _}
, u, n' l, |8 @2 ^% `解:结果: 9,11,9,10
/ h/ U* J5 o* I. O$ ~( Y% K第4章
' R$ [$ Y9 I* B2 O4.4.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
8 u5 f$ X3 W5 T6 K: @7 c; B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:
+ N7 k; m7 K6 E- ka= 3 b= 4 c= 5" D+ Q4 ~% {/ I) U
x=1.200000,y=2.400000,z=-3.600000- K. @8 G: w# Z& d) m# C
x+y= 3.60 y+z=-1.20 z+x=-2.40
- b' o* G$ l& M0 ^2 G' `0 Ru= 51274 n= 128765
- o8 k( e* @7 T7 B. _, }c1='a' or 97(ASCII)) f' G) C; T( b* V. W
c2='B' or 98(ASCII)0 L( ^9 Q$ K; d, e: k
解:/ h% B' A7 o, c$ [
main()
; @. Q: A/ ?$ P/ {5 H{. B& I1 a0 I" Q( Y/ `
int a,b,c;
3 k+ c: V r; a1 h+ Hlong int u,n;
6 t) c! H/ L* pfloat x,y,z;
* K7 K1 ]& x1 ]8 ~. mchar c1,c2;' i4 C, S) u5 }% z9 e
a=3;b=4;c=5;: Q b9 ~: W/ {
x=1.2;y=2.4;z=-3.6;- y9 B9 n! H9 N' D" x
u=51274;n=128765;; P" B, a8 I; u7 Y
c1='a';c2='b';8 _2 [- a( \/ t
printf("\n");4 ^1 [- c1 ^9 R. n6 Z' e4 y
printf("a=%2d b=%2d c=%2d\n",a,b,c);
% Q# v& ^- T9 P* E0 u/ vprintf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);, P7 ^, F& F# t" ^9 U% z- J4 c
printf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);" H7 I. x) q5 O& [% Y
printf("u=%6ld n=%9ld\n",u,n);1 y+ `* Z* ]2 k- V
printf("c1='%c' or %d(ASCII)\n",c1,c2);
; S" v: U G" \3 vprintf("c2='%c' or %d(ASCII)\n",c2,c2);
1 E( _- k/ i& O; Z8 ^}$ W* F6 @8 q( |; M9 P
4.5请写出下面程序的输出结果.
! |8 ]7 N/ M4 ~ f2 ^# |2 q5 L0 I结果:
. k3 k9 \# m7 S$ v3 | r* O57
, ?5 o( |7 J! _, n v( J1 k 5 7
! a' h6 Q$ d0 A/ V8 ~6 S5 ~# T6 C67.856400,-789.123962: x# X6 |6 H, ~8 P9 [
67.856400 ,-789.1239628 \; A6 T$ F1 u* `( i3 I
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962# Y8 a; Q# o, Z. r9 S6 j& F
6.785640e+001,-7.89e+002
# [, x( q9 w$ X# F. Y6 c4 [A,65,101,41
4 k L4 a( ]( ^8 H* [: F: b$ j1234567,4553207,d687
9 T; ~8 @; d7 |0 t3 w3 W- l# R65535,17777,ffff,-17 }, G* p" w9 h5 v( G( ?
COMPUTER, COM1 J% W. A x9 J, c
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',
% `, W! ?& O% {7 g问在键盘上如何输入?
0 d4 j( P8 L- F2 ~7 O3 {5 smain()
9 J7 b: Z7 Y2 o5 v/ z8 p{. A, H+ J, V6 n2 n
int a,b;
4 G x0 t5 [, G: h9 yfloat x,y;
e$ J. |. D. J6 x, _. Wchar c1,c2;: |( ~' L6 @( X* \" s0 j
scanf("a=%d b=%d,&a,&b);
% s2 H; K" @( N4 U& Q/ kscanf(" x=%f y=%e",&x,&y);
. U1 W& p2 c; g, J2 y8 Tscanf(" c1=%c c2=%c",&c1,&c2);
& I: E% M) d, c3 N+ J- Y4 Q}
' B4 E6 o l% q8 X- E3 E: _9 G5 ~解:可按如下方式在键盘上输入:. ~" H# R/ @7 G8 [2 q& ^/ D6 w
a=3 b=7: u) x4 B* ]' y$ Q7 \1 s
x=8.5 y=71.82' e; k D3 H0 W" S
c1=A c2=a
9 D; i- |* }" T9 u0 ?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"被第二5 t7 i1 a* O9 c( h6 s
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
- V4 V: j E- D% V% C& r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: Y$ {$ s. l1 z/ T5 W9 y
初步概念即可,以后再进一步深入理解.% ?7 a: l9 Q0 w9 q3 _4 g
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=-
6 P# G5 v+ K3 Y$ e! N5 ]3.75,z=57.8,请问* @0 _4 Q/ _+ @8 s. U& b
在键盘上如何输入数据?
2 f6 e+ w' `" {scanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z); `/ Y) ]# S4 G, k
解:5 j# [# Z) Y1 v/ w# r
main()
; g0 m! x" B7 H$ r; J5 b/ k6 W{: {/ e- ]9 b6 [5 Y/ t1 r9 z2 t
int a,b;: x6 i8 v2 f( l0 X
float x,y,z;
* |9 D1 O4 m8 }9 L4 Uchar c1,c2;
% e9 I1 N7 j% u. f& mscanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
$ ]& R, v7 i& L$ g3 N& d}
) D& _1 `& V2 `+ ^1 c" V3 c* T, C0 G( m运行时输入:
# ]# |3 z9 N4 s7 z* ]10 20Aa1.5 -3.75 +1.5,67.83 h6 ~7 w6 }" @3 K0 J5 K
注解:按%5d格式的要求输入a与b时,要先键入三个空格,而后再打入10与20。%*f+ p- l5 j' r% U- U; p& X
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+ A8 \6 S3 s7 j o# C
会赋给任何变量。
" f- r5 H/ p3 H$ I# f4.8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,0 E5 a; c, ~ \* b% 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9 p, \/ r& ], F5 L' f
程.
T9 X1 c- |4 {( ?% K; {( q解:main()! @) d. P( ]4 w! O5 C
{
+ P. N% C( C5 e4 c. _) d/ Pfloat pi,h,r,l,s,sq,vq,vz;' |4 m$ N1 [4 M7 \* Q. c8 a( S
pi=3.1415926;+ \) n, ]# k; Y& z
printf("请输入圆半径r圆柱高h:\n");; e4 l. P$ f6 U+ @. v3 O
scanf("%f,%f",&r,&h);/ |$ w" S, {6 e0 l3 a3 t
l=2*pi*r;
5 p% f4 P C7 T' Os=r*r*pi;' C& g* U/ R) a
sq=4*pi*r*r;
: Q$ } l" b! U! A; f; hvq=4.0/3.0*pi*r*r*r;
; m+ i( p; `: Cvz=pi*r*r*h;8 \$ l0 A+ u2 |6 S3 i N6 g
printf("圆周长为: =%6.2f\n",l);& a& R3 S) d [# z6 K, i; _: _
printf("圆面积为: =%6.2f\n",s);9 M' z- h8 q8 m* e3 B' y5 R$ Z, X
printf("圆球表面积为: =%6.2f\n",sq);
; m8 `: e8 m/ Y* L, xprintf("圆球体积为: =%6.2f\n",vz);+ |. d; ]" U# Q
}& z5 x9 c; u: T; A1 M; G
4.9输入一个华氏温度,要求输出摄氏温度,公式为C=5/9(F-32),输出要有文字说明,* G/ l% @1 a7 N% N& m8 a- T& ]' }
取两位小数.7 d8 V! m! j1 ?4 r' e
解: main(). ^- C/ |( E' [" a, W
{
( Q, [ T3 d7 ?8 V5 rfloat c,f;% r" G# V: p( U3 v; p7 H
printf("请输入一个华氏温度:\n");, ^( t+ w! J1 e# q% k, }: C; g2 Q
scanf("%f",&f);
7 I- R; F" I- V& zc=(5.0/9.0)*(f-32);
, Y g4 p% U* c1 Aprintf("摄氏温度为:%5.2f\n",c);
( b1 }. h; L% L/ o}
. V* @6 l9 ?; h; P; o2 v第五章 逻辑运算和判断选取结构
2 X; }' ?) ^7 Q0 Z5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.2 O7 D6 g% L" V% x! R# R& z3 c7 e
main()
$ u5 @/ ?" F9 _" T/ r: K{
% X6 n( \8 u7 L" i& ~int a,b,c;; k7 J' Q( r! A5 z9 P
printf("请输入三个数:");( {4 Z$ L7 U# n# J! d% B$ e
scanf("%d,%d,%d",&a,&b,&c);
/ `/ ^, X$ [) M2 g+ kif(a if(b printf("max=%d\n",c);
9 d `% b. T- |" I. b else; l: J4 y5 `: w* P% C* _/ v
printf("max=%d\n",b);3 N- o6 u- i4 k# [- K# {
else if(a printf("max=%d\n",c);
! ]2 M1 R$ P8 M0 P' J else
+ A, [5 G* }& V printf("max-%d\n",a);
; ]4 @& ^" s) ?9 s% r}
; Q9 z0 x4 d8 Z. K9 W方法2:使用条件表达式.7 l( j0 q- @- j: B9 m0 `/ P8 d
main()
3 J: ]- ^9 Y- F( M$ L{int a,b,c,termp,max;7 f1 | h5 c; M
printf(" 请输入 A,B,C: ");) a% k& s$ {# ~) m m: q
scanf("%d,%d,%d",&a,&b,&c);) w/ N- r0 _( b2 ]/ m8 R9 ~
printf("A=%d,B=%d,C=%d\n",a,b,c);
0 j1 z4 y* s: ^4 Z temp=(a>b)?a:b;7 r# b! i6 k4 E. ?- E2 R. B
max=(temp>c)? temp:c;( V" G$ p3 Y1 ?" d, ~, L
printf(" A,B,C中最大数是%d,",max);/ S' V0 f, N* l# m1 {. P
}
- e. F+ Y! `) D/ c8 T9 d. M5.5 main()
3 [" w- ^2 N( w! d1 B{int x,y;
1 s) m! D: r* ?: n* C1 K( D8 lprintf("输入x:");: j% t0 e; l5 b" a3 R& P
scanf("%d",&x);& Y* ]" d! c$ E4 j& w/ `
if(x<1)
# p- o4 u! O6 Q* G# J2 T- K @ {y=x;$ N" f, g' b/ H7 m& U9 j
printf("X-%d,Y=X=%d \n",x,y);
. E5 m5 J4 ]; i7 b& g }# G1 e2 s6 ?" i, H: \, c( f! O
else if(x<10)0 P/ {; m9 F h. T7 G9 A+ l
{y=2*x-1;, f, s% ]# [- Z: c* D: i, _* x
printf(" X=%d, Y=2*X-1=%d\n",x,y);
- ]% |) u+ d9 c }* c4 D4 Q! v5 P5 U7 I6 r# i
else' G6 i9 O; ~) i* M" ~% j: |
{y=3*x-11;
$ V0 N; w+ R; R' b( l$ l printf("X=5d, Y=3*x-11=%d \n",x,y);2 F' u7 p6 M9 H7 W% ~5 f% a
}
; O1 Z V$ s& h, Y0 b0 l0 [}
$ m G2 ^/ {; a/ d, @(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! " a) g6 R1 z- [! P7 ]; K" X
void main()
5 j i$ a" c7 S, c2 f( T2 E{
% ^6 c5 _4 W1 g6 `- I; B" a. yfloat s,i;
! c" ~% t1 m% \ F. \/ M( U8 _char a;/ ]1 X8 b- O8 h8 S# v1 Q
scanf("%f",&s);
9 X8 o! D1 w& j" ?( U8 bwhile(s>100||s<0)
$ S8 s# {9 Z1 s" E% A z, Y. R{
0 E' V& b* o7 Zprintf("输入错误!error!");( r- e, C0 i7 G
scanf("%f",&s);
+ A P8 C+ E5 K; j* U+ l w+ n4 N2 L}5 z0 d2 z. W4 ~+ X, [( C
i=s/10;/ A, g4 A0 P! b" ?* t+ m/ y
switch((int)i)
7 f( v; A' g9 Q* w# G2 v+ I{$ ^9 Q' h# }: u" |1 E4 n4 s- t, m
case 10:
3 _" l( d' h& F5 \7 Rcase 9: a='A';break;
) i! {! Z, `* U) {( j7 }case 8: a='B';break;+ X5 y6 E. K w
case 7: a='C';break;
1 N' u9 {2 z7 }2 {case 6: a='D';break;. H/ p+ n2 x( o7 W
case 5:' j# Y) m, B; @& @
case 4:
( k: Q% p* f' ]case 2:
* Q6 Q# T7 F; J' {case 1:! p; E* n6 B+ n9 z# o* o0 d/ Q
case 0: a='E';' s/ y+ |! N0 {9 Q0 k0 v' ^
}/ ~1 O a/ G9 N7 R8 k
printf("%c",a);# A% {$ F* g p' D6 u' W
}, ~8 z& I. g- B; J4 A e
5.7给一个不多于5位的正整数,要求:1.求它是几位数2.分别打印出每一位数字3.
e8 P) S8 B& h7 X4 h, ]6 X按逆序打印出各位数字.例如原数为321,应输出123.
. D: `" R1 ?$ w1 f. j5 p/ ~main(): N2 f# q$ K. _, s* r
{
# m! u; ]4 F; q+ u; y& o/ T long int num;
; E& f6 A/ z9 ? int indiv,ten,hundred,housand,tenthousand,place;
# g- @* t9 X# L9 L printf("请输入一个整数(0-99999):");9 H7 G! Q0 ?2 L o" d, O* R
scanf("%ld",&num);" Y- }; g5 U) S) {/ p8 k* W6 R6 N: P
if(num>9999)! E7 X. c0 t' C. ^2 ^; N' e @
place=5;+ I! c7 G f! N* X: E
else if(num>999)
# j9 p/ \2 p" |* M0 g( Y' q place=4;/ z; j- t# s& B8 U7 N! B: X
else if(num>99)
8 ?+ M' ^( W. f) C* H5 j place=3;9 w4 V2 f5 ]- T; N
else if(num>9)
- K7 z- r* `! r! Y3 F% v. P place=2;9 p9 c9 @* W5 a: h+ _9 |
else place=1;
( A! t- ^$ {" ]2 ]* uprintf("place=%d\n",place);1 L9 a+ W0 l) c9 B2 A8 |# t1 p" U; h
printf("每位数字为:");: ~ f/ G3 I" U9 Q- M& k2 G
ten_thousand=num/10000;' q- y. s6 e& w. z3 O( j
thousand=(num-tenthousand*10000)/1000;0 r: W1 l% V, K7 f
hundred=(num-tenthousand*10000-thousand*1000)/100;3 V5 ]7 o% p& ]4 J" _+ ~
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
0 E, u- i$ }4 dindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;
% q V! w1 I+ x m8 j0 Kswitch(place)
5 q1 d1 D8 Q6 N1 V/ ~7 D% Q, d1 Q* t* D' y{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);* J: e" R! v. F( z% P5 U! m; A m
printf("\n反序数字为:");' m+ x7 [+ B- F5 ?' W8 f) o6 h9 R
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);, |5 a+ R' F) v* B m
break;
, v7 O k% B1 s# H; icase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
7 G2 J/ y B2 m: C- m printf("\n反序数字为:");4 I1 ~9 Z' N: O8 ?* Y! t
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);
( V. y- k9 }* ]3 G# ~# Q break;( |+ q; F0 j/ b$ _" H* D
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);
5 N( B+ |0 ^/ y4 m printf("\n反序数字为:");3 r+ {/ z5 t' U. \$ w
printf("%d%d%d\n",indiv,ten,hundred);
8 C! b+ H4 t' v1 L# K$ O5 Icase 2:printf("%d,%d\n",ten,indiv);/ g$ L6 D5 N4 |- K3 r) c: z
printf("\n反序数字为:");; s8 T6 U- L) k, b7 ]* F- f5 V0 u
printf("%d%d\n",indiv,ten);( F# g% N, h- Y6 Q, w
case 1:printf("%d\n",indiv);
: |5 `$ Z5 z9 ~- g m7 ~1 h8 d3 a printf("\n反序数字为:");3 q' O2 t; i+ O% k# v) i
printf("%d\n",indiv);; G. s% V `! Z J% q! `
}! {* ~- m' K3 r% f% O
}" Z# g! o7 d0 t- a+ r
5.8* k- a( S" T6 J V' h
1.if语句" F# U4 P) n' k' r% S$ i0 V
main()
- M9 q, n5 Q) A- i$ x4 i7 A{long i;, U; E( ?) L2 K# Z1 ^" z! F
float bonus,bon1,bon2,bon4,bon6,bon10;
" V4 u( u/ Z4 k* f* o" t6 o- O bon1=100000*0.1;/ ?! E m9 t2 [& z( _6 N& X/ z* B
bon2=bon1+100000*0.075;
5 y* I- x! C' W bon4=bon2+200000*0.05;# |2 S: i8 W6 h4 w6 i/ B
bon6=bon4+200000*0.03;- W8 }, ?2 \# ~( P8 |
bon10=bon6+400000*0.015;
8 g2 d* l' Y4 V7 C scanf("%ld",&i);& C8 |) o. u- @1 X0 J
if(i<=1e5)bonus=i*0.1;2 c" h& z1 Y. y: v/ H, f2 P
else if(i<=2e5)bonus=bon1+(i-100000)*0.075;1 m5 f& z. X: c' R' j
else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
3 t$ _ a5 u7 @4 S2 } else if(i<=6e5)bonus=bon4+(i-400000)*0.03;
- n: T7 E+ q1 Z. Z else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
* h! y6 Q; q6 j else bonus=bon10+(i-1000000)*0.01;+ J1 h# z; o. ?2 D! `4 k' e# ?
printf("bonus=%10.2f",bonus);
2 V& R9 k0 i- [* H, X, ~5 M}
" V O# C, b6 z4 [5 k用switch语句编程序! d3 a5 O& o8 m( m
main()% E# j2 i4 _+ i! e* m8 I
{long i;
- l* E( F; {( T& b: f0 I a( V float bonus,bon1,bon2,bon4,bon6,bon10;, E+ f, _, h5 H0 C
int branch;1 T4 t2 i% a1 a9 |
bon1=100000*0.1;+ `# ^, a! |# k; f7 E
bon2=bon1+100000*0.075;
0 o8 N. D9 i# K bon4=bon2+200000*0.05;
; h/ M* X' i% A/ ] bon6=bon4+200000*0.03;
X# n, j( Z; {- X2 {1 y bon10=bon6+400000*0.015;' f- q d% f E
scanf("%ld",&i);
& a/ H+ D' U1 S8 U' ?" F3 V branch=i/100000;
4 O: C- A' T+ Q! [6 {) A1 K: E3 u& E if(branch>10)branch=10;/ f& k" Y- W. C2 W0 ~2 ^
switch(branch)
( f! t4 F1 k* a {case 0:bonus=i*0.1;break;6 H7 N: {$ K$ B* D3 ^ o P9 P
case 1:bonus=bon1+(i-100000)*0.075;break;
# Z" l& H' G: b& r% O case 2:
; [1 f) q' o7 S5 l case 3:bonus=bon2+(i-200000)*0.05;break;
1 S7 ]1 L6 L9 }6 I& [ case 4:0 J3 i' D/ \9 Z8 P
case 5:bonus=bon4+(i-400000)*0.03;break;
0 S1 d" `7 e0 o S, l case 6:3 v7 F6 G: n/ c" V
case 7
# T" e8 r/ o/ D& ?3 V case 8:1 _0 y" H* g: E- D1 A
case 9:bonus=bon6+(i-600000)*0.015;break;
" P. a' ^% I, U: K case 10:bonus=bon10+(i-1000000)*0.01;/ |$ J' w7 Y2 H5 d, y9 ^4 A/ Z
}
2 I. T. \/ g0 s A% D% a printf("bonus=%10.2f",bonus);; A* |' V6 l( a$ @+ H$ H# @9 y
}
2 [/ I8 m7 _1 j8 v; z- b2 U5.9 输入四个整数,按大小顺序输出.% k0 V* S) K/ x
main()
7 Q5 U7 D+ C, q; B5 N, l& [9 x {int t,a,b,c,d;0 @6 x& y/ ~. u1 w: g, N' g
printf("请输入四个数:");0 w$ v7 ]/ k' Z9 a0 ?
scanf("%d,%d,%d,%d",&a,&b,&c,&d);% d& F* R' `3 i6 J `
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);; C4 ~, a' H) W5 K
if(a>b)
# M1 l1 V+ n/ t( i" I! Y. Z {t=a;a=b;b=t;}
0 E5 C! K- S9 I# } C; H! g if(a>c)9 _+ l7 r9 H0 t1 X
{t=a;a=c;c=t;}* z8 m6 a& A5 B- Z/ N' L# Q+ _
if(a>d). j" h2 X! o: R) w
{t=a;a=d;d=t;}
. V# j* E# R7 `4 s if(b>c)
7 o/ ?: y* u) _# Y {t=b;b=c;c=t;}
, Q) ~' Y6 ?, i+ r5 g if(b>d)9 Z1 X# b1 T5 e" ~
{t=b;b=d;d=t;}& t" a' z8 X& c# @# B2 _
if(c>d)
% C( ?) j: C. B) O# z9 Y {t=c;c=d;d=t;}1 x/ Y% y7 n& ]# u
printf("\n 排序结果如下: \n");
: y P; V; I, C/ Y: [printf(" %d %d %d %d \n",a,b,c,d);# P# n1 H: _1 o5 S4 p* F
}
( F! ?! X6 c0 }. ~, @& _) h- |5.10塔3 m6 _1 {. ^# V
main()! }, _3 x( l1 J! }' ?
