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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
|---|
签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案, ]9 ^- U0 O) I# z6 O% q; ]
1.5请参照本章例题,编写一个C程序,输出以下信息:
; ?% G" g/ d6 ^: M1 Dmain()
: V( E; k6 h' P! I( c- v{
8 V( N4 n- I2 ]" aprintf(" ************ \n");
4 o& j% g$ F0 Eprintf("\n");
7 K9 @. ~4 p' j2 A0 R) t# x @/ H! mprintf(" Very Good! \n");- T1 l$ q1 ^0 L* c( b
printf("\n");5 U* h2 A; D! {. |, e" y
printf(" ************\n");' w0 m: ` x* k, S5 q8 N
}6 K4 g2 E$ d% w& d: [! }
1.6编写一个程序,输入a b c三个值,输出其中最大者。
1 U4 |1 d3 q' N. A3 w% f: O解:main()
+ Q* A1 {3 a) z, T1 z7 v{int a,b,c,max;9 w: y, d2 ^2 |( |" ^) \: N8 V
printf("请输入三个数a,b,c:\n");
, x5 C! p& x, Y: g2 tscanf("%d,%d,%d",&a,&b,&c);
) B& g. q+ u5 _+ P2 mmax=a;7 W2 H4 Z' x5 Z) u5 U- V
if(maxmax=b;
' g: ~+ S n- Q. b, h7 _if(maxmax=c;
% _, v- u! E5 `) K; Sprintf("最大数为:%d",max);
6 K/ v9 v; _4 k: f$ V/ a. i0 X1 C}, t0 d- l- k! L: ^6 j Q. R
第三章
2 B, o/ M4 M/ y; M3.3 请将下面各数用八进制数和十六进制数表示:+ w) w9 T1 {9 l
(1)10 (2)32 (3)75 (4)-617
0 e5 d' Z% O0 C" Q1 P+ E(5)-111 (6)2483 (7)-28654 (8)21003+ m% ~+ e/ m6 V* q" j
解:十 八 十六
8 p( E2 L" K Q2 l- W( W3 J (10)=(12)=(a)8 Y. b0 Y: e( V$ ^; Q% J
(32)=(40)=20( g; w' a3 J1 t& r$ {6 B, G
(75)=(113)=4b3 V' n3 m6 N# _- _' g! g7 ~
(-617)=(176627)=fd97
, y, S7 k7 w L0 p# I -111=177621=ff91
7 J8 I' W# Q& H) s 2483=4663=9634 E2 U! W( f% \& }- |& I; E1 Y8 M
-28654=110022=9012! c6 C1 C( q/ V$ z V4 B1 k8 P
21003=51013=520b
# W8 ?! l3 z; y0 z8 i3.5字符常量与字符串常量有什么区别?
0 r% F. m: q6 a; ^4 k/ 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5 T* U& N" ]- C" ~, v- b0 Y
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'.
2 v' t u. a8 u! X, E5 Z: m3.6写出以下程序的运行结果: n. f3 R& a% s
#include
" L' i4 h( d) d7 E) L# @void main()
* A1 E& A+ a9 d{
; a! P7 A) r( [% w+ Cchar c1='a',c2='b',c3='c',c4='\101',c5='\116';
, J& z* }$ N& ]printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);
. i e) @! ` Iprintf("\t\b%c %c\n",c4,c5);4 W/ s6 @5 J5 p4 j8 ^6 J7 h. ?( _
解:程序的运行结果为:
/ W0 b4 {" P5 @aabb cc abc* q0 }% q& l6 T. j, \/ f
A N
& ]. C; y' ?7 t; o+ C! D3.7将"China"译成密码.密码规律:用原来的字母后面第4个字母代替原来的字母,
1 h+ ?$ t* y5 y例如,字母"A"后面第4个字母是"E",用"E"代替"A".因此,"China"应译为"Glmre".. j h! I6 r- a4 l3 B& i h) o
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 t. g1 ?7 j$ p& b输出.7 P3 [ O' |! K4 h% ^3 C) C0 h
main()1 y' a/ l5 l' F% a2 ]
{char c1="C",c2="h",c3="i",c4='n',c5='a';! B7 q4 n* E* z0 ]6 C
c1+=4;) a& d9 ]; v# I B
c2+=4;
( s* b+ G1 ^3 [c3+=4;
! s( R1 j& ?3 _% Zc4+=4;
9 v1 k4 d4 Z2 R* z5 ~# bc5+=4;3 F- \& [5 D4 O6 \1 R- e
printf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);8 K3 S |1 r9 q# ^0 o* D8 F' a5 J
}* j. A& x5 O0 p5 K( s" b& ]
3.8例3.6能否改成如下:
' l3 J6 M) p9 `4 r/ U# O# H8 r& L#include: _0 n" x5 \8 [8 X5 p( t n
void main()
c; t0 w w; T* V. C- q1 \+ V( _{, ], e% Y6 I/ `. H" }
int c1,c2;(原为 char c1,c2)
' m6 U+ l. Z0 L# h* Fc1=97;6 E" W# @8 u ]/ W5 T8 k, D
c2=98;& c) ~# L6 X [
printf("%c%c\n",c1,c2);! m- d6 [2 F* O' @( `' d `3 ?
printf("%d%d\n",c1,c2);( t! @2 h1 q5 N! r/ j+ K5 [
}$ [* s, V) `2 n
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.5 `# u" F- g5 J$ y5 [/ z
3.9求下面算术表达式的值.