{; D5 f( W1 `' `/ j# C
int h=10;
* i3 ~/ O( g0 Z9 I* [/ Gfloat x,y,x0=2,y0=2,d1,d2,d3,d4;" X2 w4 n" B; r& ]5 ?: I" Y
printf("请输入一个点(x,y):");
4 V. b5 _, Y. q, n9 `scanf("%f,%f",&x,&y);
: E# X a. `8 ?4 |# R9 gd1=(x-x0)*(x-x0)+(y-y0)(y-y0);
. w# Y9 j8 @ k: _. f3 |d2=(x-x0)*(x-x0)+(y+y0)(y+y0);
: m0 ^& i: e8 i* @7 a5 Ld3=(x+x0)*(x+x0)+(y-y0)*(y-y0);1 ^! h0 f2 W- A8 |% J
d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);* W# p& X& j+ e! {; i2 I* ]& p
if(d1>1 && d2>1 && d3>1 && d4>1)
& c. k* l6 ?* v3 Y' a+ Uh=0;3 d1 O8 P" k( ^
printf("该点高度为%d",h);0 M8 g) m% _ J
}
m4 w: C: A$ `' @/ ^6 n3 w第六章 循环语句
1 f2 D; ]8 _: @2 ?, \6.1输入两个正数,求最大公约数最小公倍数.8 I X8 E3 u0 G% ^; X' ^6 _
main()
+ L9 q# m- ^0 S6 T+ m) L/ j1 W: T, H{
3 J4 L M+ l# hint a,b,num1,num2,temp;1 ^# t$ ?9 T, c
printf("请输入两个正整数:\n");
0 Q. g. O5 ]7 b p' o2 Q0 a9 Gscanf("%d,%d",&num1,&num2);
/ y& E- u& S* \& G" _/ k7 jif(num1{
# c' Z/ l- y e, ?1 itemp=num1;, P( K4 }) }& }: T/ `
num1=num2;" ~8 _1 }$ B/ x2 b
num2=temp;7 h; x5 k; N1 L6 _. H
}
8 }' j$ ^& j' x( {a=num1,b=num2;. }+ R% |4 D: H/ @4 B) s& i
while(b!=0)
& t! S% ]2 ^2 w3 Q" [2 D$ y4 { {% B' N& s4 ^; }$ |' q" W5 a
temp=a%b;
3 a1 A# L) n8 p( c0 W$ A2 F! X& ^ a=b;, z I' X+ ?+ v) M1 Y: d
b=temp;
9 t% b4 |+ W" U3 H: n% Z }6 _2 z3 E% ~0 [4 f8 \: c; C+ \( p
printf("它们的最大公约数为:%d\n",a);9 a, M9 f. f' E D* h) b& [3 U
printf("它们的最小公倍数为:%d\n",num1*num2/2);7 `$ {9 b7 K5 r/ f$ r
}
; }. c% ]% ~- D* C* z" j6.2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.
; \$ n- f5 N; ~* G3 _% U解:9 j" d4 e5 G6 _5 b, P
#include < >$ h5 s1 y2 B. x' H7 H9 m; S% Y. C s/ I
main()/ U; C+ x, f* M
{
8 T* C" I( { b, Z# \, C5 Ichar c;: _9 e2 H; i5 J! _. N9 w1 {% I: v) b3 x
int letters=0,space=0,degit=0,other=0;) H. z, d' g" M8 k/ i% e+ F' l
printf("请输入一行字符:\n");
, u7 E( T: Z( g/ c% q* qscanf("%c",&c);9 z" K, }' @, F" ]
while((c=getchar())!='\n')! S6 X0 Q0 d3 S7 s$ \. X, @- K
{/ [9 L+ `, u7 K9 ?+ h
if(c>='a'&&c<='z'||c>'A'&&c<='Z')+ F4 ^1 F: ~6 @" L2 ?
letters++;
6 l) V4 n$ w; e( |$ Z& lelse if(c==' '): r1 D9 x2 S! Y7 ^4 P7 E) c6 y* f
space++;) L9 ]/ Z" T8 G* J- ~7 U$ k3 j/ G
else if(c>='0'&&c<='9')
; b0 U* J; g9 N; ]digit++;
& f$ z% Y# ~9 J( k" m5 Zelse
1 O ?% V, M( I% sother++;9 N$ |! L, z+ b- z
}
7 v# n. U8 K g+ {' \4 `printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%% _, c4 l" R$ u4 ~1 F0 {
d\n",letters,space,2 f6 m! Q, K* @* m3 z- x
digit,other);0 Z% Q6 S- ]: Y* U4 w
}8 d1 a4 b9 s( o0 K% a, _) I
6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.$ O+ R$ N. B2 r6 w& V# A# i
解:0 f. i f, L% Q3 J
main()
4 O6 \# |2 `2 F; J- G/ p3 ~{( n b7 ?+ u* u& O* X
int a,n,count=1,sn=0,tn=0;
1 v# s# o9 E% @9 ~- {printf("请输入a和n的值:\n");
' o8 K. R! r- }! Hscanf("%d,%d",&a,&n);0 J7 S6 e0 E7 b2 s8 v
printf("a=%d n=%d \n",a,n);9 w" F. b8 P9 f# ~& H o W! m
while(count<=n)
: U, W7 u' R0 U5 H{$ j7 R1 L; m7 l2 X
tn=tn+a;' r* D; Z& M" K9 |9 T
sn=sn+tn;
' p) y) f c0 H8 g# ]" i0 Ya=a*10;$ s' V& ?4 I! C7 I! Y
++count;! J/ E& k- n5 F `. y }
}3 N: _3 ?, I7 ]
printf("a+aa+aaa+…=%d\n",sn);% {- ~' ~) d7 N) d! t9 Y
}
+ ~& k( x! Q* j" h' }: y6.4 求1+2!+3!+4!+…+20!.
. _1 K* E( C$ ^' B! E: {; bmain()
X* b; _9 i/ ^# ^' ?: K- a{# o" x' d& j; ^8 T0 {6 q
float n,s=0,t=1;) n0 |' z' M( ]1 y. h' E. N) r+ T
for(n=1;n<=20;n++)0 x( `* V8 l3 T* _2 g
{$ }. @" [# n1 F# w; ?- l* z0 K
t=t*n;# X5 h% B q! k( {4 y/ L7 s3 d4 k
s=s+t;
4 s- _' K1 X% W1 B: v$ Q% V6 Q}: d' L8 N9 e3 w& i# l0 I& h' Q
printf("1!+2!+…+20!=%e\n",s);
g/ P7 @; v' e; l}' ~$ z. x6 _4 \* F; O
6.5 main()
! o* b v4 `* ^! }3 s{
( Y/ o+ D8 P6 D J* ?int N1=100,N2=50,N3=10;' n2 T4 [& J& `1 V3 ~
float k;& Q3 P+ S+ t6 w4 b6 Z7 @, ?
float s1=0,s2=0,s3=0;
( }5 @" c$ ~' S, Z% g' R! {for(k=1;k<=N1;k++)
4 D, W9 i9 Q$ |. g3 a0 ^$ G{7 X# k2 G. z0 M' }! A% \! P
s1=s1+k;8 b& {9 }1 o; R$ F
}
# V L" f' M" v; o4 r l! bfor(k=1;k<=N2;k++)+ F' f2 I8 n% u7 G6 e+ `( z9 [
{
; Q5 h" W7 ^! z% U( j& y2 p# |' h1 v% Ks2=s2+k*k;
; a2 Q3 c- e9 A* ]0 A/ k ?}
- ]5 M/ P: x- L# @4 |7 Rfor(k=1;k<=N3;k++); g- u. K3 B+ ?
{4 j' ^0 C) R- z6 c' K) u9 f
s3=s3+1/k;
9 _1 I2 J u% b( k* y1 u}
5 P4 Z$ ~) k* U9 [( X8 Iprintf("总和=%8.2f\n",s1+s2+s3);: I2 W9 k1 M# R) z; Z0 l* ^; h
}
4 Q3 A2 E+ Y% _6.6水仙开花
* l. h+ s0 B; @% Z0 x- V: a9 t, b/ x. Umain()9 a, @# y% m. w- x
{
4 t9 M* |' X1 }9 u& v$ k6 P) O8 Bint i,j,k,n;
8 [- B0 g2 ^6 S# G4 y6 V0 I4 Sprintf(" '水仙花'数是:");
|$ X2 `' M- {% {for(n=100;n<1000;n++)$ f% G6 P, f+ N
{
* K9 ~& N) {: i; m% C' a* M- [i=n/100;
4 Q: Q" J+ R7 H$ V: O& Sj=n/10-i*10;
. s) D' Y6 V- T3 @k=n%10;
* b. ] f! C7 J! eif(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
) j( Q/ T7 X! y{
0 s- a: p; a- L" gprintf("%d",n);4 ?) m8 g$ U( X9 J4 N7 u; P
}
2 A+ {! n% u/ R% L" b2 f( F}
P8 z! h9 u! p+ K7 h1 D/ M) Kprintf("\n");* ]: X) G& P: C% j; K# F3 G, V5 C
}8 l) O0 S% M4 U( R0 q2 x/ A
6.7完数
$ x# y6 @1 h3 {6 o" P" I4 Kmain()5 z0 ^3 F" G) x: z
#include M 1000- e( L" Z4 _* M5 y% x8 u& E
main()
& L7 q; d( d" H: O{
3 l2 U5 h8 G1 F% {$ N) y8 j' Zint k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;
7 j2 }9 @- g, p" [' G ?int i,j,n,s;4 M/ L& y1 @ O( W
for(j=2;j<=M;j++)" B1 d3 q" q0 i6 y
{
; X8 a% k v8 \. N0 x$ H+ fn=0;9 ?4 M- y7 q# b8 _8 T3 k: g5 d4 f
s=j;
0 h+ @! M& A# P* ~( c, r& G) pfor(i=1;i {
' {6 Z5 @& a2 M( B' k E. g7 eif((j%i)==0)
3 a- D7 g' q' v+ A/ n. J {
8 m) s' c3 n6 S if((j%i)==0)5 ]5 ?4 Z; J8 |0 G; J/ U
{
' ?8 W+ x* n: ?) @; q n++;+ s% `) [1 l8 a, @0 v9 k7 Q
s=s-i;
2 i. m+ ?/ z/ [ switch(n)4 w2 c0 g f* y, g% n3 ?
{
4 y0 E& e& d2 O* t) _ case 1:3 L, k5 H9 V! s0 o
k0=i;+ H1 b' V- m3 x! i5 A P& V
break;" P/ Y$ A9 z7 o& N% P$ x
case 2:8 }6 U. ]% t8 `
k1=i;
& a( L, F7 e5 g* j2 b break;) J9 g; D f' {. V
case 3:
8 \3 m+ P V6 x k2=i;$ h/ M" \& ^- ^- X
break;
* ?1 y2 v* P' {5 F2 a: \ case 4:
7 u2 h7 E: w& B3 x6 v% I) }% W9 L k3=i;/ u7 Y- v- I: R. {
break;* p; c: w- ~$ X8 h3 d0 C4 l
case 5:" {7 i7 y4 v8 q& }" [
k4=i;
+ y, j* d) `# C0 W, s, e break;9 }! Z' n& R! y; w6 ]
case 6:* m8 L" B7 [, G$ I( X
k5=i;% Q* V' z) N+ ?, C6 \7 z7 H
break;
1 v' D% B1 }# L2 F, D/ I% c case 7:
) e, J6 h9 E2 B3 k2 Y2 H k6=i;
2 u; W2 F$ a3 i) ~% O ^6 n0 A- I break;+ Z7 N) e/ B/ i7 Z1 O$ \, A
case 8:/ w5 k# H; c" w& `$ g/ R/ v
k7=i;+ [5 `, S/ m5 D V: m
break;
! T+ a. ^' e% n+ G7 W' q1 S* w9 L case 9:
, l a' b( [- m* x" S* A! f k8=i;' ^9 u9 u( M+ p5 A3 r, L
break;: j# R/ {; e+ t) ]- n& u+ d
case 10:$ ?- H4 F) ?$ C3 }7 N, |, i
k9=i;% v# W6 q1 t# P7 d: K
break;. f0 I6 v8 o5 x- d3 p" Z& h
}" G9 m7 r0 z2 k7 [% O
}
6 q( t% B3 e2 j8 s: W }+ E7 a3 J! X! B8 c; a1 i; v
if(s==0)) i# i+ P% F0 }0 u8 ~2 p
{0 G" r) c+ j+ ^+ K
printf("%d是一个‘完数’,它的因子是",j);
" C% l$ z( {* k1 q8 Sif(n>1)
: J+ Y* q3 d8 q+ s printf("%d,%d",k0,k1);7 L# X3 i% c6 }9 s
if(n>2)
6 d3 u& g X- p9 e( m% c9 n printf(",%d",k2);# P( J, N1 S% F% U- V O( v
if(n>3)
" h9 z' ^% u0 R printf(",%d",k3);
1 c) {, d7 N. j2 c7 q, `) ]if(n>4)
# `$ s% W% [& P& p) C( Y! F printf(",%d",k4);
2 O; U" E& ~, V4 Z- U. dif(n>5) J6 o7 @6 H4 ?; m& L8 F
printf(",%d",k5);5 ?1 @, e; h) J5 G
if(n>6)
0 Y% q% o# O3 R6 k0 G& t; {* X0 g printf(",%d",k6);# _# g' S% d8 Q& H) V0 _0 k
if(n>7)! S8 H- y# S! s, ~7 W& O0 \* G
printf(",%d",k7);* C! @, `/ ?1 }
if(n>8)
8 n" r* @/ l% S9 q5 B printf(",%d",k8);
" u: ~2 R% m) Yif(n>9)6 X" \# D* D# K5 T8 ~
printf(",%d",k9);3 g7 A9 U6 A8 [7 J
printf("\n");
# ?( Q, ^: I" e9 M7 r( ]0 l+ A }
R ?% \% G2 v- X$ ~% k& W$ Y5 }}
& Q {- @* v* N; o方法二:此题用数组方法更为简单." H. \8 d# v+ a
main()
4 _; W8 a+ O _{3 v3 M. A6 _% m: r$ ?
static int k[10];- B2 A2 |; r7 L
int i,j,n,s;
7 }$ m4 J8 n2 @0 E: afor(j=2;j<=1000;j++)
1 C& B6 _# T2 O1 F{6 y! F& {2 E. H/ |2 Z% f5 @# b
n=-1;. R1 W0 X6 Z# e/ ]
s=j;" O' m9 |4 ~4 ~4 k) a/ j' ~
for(i=1;i{
! _: u/ k( K( Kif((j%i)==0)0 }5 G) ?) P& N
{7 G4 g; C% d8 K- K; w! |: Y1 s
n++;
, `; |7 F( S: Gs=s-i;0 m9 f5 U- a, o5 p3 ?! }
k[n]=i;9 W2 j& R3 d4 _0 e9 b; H
}
% ]5 q( r1 ?5 A; E/ d+ I T }: O: p5 k' ]' o" L
if(s==0)
" s; U; e: b* N{
1 P& p, k% o( b6 tprintf("%d是一个完数,它的因子是:",j);
6 m1 ] s! ^) A- ]; N; s. j: b/ Wfor(i=0;iprintf("%d,",k[i]);. Y y& m4 W1 p; r I
printf("%d\n",k[n]);- X' H9 F* C6 Z% ?, W- v
}
) @6 R$ K! x$ O5 z}& d6 ~* K- R$ ]( H* F4 x
6.8 有一个分数序列:2/1,3/2,5/3,8/5……求出这个数列的前20项之和.
' _# m! u: s4 ~/ W# T2 J! d" {解: main()1 O( B" f \+ x8 H, H. L9 L
{6 o! H% J9 P G
int n,t,number=20;
9 Q% a6 ~" q) D( w4 }; A; Nfloat a=2,b=1,s=0;
$ u- F& y9 n1 jfor(n=1;n<=number;n++)4 R0 W. O% I5 o' _
{) Y' u$ b4 z7 L" r! D
s=s+a/b;& M: @* `' V' |0 J
t=a,a=a+b,b=t;# N1 y1 j* h5 R$ l/ ~+ _
}
) @9 ?# r' X* t( e, }9 |4 Q( Mprintf("总和=%9.6f\n",s);8 @; b4 B5 y; y+ l! n9 K
}" |7 d# p) Q; D& b
6.9球反弹问题
+ z7 Y% j7 T+ r" [+ l; Y1 Jmain()
F! N' Z! c% n u; Z1 Y{4 o6 L( [! b8 D" {# t
float sn=100.0,hn=sn/2;
4 Z0 y& u+ Z' p+ k2 l9 qint n;
' X5 x% w' p% f9 j: Tfor(n=2;n<=10;n++)& z* D! N9 S F- Q( u4 M
{% S0 V1 k9 e, P+ X$ J
sn=sn+2*hn; s6 [" y1 R2 F
hn=hn/2;% o+ v3 `* p; W$ F& R% H
}$ K8 G: I, c7 U$ i; f
printf("第10次落地时共经过%f米 \n",sn);
) \' \% Y8 `4 c. w$ M) o, rprintf("第10次反弹%f米.\n",hn);3 p5 ?8 Z% @1 O C$ F& Q) H8 _
}
; W. |% `# d) Q- p5 A0 s6.10猴子吃桃
$ p( H4 \9 V/ F, tmain()& h+ m0 ^) Y/ L4 ^6 ]5 c h" _7 W$ y8 K
{' z; K0 n8 U7 M" I) p
int day,x1,x2;
3 |: {" K H( B8 `* pday=9;
: o/ t4 ] y" bx2=1;
0 S7 H) `4 F- ^. U5 R8 w7 s% V N/ C7 Pwhile(day>0)
3 H, G0 V5 p( l& b{9 Q6 i+ Q! V1 A3 Q
x1=(x2+1)*2;
- V8 o/ S7 d' Wx2=x1;' d9 X1 Z8 [# t7 r1 I* n2 w
day--;$ `, [" d8 P7 S1 d
}& }$ H) h/ T C) v* E& t
printf("桃子总数=%d\n",x1);7 S ^) a F# z' y
}
& ^* Z' ^7 ]# `& U
' B) k/ b1 V) P3 L* G6.128 o- @( f4 R1 Q# U' D/ S# T
#include"math.h"% r: u4 X, r' f& [- l
main()
# I- ]: K- w5 D7 D1 V& I{float x,x0,f,f1;6 K. y+ l3 K. Z2 `3 `/ O* }
x=1.5;
" w# [; h; H! L* f do# i0 z7 w; x- F* m) o) n
{x0=x;% J" f0 n+ V, B; k1 A
f=((2*x0-4)*x0+3)*x0-6;7 e( C9 k H9 }- q+ i, j' o+ l0 l
f1=(6*x0-8)*x0+3;
6 A0 b! [1 U/ ]2 E( [ x=x0-f/f1;% @$ y7 x5 g/ ^' @7 N
}
3 m$ ?$ T1 B4 w X, ?) u/ Y8 ~ while(fabs(x-x0)>=1e-5);1 s+ o5 B/ `5 L6 a
printf("x=%6.2f\n",x);
]- y1 q* }/ s ~. ^1 J+ k}/ e# Z% j: z s! ~& U6 Y9 |
. T+ n# F( D* O$ f+ b; G
6.13
, G q8 g5 H; g# [* C0 W' A% X#include"math.h"3 t8 Y# \0 k& N" v, _
main()- t. C) n. I2 ^, Z1 \' i
{float x0,x1,x2,fx0,fx1,fx2;
- R; @! V+ g- n8 n6 v/ u1 F) _ do% [- c5 U* o. S6 q
{scanf("%f,%f",&x1,&x2);
% t$ x+ w0 K+ Z fx1=x1*((2*x1-4)*x1+3)-6;
# }6 W8 o9 z) }+ s ?/ P; w+ G fx2=x2*((2*x2-4)*x2+3)-6;
( Q% W8 z' ?0 H# F7 @- Q8 U4 r } H4 \8 ]0 e0 J/ Z) {+ D
while(fx1*fx2>0);
9 y/ `; _& Y2 y8 s) O- N# ]; O do
9 J! q8 `3 c% ~1 s9 V) H1 h {x0=(x1+x2)/2;( u) s( X3 H6 I* p7 I7 o- v
fx0=x0*((2*x0-4)*x0+3)-6;
& O% y& [4 M9 @2 I4 e! Q# O& y if((fx0*fx1)<0)- A& ?7 i9 X) w) P$ Q2 F g
{x2=x0;7 g* `" |9 A6 N5 v
fx2=fx0;
+ ~: z1 n: r* y9 B, _, V }) R. ?8 z% \* w) M
else, O) D+ j |. N0 w0 H$ X( J
{x1=x0;
$ ?* i0 m: \" D! o fx1=fx0;3 j# C+ e" s. ^; X9 W6 {2 F% A8 G
}! U( r: P5 A& ~- [; v% ^
} ?5 e1 G0 x" ^$ ~0 }
while(fabs(fx0)>=1e-5);
9 g6 r& {3 R. \3 F, k6 ^ printf("x0=%6.2f\n",x0);
. o) f! l. N: Y' d* x' g% P}3 I% ` ^2 J! _$ V0 p6 N
6.14打印图案
! s% L7 t9 g! [% g0 m$ Z' N. ^main()6 Y# Q) ? Y1 S7 y! Y5 i7 ]9 F
{int i,j,k;
7 ~- Q8 e, U$ h D9 u4 R: A2 H for(i=0;i<=3;i++)& b6 p! y+ `% M7 p8 A
{for(j=0;j<=2-i;j++)1 D0 Z l# ]8 U0 D+ F3 k" e
printf(" ");
# ^& B! l" ]4 H+ g2 y" k# w for(k=0;k<=2*i;k++)
. I8 i- T) i* _! _+ m printf("*");
1 t/ V P; p% @9 f% {, e printf("\n");
3 @& v* K4 g& D+ ? }# {- u. d% V; B) b5 t% h* y
for(i=0;i<=2;i++)
- O' n! v4 \, X/ E {for(j=0;j<=i;j++)
0 `4 i2 M! i+ E2 @1 f) X) D1 h printf(" ");9 w& |$ y6 n6 ~; d. B+ @" e
for(k=0;k<=4-2*i;k++)
- J. J: |) o: |: D6 ` printf("*");
" c' E {# E4 A2 b4 k" x printf("\n");
. ~; B; [- F A9 R! h0 s- i0 s6 _ }
' e* V. i- Y' h. H}+ o( L- w* ^' y4 T! L$ q+ v) m4 S
6.15乒乓比赛
0 z& J$ w3 V) O2 z. w |main()
) e2 S5 \3 o% |* i- t; [{; h% I9 D8 u! @& p6 |. f
char i,j,k;
\6 ?6 r- } T0 n# [. Ifor(i='x';i<='z';i++)
& X0 f4 g& f5 q/ i3 |4 h* z1 y {& R% Ufor(j='x';j<='z';j++)
/ ~% ^$ t+ a9 R8 F9 q3 I {
; y% V; T% v* j2 y i8 r( h5 n3 Hif(i!=j)
5 Q# O# l9 G3 s8 r- ^8 Nfor(k='x';k<='z';k++)
: D# e( w! G& i6 }2 H# ~" U {( Z4 u$ J- g, N( X! V
if(i!=k&&j!=k)5 H( a s( r' n
{if(i!='x' && k!='x' && k! ='z')
~) E0 \7 @, y6 U" R: }& b' lprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);: S4 k" y- o6 G8 G
}
9 {& y- m/ S' Z; Q7 t; @/ U }
# E. ~$ R, O4 N5 g d1 [+ j) v' W" E5 x }
6 Z+ T7 i: X" |# p- s}5 \* R9 O% O, N- K
C语言设计谭浩强第三版的课后习题答案$ p2 `& D$ C! s4 {7 n& w+ v' E
7.1用筛选法求100之内的素数.
6 o2 l. j! v- R) d#include
+ Q% S+ {( p( o1 ~; C+ d. P#define N 101" y5 w5 R, }( Z
main()# n; R# E" H S. ^2 k# |* Y: q
{int i,j,line,a[N];( d: M9 m# y- W
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)- I- q( |3 L Z8 o, T+ w* V
if(a[j]%a[i]==0)
: S7 G1 y- E& S8 R, H/ r- g) E3 b a[j]=0;; J6 Y$ C* U! H) n
printf("\n");( B" e W# D- U9 G- t" c. G
for(i=2,line=0;i{ if(a[i]!=0)' s" Z& u7 O( i; \
{printf("%5d",a[i]);4 {$ o( x% q: ]
line++;1 ]* A2 L b$ x3 u7 b
if(line==10)8 M, ]$ l) {$ r# K0 t
{printf("\n");+ x) |1 m' X8 I/ h- ]* G! e
line=0;}/ P( ?* l. p- `. _' Y3 l2 \
}
! S3 W% j+ o& M6 ~7 O1 m}: f, z3 e, K0 M _
7.2用选择法对10个数排序.
' s& }7 g+ M1 X3 x R& _) @#define N 107 t; ]. Z1 Z) M) R+ O( W3 P
main()
1 E" g2 r- v# _' x' x{ int i,j,min,temp,a[N];
* O/ W5 E5 o& b: qprintf("请输入十个数:\n");
/ X: m4 h- S1 o) L/ \for (i=0;i{ printf("a[%d]=",i);
/ `0 D. h* c1 _' @& M/ ~ ]; ~ scanf("%d",&a[i]);
3 t3 ~# ?4 ~5 u# F1 ?" Q, v: r}, o3 [' O$ |) m! z
printf("\n");
- y. m+ |$ W1 R; H5 q7 Kfor(i=0;i printf("%5d",a[i]);; f3 U9 Z& d% F9 P( t# S/ s. L
printf("\n");0 e- e+ B) N; z* k
for (i=0;i{ min=i;: |8 p0 Y; k, M
for(j=i+1;j if(a[min]>a[j]) min=j;
3 B3 Z" }% H4 J' }! B( b1 {( S2 E temp=a[i];
1 I2 ^7 i2 @( s+ \ a[i]=a[min];! ~; M' N4 n8 ~; s0 Y: I
a[min]=temp;
; y3 o+ x+ J" Y/ e}" ^# K" n* Q, w3 q3 I
printf("\n排序结果如下:\n");- r* l. ~& k& r
for(i=0;iprintf("%5d",a[i]);7 M$ o% ]% r* r2 W8 Z& K2 |
}
& N9 N ]5 Q2 Q. B0 O% }7.3对角线和:
5 n; E8 v* o0 Mmain()& ^' A+ Q; n2 p
{
, }( |, A) _" r" A6 r$ f# kfloat a[3][3],sum=0;
" ]1 W3 @9 g5 ?6 k/ Y$ j7 wint i,j;
. D* b* b& V4 B* x# hprintf("请输入矩阵元素:\n");
4 D, x3 ]- i. Y4 r5 P u g* ffor(i=0;i<3;i++)
& k* w( e, S. n& P; T, u( t" [ for(j=0;j<3;j++)
- A4 p K; k/ ?: e# G# Z( B/ L3 U scanf("%f",&a[i][j]);4 D* H" p3 y3 H4 O
for(i=0;i<3;i++)" w3 I, T7 T( e& {8 y1 ~
sum=sum+a[i][i];( m% T1 c( {8 R. Q5 N! K
printf("对角元素之和=6.2f",sum);. \( R' r+ s( \
}: X0 G8 z$ }, g1 I7 m
7.4插入数据到数组# v8 t+ f% D) f% y4 l* z2 x* q8 }: s
main()6 v5 O0 Y! r, y8 F
{int a[11]={1,4,6,9,13,16,19,28,40,100};) H0 f$ W2 N; J7 h" q
int temp1,temp2,number,end,i,j;6 i! U! |- U* ^- [+ {, s \' p
printf("初始数组如下:");
4 J& Q' _7 N/ ^! |9 Gfor (i=0;i<10;i++); t8 W# j5 p9 _/ H8 U' H
printf("%5d",a[i]);! Q! [7 z& P# |2 l" h
printf("\n");' ^- q& a) C% c, \' B! T( l% ~
printf("输入插入数据:");6 ^* l9 S# |+ y3 v
scanf("%d",&number);6 I9 o: M! f8 ?! W3 \
end=a[9];
1 n5 V" _3 D4 Zif(number>end)! |& v; ?$ [1 `- r" O: c
a[10]=number;
# D$ s9 [2 k, y( t* c. T5 b4 o) d; ~% yelse( w5 z; q4 I5 }3 z
{for(i=0;i<10;i++)# j7 }/ b0 ^+ e& Z( C
{ if(a[i]>number)
2 T) Z7 N" H4 j {temp1=a[i];5 U% Y5 W+ ~5 W: u' M
a[i]=number;
0 E \) E3 _' T4 C# u9 [ for(j=i+1;j<11;j++)1 w. b! R/ {, T6 @$ o* [
{temp2=a[j];
. X b" I) z- A: x5 h a[j]=temp1;
6 ? K% @0 `. X! } temp1=temp2;* U y8 l4 |) F5 m! `
}
. q$ _) E4 Y* F0 z! x$ f& E break;
/ N& V( [5 o& r }& q* x! I$ m( x$ Q* w; w
}" J7 Q7 l. m0 P& ?+ }
} d6 N; o$ S% k' O' }
for(i=0;j<11;i++)) L' {" d2 M; ~# z
printf("a%6d",a[i]);: r* K/ \* u% m6 [8 B8 X9 @6 H [" y
}
( l- X0 j4 [$ P, }- z6 b7.5将一个数组逆序存放。: E* p G. f5 A; @$ x
#define N 5
2 [% z/ N* c" ~4 @- j9 f2 U' V* Smain()
+ M5 i. p X" W. s. ]) H{ int a[N]={8,6,5,4,1},i,temp;" ?4 l' Q! Z( L/ ^
printf("\n 初始数组:\n");$ w/ m( g- s/ r. ^/ G& c5 k( ?
for(i=0;iprintf("%4d",a[i]);4 H- [, P5 n7 {
for(i=0;i{ temp=a[i];
/ l2 ~: o+ z1 V. I6 M, N a[i]=a[N-i-1];
, d% E6 a$ @$ g/ S9 l$ P* q' x r( x a[N-i-1]=temp;8 i9 x+ N& p% D( F' b- [/ I/ e
}4 f9 `! L p F& Y2 N
printf("\n 交换后的数组:\n");
2 f( M: \9 d# S& y- }7 @# ofor(i=0;i printf("%4d",a[i]);* o& M% F' W' N% k" `( z1 `8 n3 m9 M
}
$ P1 u. I* F, E6 A' a5 \' G7.6杨辉三角
7 p" `( @; R6 T2 X9 r' F" N#define N 11
8 D& \; j4 X; M, n3 W. f; [* Qmain(): Z$ ^! U" I. `" C1 d$ T( T" O. b8 W2 t
{ int i,j,a[N][N];1 I8 x8 G! n B# X
for(i=1;i {a[i][i]=1;
. _+ Y7 n+ R2 x! ? y) j- P8 u a[i][1]=1;
& Q( y6 U# S- z; i! n }
5 J% i/ p, ]: g; g$ H6 P! o ^ I for(i=3;i for(j=2;j<=i-1;j++)
) \+ W* k! o- k% r' s9 c a[i][j]=a[i01][j-1]+a[i-1][j];
) N* q* l- F4 N2 d* j) \1 E$ y for(i=1;i { for(j=1;j<=i;j++)
4 X; u7 c' ~, a8 o printf("%6d",a[i][j];. f6 c0 Q- J3 r- \ f
printf("\n");, u3 h3 l# w9 f8 A, t9 @. ?& S
}1 `3 R+ i% s3 G; s, {* N% p
printf("\n");# V( G) {. ?. }4 m2 }- }5 C
}4 [; x/ q( F9 a7 B! E+ ^
7.8鞍点
2 k+ D9 k( @$ B, M1 S: Q#define N 10
( G/ s* {" G3 E% n9 y#define M 102 \; q# T2 e' r
main()# u9 T0 o, i5 x2 J+ ~: o- d" M
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;
2 W2 @$ R9 ^# i6 q% p5 D$ w printf("\n输入行数n:");
) A$ C% U" J( i; [" W/ n; i scanf("%d",&n);
$ T9 G8 Z1 I( s a t printf("\n输入列数m:");
% B( |( W5 C0 M( B a/ _ scanf("%d",&m);! D( P! U+ ]1 i% z- Y
% ~- B! f* i( f4 w+ m for(i=0;i { printf("第%d行?\n",i);
5 q$ Z, E2 w5 T6 ?0 w3 ^ for(j=0;j scanf("%d",&a[i][j];! X: Q! r1 Y, @1 p1 n4 i
}% ?8 s/ ~! F. ^4 k' C4 T
for(i=0;i { for(j=0;j printf("%5d",a[i][j]);$ ~1 l# |1 b+ Q" K
pritf("\n");8 x( w* K% @4 {! U) }" G: w
}
0 G) T& i$ k( X- ]# @4 n flag2=0;' M- {7 O* e& U, V' Q9 G+ o$ B/ |$ n
for(i=0;i { max=a[i][0];: w6 B" p; i8 o/ D
for(j=0;j if(a[i][j]>max)
5 ]5 Y$ U" m$ b' H) Z" ` { max=a[i][j];
+ R, j, S* S- V/ V1 t: z8 C+ g maxj=j;
# Q/ w" J* ^1 R' v; ]/ X }
2 G) X7 G& T9 j a/ G* } for (k=0,flag1=1;k if(max>a[k][max])! N( [' c+ r7 O/ F9 K
flag1=0;4 V/ W3 l+ r# d' i0 `( @
if(flag1)
: f$ H0 F* |' p) l4 w { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max); B6 m* {) R! d9 \# r$ W: P# T: S
flag2=1;
! |0 o9 A) }% q2 Q* M5 |6 M# j }
7 O: H, @) I6 P( i5 m}
1 d' x. F$ ]1 e, ~9 Kif(!flag2)
( {; a9 p4 S, A+ d0 G2 |; s) d ~ printf("\n 矩阵中无鞍点! \n");. M- ~: l% d6 P% d# D
}6 Y( l! ^% R6 j/ x+ W! I1 W
1 c% G' n6 x0 O" L1 [
7.9变量说明:top,bott:查找区间两端点的下标;loca:查找成功与否的开关变量.4 C% x) n+ G! r1 ]) t
#include
* u! [2 B: G+ d. L#define N 15
9 Q0 ^9 u( m# [$ [ |main()
o. ~ o8 ]# P( \{ int i,j,number,top,bott,min,loca,a[N],flag;* M$ h# x* C& t( j$ n4 R
char c;3 k% u, S; Y" Y( ?! d4 N- E( X. |
printf("输入15个数(a[i]>[i-1])\n);
+ l$ I- G$ R* C$ p- t- J scanf("%d",&a[0]);
" B: l0 {2 t6 T5 V5 m j5 \4 X i=1;
+ b& _4 O) N3 y' s! e while(i { scanf("%d",&a[i]);
h3 L8 w) v0 |9 ]/ k9 @+ f* m if(a[i]>=a[i-1])
: _: A7 {, R' x" g* |' V/ w i++;
# ?' n& F- X3 m, L esle
7 A0 n) w( t1 E2 ]: y1 F( W {printf("请重输入a[i]");. `- A9 l/ d& U. A% P/ W
printf("必须大于%d\n",a[i-1]);& C9 x# h7 x" s/ C* C! C8 X
}, H: e3 ?6 E2 G2 h! U# I) W
}
, v4 h% H( B9 C0 A. Z1 J1 L* h, i7 g printf("\n");8 q# p- i) n* A' A% A3 S4 m( A
for(i=0;i printf("%4d",a[i]);
1 e. g2 V0 y+ n4 l3 S printf("\n");1 c8 \! \- Y0 c: r4 f
8 O n2 q ^/ [+ E
flag=1;
; h2 M) H; {. i0 z/ t$ E9 [ while(flag)! Z# K/ |, y |; z* ?2 r
{
8 d0 y) f; i6 @# M" \/ N printf("请输入查找数据:");" B. g- S: x8 X U P& ]) g" A
scanf("%d",&number);) Z, w, O8 T. a: m( Y
loca=0;
# j }9 v9 H- j! A, z# z- f top=0;
; Q3 A5 k* y2 k/ N3 P0 v bott=N-1;
& N4 F4 P2 P: |; t if((numbera[N-1]))3 P" G I( m3 _
loca=-1;
, [$ _8 u3 Z# f& _' { while((loca==0)&&(top<=bott))
" b) ]6 \/ @3 W { min=(bott+top)/2;
. {6 W) e5 x4 X# ^ r9 { if(number==a[min])
& M1 G% t2 o2 g8 v7 u/ E; U { loca=min;
6 K$ ]+ |- J+ j: K% \. v' Q printf("%d位于表中第%d个数\n",number,loca+1);
: ]5 x# q/ |9 `; S) j* ] }9 J& }) z+ k( v# v
else if(number bott=min-1;9 n0 v5 y" t0 t. K- x1 b* L
else& j& k! H8 ~. w5 s& R
top=min+1;' T) l3 x+ b- v- S; M- t# z) n
}
/ Q. M/ |; S. K' M! f/ o- }: e if(loca==0||loca==-1)0 V& i3 p, W8 i3 _1 x
printf("%d不在表中\n",number);
& |; S/ ?9 c5 f7 H5 Y7 R printf("是否继续查找?Y/N!\n");
- J2 b {+ ?" j! X7 V: S: q1 z1 ] c=getchar();1 F& k+ W) x/ U( Z3 }
if(c=='N'||c=='n') D6 \* ~5 ?+ d; b. m0 M0 t$ {9 l j
flag=0;! ]6 E/ t0 @/ \
}
- ]& R, M! U. B$ E+ p% D t}0 D0 M. l8 H6 i% h
+ \% \6 I3 n9 X$ `) j& p) C/ f
7.10+ g8 @5 j; y5 {8 {/ o
main()
# w" a) j) k1 l. Y: b( h{ int i,j,uppn,lown,dign,span,othn;
' x Q4 W* E/ Q8 R# o7 d6 y char text[3][80];
0 X/ i$ g$ J4 Z+ W! o' S+ A uppn=lown=dign=span=othn=0;2 p5 v6 z! o. m1 ~1 s4 ]
for(i=0;i<3;i++)
5 p1 } i, z$ m, O1 s; K2 ~ { printf("\n请输入第%d行:\n",i);$ n# T5 N# i8 X. n9 b0 J
gets(text[i]);5 C# a5 t2 B8 e, H# T
for(j=0;j<80 && text[i][j]!='\0';j++)
7 u% g. G6 Q9 `! @7 q8 x {if(text[i][j]>='A' && text[i][j]<='Z')9 t0 J$ r: [) A% u6 f$ n+ ?
uppn+=1;3 }( ^ J, @5 `
else if(text[i][j]>='a' && text[i][j]<='z')) P( c4 M6 w" V P
lown+=1;
0 u- m- ~5 ]# O, W5 `* N else if(text[i][j]>='1' && text[i][j]<='9')
( Q# V, L4 W% s1 \* X2 Z3 r# | dign+=1;& h% m' F& k7 p, I u: n+ v
else if(text[i][j]=' ')
' z3 v+ g4 q% D4 t span+=1;1 t$ k% z! \ B& y- r
else
$ Z# A6 R6 L% [4 Q1 L othn+=1;
- A2 f: l o! W# ]: N0 d# m3 E1 N }
, m$ @7 T3 D" i% s! K; ]; _3 L }
( f' ]5 z6 r7 [* ^( V for(i=0;i<3;i++)
) B4 s$ o" t! K! |9 E: ~ printf("%s=n",text[i]);
; k* ]& p3 K, V; H ?% x printf("大写字母数:%d\n",uppn);
$ y% r, b* w" j z1 W printf("小写字母数:%d\n",lown);8 u' V7 k. K' u( |7 d
printf("数字个数:%d\n",dign); g: b' h5 M; L' G
printf("空格个数:%d\n",span);$ l: n: q; j7 L# o
printf("其它字符:%d\n",othn);
1 q% j* N, T$ n2 \( Z}
0 p3 e L& J6 o0 R& E+ j! k5 ]' m/ h( L3 x% T& J! h
W! E9 e6 y/ p- j Q' H7 T5 J7.11
+ b2 |; @% l8 S' [) Z _- Bmain(): \+ J2 n2 D' n6 [
{static char a[5]={'*','*','*','*','*'};
7 y% j0 D1 N; @/ F4 m7 M, K' Q3 L int i,j,k; r( r6 B' D- Q, A0 k0 V. A# V
char space=' ';
% W1 J d- r {; N for(i=0;i<=5;i++)7 Q' ~6 c( ^: n+ ?