: O- W# A1 ^( g(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)8 t- N7 x: Q2 H- C9 ~( H' l. S
(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)) U0 E7 [: U* v3 `1 I$ [; u
3.10写出下面程序的运行结果:! C9 v6 i' z) [2 f! J* g
#include( S+ F6 ^. U: ~# `/ o
void main()
3 Q; l' h1 _. v' x1 G8 I{7 n% s+ s. G- N ~5 o Q
int i,j,m,n;
) k' b4 s2 P/ Z* E7 C' ^i=8;, ]) g, V" i3 z
j=10;! Y3 r! {& _$ d5 z2 q' M9 _$ f
m=++i;; i3 M! H- `) ^$ S6 m$ d
n=j++;& n/ Z2 ^; S/ ~5 K0 v7 N8 ?0 C
printf("%d,%d,%d,%d\n",i,j,m,n);
4 d' V* K, }4 o9 C2 m w$ o}9 w8 w Q/ Y1 t) s+ M9 j5 L
解:结果: 9,11,9,10 O8 S2 c; q% J" W) e
第4章
$ ?- E6 c. a9 g% N3 G; J; c4.4.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' _. G# T5 ]8 z8 E1 p2 B( P4 ^* 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:8 v1 S9 m7 d2 R( {3 ^6 _
a= 3 b= 4 c= 5
. h1 \. D/ }, b. s& _; Fx=1.200000,y=2.400000,z=-3.6000003 l5 e$ H4 N# X3 J. O) b
x+y= 3.60 y+z=-1.20 z+x=-2.40! Q" B0 ~( ?, r4 D
u= 51274 n= 128765" C0 c1 S: p/ \: @" O* b" M5 K4 U
c1='a' or 97(ASCII)4 m' }7 i2 A. Y7 ~! ?0 @
c2='B' or 98(ASCII)& ~$ z0 J0 R+ u
解:
6 ]3 b0 I. T* R- m" Emain()( E4 ^) p# t1 R* U9 G, e( H
{
. o5 m' ]6 Y" s3 z2 Qint a,b,c;; `( [) U( |/ @/ T: T1 [$ S
long int u,n;
; w/ s7 n' Z$ Y _float x,y,z;4 N) o& \7 ^0 j: p. J4 S& A" K6 [! p
char c1,c2;$ L6 P& _! z5 u2 n
a=3;b=4;c=5;
& _) j1 t) n( W& qx=1.2;y=2.4;z=-3.6;
0 z! ~- ~) @% iu=51274;n=128765;
8 Z" k6 U& Y8 s. qc1='a';c2='b';
: b1 s' s: C9 b: aprintf("\n");2 C$ G7 N, j& C- ]& a/ P- W
printf("a=%2d b=%2d c=%2d\n",a,b,c);( r3 P3 G: x- v
printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
5 a' o5 S" G- u, n% y! N' }" X4 Bprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);
8 _1 }7 ]2 o8 ~# I' Xprintf("u=%6ld n=%9ld\n",u,n);8 O# z6 r( B7 E3 L" H
printf("c1='%c' or %d(ASCII)\n",c1,c2);
. V6 X4 I. [( yprintf("c2='%c' or %d(ASCII)\n",c2,c2);' q# S( j- ]/ a6 B2 s! I2 a7 [ f
}
/ I7 H' _: ?. Z% P9 g' l4.5请写出下面程序的输出结果.! v& E2 u! n+ Z& V5 A8 n0 q
结果:1 w. N( L# @) t7 u4 O# P
57: z: Y+ z0 A- Z" [
5 77 P' T" y: \# r" D7 \
67.856400,-789.123962$ X N' P) I; c( ~8 |1 v5 b' _! l
67.856400 ,-789.1239621 e* D# z+ J( \0 u
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962
$ W" |$ h6 D, o \4 f& g% P1 R/ d6.785640e+001,-7.89e+002/ s6 w, w: q% m
A,65,101,41
+ Y& b4 R0 B1 G1234567,4553207,d687
! o2 R$ }+ h6 n, x65535,17777,ffff,-18 g5 y! K- |2 k4 Y- v
COMPUTER, COM0 q2 v: e3 O, d ?5 W4 x* l
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',' A K/ F# w6 t! O
问在键盘上如何输入?
+ W/ V% p' F3 W9 [1 V" Vmain()5 X! K. Q3 w, o# T% n! C
{$ }! h8 Q1 H/ H+ k+ p8 R& D) A( O
int a,b;
5 @8 v+ j$ B8 f" p$ R6 Z8 f9 zfloat x,y;
- f2 N& q( \1 Q8 L' U- l, |7 \+ S; Lchar c1,c2;& Q! i! B' q( q% A' X8 _* a
scanf("a=%d b=%d,&a,&b);
$ H5 A: s5 ]& g/ ^2 B8 I3 Zscanf(" x=%f y=%e",&x,&y);' W/ I- ~) a# ~" v' D/ Z! x
scanf(" c1=%c c2=%c",&c1,&c2);0 i3 h" {2 }8 T9 @) O
}
& \$ A, d7 |& C2 B" w6 Z解:可按如下方式在键盘上输入:
, F* a# a. q! T8 Ia=3 b=7& {: b3 Q7 L! o6 J2 q+ D
x=8.5 y=71.82
; G- I; \3 y8 L7 L) ?; z6 bc1=A c2=a
3 S m) W% |, @1 `3 e/ f" I9 y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* a) J" ?- W" b. M! K% 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4 C( Z: _) L# M& b/ }; T! Y
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一0 y: _6 _6 L5 d) |/ @
初步概念即可,以后再进一步深入理解.