{printf("\n");
~8 `, F. T( X* _+ G for(j=1;j<=3*i;j++)5 Y; L5 i# W6 z
printf("%lc",space);& _/ f: I3 b* K7 }+ I$ i
for(k=0;k<=5;k++)) m x6 L% u1 B, J/ y+ J
printf("%3c",a[k];
3 `3 o. l i# x: ~. q5 a3 s/ q }
9 K* f" x: d1 X+ T, Q! R}
, ~9 W* P( f3 q$ Z( ~7.12 Y$ u# ?& J3 [' S! I7 I' C1 w
#include2 F) U: Z, `$ ?( q% i. L h
main()
% }* T) T" q6 ` E% L{int i,n;/ P9 C; z, c5 }9 y
char ch[80],tran[80];$ `) J! B! D) b" g9 y9 I5 N
printf("请输入字符:");
5 N* n- j/ W- c: Y( E8 s" f gets(ch);
( |7 a& y4 p6 R, g0 fprintf("\n密码是%c",ch);
( x+ h- v3 D% H0 G9 Zi=0;
8 b C" t/ q$ c2 qwhile(ch[i]!='\0')- h3 h& }( o8 \0 L0 x
{if((ch[i]>='A')&&(ch[i]<='Z'))& _4 E9 n6 ~6 a) U0 O- C# t5 }* ], }
tran[i]=26+64-ch[i]+1+64;
! l1 g) g7 `+ V8 d; P2 jelse if((ch[i]>='a')&&(ch[i]<='z'))0 F( U8 b0 Y6 m' f$ Q1 w
tran[i]=26+96-ch[i]+1+96;( D% M& |$ W- i
else% G5 [! o" s' E1 J6 J: O: B
tran[i]=ch[i];% }1 U, { O' g, r6 b, |
i++;# r2 l9 E M8 U. b6 Z2 [$ l
}6 M0 s! k5 V1 O; _. {- I
n=i; _* D# D" K& \* N/ }5 `6 `& `
printf("\n原文是:");4 A% Q8 z7 F( i5 e- X
for(i=0;iputchar(tran[i]);
* T% q$ l8 c. @+ s! m y& C} O$ C$ v: m/ L% v9 G
7.13. C+ ^% T) u3 h3 {$ [' d7 a
main()' K# \& D8 G, \, n
{
( Q& ]+ x# d9 L char s1[80],s2[40];
3 a9 x% X7 \' P9 m3 A# i, Q int i=0,j=0;
+ E* Q% b: B% l printf("\n请输入字符串1:");2 b0 f! ^# v3 u9 o' [) t! x2 p
scanf("%s",s1);
3 u- N& ~* h0 w9 H$ m printf("\n请输入字符串2:");. ^6 o$ b% P# R, H
scanf("%s",s2);5 p3 I' R) {2 P4 ~- Z* u
while(s1[i]!='\0')# a$ X% B0 }* D3 I% Y! k, B
i++;
) \& k8 W K/ B U! n/ kwhile(s2[j]!='\0')5 ^% o3 c( g' ~( Z" o9 Q' `
s1[i++]=s2[j++];
* s! F% K; E; S" q7 ts1[i]='\0';- h ]/ b+ t( q# X" G. @
printf("\n连接后字符串为:%s",s1);
! x* G9 b/ ?: b9 j u" q+ D0 M) G. {( p }
$ F8 K' @" w2 q
# t/ @8 u% d. L& \# n: f* J6 F
5 _% N. O, z3 I4 A# ^9 }7.14 ]* F1 c2 T4 c' a
#include
+ N- m; p9 R8 F" A% t" hmain()
( V; V/ v! c( y{int i,resu;! h: e# o8 d5 ]. w! } ^; r5 z4 g
char s1[100],s2[100];
: Z& F L$ P+ m' P; g printf("请输入字符串1:\n");
2 g, ?+ R: M t gets(s1);
! L' a) r ^1 }7 u0 y% Y( _1 [7 u printf("\n 请输入字符串2:\n");
* p0 u" w) B. _ gets(s2);
0 V3 e9 q: I8 p- i* |3 _ i=0;
. b0 E, V I7 Q: o" M while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;
. _2 G7 u6 o& P% ?1 M; f: n if(s1[i]=='\0' && s2[i]=='\0')resu=0;
% e$ M2 [6 M/ a2 X8 g7 W else
* X; M3 y7 \ L& a9 C' @: f resu=s1[i]-s2[i];3 J4 c+ ~- ]& L3 y, J* p
printf(" %s与%s比较结果是%d",s1,s2,resu);
$ H+ v$ @7 q3 m/ M4 p}
8 C9 B& i& f$ k1 `7.15
/ e6 C& \, z5 i4 o3 U#include6 r/ q5 C& ^) o' P! f6 l$ `
main()9 g4 A+ Y. A8 W+ D, q6 h
{
7 c' B3 |: ]6 D) d% P( G2 e& E8 I3 ] char from[80],to[80];6 A4 j* G L7 b7 I Z
int i;
# C. B5 O5 S' c4 Y) ~6 C9 G& o printf("请输入字符串");# x+ I0 k) a3 j3 j" ^# w
scanf("%s",from);
2 }, E" T$ c8 @* z for(i=0;i<=strlen(from);i++). t) X1 Y; D% }0 x6 e- G
to[i]=from[i];6 i) z b1 L1 Q8 i; w$ b Z' N
printf("复制字符串为:%s\n",to);
0 y5 N( Z% t, U7 s8 p- A6 Q }6 R. u; P1 _! ~
; s+ e& h* t2 N* f
3 ?7 ]) X% U9 O8 r! _第八章 函数
j. U$ q; q5 t. C8 L* W$ g' u8.1(最小公倍数=u*v/最大公约数.)" w7 \/ U) H) L+ u( \4 V5 q3 B" Q
hcf(u,v)) X9 n" F6 \0 U7 {+ J k
int u,v;6 i5 D( y+ {! O& w: p7 y
(int a,b,t,r;
8 A& @7 |- ^% r" B9 ^; P2 h if(u>v)
: n) I1 K! J) D- n! V" A {t=u;u=v;v=t;}
4 R( ~0 `- l8 E8 V" R a=u;b=v;- I4 v; H1 Y7 i. W
while((r=b%a)!=0)+ K c4 m; I a1 {8 m
{b=a;a=r;}3 _5 u: q' ~- e4 k' ^
return(a);% Y9 g0 e( ?' Z
}
8 m* T& s! a9 v# e% q0 r; x, v0 C8 g lcd(u,v,h)
0 N9 ^$ R2 k6 | int u,v,h;
1 ~* g" k+ s- i+ L9 U! D# ^ {int u,v,h,l;- h4 T6 g" M9 h5 G' {5 x# G& @
scanf("%d,%d",&u,&v);
z4 d, l" a! _/ s7 D7 ~5 t h=hcf(u,v);/ q2 W. j( I( e. @$ k& x/ H! Q
printf("H.C.F=%d\n",h);' D* x: L0 h; ~2 G, K8 _5 J
l=lcd(u,v,h);, ~- u0 _6 N* f% [" a
printf("L.C.d=%d\n",l);, R! W8 R. w" `0 ^. e
}
# V S8 F3 w4 [. N2 N {return(u*v/h);}$ D. }+ o/ ]2 t2 u3 S& O
main()
, l J' P2 i+ F {int u,v,h,l;3 @* b* s! @6 ?% e& P( w, t: z
scanf("%d,%d",&u,&v);9 m& \2 z: s* ^' F& `
h=hcf(u,v);# w$ k& m1 k9 P" I$ p- {: B
printf("H.C.F=%d\n",h);
2 g- A* L% K5 X3 u: C l=lcd(u,v,h);8 r+ d/ v3 O7 W3 _
printf("L.C.D=%d\n",l);) w; {( [& ^) q G# T4 ^( V# N1 ^
}. e; L% U: p; ?; w; R2 v
. f$ R! F: H. J4 z3 ]
3 q1 g/ R4 V. d5 I/ W
2 W# f* L8 a: V$ R- g; l) M8.2求方程根
& [' b& x. Y7 q$ b% S6 ?# z#include
7 K# }3 ~( j/ Yfloat x1,x2,disc,p,q;" Z" x% } I1 ~6 l% n$ |$ |
greater_than_zero(a,b)
$ z/ I# n6 G+ O/ p; vfloat a,b;
% T( `0 \5 i& |9 y{
) E& ~; F/ U" L. ?$ E; b; x+ e. ex1=(-b+sqrt(disc))/(2*a);
/ `; \6 k- |1 v6 A- c# L0 \x2=(-b-sqrt(disc))/(2*a);0 N# [( |9 x: y
}
7 V' f W3 G, Z6 V% B: D( Iequal_to_zero(a,b)' C8 T) [9 v$ m% O# \* Y* b% z
float a,b;
) D7 \! y; T* |4 z3 ^7 m{x1=x2=(-b)/(2*a);}
; ]3 T7 d! j4 x7 v& k5 f8 e' u3 Asmaller_than_zero(a,b)
+ Z' f, J5 f: m5 [ ~% ?float a,b;
0 p. L6 B' {3 }0 Z# i/ j+ d{p=-b/(2*a);
2 C0 h \' O5 H9 dq=sqrt(disc)/(2*a); q/ o# ^, M' q6 o
}9 t5 g/ l. @* }; ]4 j, @$ O6 u4 i
main()5 G8 M, ^: ?4 D( ]4 b- r0 p
{
: n: R9 g- D1 K9 f) Y' O# Nfloat a,b,c;
+ j& c! ?0 [! O* G. I0 Xprintf("\n输入方程的系数a,b,c:\n");
7 v! p% d8 Y* {scanf("%f,%f,%f",&a,&b,&c);
0 P9 u6 h+ F- h- f" l' Zprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);
. A% m4 z3 g+ F; }& w0 fdisc=b*b-4*a*c;& u8 q% m' D6 j, d
printf("方程的解是:\n");! ?/ H5 u) ^8 o* _& K
if(disc>0)
" N _# E& d1 d- }) v3 L% p3 B* [2 ~$ F{great_than_zero(a,b);
- p$ c3 l* z3 p Cprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
3 g7 E0 p* I! c `1 o3 X7 ]2 u$ A}3 Z2 C3 E! u5 k6 k7 D: t; X. p& y
else if(disc==0)% F) a8 p- ]8 I# N' W
{* `3 T) P! ~9 Z4 W6 E/ ?
zero(a,b);
& b8 G3 \. t- d3 z9 Kprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
! v4 S. l h6 h$ h9 Z( b }( `$ S9 d6 [1 [, @* `
else8 C6 z; D5 N4 W W3 g2 L
{$ k8 R/ G4 `$ x& n
small_than_zero(a,b,c);( \8 ~6 b: f. l6 F5 R) E. J; n+ l
printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);" e5 W# W/ c+ i$ h7 X( Y! ?2 u4 L
}
$ d4 P7 Q% R: h}8 L* S: _: x% P; f) G& \5 [
8.3素数
6 P2 g* `6 R9 k! h7 x1 x h#include"math.h"
, h9 b" H F2 \2 Pmain()
9 |7 {9 ?% v9 H( X: n4 a1 p7 q{int number;
- p! `# p; a7 w9 v# }5 Q- i scanf("%d",&number);' r- u3 T- {$ E; f9 m
if(prime(number)), d U6 ]/ f5 o
printf("yes");
( t. E- L) [# b, H9 _" N else0 {- c' t+ C2 S
printf("no");
L6 [+ i, d( J; Y- ?; h}
3 J+ ?8 W2 X4 K; A. ]* f8 U0 Fint prime(number)) Y1 E% |( `* F1 a
int number;
; h+ g7 m' W5 H. _{int flag=1,n; F) g; l! o3 m: H7 }
for(n=2;n if(number%n==0)
D) v6 W" T2 Z; q2 i3 A4 g! x7 M+ y flag=0;3 R- X9 j* N0 [
return(flag);
+ m c1 A& M6 ^& m# ?" a2 F9 w9 b}3 O! N6 S6 R' I/ C( U' c
2 x k# [% l9 S. Y' Y& k
, G* }6 Q3 X% q# u' g7 j' x6 s4 w* {' ?$ |: N$ {9 x: R
8.4
. q9 u. z9 P( w0 V#define N 3
" Z% v. x/ V7 S: u/ Aint array[N][N];- a( Z! H9 G ] G. k: k
convert(array)
. E; X+ p4 h9 {- r2 Rint array[3][3];7 [* {+ H( S! Q7 e
{ int i,j,t;, H [& D! O E% `0 ]; X. P
for(i=0;i for(j=i+1;j { t=array[i][j];' Y! Y \+ B2 m: B, Q" T
array[i][j]=array[j][i];2 _5 ]3 B3 L( D" d
array[j][i]=t;- K$ F6 R" P- ?& |6 R
}. G0 Y8 e6 @7 N- a9 l/ I
}) d l8 X! u7 C: Q4 `
main()/ N$ z8 W0 F, G& a* O, r
{! k0 P* ?2 u" B# P7 I+ Q! r | A% h
int i,j;
5 Z. U2 b1 O/ B$ ?4 y6 e printf("输入数组元素:\n");5 X' a* g7 D' v& d$ E- z# @0 s
for(i=0;i for(j=0;j scanf("%d",&array[i][j];2 U' @; b: Q' j, |+ f! a9 |# Q! a
printf("\n数组是:\n");
% p1 W0 b9 @, f. k* vfor(i=0;i { for(j=0;j printf("%5d",array[i][j]);( f( \+ A. t {7 D! X
printf("\n");' F8 L1 R. `* }( B
}# U, m1 T0 R1 a5 J& E8 t) g. l
convert(array);3 Z0 y+ \7 |+ u, Q
printf("转置数组是:\n");3 t1 t# [* |' Y7 i' D0 Y: y
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
3 y5 } b, E) n: F9 [9 V; ~ printf("\n");
9 j- Y: l; k% l# b+ M' T) }+ Q }/ y/ u4 m; P/ d1 ~% ^# D
}/ Q; ]; h" K& h; q/ K0 ?% t
/ X+ A" P. W) P& R6 A" ~0 p! F
1 v4 J) ]6 r' {
+ Y5 l8 x; y2 ]- [1 K2 {5 d8.5! l6 [$ {2 Q9 b9 d d
main()2 Z4 x) t) t) t1 y# F* m3 [
{2 s# \2 Q; H6 `( G6 v6 J' \
char str[100]; c' V8 N4 F% T* u" J
printf("输入字符串:\n");
) d7 O$ Q' }. Z4 {" v; u scanf("%s",str);
/ L5 P. Q3 r c. H inverse(str);. C/ v$ n1 g2 v; p
printf("转换后的字符串是: %s\n",str);& t; A1 _ h+ F1 s' O1 E) A
}
& T' B3 q1 I3 t H( Oinverse(str): k9 A0 G0 J6 b9 s
char str[];4 F: t7 K, q; e9 F9 x
{
5 b, Q2 T$ g* o- ?! E char t;
* F" n3 W* P% u2 ~* j4 R int i,j;( B( V4 w! o6 F3 G( h
for(i=0,j=strlen(str);i {" X2 b7 ]5 ^( s+ m1 L2 H
t=str[i];! D9 s7 c+ r" y) x5 E( {
str[i]=str[i-1];
0 |7 L8 l/ }; x# X; y* j5 S5 s/ d" l str[i-1]=t;
! w) p1 X s& n }+ v) ^* \' G* w/ y1 }
}) m3 l' H& s2 F& Z- o: ?% \
: \# I! n* {: O) N7 I1 i0 u7 r3 c9 J1 Q& P0 q: h& B
6 J7 a& _1 ~6 n1 W) u+ k: n* r8.6
5 l9 _ D/ I8 c% J/ t# @2 Xchar concatenate(string1,string2,string);
2 D. L: Z* E6 A$ ?* d0 b3 M! |char string1[],string2[],string[];
/ A( j4 P' b" R. K$ f) y. q! ~{
7 W! y2 G+ V* \int i,j;
, s8 {, z: k/ t$ Lfor(i=0;string1[i]!='\0';i++)
5 T* K9 F# E( q9 a string[i]=string1[i];- ]8 Y$ i- V, @
for(j=0;string2[j]!='\0';j++)
4 i& X7 v$ [, f; N/ X$ Z string[i+j]=string2[j];
* K8 K. W' B7 s string[i+j]='\0';4 ]/ b M+ V/ \; H4 ^' b6 d% \
}" M1 |/ r$ n. y' {
main()
) ?9 I/ |5 F6 M! {8 y{
5 F) z. v( b4 X& t6 R: I0 c char s1[100],s2[100],s[100];) S' J ^" H0 E( B: J
printf("\n输入字符串1:\n");
) E h/ t$ m/ B scanf("%s",s1);
' I1 f/ b8 p0 e' _$ @: y) c0 w printf("输入字符串2:\n");0 ?% H! d8 ~+ B; d4 J
scanf("%s",s2);
$ _* s. z" }2 C: z- Q8 C$ Z+ N concatenate(s1,s2,s);
b& K1 E; P; Z; A+ o: X7 [' Q printf("连接后的字符串:%s\n",s);: J+ P' g9 ~% x$ J, T
}
4 f* g" B3 N5 B/ q% h$ k& e
: A$ k% ^! Z' B/ v$ V& [* T5 u$ w
* L; C% k6 f% u+ Y8.8& H; U+ T9 H5 z, M: j5 B
main()$ w& `3 A4 X" I1 l
{
" `1 q: A2 p! T, h g5 J" U2 p char str[80];7 }. m# C$ z, v3 X0 k; x x
printf("请输入含有四个数字的字符串:\n");
% l* W# Z7 I4 W8 @. r6 V scanf("%s",str);% S* i% ]! o2 p1 ]! p
insert(str);
7 W! Q# b O {2 a/ Z}
, y2 W x* H$ \5 i8 Y6 n" Sinsert(str)$ F5 g! k3 A% D- V
char str[];
) u- J" i. k7 X9 y" V{" V |, c' ]$ w& a7 u4 @( {* q
int i;, H. }) b& g' l/ E8 a) n3 v4 X" _
for(i=strlen(str);i>0;i--)
; ?- W P; _" l7 F4 M1 m9 T9 s6 E { str[2*i]=str[i];
! ^1 C; @: d5 j' q% z# R% y2 k str[2*i-1]=' ';
- c9 K. q4 t) [1 a. X }: V5 j* \; K6 _& q5 ~: u
printf("\n 结果是:\n %s",str);
2 t. ^! {; [/ W' T }
; ]% `/ S4 f1 d9 f- n! b% }/ ]/ E) h5 O4 V8 `1 w" U
$ _/ \8 q5 U2 H) ` q
# l X3 t% O" o! r2 D1 D8.9! i0 \& E$ P) p# a! n
#include"math.h"
% [- g: }8 P& A; gint alph,digit,space,others;6 Y2 K1 {% R- t; Q) s3 ?* s9 V9 ]& y5 ~
main()
* S) `/ h C/ u6 I{char text[80];" }5 |6 }& j4 w3 A
gets(text);. t4 q' Z& ]& w0 F* H
alph=0,digit=0,space=0,others=0;) P4 R0 d+ ?/ v* r& d$ s! F
count(text);2 l7 l% k1 z: c3 s
printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);: t8 k* i& \& m) @% K
}* c- q# @' D; P/ Q! i9 M5 C
count(str)
/ w/ A$ ?# k" @ c5 B' | o& schar str[];2 D. F1 T' m% n
{int i;* w8 Q; u1 L9 P
for(i=0;str[i]!='\0';i++)/ r8 c! f/ d8 \0 g
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))) o8 G! @7 @4 z) S: M% R) F
alph++;, e& q3 R5 L1 w9 C1 W; R
else if(str[i]>='0'&&str[i]<='9')
# \% ?# E$ \, D$ E digit++;
9 H# O6 G3 M/ j- @! n* {2 l else if(strcmp(str[i],' ')==0)
1 |* W" t, u' r, D3 W7 k' O space++;. L5 b8 A! A3 ?. ]% n5 j$ k
else
[/ B$ c+ X0 V' |. I6 I others++;
& e" M9 V. C* Y3 d* q* l}
; K/ [% ^, t) L8 \
% ?( Q; |, R5 _2 \, U$ g3 R, ?2 @8 _# N6 i* `! T; D; a! X
8.10
& q! Q' O9 ~) A; _/ `1 _4 n+ m' lint alphabetic(c);
6 Y( A) m3 z- P( ychar c;# C: e" K) E4 U6 f3 P& s
{
& @5 B+ ?" f' K$ s& A, G9 g- _ if((c>='a' && c<='z'||(c>='A' && c<='Z'))
8 K* c$ E+ W0 {- }0 ~2 m# w- r) l return(1);1 ]9 P0 g# `! Q7 U- f. c
else, |: j, g$ R0 d) E& R# {
return(0);
# o+ T* f9 a- [5 S}* t# d2 A6 ? a# @( [2 _; v
! q& \) u' ^" b2 U; ]int longest (string)
$ S' X4 Q4 F" ?* ychar string[]; a' y7 T0 i K& N
{3 u) A `' }, R/ V: n6 F" n- J: a
int len=0,i,length=0,flag=1,place,point;
+ b7 `# L, ~5 H! D for(i=0;i<=strlen(string);i++), ` U9 A$ Z* [" r
if(alphabctic(string[i])) p* N5 x( l6 Q& f
if(flag)
1 ]' q* V' |* P4 i( H {, S8 p" N! I2 D" y8 p, h
point=i;
/ v7 M! F, u# l; [+ b flag=0;5 X* X( t/ z# O2 f. P
}5 A/ U5 S& H7 C* ^3 A
else. S. a3 q6 o- L( K# a
len++;
9 N) o9 V* d$ L7 X+ P else; _! u' K5 C; W) u" W d
{ flag=1;
1 m2 s3 N2 m* f if len>length)
( I0 u T P0 Y' g; ^# L0 N {length=len;2 s& ?9 |6 [% p- v7 L. Z5 _3 i
place=point;
7 n4 V* t% m- K& k len=0;
4 G- a7 @1 i0 x5 z" U }
* ?9 o |% \5 _$ ^8 @$ p/ c& y9 I }( [' H3 ~9 K6 o) B/ N2 c
return(place);
8 N$ s: L8 u# y9 Y& O }
5 N% `1 \2 c- o- mmain(), r1 s7 |4 M- \+ H/ x/ {/ [
{4 M+ g8 L3 C3 q0 F
int i;
6 v9 _! e; W# ochar line[100];
@5 ^2 p& j: ~7 H8 x: U9 Yprintf("输入一行文本\n");
0 ^6 U: J3 Z% H4 x6 Pgets(line);5 w7 ~/ I: {: w" A, B+ c
printf("\n最长的单词是:");2 J9 w$ i' F3 p, u- s0 c
for(i=longest(line);alphabctic(line[i]);i++)
/ t K- s8 H$ O$ h printf("%c",line[i];4 l7 R6 \- E6 |$ J' a
printf("\n");
% {( q3 o) j% G4 W3 I}% l2 ?$ q; u- k4 k7 K* A" e$ W
0 G* ~+ d% c7 t1 W8 ]. M8 x, Y
u! v5 d; }" J: m
' c3 [+ _- l& K% @8.11. k7 F) Y+ y) \9 `
#include
1 T+ d6 z4 I* L6 w1 C% v7 p. ]8 ~) N5 H* h8 R. G
#define N 10" _3 y) d( u( E" s
char str[N];! l1 G! r0 L( i
main()6 B2 L4 w$ {, R# H9 R, m
{/ s& j5 B: w3 E. g$ `3 t/ V
int i,flag;
9 ~" I4 h: u0 A6 bfor(flag=1;flag==1;)
" y( N: H7 S9 h! g# l{
# V" d+ l& v/ |" p& e: { D' E* [ printf("\n输入字符串,长度为10:\n");5 m) \3 z K% s6 r3 ]) N q% Q
scanf("%s",&str);, ~8 u/ Y/ B v9 j2 v7 Y e
if(strlen(str)>N)
/ a9 N4 B" c5 Y! E5 m& M( a4 h4 f printf("超过长度,请重输!");
. C+ y/ Y- b2 B6 c) t) G else
4 l- H* U5 y8 V5 _' z flag=0;
4 Q7 e9 l+ W% a0 e}
+ F4 Q9 W5 ]6 B1 \sort(str);
4 u- a: H2 B& B% Z8 eprintf("\n 排序结果:");% h" w' ~; w" R* t9 `9 Z2 R
for(i=0;i printf("%c",str[i]);
; O M' ^% @5 A! Y( [ L}. Y3 X, O5 C8 q+ n" L5 ~2 t
sort(str)' Q! x% k+ |; ~$ v5 v# ?1 U
char str[N];4 ~) k4 {7 O: n# }4 ~$ l6 S+ o
{
* j2 _6 Z) D w* }5 f) q1 Pint i,j;
* ?5 o4 b1 y! {: J$ jchar t;0 `7 w, u% N5 U/ L1 I: |
for(j=1;j for(i=0;(i if(str[i]>str[i+1])
9 |0 d2 M$ l% L$ H/ Y { t=str[i];
6 D! c+ I1 r+ s! ^4 m str[i]=str[i+1];
! J; f: y. k+ Y' W) i& R/ n str[i+1]=t;
( G2 a5 F2 e9 w T2 m0 {' o }) w% V/ b! u' p u
}
; G4 F- F5 m& }9 P- `- B8.12
' P( Q b/ U, x% s#include
) { \) S1 S5 G1 Y# z: j#include
) k0 n# t& ?1 k6 }, s0 Nfloat solut(a,b,c,d)8 `9 Z7 u* J3 M5 ]2 ~6 M# A7 W1 G& v
float a,b,c,d;
! A1 I" {6 q D! E3 s: q5 w: A{float x=1,x0,f,f1;: K' p6 P/ t' e) j9 p- M
do- Y$ M' z) l0 y$ R+ H
{x0=x;: ~1 X6 D8 B; ?
f=((a*x0+b)*x0+c)*x0+d;( D; [) R K+ c3 a+ a6 {6 Y0 i
f1=(3*a*x0+2*b)*x0+c;* y& w9 f6 Y" O" F6 c8 `: Y H0 e; F
x=x0-f/f1;
8 ?9 W( r x2 R3 e. x% j0 T+ u% a }3 S- M8 Y0 Z* z/ T
while(fabs(x-x0)>=1e-5);* R% x G# C) [$ b. Y5 o
return(x);4 K8 c% d7 W/ `+ i, k8 Z& j& r
}6 s$ }4 Z" J4 I5 N# {7 k& T2 ~# t
main(), ?6 j8 B( m6 y( `
{float a,b,c,d;
# u. |8 h3 H& h: ~ scanf("%f,%f,%f,%f",&a,&b,&c,&d);
1 b) `$ n" i3 z$ t; ? printf("x=%10.7f\n",solut(a,b,c,d));# i8 {2 W3 l) l8 B
}
, F; {, m+ I3 v# ^( v9 i8.13% g' z6 |+ n. W6 |% s8 e/ u
#include) M/ Z0 F( Y7 N: ]$ U
main()
3 N: d/ V$ d* M- j" u/ H{int x,n;
3 \) X2 H+ Q" l7 E5 ] float p();; J B+ z+ h" M1 A
scanf("%d,%d",&n,&x);5 m4 Z1 r; l; x( b8 I
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
# f T) R) g( e. D. L}4 B! f) I; x9 i
float p(tn,tx)
. _1 Z/ x9 p% Z' l( @2 \int tn,tx;
y7 d1 B4 w9 y: X( k" [5 Z{if(tn==0)( ~$ g0 e# q! v, D
return(1);
/ N; f& G' Q- g" h7 [! e else if(tn==1)2 e- `; S7 ]% y8 k
return(tx);
; B9 @6 a* @2 T else1 G' C9 C' } r# H7 T
return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);
6 W9 G! \' {" U$ Z}* U: D& u1 B( E4 `5 ~' d
8.148 N" ?* l- Y6 M( V& f" a2 Y
#include "stdio.h"
. S' Y( I" o, x- V# L# S: l2 Z( S#define N 10" `. @6 H( d: A* x. Y
#define M 5
5 I' e: i: S" xfloat score[N][M];- @" Z5 P: ]! q7 N- O. B
float a_stu[N],a_cor[M];
( Q% E9 [/ n+ {$ }. p) y, D) cmain()
" }* f# ]8 Y3 v5 R% c{int i,j,r,c;
# i% ~( ^4 l# l float h;% n8 D% F: l7 }) a
float s_diff();
5 p5 f* E' o- @+ \, ^) c float highest();
' t; Y2 v5 K$ ?) `6 F r=0;8 R" x j+ @6 [# S* W
c=1;+ M# c! \8 A; N# J
input_stu();' l: V( y; \" Y
avr_stu();
- y+ q( T/ n# O8 E+ S avr_cor();
5 ?, i2 C1 F9 ]$ p# a' } printf("\n number class 1 2 3 4 5 avr");. Q: |& J) Q1 Y; ?5 [! h4 B o
for(i=0;i {printf("\nNO%2d",i+1);
1 b, q" Y& u# ] for(j=0;j printf("%8.2f",score[i][j]);) J0 z3 t0 j: J% v6 L0 Q! O
printf("%8.2f",a_stu[i]);
2 w8 U1 c6 s( W5 r& Z/ t( N }
5 G B i; [# U% g' u8 Q printf("\nclassavr");5 W, i0 ?) U9 h6 b! q4 u0 M
for(j=0;j printf("%8.2f",a_cor[j]);* T/ P) f4 @' v) d
h=highest(&r,&c);9 o& d$ e( x% i/ ]2 A
printf("\n\n%8.2f %d %d\n",h,r,c);3 s9 K4 [. l/ z ?4 u
printf("\n %8.2f\n",s_diff());4 m7 |0 L* J/ h( E) q0 [9 s
}
( T5 `$ n1 ?2 U1 q* [1 k, P9 \$ k+ W) Linput_stu()4 c% D' p3 V1 _+ k1 C
{int i,j;7 S5 _, o+ \$ v$ ]2 ^' u- \: L
float x;1 V2 k9 |- k8 B2 j D7 C! M5 ]
for(i=0;i {for(j=0;j {scanf("%f",&x);" J, ~# L! R( h* ?: C% d! ]& K/ u
score[i][j]=x;5 h$ j% G9 t3 \0 L: ~4 o) ], p
}. L" u( b7 a: ?
}( \ u, N; l9 R* k: Z7 B
}
1 X6 } A6 A% c" B' D' Q5 Havr_stu()
0 [" Z' t3 S) _6 Y ]% D8 [. o: f{int i,j;
4 p& g( w7 d4 D: w @" _5 ? t float s;3 ~6 f7 p1 J2 W# S
for(i=0;i {for(j=0,s=0;j s+=score[i][j];
0 J% k6 z# I/ V- n3 V* P; E& Q a_stu[i]=s/5.0;) C: h; L% ]3 i- h$ b. p1 M. V
}' y& f, Y: e+ C. N1 `* S. {7 z
}4 S; e0 j; G( t# H; P
avr_cor()
2 C& M5 J# l) Q! u{int i,j;
& b# u, {, W \! F. ]) z9 ?3 ?! R. U float s;) U1 U6 N( W* m6 u( d1 [1 y
for(j=0;j {for(i=0,s=0;i s+=score[i][j];% `) u$ F* o5 I9 o
a_cor[j]=s/(float)N;
' {9 T7 ]5 }% C3 a( o1 F( j" R. X }
; O* ^+ j) L. l6 C& |2 x! d}
$ S9 B9 R. j* ]& C% I: c* Mfloat highest(r,c)( X9 b# g' U& D, h5 l; B
int *r,*c;
3 {& \- n3 G) I+ a% O, z! h0 H{float high;
' A0 g2 [3 ], ?. R( l2 W3 P int i,j;9 M0 p/ V* ~3 g E" C+ q5 n0 C. D5 ~3 O' d
high=score[0][0];: ~* f( J1 b, k9 ~- V3 J, X1 R) H
for(i=0;i for(j=0;j if(score[i][j]>high)
/ r3 t2 e3 G3 M9 _# p {high=score[i][j];
% H( G/ q& n* |- y *r=i+1;! ~! [& x7 F; F3 h7 C6 A9 c+ ^7 f1 L
*c=j+1;
8 a7 g1 p" m V2 M( Y1 |; N }
: P3 e6 j W) F4 _ return(high);
) }/ i% G3 Y/ T9 I1 z, ~/ I. K}
* J- u% e. u) X2 j4 Vfloat s_diff()
; J4 h5 S" M+ G$ L& i{int i,j;
& E2 ?) z6 n- @; h" A F float sumx=0.0,sumxn=0.0;" X: ]% A4 ^% z) [) w1 q
for(i=0;i {sumx+=a_stu[i]*a_stu[i];8 M) @; g1 T; z% Z7 g
sumxn+=a_stu[i];, }) q1 x' l% L7 y9 M
}3 A) C. t o! k) n! l
return(sumx/N-(sumxn/N)*(sumxn/N));
% [( f, ^& z* F7 d}3 n! Y; }$ F. t
8.15
2 {, |: }: m; I# K! r0 \) i#include! X+ M C y; J5 C1 x
#define N 10' { G# W8 L4 T/ A
void input_e(num,name)* y$ U+ c& H3 b' Z. J T
int num[];
! t; w5 S% A' Mchar name[N][8];
- l1 M( B% B; M4 z4 v) o{int i;' |; H1 M X+ W
for(i=0;i {scanf("%d",&num[i]);
. U+ g- Y( D9 `9 t4 Z/ ~ gets(name[i]);5 l: o, b3 Z7 w, u5 B- ^
}5 \6 J' y" U7 X' y2 F9 q1 Q$ t# V
}
! ?8 B. g% Y: K% X6 c, {void sort(num,name)" Z; d- \% b& G: }3 f7 ~ N* s
int num[];
2 R, R& Z* n& f* dchar name[N][8];& ^; u$ Z6 X& g' w4 \
{int i,j,min,temp1;) y6 Q0 L) q) W% P) j1 S
char temp2[8];: ^* o! S6 C- |3 P
for(i=0;i {min=i;
$ b0 ]+ L3 X/ d1 V; {& p2 x) P/ R for(j=i;j if(num[min]>num[j])min=j;& w) k! O; X7 Y! F& I0 W5 U2 ]
temp1=num[i];3 b- H; p$ T* s: |
num[i]=num[min];7 l' ^; P6 u2 y v
num[min]=temp1;
! d3 h: {9 d x" V strcpy(temp2,name[i]);* F3 ~) n" b7 J" D2 v6 G
strcpy(name[i],name[min]);
! B a3 ?) i. b1 X strcpy(name[min],temp2);# r3 [/ C( _( B# L7 W5 @
}, L' W; _) C3 k" }2 H, i. T
for(i=0;i printf("\n%5d%10s",num[i],name[i]);
- V& V+ H, u4 f2 `8 I. F3 S}
* s) C8 ?0 Y+ x; \' m& y1 Svoid search(n,num,name)
& B* @4 e- h c z) h8 v5 Zint n,num[];
( J, b, P9 j" E: zchar name[N][8];
9 J. s1 I t$ N. m) U9 V$ O8 u{int top,bott,min,loca;
9 f6 N/ l' n3 C- r' F+ ` loca=0;
2 V, T3 @: h% |9 t4 h+ D top=0;
7 s" o! ?9 `4 z5 r, s! b bott=N-1;" a/ x3 l" ~1 ?! F- G' t- }+ ?. e) v5 h
if((nnum[N-1]))
, s+ t" Y# v( n* |! F% ^. s* u loca=-1;3 z! r. O9 X, F2 }
while((loca==0)&&(top<=bott))
$ h4 c3 d. n. l% p! r {min=(bott+top)/2;
. W4 X# W% n* l w9 x: j2 M if(n==num[min])
, q( t9 Y5 {2 O {loca=min;
/ D; c$ T% m. I7 ] printf("number=%d,name=%s\n",n,name[loca]);2 ^8 ^- @, T5 v2 C. ~: G1 a# y
}; V( e& ]! |$ t% l% N& u/ p; I
else if(n bott=min-1;9 u0 _9 {$ ]& E' _. e# D2 [
else
: `/ s$ r- G# s; B, h. O2 l1 O top=min+1;- j7 ?( J7 C1 ~* F+ R
}
2 W, {& K/ E( J% L6 t if(loca==0||loca==-1)
/ D) H) P/ i2 Q9 { Z+ D printf("number=%d is not in table\n",n);& \6 h! T" O! C, ]
}1 h# l9 y9 P% X8 C2 h5 O# {
main()
/ x7 u0 X1 G0 w/ c{int num[N],number,flag,c,n;
; @' D6 X0 l' w' |7 X* w char name[N][8];3 w& m5 P8 y' R) X+ Y
input_e(num,name);0 A0 p. ^) K; v) r* ?