( \% z5 \' d2 H+ L h) P9 O4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-( v) [4 s6 k0 W- X9 y' m$ M( E) A
3.75,z=57.8,请问
8 |) ^. O8 C2 ^6 |, L在键盘上如何输入数据?4 g+ i/ j2 e5 W! }1 e
scanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);
, T- q# h0 q% d F" [解:# K+ y! I) A5 K3 A
main()
( }: `& Z9 U# C& D/ v7 h{( @7 `& I" k' t7 e7 p
int a,b;
8 {* ^+ s% M9 a: g2 Dfloat x,y,z;
8 M1 B# J! }) S3 E4 K$ ^char c1,c2;
$ \* W, f5 o b) R5 \2 Qscanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
, @4 L$ j. Z4 Y/ i! p; I- l2 C: D {}
1 Z8 [# d% j Y `运行时输入:$ J% n& }" j" ]# f+ a: B) n, d
10 20Aa1.5 -3.75 +1.5,67.8
2 h( u' `# V% M' R5 B5 C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' p7 [% W3 P) @; }2 c1 J
是用来禁止赋值的。在输入时,对应于%*f的地方,随意打入了一个数1.5,该值不. c8 n3 A+ u6 x$ a, v1 n3 G O( q
会赋给任何变量。
# E4 a, J. b4 ^( ~3 [4.8设圆半径r=1.5,圆柱高h=3,求圆周长,圆面积,圆球表面积,圆球体积,圆柱体积,
: v5 x1 \7 ?" D2 N5 k2 P+ r& R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. M# D$ c i, F! }2 g
程. s; Y9 P' \! m
解:main()1 v& y* _) q( ?" |8 c9 s# j2 w
{
j2 s6 ?+ e# i- {float pi,h,r,l,s,sq,vq,vz;
! ~3 ]! p3 T3 R' S6 [4 vpi=3.1415926;( s/ r5 w+ Z: Q$ \/ a/ v& }
printf("请输入圆半径r圆柱高h:\n");, }3 G' S/ F/ n; B, o6 _
scanf("%f,%f",&r,&h);
2 ]% V% a8 s. I {6 y7 |l=2*pi*r;
4 D. S8 W' N% N' {s=r*r*pi;
; \/ w8 X& n# bsq=4*pi*r*r;8 E, B! L* M) R. T& V- K2 m
vq=4.0/3.0*pi*r*r*r;; d8 V1 M; h9 x' w
vz=pi*r*r*h;
+ U: C, s# X, ?. vprintf("圆周长为: =%6.2f\n",l);; [) p# `! [: J8 T8 a8 ? o
printf("圆面积为: =%6.2f\n",s);" C; L w! o7 q7 K1 I$ G
printf("圆球表面积为: =%6.2f\n",sq);" x$ a* L! ?+ A& {; I
printf("圆球体积为: =%6.2f\n",vz);6 n* V5 c. D \. J- z
}
1 I7 M( y1 Y; x1 o+ e% z4.9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,
" o* B) F" }1 r0 J取两位小数.1 D) u: i5 _6 V- b! a3 Y: r
解: main()
) B9 J6 K7 a; p- v! U{
3 c1 K- w3 s7 [; v/ rfloat c,f;
+ `0 y# t! o0 U7 l1 hprintf("请输入一个华氏温度:\n");
+ m6 W7 W" B8 ^* N, h+ sscanf("%f",&f);+ z' v- N+ y9 N$ K. M" e
c=(5.0/9.0)*(f-32);! n' J! q. d/ V0 ~: `3 u
printf("摄氏温度为:%5.2f\n",c);
/ B: L& I- z) c. g9 Z1 W; u}
# U+ B* r( J2 A# `7 n第五章 逻辑运算和判断选取结构
, k# {+ U' f: F* `! ]) M5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
+ K5 {1 [7 Q0 c# G4 zmain()! t+ z3 f. y, @( J1 B* A- W2 s+ b
{
3 l$ L" K6 W1 a0 o0 e0 Aint a,b,c;. w- {5 M' ?/ _ z5 K
printf("请输入三个数:");- ?2 f+ ~' V( a! ^' C
scanf("%d,%d,%d",&a,&b,&c);
+ ^2 e& J) [# p9 g+ Z" B+ fif(a if(b printf("max=%d\n",c);
- T, N* v$ x! B- p; Z) B* Q" E! r0 { else; v' T, A* s, @& y9 s
printf("max=%d\n",b);
* y6 V* M' ]! Z3 Felse if(a printf("max=%d\n",c);: r; I% C4 q2 H1 ]) s
else
( x0 ~- M# V1 n printf("max-%d\n",a);" C. c0 O( i0 e! V3 M3 Z3 b
}1 M5 {' A; M1 |: G& T! N
方法2:使用条件表达式.% S# Q% l; o% A" L* [5 S2 a
main()
& Q, _2 U1 g9 w9 p+ x- C0 J2 _{int a,b,c,termp,max;
/ Y5 e+ l% R# `4 R7 I# Y: T printf(" 请输入 A,B,C: ");
3 M: m ^( t1 V$ e scanf("%d,%d,%d",&a,&b,&c);; L0 r7 o* W; }0 ~6 K) L
printf("A=%d,B=%d,C=%d\n",a,b,c);' a u ~) H9 u9 {, T
temp=(a>b)?a:b;/ N) x; q/ w/ y7 v; T" l7 @
max=(temp>c)? temp:c;
- a! O+ y @7 H% y l printf(" A,B,C中最大数是%d,",max);
; G( @& f3 p9 P) T% O p: `! p}
+ ^/ \- j0 M$ r( e# W6 ^" J+ ~5 g( O5.5 main()
3 W7 |* `+ y/ a2 ?, g* J! s& z{int x,y;0 b C3 O2 q% D9 m
printf("输入x:");: b: t+ N$ B: Y5 A4 Z& N4 l
scanf("%d",&x);
- B2 l O: J9 x; _if(x<1)4 Z8 ~ a1 N* S% r" G/ J5 ?