sort(num,name);! Z% Y& V6 R# J6 @; x/ X1 b
for(flag=1;flag;)+ b/ s0 D" k4 c4 c
{scanf("%d",&number);( Q# E# ]% }' L4 a
search(number,num,name);
3 _, h0 Z9 R% E) J" | n- q. E- U! d printf("continue?Y/N!");
' ?$ M7 q2 l7 }5 t: I c=getchar();4 D1 C& g; b, w q5 I7 l3 f
if(c=='N'||c=='n')8 \$ V) g7 g& S! W
flag=0;
. E! J7 t! K# v. c! c/ U2 ` }
( \/ O, ]8 a- F}
2 S0 Y! X+ T' Z7 m0 }4 a& G2 }
" M2 \( m2 N6 K+ \8.16 f- H' _& _( M, `" x, c o
#include
7 C2 W1 y2 ]. S' o9 l#define MAX 1000
2 y" ~" F+ ]6 J, C7 fmain()! e. G3 V* e& u
{ int c,i,flag,flag1;* }- V4 V9 z i
char t[MAX];! M. a" `7 z3 k. S7 V+ b$ ~
i=0;
8 I1 {; e7 B2 H v N7 [ flag=0;* m5 m* X/ h I- S/ N
flag1=1;3 m. T9 Z- R+ R6 y) L/ K1 i- O
printf("\n输入十六进制数:");1 s2 U& Y: l* Z0 G7 Q, W
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
$ E5 [$ u2 Z7 B5 }/ _ {flag=1;
& j1 X0 T8 c1 `; l t[i++]=c;
1 c% Z: V6 F0 r& y4 B }) ?8 {. e/ A- K! ~4 U
else if(flag)
/ J- h: u1 ]1 p9 t) D {
8 H- m2 c5 C9 w! K( E( F% C/ F t[i]='\0';+ s( X1 M" h% I7 D3 ^
printf("\n 十进制数%d\n",htoi(t));
, }4 @( f% _4 F/ p. s4 L printf("继续吗?");
0 C9 M. U- @! K# g# V3 x4 c c=getchar();7 t3 D9 {$ }, Q: \, C
if(c=='N'||c=='n')
?# o& s0 v* o, h flag1=0;
9 I- N: B5 J3 C! w6 F& L7 E1 ] else' i2 ~+ S, N0 q
{flag=0;: A3 |" F8 `9 Z3 f9 m3 _- k3 C; C
i=0;3 n, p8 |' E i: J8 ?% k, d1 D
printf("\n 输入十六进制数:");( c: s7 \/ H2 ~, V' @1 I
}
U3 x' K3 n1 |: C7 ^}
8 {6 b8 k! v1 K0 k. T}( i7 T9 Z* r4 p4 ^! n
}
: a( v1 |* G0 l4 i3 k2 J! rhtoi(s), P' J, M! ^2 G, U& ?4 ]4 \
char s[];
+ i6 B8 w. x# A u% _( }: b0 b{ int i,n;
/ n( K# S" a: K u f& h n=0;
% a' }5 ^7 m- D7 B! e& | for(i=0;s[i]!='\0';i++); d- L7 F9 }4 B5 i
{if(s[i]>='0'&&s[i]<='9')
$ U# `& E$ m. Q# u6 F n=n*16+s[i]-'0';
, B4 A, K# b0 ?" Q if(s[i]>='a'&&s[i]<='f'), f5 F/ o( T4 ?/ L0 g, a" i2 m8 x
n=n*16+s[i]-'a'+10;
( ~! u. Q5 B/ b& F" Q. p& ^ if(s[i]>='A'&&s[i]<='F')
3 d3 \3 b1 T/ ~# X# _3 e n=n*16+s[i]-'A'+10;6 J; y! R8 o+ G8 p5 I
}
; D5 g3 d. D" x' y- E, E: h return(n);
+ U4 } D4 j. @}
' p( } |% j7 u2 E$ [
7 {# u& U6 O/ v- e+ p( P' v3 v' z! [
9 G6 H! \) }% A# z" n
8.17
6 W& P2 k/ H6 Z# o5 |0 Y* [#include
1 k& |1 B# Q. ?7 \- M5 Pvoid counvert(n)
! M F- [' c2 j uint n;
1 p- R, v- L/ E" n. \# j% ?{ int i;
9 p6 s+ E2 x3 z if((i=n/10)!=0)1 D+ I3 c& t0 V# J/ a6 U
convert(i);) p1 o5 H3 m+ V# d
putchar(n%10+'0');
7 f" l( f0 A! n) Q}
3 g, d1 N5 i; h! Tmain()+ p1 x: P6 M' E+ o4 g8 d5 Q9 c
{ int number;4 x! O8 Y$ T& B# ]/ N/ e9 u& P$ g
printf("\n 输入整数:");/ _7 A3 O( N0 g5 Z8 S5 d; e3 r
scanf("%d",&number);# @8 `, L) u' ^! r$ O
printf("\n 输出是: ");! Z' s( K" T" B; y9 s
if(number<0)
/ w- [2 C8 L' T4 z8 ~ g$ f { putchar('-');
0 \ a7 `. R* A: l& ~ number=-number;
3 c2 k% Y) D$ h }6 A N; [, |, e: r
convert(number);" `; ^4 x B5 `/ I4 v
}
, T1 ^' {/ I6 H0 H8 J/ \
( ?" K4 r* D4 m+ K- E% S0 V. [+ R* y3 K' o: ]
@5 @% ^6 Y4 W- p) u. u1 a8.18
4 g$ K( X( V! C) E: q# x0 amain()
3 z; u3 P5 _$ M* o$ c% n{
) g: D+ }' j3 N8 B2 o int year,month,day;
" M2 {+ k3 {8 i7 v) N int days;
8 V; Z$ N7 F. @6 I printf("\n 请输入日期(年,月,日)\n");8 K# J4 E) j( x' b
scanf("%d,%d,%d",&year,&month,&day);$ l$ w, w3 E/ O7 v$ Z& G+ h. S. b
printf("\n %d年%d月%d日",year,month,day);# ~! P# \5 [, K# N
days=sum_day(month,day); k2 G0 @' R+ @; @- z
if(leap(year)&&month>=3)
4 s8 ^& \. _+ A* G1 d, d3 D days=days+1;
" a. V" U9 V+ A; z Q M ? printf("是该年的%d天.\n",days);$ }( {- S2 J5 z; t \! G% M
}+ D/ o- C; v* k( Y& D& d$ p
static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}$ d+ Y1 M1 P: n* x* a
int(sum_day(month,day)2 Y7 a! T( \' B4 B: [. U6 s) r
int month,day;6 h$ C- D( t4 g9 i
{2 l' x& U, F& Q
int i;% j7 t" t8 \- k8 O
for(i=1;i day+=day_tab[i];
, |, S7 Q) R2 h l return(day);* K# S# x. z) q6 a- O4 Z( X, n$ v2 c
}
5 c, a# ~" r; f( ~: P. V int leap(year)5 @, M, D: c/ B4 `7 [6 n4 E8 k* i
int year;8 ]5 D7 B, n6 c6 r; G, x6 V, B5 ~
{: E5 z$ Q+ _, j
int leap;# g+ F; I/ L, x) h- w. |" J; n- n
leap=year%4==0&&year%100!=0||year%400==0;) j2 t$ _" n) d! {8 ?) E
return(leap);. |# z0 |3 F) W2 r! W
}: N% S% x9 L3 X% H
第九章 编译预处理 S, I# B9 k4 l6 M+ M: D/ E
9.1" j# F& w7 j' t- q
#define SWAP(a,b) t=b;b=a;a=t4 B+ f) A4 X5 H9 S! W5 n$ v
main()
: X) ^5 @& P5 p# v4 z" g{
0 i3 N+ ~6 ~% f# O8 jint a,b,t;
6 i/ l/ u* }9 M& [( \printf("请输入两个整数 a,b:");
C' G4 G; A& Q3 T {" Mscanf("%d,%d",&a,&b);# ^/ a. \$ f7 A7 @
SWAP(a,b);( M2 M" I& f# l% n
printf("交换结果为:a=%d,b=%d\n",a,b);% i& f* @' Y4 I9 N- ~7 C& M7 u+ q
}
. ~2 ?: g( U4 H* L0 u4 p / O: @$ w: H' g1 t% |
7 P# x6 R/ d! @# K
9.20 C" T; S" r4 V; M) f
#define SURPLUS(a,b) ((a)%(b)), v: F4 B' A* U1 y4 P8 M
main()6 j. r' _8 i+ c8 M9 T9 U* B9 }8 N
{
% q+ ]7 \& B/ a3 k& J. j int a,b;) C6 T: P- w8 S, [9 f' b2 k
printf(" 请输入两个整数 a,b:");( O! L" U( z: X. |9 I' y8 a' E+ o
scanf("%d,%d",&a,&b);
" C% Q, q6 j& r) o! fprintf("a,b相除的余数为:%d\n",SURPLUS(a,b));
) \5 X0 \; |3 \$ H" Y }
5 O" W! b# v" i
3 m6 w- f" p' o, ` R! C5 C4 F) d; |* }' i; y6 [
9.3
! K: M% V5 d* B$ A. A* f#include
) {3 I1 |3 ^9 w- x6 h- m: L% N#defin S(a,b,c) ((a+b+c)/2)
2 w% F1 d1 T1 f0 E( i" C#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-4 v% k. d8 H# y1 P9 K
c))); y. u& d! L: B7 z# E4 E$ W# k, `/ G
main()
* W- g6 Y' ~* y" p Q! k {. C. Q5 E% {- q8 e" S- o, ?
float a,b,c;0 T7 `, G5 |: v% Y3 Y+ |" M+ \* b
printf("请输入三角形的三条边:");
8 I5 r% h3 e+ P6 v7 e& X" L scanf("%f,%f,%f",&a,&b,&c);- X* Y/ d0 t. Z7 { R
if(a+b>c && a+c>b && b+c>a)1 p1 R( w' `- x( f& F3 A& H
printf("其面积为:%8.2f.\n",AREA(a,b,c));
* s/ c; n& p+ |: p+ [ else
, N) V: n) Y/ o5 M. \9 ` printf("不能构成三角形!");5 E* ^. W# l1 D6 F# E6 Q/ ]' n% ^
}, P1 K! M1 t- t, O
) }. Q7 K" X3 R, T' H) q* y0 }; P. l0 p
. P+ ^' h# z8 x/ X/ z* k. ^) ^9.43 _5 i3 w P) p: P& O: x
#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)
; N( B0 U4 q! {* F4 Q/ d0 imain()
1 W& @# i. @3 k9 a4 D {
% x4 i# O- ~ M; s% I int year;
5 `$ R; L5 b: n2 z% R6 {+ ~' l printf("\n请输入某一年:");* B4 @5 ` d) O6 w! D
scanf("%d",&year);+ f3 _+ q B& m% W: {/ l5 V
if(LEAP_YEAR(year))$ m$ j( H: t+ l6 V$ {0 ^
printf("%d 是闰年.\n",year);7 p/ _0 \. I$ S6 z9 L) t
else
( E6 i' z4 p9 D; q- P' s/ t printf("%d 不是闰年.\n",year);
7 C8 f" l4 f' h& z# n' F" N }- ^* |% E# s3 {% L
- T" u3 \) i+ ]9 X1 R; c4 r% C
1 t1 J, k6 Q% ^7 a$ {9 D9.5解:展开后:- u) q" f* R, I1 \" h: ?$ O7 D
printf("&#118alue=%format\t",x);1 S4 s+ E+ g. B; {% @
printf("&#118alue=%format\t",x);putchar('\n');1 k' Z* T/ F8 A; ?1 K
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');: e8 X8 x. _ c. u! D0 [0 t3 ?
输出结果:
9 }0 ]0 n8 i+ a5 k&#118alue=5.000000ormat &#118alue=5.000000ormat
' ]; v! \# q% w&#118alue=3.000000ormat &#118alue=8.000000ormat6 Q' z: Q, N$ h( w6 e2 G+ P' _
. X* C: p7 i$ z# E' T; I6 Y+ a: A$ W) `) I
9.85 w+ o) r* N# J
main()
% g0 o$ v9 j9 ^, x: @ {
; H5 J1 l, _5 J& H' w! {# c* a+ A, p int a,b,c;% T" b; g) F; k/ K) i' r; ~0 [0 d
printf("请输入三个整数:");) {& r9 o. X: W& \6 L% C
scanf("%d,%d,%d",&a,&b,&c);
& Y% o* O7 T& `% x, I printf("三个之中最大值为:%d\n",max(a,b,c));
Z* @+ t- f! n1 Y! b( r }
" a9 c' ^$ I t* v! S max(x,y,z)" A) K: h) K- I! u) w1 [* q+ g
int x,y,z;
4 O7 D5 z& m L. r6 e/ S {! y) n5 b! g3 ^. }$ l
int t;
1 \" z/ ^" A! r3 t t=(x>y? x:y);$ E" ]8 L6 {1 c: F+ x. b2 {
return(t>z? t:z);1 R% Q: n$ x3 z4 w8 Z3 o
}
/ _6 [* M, q2 ^# `3 ?8 ], g7 W+ R2 n L+ h" F# \0 k& l4 n
5 p z0 W5 D6 a, y0 y/ z# V# c" N1 A6 U& m
9.10
1 F" s. ?0 `. e2 C7 H5 K- p; Q#include& F$ k( u& C. U
#define MAX 80 z3 s3 X0 x0 A+ P
#define CHANGE 14 M5 |- E' g$ V I. \7 \
main()
6 T7 C, l5 j, T6 V, E7 y+ M8 I{) N4 X( ^/ l: Y& r1 @: h# R
char str[MAX];9 h. m' B% N& n" [/ i ?" E) V3 W
int i;
, G/ h' A" c+ M+ t' F# h; Q) h' r printf("请输入文本行:\n");5 ]! R) n A- f, ^, N
scanf("%s",str);5 ?' A6 [ K* {4 }: Q
#if(CHANGE)
. k4 K, Z1 |5 I( U7 Z: W {
# \! q, W" y) H7 U0 n5 ~ for (i=0;i {! f' V ~7 u8 M/ z0 n# a
if(str[i]!='\0'
" [$ f5 v% e6 m' ^+ V3 C; X5 U7 P if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
5 M ?+ j ^ }2 Z str[i]+=1;0 Q7 d" _6 h: n' g. P1 w6 E8 g
else if(str[i]=='z' || str[i]=='Z')1 B/ P/ E: s3 P. w* L
str[i]-=25;
' q$ t+ C3 g! ^ }8 ~9 p' Z* ]3 ~( j
}3 l# J0 l0 F2 K- ~
#endif2 F1 l: e( M9 Q$ y! X+ |. H: u0 |6 v
printf("输出电码为:\n%s",str);& M& o9 J/ [( O# O) D% ]+ P2 V
}
/ h7 c& h, X8 D0 b- v) R% k第十章 指针
) \$ X5 |5 w6 s4 M* q, c10.1& M4 q1 P M' d: D5 i- p
main()
) ^- q$ f7 r, ]9 R% }: `{int n1,n2,n3;
0 {) Z r G" x h; F1 b int *p1,*p2,*p3;
% P1 r9 x& E2 {8 y% N7 j scanf("%d,%d,%d",&n1,&n2,&n3);
- h+ m& b2 D9 p$ x p1=&n1;. x* }- \% w$ H1 l' L3 J) X
p2=&n2;
* `/ {# K* | {9 y) _ p3=&n3;
1 O& y% T+ ?1 b( ~7 C+ i o- L5 \ if(n1>n2)swap(p1,p2);! B8 \2 `3 m/ e' D$ {; S
if(n1>n3)swap(p1,p3);
. U( ?, L: `4 _# d/ E if(n2>n3)swap(p2,p3);
4 u5 l% v4 l q3 j+ g) G printf("%d,%d,%d\n",n1,n2,n3);
( m9 m6 K$ W9 C}* b6 I; l+ x7 H( I$ X) K, x
swap(p1,p2)
) k# |+ Q) P4 b; P5 zint *p1,*p2;
$ x5 |7 R3 G5 A{int p;
/ W- I& ^/ Y' l4 N% n+ Z" T4 E9 Z p=*p1;*p1=*p2;*p2=p;
1 Q- Q. ^# T( m% S. I}
6 I* O3 c4 v# `5 @10.2$ G7 q7 `/ b4 B- P
main()4 Q+ `7 H! k8 |7 @% Z+ `+ A8 |
{char *str1[20],*str2[20],*str3[20];
1 S" [' J5 b7 o3 r char swap();% o' A% S3 S! L
scanf("%s",str1);9 }8 `+ P, T, H3 V/ v! B
scanf("%s",str2);2 ]) f$ {" }" L) c, l+ y
scanf("%s",str3);3 h7 @, i& a# G" R# `' l |; M6 Z6 F
if(strcmp(str1,str2)>0)swap(str1,str2);
4 U9 S) J( }) ~' N9 i6 | if(strcmp(str1,str3)>0)swap(str1,str3);0 [/ ?1 L, ~9 s1 e) N8 ?) x
if(strcmp(str2,str3)>0)swap(str2,str3);$ j# g! u3 q$ h
printf("%s\n%s\n%s\n",str1,str2,str3);
2 a- N& O6 b2 U: v}
0 Q* f# b2 ?$ N" ?4 Q' `2 L: b4 Wchar swap(p1,p2)8 d* S% U+ f* `" Z! L
char *p1,*p2;% V- k3 W2 _7 P5 y
{char *p[20];* X) {) z4 Y3 m0 k. w
strcpy(p,p1);. z$ [* l6 X3 A E
strcpy(p1,p2); R9 c/ `. h+ B! z, B# K+ [
strcpy(p2,p);# t" v' p" D f& d1 ^
}
1 g) e: k: o0 Y10.3/ O3 B+ _6 `/ \. }
main(): S0 Z+ _7 L( A2 K4 U( X0 b
{int number[10];: F5 \' [: h$ I0 Y3 a4 d1 r- G
input(number);0 o$ y* Z4 d+ y4 v1 Y( M
max_min_&#118alue(number);
' n; c$ P' d6 `& |3 a9 G- l, j output(number);, a3 E! W2 }$ f
}
' o o! Q1 n+ U9 o8 i- Qinput(number)2 K, e3 G) W; @- m& U( @
int number[10];
3 h2 I$ V. Z# @8 t1 ^{int i;6 J) _' Y5 }2 W {0 p" k
for(i=0;i<10;i++): p- \, N1 W! z) x) E
scanf("%d",&number[i]);
- z/ X/ K% Q# o# X+ {}
2 {$ M8 R, b% f5 S/ {" nmax_min_&#118alue(number)
8 S2 u! _9 w3 \int number[10];: r6 X3 D: \! ?
{int *max,*min;- Z; c' @% k# U9 f) B
int *p,*end;, g& y" C9 N, S
end=number+10;
6 m' `' Y, ]1 {+ A" S4 q& t7 [ max=min=number;
8 n l, j+ l- D- \" U0 }2 A2 T1 v5 r; y for(p=number+1;p if(*p>*max)max=p;
h$ z0 H% j* A; Y else if(*p<*min)min=p;5 N' Z2 M5 U5 x: T
*p=number[0];
+ R; v3 x- s+ C$ ~! f$ Z3 E number[0]=*min;0 o/ ]( r$ I7 s) H) E
*min=*p;
8 t7 P- Q3 u% @" c! r+ X7 i T2 t *p=number[9];5 B. F! W8 K4 K+ d& ]
number[9]=*max;
/ w+ x# I# I' S6 P! z& V *max=*p;) _& g2 j4 @) [8 [0 p9 ~( x# s
return;! s* F9 z0 ^9 e% _6 S. Z
}
0 y- T @5 F- y# d# s; Q$ u" p% i2 voutput(number), `; |. C; Z3 \/ R6 ]
int number[10];) x4 ? A" W4 l1 Z9 i/ q" r
{int *p;9 J( z* X- J& M0 r% N* Y* ^" R4 W
for(p=number;p printf("%d,",*p);
8 o2 y8 h& t6 d$ `! Z: A printf("%d\n",*p);( Q& o0 ]9 @$ D3 B4 P
}
# h2 k% [1 J5 o1 e& O10.45 w) L( L, j8 _7 E/ v! p, y! h
main()- v. j& j0 b( x, C
{int number[20],n,m,i;3 f& t' m7 m0 d, K& y
scanf("%d",&n);
# Z, o2 F5 R, q! B' y( F) D1 e scanf("%d",&m);8 b% |, X0 i! u$ S' O4 e( i" u
for(i=0;i scanf("%d",&number[i]);
; `& S2 F. ~9 } move(number,n,m);( }- y: b. O' Z# Y2 B& E
for(i=0;i printf("%8d",number[i]);) |% w# A& F% a; Z
}. E; A8 P2 n* |
move(array,n,m)
6 e2 h$ g- h$ rint array[20],n,m;
/ q: ~+ j0 X: G6 V{int *p,end;
* C C2 q/ r( @; N( r end=*(array+n-1);7 v; a5 d0 W8 X# U: f8 I2 K
for(p=array+n-1;p>array;p--)' q* X; k6 m, A/ e* S8 w, j
*p=*(p-1);* I6 x5 W4 m; F. }! k! q9 `
*array=end;
' H* K) i+ Q# ^3 ~& o m--;0 Z) e( o( Z% \( r( w4 c( s1 H) d
if(m>0)move(array,n,m);) z" d- {! t0 O( V8 n+ d, e G |
}
7 E8 A1 ]! E2 h8 y$ V$ U10.5
! x( s2 `1 D, Z) N- n#define nmax 50" a, F' Y. Q b$ s3 K) d
main()4 p' m5 k) ~5 G$ N
{int i,k,m,n,num[nmax],*p;
) z1 p) _' E! t5 u) J" a: _ scanf("%d",&n);
/ g D& n9 b( z4 Q p=num;
# r# e9 N9 R) i" s for(i=0;i *(p+i)=i+1;) Y, A; [4 u0 g9 U) h, F% G( [ T) S
i=k=m=0;
3 E C8 j7 K: g% {# I. ^ while(m {if(*(p+i)!=0)k++;
! _" M( V2 Y4 r if(k==3), ^2 o$ E# B2 Y, Y. e% P" c& A/ f2 s% l
{*(p+i)=0;5 j0 G7 y$ v' f L
k=0;& Z o! O( a. Q0 M& f" L* |" Q
m++;! u1 \& a8 o: b A: y( \
}, [# V- j% \ O, p
i++;
7 T9 D/ H) w8 s! r9 O if(i==n)i=0;* ^9 b' Q# f# B {% \
}
9 R; {7 s8 |5 ]4 N" H6 ~1 u$ @9 }+ | while(*p==0)p++;& c8 H& r3 { |7 P3 b) d
printf("%d",*p);0 b' @+ e* Y: A, r h
}
. m# R9 w* { L2 O3 N# Z10.6$ s) Q% F. l, Q) `
main()
$ T D+ k( D+ |3 Z/ o; h{int len;" `* ]4 ?$ s% Y& u U9 B
char *str[20];7 k, ~) |; e: K/ F
scanf("%s",str);: |4 q% n7 y# h# x0 A9 s/ w
len=length(str);+ M% H B! @! Q4 |1 `
printf("\nlen=%d\n",len);
! W- H6 t% ^0 `$ I- M* A}. k, r4 Z* @0 u+ d1 S( O& ?