{y=x;
$ z. X' _. O0 ^; E, c) P printf("X-%d,Y=X=%d \n",x,y);' Q" }6 G8 b) e7 y3 E' Y
}
0 [" o( C" u) Y9 n. \8 H- _else if(x<10)
, S. Z4 t% {; S$ h {y=2*x-1;" Q% j; _6 O, G+ {4 k" i1 U
printf(" X=%d, Y=2*X-1=%d\n",x,y);0 ?% ]7 q# j' @. [. `* F; L
}
7 `( r3 z% L% Q/ e. nelse: X& r) t6 |, `8 [' F
{y=3*x-11;' B$ F; u+ e7 M
printf("X=5d, Y=3*x-11=%d \n",x,y);# W3 e* x8 H5 i6 l
}2 u( r% C1 J; K$ z- v0 J
}# r O: m4 B" C9 l' k
(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!
4 s+ J- f/ Z- G) l/ \" Xvoid main()0 Q5 r, c. `* O3 U/ i) U
{
- |: m5 ~3 P( }3 B I. Zfloat s,i;( L& E8 d f3 M, `. H+ {
char a;9 ^* r9 e% J2 H/ @3 X6 Q' L
scanf("%f",&s);
8 }( ?) x3 r1 t9 T, K: wwhile(s>100||s<0)
) l8 S/ j0 o' n1 g% U' c{
# w/ Y! a* c7 o1 R9 j& `9 {6 D! ~printf("输入错误!error!");/ f" h# E/ c" e) `
scanf("%f",&s);" B( X/ i- y6 ?: P
}
8 X$ X' i& Z9 n. f7 K, Wi=s/10;3 K R/ V" m; s& B
switch((int)i)5 F, n/ b; [# X" t3 `% i/ H ~1 V& y, m
{
! `" y6 u: B7 W2 c5 K3 n$ o" Y- ccase 10:' @# w) B) x0 T7 @ v
case 9: a='A';break;
$ W# L1 m1 V$ m+ {7 ucase 8: a='B';break;
" w3 C {7 R6 t8 }case 7: a='C';break;
' `3 m. P1 z8 b# B5 ~' x8 |case 6: a='D';break;
, x0 e0 i9 Q4 C* Vcase 5:
1 p2 y1 g, L/ v P2 D" ? U( A8 pcase 4:4 E) Z8 D3 W/ K/ P! B5 _
case 2:0 Y4 U% s: Y) X/ b. o# w
case 1:. A7 q/ x; |' y$ e W7 h
case 0: a='E';
8 s4 D! c6 \& c! ?5 N}
9 F3 i, o' V0 z0 L! Z, `printf("%c",a);3 ]( l/ d7 Y$ j0 y& a2 n
}# {6 C! j8 x/ s( Q
5.7给一个不多于5位的正整数,要求:1.求它是几位数2.分别打印出每一位数字3.' X! |; N! z( b% M1 t3 Q( ~6 {& j8 W
按逆序打印出各位数字.例如原数为321,应输出123.& \. T1 ?0 @2 [. k/ I6 ~' N
main()
1 m @; J( I+ u- T5 Z6 E {3 p! g2 A' |2 a! v' {2 x8 Y
long int num;
1 z5 \7 Q! Y0 O- y# e int indiv,ten,hundred,housand,tenthousand,place;
, R7 N% m! y1 l/ k printf("请输入一个整数(0-99999):");4 x4 y8 {* r! r2 P3 r0 [' {+ `
scanf("%ld",&num);
5 a& I" Q+ K u2 k5 D if(num>9999). w, M" x$ D; h' w6 @) j8 l
place=5;0 l0 r8 X5 C/ b8 n
else if(num>999)1 g7 m7 W! m2 @# E' e
place=4;0 u' U8 z+ d- S, S8 Z9 A
else if(num>99)4 I/ ~9 r6 G& u
place=3;8 `% p, _# ?+ l2 n V
else if(num>9)8 D0 n: }. |' x2 l. ?