length(p)# g5 ^' m, F7 p6 h
char *p;
# Z0 J% Y6 _9 X( C$ Z7 G7 r{int n=0;
0 N" u, l( r, ~( h1 d% j! D while(*p!='\0')
9 ?. Q: X+ L$ r8 c+ q {n++;p++;}4 ~4 Y% w9 `" K8 C
return(n);
. T6 N! \/ N4 u) ~2 W1 W8 W}
" q5 \4 v: L8 } x7 s" D+ F$ D) ?10.7
( s& g. |8 P( }3 U7 j0 ~% _. o# mmain()
; y6 Y& I& Z q0 n4 V0 X" f7 z{int m;( W8 d& F( C: _% t" z1 v0 W9 C7 f2 G, C
char *str1[20],*str2[20];5 O% ]1 `, y* Z
scanf("%s",str1);, n/ L: h( `7 \' G
scanf("%d",&m);6 c$ R. Q8 i% ^: ~3 X5 ?" v
if(strlen(str1) printf("error");
5 T; X9 @' Q4 R' z% _- ^ else
( |! A1 N: R9 E( i0 g {copystr(str1,str2,m);
! u& |7 h% L7 F( U0 J9 E printf("%s",str2);
* o7 t) ^2 O4 ^/ n( G' j }* u7 q7 r- @5 d- T
}& c5 x& K- x! T
copystr(p1,p2,m)
- ^, S* \" L$ P, s- \: Q) fchar *p1,*p2;
s' m. ^6 Q0 t3 b, R/ Y | {4 @int m;
" k' r1 @- L$ k% T, _{int n=0;
8 ] E% }: u# Y8 D- A while(n {n++;p1++;}( |# H' T( ?+ @3 m ], J
while(*p1!='\0')6 Y# H: T( ~3 g/ u Z. S
{*p2=*p1;
/ L, n4 E% d4 I4 X; D p1++;+ r& P7 h3 f7 Z& H- h* i
p2++;6 Q, G+ T" M7 V) k# Z/ X8 Z
}) S0 ]' z) S+ [+ O) Q+ o" t( H1 S
*p2='\0';
' L3 n% @2 R0 u; C1 u0 v}9 t* E7 }. R" g4 i; H( `
10.8
0 n- E& i& e7 f9 i#include"stdio.h": N* g# q( Y4 |4 x
main()
1 h! ~$ J: ~7 N9 w: y6 ^/ K{int cle=0,sle=0,di=0,wsp=0,ot=0,i;
8 I- W9 @5 o' ] char *p,s[20];3 C/ T* P4 l$ W+ t
for(i=0;i<20;i++)s[i]=0;
$ A( c; z% @ `1 P k* w, N# ~ i=0;
( s, H; D& N3 D! M ?0 ? while((s[i]=getchar())!='\n')i++;
' e3 I9 L5 ], D! k3 j. [# n p=s;
' c, X& o: y% f: k while(*p!='\n'), `& e' g( L' t7 S" h
{if(*p>='a'&&*p<='z')
( K8 N+ {; D( \4 w9 E ++sle;
( D7 @ l* r( l3 V3 i+ q5 {; ^ else if(*p>='A'&&*p<='Z')1 A9 }. V4 U$ Y6 B7 S1 P, I$ O3 _
++cle;
: I5 a6 w! E" L1 f else if(*p==' ')
$ K2 S# V8 K2 v% t3 O! s U5 Q ++wsp;; L' M* F `6 z: L7 _
else if(*p>='0'&&*p<='9')( q- v5 I4 F( ]9 B6 u, D3 i
++di;
9 b. _) N' o# e( j: w else
1 k, ?" H# b+ d5 Q: B4 E' c7 R, f ++ot;( H. p, @0 Y0 f) G; C h
p++;; v+ ~$ _8 | B! U. D/ J
}: G+ K9 s1 }! u4 R. u
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);8 j( O4 ^1 N+ r" f6 G/ T
}, G2 } F! N0 G% a( u% J
10.9
$ u: L O8 Z4 }- i7 g _& f& O" xmain()& v7 G* ~: E4 ]2 X4 y
{int a[3][3],*p,i;
+ _; z. g- u; h% ^. O for(i=0;i<3;i++)6 {6 W' F: m# V+ M9 }
scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);
3 p- u( |% U5 X7 p& N p=a;3 X+ b9 P9 h5 V- C
move(p);! r$ `3 I% z; `
for(i=0;i<3;i++)
+ T- R+ k! ^' D" R" c$ F' ] printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);
& o+ i Z3 k+ T2 m, W- t/ q! ?}- G0 q, G' l. p3 X o" U
move(pointer)
" [: {( O) F3 @6 c& ?int *pointer;" \$ Z% u; H, y" ~8 Z/ u
{int i,j,t;
3 g' J0 r% |7 w, O for(i=0;i<2;i++)
( f: o% { S+ V4 Q2 B# H0 X/ ^& h for(j=i+1;j<3;j++)
8 k9 v$ a2 k' G/ @% r5 C- h {t=*(pointer+3*i+j);
G8 s9 m+ `2 y* @# s *(pointer+3*i+j)=*(pointer+3*j+i);
. J0 T7 t" G Z *(pointer+3*j+i)=t;6 q+ O l$ P+ q( \
}, t0 v' E: H# ?3 Y2 w3 D
}& z. B7 h1 r! p( V
10.10
) a- D ~1 [ G7 p8 a$ jmain()# ?3 e( M0 L8 q& Y1 ]) ~& c( w
{int a[5][5],*p,i,j;
( r# }7 L! l# x8 g! J9 ^7 J: L for(i=0;i<5;i++). L, d! ?) ]' a0 C. ]
for(j=0;j<5;j++)! T- C7 L' C9 f% O v/ i, y1 @
scanf("%d",&a[i][j]);
# C5 x5 z0 q+ v( D" R+ Y p=a;
4 d" B4 J) U1 x O0 i change(p);
8 `& l8 J" D* p" F- j$ _2 O- m8 x for(i=0;i<5;i++)' i Q0 E8 O, o1 r9 r
{printf("\n");! b S; `' O0 S" s' t& L, [3 M
for(j=0;j<5;j++)' L _6 K: ^" w* C+ g. K, M
printf("%8d",a[i][j]);+ l! z% @! o2 V# f& [" A, C& |9 w! M
}
' A* E; `4 c6 w% _! ]" n+ Y' |}
X/ }$ j' U1 S- u2 h% i1 Mchange(p)8 F% A. V; i: K( @' f5 q: ?
int *p;# `$ L" z" i! Q; Q' r
{int i,j,change;4 H; p7 Z5 s! C* Q
int *pmax,*pmin;
1 @% c1 ?( W2 K' j) K pmax=p;3 H, z' N$ w+ b* y |
pmin=p;
3 R+ N; d" |# E* \ for(i=0;i<5;i++)
D0 m& z" m0 @3 ~* Z for(j=0;j<5;j++)
3 O5 b% W5 u, S& p {if(*pmax<*(p+5*i+j))pmax=p+5*i+j;5 u: Z, p* h/ y i
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;, g1 m% m# g* O" ^) g9 \4 x y# g
}
4 w& w8 b, f0 ?' {! O change=*(p+12);
b g( O2 t8 G5 @# ^* a *(p+12)=*pmax;
0 O1 S* a+ L9 m6 k *pmax=change;
- [, [! Q6 ^5 H6 z" X1 Y' V! ? change=*p;2 y- T* w6 U, W4 B& M8 A, ^6 m
*p=*pmin;: O$ h- L. ?+ M+ A
*pmin=change;9 U) H& J j) f' Y/ `2 k' f. S
pmin=p+1;
$ R9 X- Z4 w* K% Y8 c for(i=0;i<5;i++)
7 n6 e$ w/ Q$ q for(j=0;j<5;j++)) q$ ?$ B/ J: d
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
0 |* A) M! H; X& `$ D* C change=*(p+4);
5 b1 w* M8 S; ^4 n6 u6 U+ h2 Y3 S *(p+4)=*pmin;5 v+ D% M5 K6 q, j
*pmin=change;, D6 h* e; t4 R C
pmin=p+1;
3 M) c" {, g, e9 ]' O2 b% M for(i=0;i<5;i++)- j' [) p! B& D7 b; g9 P8 V7 b
for(j=0;j<5;j++)6 n. Q T- W+ H; ?( L/ Z* c+ e
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
6 X3 ]8 v) C& `3 { pmin=p+5*i+j;
N+ H) H& y: D+ |5 X: [ change=*(p+20);
! |, l5 O3 j2 m! T *(p+20)=*pmin;5 b+ F3 e2 {, Z
*pmin=change;1 K; C7 e* u1 N3 i& v
pmin=p+1;# `. I- B+ R' u
for(i=0;i<5;i++)8 W, `) o2 _1 h, U
for(j=0;j<5;j++)
+ n) f* Z9 `" y- m x8 Z D if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))% E' b# n1 U3 W& ]% b3 A
&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
# i; h- u& I2 x) n change=*(p+24);+ r* Y7 O! z. D' x7 V
*(p+24)=*pmin;
% _ w L- _) O *pmin=change;6 k5 _8 L0 M3 p8 H% s/ C/ x
}: r! |- `) G) z
10.11
7 I$ W5 W/ C- {# b% l Pmain()& W# j, {' D9 y2 J
{int i;6 S2 r! n+ M m/ ]8 V1 X
char *p,str[10][10];
' r# p" g4 o. N! p/ i, d for(i=0;i<10;i++); A* f. u0 B, X; K
scanf("%s",str[i]);
8 l7 R8 G7 K, G! S p=str;, Q% e `. I- q) u( g
sort(p);
& V5 Q/ w! V! \5 \$ H9 F for(i=0;i<10;i++)6 v' T$ V' ]1 y& Y1 W' C1 |
printf("%s\n",str[i]);
, A. Q- ?- i7 ^ W4 W}
5 T" y) ]6 T( F9 U/ C9 W8 G2 esort(p)& Z+ ^& z7 a* I6 y2 f
char *p;
2 q5 G0 g& Q- M8 n{int i,j;
8 B: D. i. }0 Q2 ]6 `, Y char s[10],*smax,*smin;6 W' C8 p( x/ H+ i8 Z8 i
for(i=0;i<10;i++)
2 ]/ o7 e# l% o' D+ r2 @5 o {smax=p+10*i;
/ J' x) U5 f( o$ f9 a+ n for(j=i+1;j<10;j++)
5 M7 w0 ~& Z- b2 z4 N, e/ Z {smin=p+10*j;0 H/ y4 f0 S; f& o' u
if(strcmp(smax,smin)>0)
$ X5 Y7 k1 M" b) k6 i {strcpy(s,smin);
$ q0 P4 p* B) o strcpy(smin,smax);' X7 ^' h6 s, N" \
strcpy(smax,s);
+ ]" w. z0 L: s9 Q9 l }
: D% A& h# T6 Z }: _4 C+ G" Y; I8 Q: k
}8 n% Q" @7 u8 I$ Z, T0 e/ [$ R
}
; Z5 J7 r' g* u' }+ Z \8 ~10.121 q5 H6 `0 R/ F1 g/ W
#define MAX 20
/ a! i6 U+ O& a' B$ Z! c; T+ jmain()
7 n% ~ A5 p" [" f{int i;
3 s l8 M( a+ X+ }/ t7 i char *pstr[10],str[10][MAX];
5 a9 Q0 k+ Y; k& k7 r: s for(i=0;i<10;i++)
* ]$ N1 X+ F7 F+ J" u pstr[i]=str[i];
) V; h& ~- i f* Q: I; b' Q8 a& r for(i=0;i<10;i++)* q% t/ J9 r" B
scanf("%s",pstr[i]);
: z( U4 E5 v8 ?9 ^- O% M# _ sort(pstr);$ K9 O" u2 k" R y8 z
for(i=0;i<10;i++)& L: X9 U) B9 o; ?( s ~ q
printf("%s\n",pstr[i]);3 Z1 R; o6 H4 f8 Z' K0 K
}
; U, K+ {& u: x2 b9 _( i) |7 B, ^& W( xsort(pstr)% ?1 T' f4 ~; ~9 n1 X# ?2 X+ }% g) H \
char *pstr[10];
1 M; r3 I! `9 J5 Y) m/ h{int i,j;# x. }- [7 b' C+ d1 q
char *p;
1 X9 {7 L! e/ s' f0 A for(i=0;i<10;i++)
! j1 X. X; ]# q$ U {for(j=i+1;j<10;j++)
/ T. F& z( m( ~0 Q6 y {if(strcmp(*(pstr+i),*(pstr+j))>0)+ j% k6 p6 O3 y1 d3 M4 p8 H
{p=*(pstr+i);
& E, f% s- y, P8 q. |) ? *(pstr+i)=*(pstr+j);
* h* j5 A; c- `% P+ m! w* x *(pstr+j)=p;+ {1 D2 {" I) m6 u) z
}
. h/ i; j6 {) x: _1 }1 V }
- x, w3 h! l5 } }
6 y8 q$ v3 Z U}* a" `4 H. m6 S ^5 ]/ c( _# H) o0 r
10.132 E8 O o4 E: W+ e+ V3 T; L
#include"math.h" k* A7 T' r/ s$ O E ]
main()9 ?( f9 b3 u: h+ j6 I1 m
{int n=20;
1 L+ s( i0 @- `/ u- Q+ V float a,b,a1,b1,a2,b2,c,(*p)(),jiff();4 z5 J7 n+ \3 Q; `) ` J2 J" q
scanf("%f,%f",&a,&b);1 T/ Y* o% g, M/ B
scanf("%f,%f",&a1,&b1);' v) z% X9 H2 H3 k
scanf("%f,%f",&a2,&b2);( Q! v# @* X9 M
p=sin;5 j7 v; O" l* \
c=jiff(a,b,n,p);0 R0 U/ I/ Z% K; Y( |( Q1 @/ X! T; V
printf("sin=%f\n",c);
' T" }( X3 i! z4 h q p=cos;6 ~( U1 P, K9 E; {, g* u1 o6 _
c=jiff(a1,b1,n,p);
' a0 i9 i% Y0 \ C6 ~+ N. q printf("cos=%f\n",c);2 B9 }+ @% ]2 {: P0 |
p=exp;
3 L6 T/ C# G" C6 S5 e c=jiff(a2,b2,n,p);) J( I5 h# |7 h! W5 l
printf("exp=%f\n",c);
1 h2 n. P9 }8 D: Q [}) r, A, Q% ?3 Q7 E) U4 w
float jiff(a,b,n,p)
3 u: z, }; b4 x2 Y1 p; [float a,b,(*p)();' ^9 d( B. ?# B2 p e8 A
int n; c" }9 b9 m' N" `. k+ \( S
{int i;
/ e6 R9 E! k2 A3 E: u% m float x,f,h,area;
; o3 _5 }2 m9 @9 [) f' M- L h=(b-a)/n;' I8 V/ M0 o# F! B6 A
x=a;! d E( t" e0 q8 r6 [) R
area=0;
1 c0 s; `! G0 _; p for(i=1;i<=n;i++)
$ o2 p! Z; q# ?9 Q. a6 t- D5 U$ \ {x=x+h;, H1 ^# g3 W! d1 C) E" B9 p
area=area+(*p)(x)*h;
3 K, D; G3 p( T& G: Z2 `) e, k }
3 [+ d) N8 e, n) M$ P% G return(area);* z3 H# G8 h2 ?( t n1 t
}, P8 J% G3 K8 r' m
10.14
) I! P! i4 ]. ]main(); I. A8 r8 x2 C9 R6 r
{int i,n,num[20];! y- p" G4 r% ]; k$ s$ ~6 l: T
char *p;: m- j" M) K+ H7 y6 L6 _
scanf("%d",&n);' _# y* W; W0 ?# _, q/ G6 z
for(i=0;i scanf("%d",&num[i]);" e* N! Y' b/ n p# L
p=num;
( f2 R j0 D7 w7 B2 J0 J4 k sort(p,n);, [2 S+ g8 @- i. k
for(i=0;i printf("%8d",num[i]);3 f* `! Z6 ?3 `9 |4 _# B1 I) R
}
& U* K/ k& @7 ]! jsort(p,m); p$ y; |7 Z7 l$ O% y: n
char *p;
4 `* ~% h; ^' b. G( @int m;3 k9 M; d" `8 C. y4 i6 L* `) \( O
{int i;, d# @& p! o2 C% n
char change,*p1,*p2;& c& Y/ o+ W7 ?2 E
for(i=0;i {p1=p+i;
+ M' F" s7 c9 O# }+ k0 l u0 i% f p2=p+(m-1-i);
& U) {2 ^; a v6 y* F change=*p1;* I7 @+ k6 p* q+ J) V) y; ?, c
*p1=*p2;
! a% U5 x$ {2 Q* }. H9 t0 f *p2=change;- S7 p5 R+ T+ h" r( T I4 @
}
; `( Z# K4 e! g9 i# f* L: O+ b}0 |- w" A' m" r! ^6 L
10.15
/ Q; |" @+ T2 D- s- z) J+ k( `main()% u5 g2 b! {! h0 K1 F
{int i,j,*pnum,num[4];
5 [6 H2 ]" B( L, r) J float score[4][5],aver[4],*psco,*pave;
" j" a k( R; J) k. O$ _ char course[5][10],*pcou;6 h& ?/ k2 f1 C: I
pcou=course[0];. O" ]/ v& {9 i1 R; o
for(i=0;i<5;i++)$ t' g6 C$ ]6 ~3 i p
scanf("%s",pcou+10*i);+ P$ i+ s$ Y2 ~( |
printf("number");' m8 d) \# O2 x @# r
for(i=0;i<5;i++)) T! t# F3 _) G+ A9 i+ r
printf(",%s",pcou+10*i);, q3 X8 W1 |4 _% ~) K
printf("\n");
; k+ s! C- H3 H' H8 n7 e psco=score;
& W! i' ^2 @! s pnum=num;
! Q8 h/ S! L7 O3 r# D9 [4 C for(i=0;i<4;i++)
- n& w/ B' {& u" U& C {scanf("%d",pnum+i);
2 J: z5 J' o; ? for(j=0;j<5;j++)* C0 l/ }: E1 Q" S$ t3 b
scanf(",%f",psco+5*i+j);' v: c5 y) x& Z% H( e0 E4 v P/ `
}' A+ r) ~( r* f! n- A( d5 Y
pave=aver;7 z$ O8 _: d) b! E L" }$ \% V
printf("\n");
+ Y( N0 u' L! {: c, O avsco(psco,pave);) B" c. }7 [9 b' {8 h+ P
avcour1(pcou,psco);; Q7 k7 r# [- M% d
printf("\n");: f1 P* o4 j6 K; i: o. t& y
fali2(pcou,pnum,psco,pave);9 }$ k4 ~. k/ ]; D$ o
printf("\n");. I! O h" i2 @& s% [/ i8 F
good(pcou,pnum,psco,pave);
* ?; G9 e/ L& j+ a/ x: a5 m$ q}% T+ q5 w0 g# V3 V" ~/ z7 H& V
avsco(psco,pave)
# e4 D# M0 y. v- afloat *psco,*pave;+ e3 S% _6 q( } P2 P) p: A6 U
{int i,j;
! n* }8 T$ k' C& P7 s1 V2 j4 s float sum,average;2 k& r1 _8 t5 I- D) ?: t7 l
for(i=0;i<4;i++)
1 q# @' a5 x6 g1 v" x- x {sum=0;4 i& _& l; \# `) j
for(j=0;j<5;j+)
& K n* s2 ?$ q8 E& r+ O. Z sum+=(*(psco+5*i+j));/ A2 R& w" E/ l2 }8 A
average=sum/5;
$ ?3 H- G( v5 S ^2 d *(pave+i)=average;3 c$ n7 g N" f* [
}
( O8 I+ z/ l8 q8 W* B4 [3 F' a' O}' ?3 {, g, |5 p
avcour1(pcou,psco)# ?) a+ }( x) F# b! d o5 s
char *pcou;/ X3 R- w2 N! b2 ~; N8 i% O
float *psco;
( ~9 F" P$ o4 y5 \+ E{int i;
9 A- z% @. \" ?% w) b4 U6 J float sum,average1;9 U7 p; x E9 T' }8 w$ [
sum=0; w+ x5 W- C& n- z
for(i=0;i<4;i++)
( t* V9 Z& f2 n% h sum+=(*(psco+5*i))
, h+ C5 q& w" ]# T% h average1=sum/4;" t2 j% d$ E7 T q! m3 q7 i) z$ n8 f! F
printf("%s %5.2f\n",pcou,average1);
* g5 ?8 Q/ }5 }% d}7 l$ {( B. n$ s. e: `
fali2(pcou,pnum,psco,pave)
' V/ }0 e0 N/ L n) F8 A% |: pchar *pcou;
: V6 K7 E' `6 `! Z3 W1 Qint *pnum;( }- z. N1 s! t8 v; x/ U' O; @$ F
float *psco,*pave;
* W8 t; m+ [) j U* z; y{int i,j,k,label;" \1 r3 I, e) b& M& ~; m
printf("\nnumber\n");
& \, j1 x) N6 D$ R+ i for(i=0;i<5;i++)' l! `, P: C+ W% s$ ?3 Z2 E0 u0 t
printf("%-8s",pcou+10*i);3 ]$ s6 q; @+ b" B
printf("\naverage\n");
( a" C- K4 ^; v+ s: n; W$ \ for(i=0;i<4;i++)
: p% H; Q1 t8 e g9 b% s0 |8 J {label=0;: b" [5 ~5 n& `& m1 J9 u3 @ i
for(j=0;j<5;j++)* K1 S. ]6 ], f' U
if(*(psco+5*i+j)<60.0)label++;
1 W. p! k. I$ z' C; h! I. n3 O9 A) V if(label>=2), ~1 |8 |) R9 `% k$ F1 f
{printf("%-8d",*(pnum+i));' y$ Q/ @8 v* n& u+ O( Y5 b) @
for(k=0;k<5;k++)
* u" [5 i8 ?- f/ ^5 O printf("%-8.2f",*(psco+5*i+k));2 c. ~# \% B* N
printf("%-8.2f",*(pave+i));5 y( c) ? q) h3 v! O& q7 ?