place=2;
2 r* B0 p; E& J2 Y1 |! |else place=1;
5 u/ m3 X4 @4 b/ v4 f: Z6 iprintf("place=%d\n",place);
$ J) p9 U& j/ `- o1 y1 Pprintf("每位数字为:");
" E+ |5 W' {2 o: y2 B# Qten_thousand=num/10000;" M, e2 p% l: Y- |; f
thousand=(num-tenthousand*10000)/1000;
9 M0 ?% G0 S" K- ]3 L: i; zhundred=(num-tenthousand*10000-thousand*1000)/100;
/ a' M7 z) S6 Iten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;0 d% f% {" w' Z) H4 p# b: g
indiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;9 z; n( y+ T W* A
switch(place)
# G. K' s% U5 g% Q* p1 s# Z{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
) b; n! Z, ]" \# v6 z+ M" V' C printf("\n反序数字为:");
( W, Q; Y5 [! s m8 j E: Z printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);* |+ O. E; D& l- m \' g
break;
: O" B5 B% r' hcase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);1 d6 M7 Q. _/ B! G
printf("\n反序数字为:");
" f! _8 F$ b: U- q4 W printf("%d%d%d%d\n",indiv,ten,hundred,thousand);
7 n' @% x/ R! e3 | break;
+ |! w& t" A, U, i. ]; Jcase 3:printf("%d,%d,%d\n",hundred,ten,indiv);) r4 b( J0 Y) v
printf("\n反序数字为:");8 Q" ^( ?, ]% C8 Y( J6 S
printf("%d%d%d\n",indiv,ten,hundred);
c$ U) j. T" s2 w1 q, k- s1 {) I( Bcase 2:printf("%d,%d\n",ten,indiv);; n( @9 n' b4 P6 R5 I2 w! F+ [
printf("\n反序数字为:");
: I4 { i2 R: U0 |# ^% k' s7 ] printf("%d%d\n",indiv,ten);
: e" Z0 ^5 {. W& Ocase 1:printf("%d\n",indiv);( i; N) [' f2 M+ e
printf("\n反序数字为:");
$ U* T' Q7 D7 G6 R* R6 A3 o* r0 t printf("%d\n",indiv);' K. I8 l; w A0 [. W
}
/ y- S7 u( e8 E1 x+ _}
0 Z! [- N5 [; d6 b) Q0 G5.8
6 Y( M: T( D q3 `' V- O1.if语句
|+ w4 b& R7 [; Y' k7 m4 G$ |main()$ O' M3 K5 w2 |8 w& c; K, t
{long i;( f7 U! c6 N1 @6 J0 b' o9 R
float bonus,bon1,bon2,bon4,bon6,bon10;
8 H( H3 Q" `' m3 p bon1=100000*0.1;
`% P5 h0 I* R1 W3 F bon2=bon1+100000*0.075;
* H; a! P( r6 P bon4=bon2+200000*0.05;
) @! N2 G7 p7 S2 t$ k7 t bon6=bon4+200000*0.03;9 o9 B2 m* N, s0 h
bon10=bon6+400000*0.015;1 Z) @' J* j u$ M
scanf("%ld",&i);
% D7 [2 z& h* Y& u' x if(i<=1e5)bonus=i*0.1;
% J0 F* v8 I) n% M' ] else if(i<=2e5)bonus=bon1+(i-100000)*0.075;( d3 m7 P% X) k7 R7 j- S& x' q K
else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
9 E% Z) J/ @7 V+ n else if(i<=6e5)bonus=bon4+(i-400000)*0.03;7 w& J3 V7 r% b9 Y- t
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;* l2 r! M* e z
else bonus=bon10+(i-1000000)*0.01;
+ ^ w' \ Q" E* M; K) K( t printf("bonus=%10.2f",bonus);& M L& `/ X( s3 g
}' l9 ? h# r( e9 z& D/ q
用switch语句编程序
* B# A) u& I+ e! Cmain()1 [+ |; r; X5 C! m
{long i;! Q# {5 {6 F, Q1 {+ w0 f
float bonus,bon1,bon2,bon4,bon6,bon10;
& Q+ n- G1 D- ]/ v# b3 C+ O int branch;
/ S! `* F+ ]5 [% U t bon1=100000*0.1;0 @1 V, J" [" V* m: R9 y
bon2=bon1+100000*0.075;
% l) c# R# p% @4 H bon4=bon2+200000*0.05;
( Q/ A/ f8 y: B1 ~ bon6=bon4+200000*0.03;6 x" P8 B$ J8 v7 X' H) [: W: ^
bon10=bon6+400000*0.015;; H3 s, E1 D* I; }& |% @: I0 {
scanf("%ld",&i);* o1 n% @ J0 h6 v
branch=i/100000;
+ D3 J9 L5 _: l% q+ V4 r if(branch>10)branch=10;
: L( P# i, Y, ]9 E: S switch(branch)
" s" I! h8 P) A1 @ {case 0:bonus=i*0.1;break;' N- ]0 A) y' Z4 ~5 H+ C
case 1:bonus=bon1+(i-100000)*0.075;break;
. Z; r" r" ?, `2 Q& y case 2:- {# Z; s: \, I4 F0 a4 g3 b( e
case 3:bonus=bon2+(i-200000)*0.05;break;
* O, f' {8 S# C6 {+ G case 4:
! N0 R5 W1 L% M* d6 ^1 n case 5:bonus=bon4+(i-400000)*0.03;break;
T1 w$ o, ?: q: p case 6:; v$ S; U% v# K8 P5 h$ I$ H- w
case 7
/ u* I' E, l, p& r; Q1 V8 F+ o case 8:
$ q$ [+ _9 n# N& S; O& t case 9:bonus=bon6+(i-600000)*0.015;break;1 g$ r6 P6 y4 a7 q; Q
case 10:bonus=bon10+(i-1000000)*0.01;" {& u4 ^0 H% a2 K$ n: E4 J4 E j
}5 e, `5 ^/ X3 v( P
printf("bonus=%10.2f",bonus);
$ K3 {* h: v8 ]* G/ Z: ~! N} % f0 p" P4 _3 E' r" o/ U& |1 N
5.9 输入四个整数,按大小顺序输出.: L3 C8 {6 O, e% h& q
main()
! C& |2 n& d4 _7 y0 I3 Z* c {int t,a,b,c,d;1 }( i% d8 r# N) d
printf("请输入四个数:");( C$ _. A; V; t$ Y. H) l. [* C
scanf("%d,%d,%d,%d",&a,&b,&c,&d);- P3 m8 _0 F' W8 m) o1 A
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);% J& t: q2 p. d+ Z; M
if(a>b), f" h6 H, Y/ E6 y* S# Z+ z. d! I6 |" S
{t=a;a=b;b=t;}7 {& Y# m3 v4 |2 y' M& s) l) P
if(a>c); ^6 \; ]9 q! {" L
{t=a;a=c;c=t;}
+ g" W3 x* t. G m! S" m, S _ if(a>d)
0 L3 d: s# [( A' [% B {t=a;a=d;d=t;}
9 h0 U5 _: x0 S1 k9 w+ |6 [2 s if(b>c)
' _8 u* Z# }% u' Q; Y" j {t=b;b=c;c=t;}' G' j) E: F8 W y, F4 [1 W, O
if(b>d)+ S0 Q( v5 r$ W
{t=b;b=d;d=t;}( d2 [' V- s+ K4 a( u- `- f
if(c>d)+ E% H& D* ?* e. j, J% P* B9 }. v
{t=c;c=d;d=t;}
9 d) ?, @0 n1 x& Nprintf("\n 排序结果如下: \n");8 f3 t7 x3 m# Q5 ?# K+ L- ^
printf(" %d %d %d %d \n",a,b,c,d);# c. }9 Y# f( Q8 L' H. ~
}& |/ y: U3 O+ d8 N; O# Z7 S) u/ y
5.10塔: p3 U6 Q& X+ h5 D2 c6 ~3 B7 m' Y
main()
# M1 R, }5 h0 P. l+ `- O# m{# N- g, s( Q9 e* Y* B" s) ]# ?% U
int h=10;& {3 E/ I) X- P
float x,y,x0=2,y0=2,d1,d2,d3,d4; z; j5 I% w# {" S$ C' L
printf("请输入一个点(x,y):");7 T$ d. E% O0 K$ u- W' S
scanf("%f,%f",&x,&y);
I9 a( q& q* K: {3 i9 ?. W u" k( pd1=(x-x0)*(x-x0)+(y-y0)(y-y0);
, a) O$ |2 H4 ]/ X5 td2=(x-x0)*(x-x0)+(y+y0)(y+y0);
) X- s% \6 Z, F# ld3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
4 O/ ?1 O( h* c$ |5 m+ C5 G; }d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);
( P1 ^- Z5 d8 i0 Y' ~( b+ qif(d1>1 && d2>1 && d3>1 && d4>1)
6 o) w: t+ V1 gh=0;
+ ^ j" G) D5 o! f( Mprintf("该点高度为%d",h);) T& ^% z0 D! ?* M1 w, }, k# z% }/ ^
}: H. S7 Q. Z% e8 ?
第六章 循环语句
+ W. p% q6 \ x G o6.1输入两个正数,求最大公约数最小公倍数.
4 }3 R! Q; |+ ~2 jmain()
z/ i5 v9 q4 {5 j6 V. K* j. v6 u{! `% T- U7 l) b# \3 d7 @9 o# d
int a,b,num1,num2,temp;2 i) }+ x3 K) h$ X1 c" _
printf("请输入两个正整数:\n");
D5 Q0 L! _9 ?scanf("%d,%d",&num1,&num2);. [1 W1 a% j/ Q* g
if(num1{/ y1 u1 M7 g M; m% z. L( Y. P
temp=num1;
7 t& z4 c$ M' }2 s6 U. j! J1 qnum1=num2;
# C. P, S# c8 E" V% Vnum2=temp;
+ t# I& ], F% Q* k}! c9 Q2 `3 x% j m
a=num1,b=num2; a2 }9 o3 ]* j) O: {% p9 a
while(b!=0)4 d; H5 [1 V4 D6 b( q0 h, t( s$ E
{- R/ f% i8 z% [0 u5 e) X- `# w
temp=a%b;; `4 E+ Q Y+ ?; h# t. N% E
a=b;
1 _0 \- [" j! P+ k/ j b=temp;9 F2 Y0 u" _7 I1 a; W: g
}
4 x8 s7 u% r: u$ }2 b* V8 H, x1 wprintf("它们的最大公约数为:%d\n",a);
; X8 h6 O- |! t# D$ O: [printf("它们的最小公倍数为:%d\n",num1*num2/2);
2 f/ `& X- s& V9 O% D- J}
& t# F) k. p t* O, Y* J6 a6.2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.0 o) w l* m* V* k0 j
解:
; r9 w4 U1 Q6 w9 p. m) H#include < >
' p4 y4 P- h+ W9 @- C fmain()
5 y( N8 L1 _6 M1 J, I9 K, }{! {; v3 f: I6 J9 o3 [; m
char c;
) r9 H6 S! E# ?int letters=0,space=0,degit=0,other=0;
% q2 c7 H8 }: F* n, B; P2 Bprintf("请输入一行字符:\n");# }9 f" X" F' H* F* M
scanf("%c",&c);9 j3 N9 n# l9 }( w: U( l" x
while((c=getchar())!='\n') C; h. {1 L8 M$ _) w
{) @$ M) }7 l" b
if(c>='a'&&c<='z'||c>'A'&&c<='Z')
/ M1 h9 F+ t7 \ x& {letters++;
2 E& @# E2 O/ M- p2 [else if(c==' ')
R2 B" i, [% w* \& Cspace++;
' x" s% @/ P+ G" jelse if(c>='0'&&c<='9')4 S; Z3 ^3 l- h! [* w0 c
digit++;
2 n5 X. ~# D- B1 v5 S7 xelse- K u# B. V, a N/ k
other++;
2 _) O, c# F7 n: B$ T}& ?( k: A; s+ I& o; X" ?