}* @+ U- x: j' C, K
}
( T3 a+ i* F' s7 `}7 y3 d% u* U' E2 L% O6 T
good(pcou,pnum,psco,pave)
6 S( k6 S7 V. f2 k# @char *pcou;; s: a! o d) R
int *pnum;
/ R$ U j# t9 K) A( J# ^float *psco,*pave;
1 ` R2 Q, S, [) Z. P& h6 I{int i,j,k,label;5 E/ D0 T' ]% K2 F! E! g/ A
printf("number");
. w- s' b; C! `$ `+ | for(i=0;i<5;i++)5 `4 d+ \1 s# G* \
printf("%-8s",pcou+10*i);* l. B/ r; D$ O* R0 G; \
printf("average");
+ ?; O+ N! l# Z' Y ^0 K for(i=0;i<4;i++)' n0 b; m3 e7 L2 ?" z8 b- h# d6 n! a
{label=0;, D" g5 Z; d9 W, y
for(j=0;j<5;j++)
* ]$ G. R$ u+ g% K if(*(psco+5*i+j)>=85.0)label++;
4 g5 [; H( `% m) r5 N! d if((label>=5)||(*(pave+i)>=90))8 ?! F9 y" x/ B' P8 `
{printf("%-8d",*(pnum+i));0 j9 ~$ ?7 l; K0 q9 B# `; w4 w7 E
for(k=0;k<5;k++)* w+ |- L4 ?+ D5 A
printf("%-8.2f",*(psco+5*i+k));7 C$ u6 Y+ }, G/ k. J/ c
printf("%-8.2f",*(pave+i));! d. Q! v3 `: N4 J
}
6 A% u4 O& {$ b }
2 g& o. l. @' H# v}
* D* @+ u- i: m10.16
+ ^5 F4 K4 o+ X& E. o#include"stdio.h"
4 r8 `9 G9 h! ~main()3 k4 W% J) g& U F/ P( t
{char str[50],*pstr;
7 |$ u2 R# s0 ?% H9 r6 b8 a* C int i,j,k,m,e10,digit,ndigit,a[10],*pa;
1 j h7 a ?! N) y) i. W gets(str);# p/ t( M6 M1 a2 h4 E4 m
pstr=str;
7 l! }1 R1 e+ J# |* l: k5 I& {$ Y& z pa=a;! c G3 o. N. e' y
ndigit=0;1 {0 K' j) X' j3 L, A
i=j=0;& h( B: E6 q8 v
while(*(pstr+i)!='\0')
$ f. V& A; l' P: {+ `& O5 B {if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))- b& y/ q1 K* m% K% |2 w
j++;
7 C8 `4 U5 {8 D. ^ `8 a5 L2 `6 \ else4 d8 u; J9 Z7 \1 G
{if(j>0)8 Z8 n" Z. F# l0 H, r
{digit=*(pstr+i-1)-48;
' p+ i- p6 L/ G4 S" B$ ? k=1;- l/ n( c& ^# Z- k
while(k {e10=1;( y& _5 i9 H0 r6 s% z
for(m=1;m<=k;m++)* _1 e& A, V: `% N6 k) R
e10=e10*10;
! A. d5 }) a/ o: B! X+ w digit+=(*(pstr+i-1-k)-48)*e10;& t% y( b7 O1 F& Z$ s/ [
k++;$ ^: c; A2 f( y! W& u! v
}+ T3 B9 j. D' y' T6 T
*pa=digit;
% i# w0 S8 t8 N2 K ndigit++;+ j6 E4 V5 E- C1 j2 U d% Y1 e
pa++;+ T% C J: Y* x/ w1 s i. w; m
j=0;9 y F; ?2 x/ }- ]% @1 q
}- p! P1 m8 N3 k
}0 Z' L. v c, A, J: W1 ^ j
i++;
0 M' V" \% v, c3 [$ ^. t }$ b7 C2 c( o& U7 ?/ _$ D4 w
if(j>0), d+ u- {$ l5 K1 z' n; [+ b) x
{digit=*(pstr+i-1)-48;
6 f2 Y3 a, t- f9 u k=1;/ }) e( ?. H* o) y( X
while(k {e10=1;
9 P" H' D j- m+ f for(m=1;m<=k;m++)6 `0 t% T) Y" h3 T
e10=e10*10;3 B* f6 t1 v8 g
digit+=(*(pstr+i-1-k)-48)*e10;
2 s. l5 G, H6 n: V% ~ k++;
5 Y# C' L" d, z5 z } R6 m; m* ]0 i; X- o: n# G
*pa=digit;
- I+ E. W7 b, Z ndigit++;1 u% _/ ?$ [* a9 ]& Q9 J4 L$ c0 c
j=0;
! K' q9 ~. V9 n* ] }
- _( J" a! V& ^2 ~* r2 `# h" c printf("ndigit=%d\n",ndigit);
* [" z8 T" i7 Y* |% d j=0;9 |3 p0 J+ u1 n6 }! e5 E, Q* ]9 H
pa=a;
9 n* t# q7 k0 X1 S3 B. i for(j=0;j printf("%d",*(pa+j));
6 M4 d$ n) ?1 o- u3 ?* N2 a}
8 k, ]/ I1 t: i/ K10.17/ j' I6 M( C: v
main()" }6 E: D( z0 Z* g
{int m;2 Q G. w& ~" W# y7 a* \
char str1[20],str2[20],*p1,*p2;7 ^- i, n+ }8 @7 r( c
scanf("%s",str1);' c: G- O2 e- V6 q
scanf("%s",str2);
- L& } ^2 Z( `4 i# K p1=str1;
7 r, {3 I+ P- x: g) c; X p2=str2;
& b1 F2 @! A* ?$ U n/ H( n5 N m=strcmp(p1,p2);/ z# q* p$ c( c7 \$ L, w
printf("%d\n",m);
: K! J @2 ], A3 L: J}
: j3 W" `6 r% y* m2 _strcmp(p1,p2)9 n2 a4 C# w/ V( B G
char *p1,*p2; G! }. w2 p( R0 w' l
{int i=0;
, O R1 y& O/ U while(*(p1+i)==*(p2+i))
9 E7 x7 q# T. z# e) U if(*(p+i++)=='\0')return(0);
5 _$ ?* b- w$ { b$ \* o6 j7 d return(*(p1+i)-*(p2+i));
% U! [9 A7 |1 {7 G+ }}+ [7 w, \1 W$ [
10.18! {$ t2 ~, Y+ ~3 w% e9 _0 K+ D
main()
" H; k! I& d" ^3 M' {# A! g{static char *mname[13]={"illeagl","January","February","March",% p0 N+ y' K3 L) l" x
"April","May","June","July","August","September","October",
/ m: n" r6 p/ c9 ^0 s "November","December"};
$ o1 r8 X5 t. M- n; \8 W' r# }- t int n;$ k2 P" `3 h/ h. a8 x2 I
scanf("%d",&n);& {1 Q8 i' J& A% ~7 R1 K
if((n>=1)&&(n<=12))
# c1 [ S0 t/ k2 l printf("%s\n",*(mname+n));, B8 A5 Q1 b5 p
else/ F( e' k' y+ A6 ]% Z4 Z0 ]$ z
printf("error");8 X1 \/ q- W- i7 T4 l
}4 p0 E& i) E/ C' a
10.20
3 W9 H G" {% {/ N9 Emain()( \2 L" w& `: x: s$ p7 v% o
{int i;
+ x8 R8 [; q4 k+ ? char **p,*pstr[5],str[5][10];
- s- N% Q* m6 V) q" m5 F& B$ \4 k for(i=0;i<5;i++)0 [) B8 a2 N% J6 F; Y0 S! X
pstr[i]=str[i];
0 [+ h' b! ~" F9 p for(i=0;i<5;i++)
9 t0 ], A3 n! Z, g scanf("%s",pstr[i]);0 ~9 G2 x: U3 U8 |
p=pstr;
4 C- p' f! ?( G1 q) k sort(p);
2 z$ b8 s4 R6 B for(i=0;i<5;i++)
5 S s+ W5 R+ l+ f- i" `( \ printf("%s\n",pstr[i]);2 M, w1 t o! e, [4 [
}3 l' ]5 { s) T! B$ k, D
sort(p); a/ w# E( Q. ?8 }1 z9 N0 Y5 l0 f
char **P;% J# L+ ^5 F& w$ N/ O* l4 \6 v
{int i,j;- ?' X+ f" m3 a+ @9 c# w. A
char *pchange; A1 \6 B# r# k
for(i=0;i<5;i++)
8 i; @& g$ X: N1 i3 I {for(j=i+1;j<5;j++)
3 U: |6 B3 t: |0 [- n2 e {if(strcmp(*(p+i),*(p+j))>0)7 r9 Z* c7 B, [ C. w9 }, i$ V
{pchange=*(p+i);( ^; K9 m1 k6 h9 G' z( L0 f
*(p+i)=*(p+j);
8 ^' q- q) b1 ?5 r *(p+j)=pchange;. A' d; c* Y( l# G; c/ F
}
& G& w; T3 T! N& J* B- o6 m( L }! n9 \, x0 Y5 p0 t# o, S0 k( Z
}4 B( p& f3 \% r. o: R
}6 R! ]8 ^! R; r
10.21. A6 p# R: C. k' M4 X
main()
5 @' b* R( \- H- y9 G{int i,n,digit[20],**p,*pstr[20];; ^# y: b) S; p/ Q) e& n
scanf("%d",&n);
- v' `5 u( _: `1 Q5 F; c: R' G6 Y$ Q for(i=0;i pstr[i]=&digit[i];3 l+ k+ F4 z+ c! I, O
for(i=0;i scanf("%d",pstr[i]);
' C, ^$ D+ y- `1 q# b. n p=pstr;
2 x! S B# v' D/ k- n sort(p,n);
/ ]0 R! N1 N- ]* ?7 \- k% N for(i=0;i printf("%d ",*pstr[i]);9 Q! c% b' x5 ^7 b3 }0 O
}* Q4 v& @ m5 E: w# H- n2 n
sort(p,n)
3 e: A# d# a+ Bint **p,n;
( _6 l# C. o; ?3 J2 `& ^ e. k{int i,j,*pchange;. g; ]" F2 C: h4 J3 S
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
& u9 a# A& x4 K {pchange=*(p+i);
# Q8 F5 L( o! _0 I, g8 F *(p+i)=*(p+j); I0 @; y* p6 F! u. j: P: w8 `; ~
*(p+j)=pchange;6 B' h. @6 X3 m6 p( S0 t
}
8 ?( d* K% }7 V }$ [& l; F v0 g* M+ I1 B* n; `- r! e
}
( f+ \0 d" t s}7 t: B+ I+ t, V% ^0 u, e
第十一章 结构体与共用体
* J- @% h$ Q R6 i/ j- D& x11.1" S+ K) L* q0 [1 S
struct
$ S) @" _6 W- A1 j0 ~6 f1 [4 Y2 w {int year;
8 e, [1 k' ~! H1 H% N# Q" E# j- Z int month;+ |- x" t3 z I( X. X5 R
int day;+ T, ^9 w* V, v" k% V
}date;
: {: r o! a2 q+ q4 ~$ umain()
' _/ K* ^. e& Q: r6 L{int days;: f3 `' k& Q9 w, m" b
scanf("%d,%d,%d",&date.year,&date.month,&date.day);* u0 t& B" ?5 ]+ W( V8 S7 W, S
switch(date.month)- d9 |$ z4 T2 W8 Q" @
{case 1:days=date.day;break;
1 r. D1 t" H" O% ~ case 2:days=date.day+31;break;! K* p% }8 A) L" v
case 3:days=date.day+59;break;& r }, G) }3 Q6 t
case 4:days=date.day+90;break;
. c8 \! o$ H; s case 5:days=date.day+120;break;5 o; w' ]: r( B4 ^
case 6:days=date.day+151;break;9 |) t4 o( E/ |0 ]! E) G
case 7:days=date.day+181;break;; b. A1 ^& ]: G: p c8 M n
case 8:days=date.day+212;break;
6 [1 O9 S- e0 H. j5 t case 9:days=date.day+243;break;
& J4 |5 W4 \3 |0 ?) }( l) H case 10:days=date.day+273;break;& e7 A3 j2 T& W; s6 v) G
case 11:days=date.day+304;break;
6 m3 L7 T9 H |! s7 \ case 12:days=date.day+334;break;" h1 @, F; S" ]1 D! m/ O
}/ M, x( s9 o- n& I7 j/ d
if((date.year%4==0&&date.year%100!=0||date.year%400==0)( k1 E% c# L+ `! L0 f r
&&date.month>=3)
: S' s( S4 w5 y Z/ k% h# v/ \0 K days+=1;
3 n5 D6 J8 y' ]% y( k printf("days=%d\n",days);
9 o) b! ]7 s2 E R}
8 s4 x" f! Y7 ^11.2
( d- Z5 L5 i7 b0 f5 t a, Gstruct dt( t! `2 V- D0 W% m
{int year;! o" _7 F1 s+ {0 u2 N# A
int month;+ i$ C! k& }; q% M
int day;
( A& `' t9 i+ ~, n }date;
% S1 f4 R+ b$ I1 Q6 T2 Nmain()6 K* G7 E: I: p; C" A. b `
{3 N& V b5 a5 m0 v S* S
scanf("%d,%d,%d",&date.year,&date.month,&date.day);6 N/ W. o W7 k1 J& { A4 u5 C
printf("\n%d\n",days(date.year,date.month,date.day));
) j* J$ B! [$ d' `: F}( w6 n, J. Z6 p. C5 }2 `2 K
days(year,month,day)6 ~, h% l' a. `# }
int year,month,day;- j: r5 f) j1 _
{int daysum=0,i;
; _) ?; v! A& _1 ^! O static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
+ }3 ?* i2 l2 z+ b( } for(i=1;i daysum+=daytab[i];' J" L7 v9 n1 L0 ?# S
daysum+=day;
# M7 m% _3 d9 Y' [" n if((year%4==0&&year%100!=0||year%400==0)&&month>=3)
8 Z' d/ H n+ i daysum+=1;
" \7 r1 {+ n) i7 a, ~8 z return(daysum);7 ]" y7 V) u% M. N
}
/ J" ~2 w& J6 O0 T11.3
8 L o5 D% {/ I) f' z11.4' [8 ]8 r& E: U
#define N 5
' T3 J- Z* `9 N$ q2 M% Zstruct student9 k$ G9 S- K0 B9 V
{char num[6];, F H4 X% G& |
char name[8];
2 l @5 b1 F- d0 f$ P0 n int score[4];0 h$ l, J, q# D: ?* S* V
}stu[N];/ J3 K1 Q# s% S9 O- A7 h n( q9 P
main()
7 T; }, k3 B8 X1 o" {9 R{3 M S/ I0 t* f( u0 `
input(stu);
8 d2 S9 A0 r b9 T) r& y print(stu);8 {: g- y2 q3 {6 Q W3 c
}
( e0 l* {7 L/ A }9 P( Linput(stu)) e) I% {% i7 B6 o% ~! {' A. a
struct student stu[];
- b9 J2 N @ i{int i,j;
* ?( Y8 F1 _( T$ c/ N: [# h/ ~ for(i=0;i {printf("number");
" B/ F% ]. B: ?& r scanf("%s",stu[i].num);
% C6 U! V& S9 D& C! M printf("name");
Q) X" Q9 \% R, H# g4 N scanf("%s",stu[i].name);
+ b" [9 Y ]" [4 u j for(j=0;j<3;j++)
e t! O! w* k0 s. @$ F {printf("\nscore\n");
4 f2 a. @4 j: E1 B7 i scanf("%d",&stu[i].score[j]);
7 T: W. V% B: m: f+ R0 F' f+ e4 p* d }; C7 k+ b$ A1 E2 V; ~; Z( g
printf("\n");
9 v/ A+ j! l' U }0 [" ?- U7 F* @6 D6 ]
}
0 K) f1 |4 e9 iprint(stu)5 H* i8 A0 P' N" ~( v: a8 V
struct student stu[];- U- o; @8 R. p# m5 r
{int i,j; I1 A- ^. i% L/ u! J$ f0 ^, ?
printf("\nnumber name score1 score2 score3 \n");# A3 m& C& O+ w
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);9 Z+ w7 B* p2 `
for(j=0;j<3;j++)
& F+ E$ K4 ^; R2 z$ b8 ~# G printf("%7d",stu[i].score[j]);, I- J( `1 |" I0 ~. l- ^4 h1 r
printf("\n");6 h9 b6 ]/ p6 I4 ?
}: u# n& u+ s3 F$ R" o
}
% l' h' Z% P0 y0 {' C11.5, ?: D& o0 `' C5 w& V' ~
struct student) ?% q! e* t4 x* @
{char num[6];; B3 S9 j0 p" G& K: r
char name[8];
' [) Q- t$ a [; }7 P; m: c! O int score[4];& m8 z/ d; @8 }1 M: a* L
float avr;
1 n7 c7 ^7 t, k( \3 ?6 K }stu[5];# L) j+ U* y# ~1 y
main()
" O3 q/ i4 X a# N4 x2 e4 |{int i,j,max,maxi,sum;
# I y1 m+ F% X( U% }, n6 o float average;8 a3 H& B' T! g: b) \
for(i=0;i<5;i++)
}1 T/ Y/ T/ Y9 ] {printf("number");
; f' ^( r! t' ^! [ scanf("%s",stu[i].num);
9 I2 @5 A1 [! Q1 l0 v printf("name");
) v9 f; h) q- { scanf("%s",stu[i].name);
, u* b. Y: H; C9 ?7 E4 ] for(j=0;j<3;j++)
& ~$ r( p4 m4 k7 K4 O6 o# D3 t& e& L$ N {printf("\nscore\n");$ M% Q e8 V9 ^
scanf("%d",&stu[i].score[j]); |1 C0 N) v* f1 y' e" |
}
- A: A' g9 ~2 _2 H# r( f! \ }& x n8 c$ @. Y( D! p
average=0;
9 h1 j5 z- h& s, w6 Y# g( y max=0;1 k( a) p1 e: m' ~& l* h
maxi=0;
% N3 A0 k( z9 e7 { for(i=0;i<5;i++)
# ?# H8 E1 e4 G$ G {sum=0;" ~, S9 u+ x* s% o9 h3 u5 d1 L
for(j=0;j<3;j++)5 d j! E/ D* s2 w# O
sum+=stu[i].score[j];2 [8 z; [2 A+ e) s+ Q
stu[i].avr=sum/3.0;( v6 S4 t/ F5 G6 x, v1 N5 s
average+=stu[i].avr;
L: J! Q' [+ X- w2 ] if(sum>max)% g4 \! x- D* L9 w: C
{max=sum;
1 w; ?5 d. j- U3 O: A maxi=i;
- E7 h6 P2 M9 D1 z; x: \ }
h) h, \2 j; G. @& `+ _ }
5 v! s! f- U3 n average/=5;. W: h6 g2 {0 |: H
printf("number name score1 score2 score3 average\n");
8 J2 S* i- N5 e) s for(i=0;i<5;i++)
2 p V3 T0 b( x& W" N1 J. t( W l {printf("%8s%10s",stu[i].num,stu[i].name);
/ u) z/ W9 e4 }8 a) O3 k for(j=0;j<3;j++)- [% p! B# C# u& n. q& T3 L
printf("%7d",stu[i].score[j]);
4 P* b8 r" {) Z/ W7 U$ A, c printf("%6.2f\n",stu[i].avr);: S4 a3 Z7 m# ]& r) ~
}9 S% }3 J$ Z; e i& E- r; s
printf("average=%5.2f\n",average);
' b- l( \. s9 d4 l( B2 j printf("The best student is %s,sum=%d\n",stu[maxi].name,max);
# _: \& X. s, N}' a) b, ?& S3 O4 o; {: {7 ]2 T
5 x' S& B! Y& Q7 } |
zan
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