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%1 s7 `/ w7 F* E' ?7 N( P' G% G
d\n",letters,space,& O* d& s7 b0 o+ R" C0 i5 q
digit,other);, y: ^. r; [- r
}
/ f( q1 _6 J8 Y6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.
* A# L) a+ O+ [2 i1 F$ g& Z解:
0 Q$ j: J( _! K6 J* n4 L& Emain()
/ k' K. w. A* o/ [3 x3 r- Z& `{3 X+ Q4 p9 k- j! y, g& h; J1 o
int a,n,count=1,sn=0,tn=0;
2 J. P; u- G" ], I gprintf("请输入a和n的值:\n");
% C b/ ^& }5 {( P5 _2 t+ v# Vscanf("%d,%d",&a,&n);8 [, N0 V0 }1 C' y. g
printf("a=%d n=%d \n",a,n);( O4 R* U6 \: e. g9 ^& _
while(count<=n)* b; ]+ y3 M Z' H# ]
{! i0 v: c, j% t N) ^/ U. f
tn=tn+a;) ?: D H/ \4 {8 V/ e0 B+ x, U
sn=sn+tn;
, L' `+ _2 c5 }; J7 x, A. {a=a*10;6 e% l. F3 e7 i
++count;
$ h2 \+ t! y) L0 m9 S5 o}
! L2 b0 J; E0 e) L) I- Oprintf("a+aa+aaa+…=%d\n",sn);' D6 ]2 X4 Y. l7 M
}
2 h* C( }" \- F4 e: p' E1 g6.4 求1+2!+3!+4!+…+20!.0 [+ w! x0 c+ T! t
main()
& ]4 U# D) P8 r+ R7 i4 s8 Z& W% W{
. b* l5 `3 j2 M( _! ~" ]float n,s=0,t=1;
$ \7 t. y! l( c* t, X' r2 f! R$ B+ ?$ N, pfor(n=1;n<=20;n++)
" R! O& S$ k7 z/ i1 P{
& n( j4 g$ l3 N1 F$ xt=t*n;
, @5 F& J! p# F% rs=s+t;
5 M, \$ z2 H$ J6 }$ S" J1 j}
& e- w4 c( K, c% gprintf("1!+2!+…+20!=%e\n",s); N; e( k+ f: {$ d+ f ]; s
}0 T2 I/ O# n. z7 t, ~/ e; P
6.5 main()
. `% y7 I$ \0 Z% ^{, w3 b2 s2 @* H
int N1=100,N2=50,N3=10;+ S! }( E- T+ n1 @: k/ `
float k;
- \/ Y: N7 g+ n8 h# Nfloat s1=0,s2=0,s3=0;* ^0 L$ v7 Y! Y4 H# l
for(k=1;k<=N1;k++)4 w* s0 l2 v$ ]+ }. m2 G, Z5 g" o2 G
{9 q7 b( \' Q! U. O+ u
s1=s1+k;2 Q& P: O8 R+ z6 `( J- N% p
}
' A6 p, x4 @' a& L( v- a1 ?! zfor(k=1;k<=N2;k++)6 T! V# \3 w2 g ]
{: _2 h1 J8 _6 Y3 {
s2=s2+k*k;
5 ~ k9 |4 n8 T A0 z, N. ?}
5 e% e5 x& _+ ^3 H, C# Ufor(k=1;k<=N3;k++)' n) Q: m, Z, |4 u+ C
{
; l2 h1 J+ q; C% }s3=s3+1/k;0 `1 [' J" L! k
}
% Z" f. c( g1 ~! |8 u t- hprintf("总和=%8.2f\n",s1+s2+s3);
# F0 g/ {; P8 d7 K# q}0 g5 o/ `1 ~0 w! S2 f
6.6水仙开花) ? j( W5 o2 S4 E% K2 a2 m6 z* p
main()& j6 j" u6 s3 \ V. A" { l1 U
{+ A# u3 g4 i% p. s! U0 B2 O
int i,j,k,n;! b1 l5 Y& p2 J7 ~ o: s
printf(" '水仙花'数是:");/ h% s5 w9 T9 m& p* M7 l2 L7 P
for(n=100;n<1000;n++)
% k( F' ^4 q8 C7 K9 l{
$ N. }5 {' z# |* p9 b4 _0 c7 h3 vi=n/100;
: c) N6 Q# M* t) _+ f+ Z" A, cj=n/10-i*10;8 G* T" y0 B. d' e* w# l. ?5 n
k=n%10;
- v3 n* }$ J3 F( n; P9 k( l- bif(i*100+j*10+k==i*i*i+j*j*j+k*k*k)4 w2 F( A8 K- U7 a8 k( ~& S" M2 A
{* b1 b& f) T* D
printf("%d",n);8 P- F/ ?+ v/ c' Z' H" I
}+ G) ?0 P3 b7 ^4 Y% t) A" P% d# q
}6 w! o. C. Z' p8 i5 u! y" b
printf("\n");
+ l' b, N: Z6 o: D}& r4 ?1 l* f( `; z/ c5 H4 H# P
6.7完数
% }0 f2 }& L7 jmain()) s% |1 B2 _7 I" m: e6 B
#include M 1000
+ \. ~* J. ^) p6 \1 `$ Pmain()
1 Q+ [$ a. M' E4 {7 V5 R+ P{
6 F% F' Q8 O0 }" e7 ]- z1 Tint k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;
K/ J$ l% K5 w$ ?. z0 eint i,j,n,s;3 ]; L8 f8 j2 E4 e; f
for(j=2;j<=M;j++)
8 w6 {" @4 `! c7 } V{; q0 q1 f6 ?/ w) ` i, i7 m+ w
n=0;! c! B4 f* Z1 R
s=j;
5 m( Z" U, A3 O7 _/ `6 M6 ^for(i=1;i {
, Z0 p% g: Q" U3 Y; M, ~if((j%i)==0)' P$ Z+ L+ n% ~* l9 }* e
{
5 H( Q$ v( i' c3 L if((j%i)==0): {5 o3 }- W* D! H4 r2 K
{" @$ M8 {7 S( d3 L5 p9 W
n++;
" K0 q# q& t' o) O8 S s=s-i;4 F; l. b! [5 F+ q; z
switch(n)- i% @' s" s8 S' ]! @ |
{
: l/ B8 p: U5 Y( e- s case 1:
# l8 H6 X! `5 u9 `- S# b$ r | k0=i;) x4 k' G+ h6 f; H+ \9 |! ~
break;
' p8 O* L; r, p: C. h* ?$ ^0 Q case 2:
$ S l6 A% q) D( }! x k1=i;
& W6 a L: J: @6 c break;+ Q; [ L: e2 [8 t, L- r v7 d
case 3:
$ Q- p# q) V% j! h) H- w a k2=i;5 F0 ` f" [8 b# j
break;
; H* z* V5 u1 D+ R0 o2 n9 }5 _- A case 4:
( S' ^: Z/ x `1 } k3=i;
% ^' R+ ?7 b U, E7 x break;' C8 ^+ y7 l( y
case 5:
6 P) @: {& X% U/ h: U- L+ `( ? k4=i;2 y. m0 u5 T ?- D$ g! P
break;; q0 g8 ^* }1 v1 V5 ~
case 6:
# M5 J) z" `7 g7 J k5=i;: V0 d" t4 M+ a; B7 H( V, l
break;
% b5 @! b0 N& J$ H; y case 7:
+ n3 Q+ l7 Z9 c; {) F7 }1 u k6=i;: d9 W1 c% r1 l( i5 r; v9 F
break;
- u! C# q* ~- b case 8:6 F' S0 ?1 y4 L. ` s
k7=i;
5 X' n( r. t- r( Y; l break;" |+ K- _0 w: { n3 l. ~6 ]; o
case 9:
* F% k5 @: D7 [/ k g6 T# \) O k8=i;4 e' _; M/ h% m3 j
break;" J/ p8 c- C, `$ _; w
case 10:7 l4 O# U. ]* w7 b$ M4 R; h
k9=i;
6 w. J5 L& p( |8 [& s6 p6 s. {2 u break; @) P d" g; `. B0 D8 e `5 z
}
' s. L& E7 f. U+ N0 J" t' i }
& e+ t: Y- ?; X1 h+ l }' ?) a( x$ @. t) p. |3 |
if(s==0)
/ G/ V6 R, g& p' u8 w4 Q4 V0 Q; ~ {' T2 H I5 V! l! c
printf("%d是一个‘完数’,它的因子是",j);( J5 j5 X; W) N9 h, k+ T% k; W
if(n>1)9 W; j1 y1 o4 ~3 w& h' L
printf("%d,%d",k0,k1);& _! J) `+ r3 {% ]5 v
if(n>2)
/ Z4 f' {8 u S t7 k m, H printf(",%d",k2);+ ^4 i0 J7 a/ ~) u( q9 o
if(n>3)2 ^& I+ z* f. D5 H
printf(",%d",k3);
+ g7 N6 i' @3 K# lif(n>4)
. T* _6 a& `. n$ i, n printf(",%d",k4);! Z5 y2 H G" @7 C9 [; t
if(n>5)8 y2 Z( H- @7 `3 e6 x+ d7 @
printf(",%d",k5);$ Q f% b$ `, P% W
if(n>6). C/ S, t- o! }! n$ [5 K
printf(",%d",k6);
$ `" r1 D, v* e/ qif(n>7)8 [4 b4 D- m1 c3 b
printf(",%d",k7);6 h3 m" \0 y2 W e8 e8 i' L
if(n>8)
8 `6 g$ X4 X7 ^+ Z2 S- g" y printf(",%d",k8);
9 Y) z& J0 `6 e! wif(n>9)5 i0 s4 y' r( _$ O5 u$ B6 m
printf(",%d",k9);
9 x% j: x. ~3 N. Q1 r0 l, Lprintf("\n");9 Y2 ^: m. z9 H: ~# V0 C
}/ ^1 o0 d7 V# S% Q7 G( B! n
}
' p" _8 x* x* W# ~" K( s2 U方法二:此题用数组方法更为简单.& h% P9 r6 b" X8 U
main()
6 ^ m1 a+ t. E l5 b{
7 H4 I$ Q' s$ ]+ hstatic int k[10];
2 ?. d) J9 P* |" R5 i( O6 i7 bint i,j,n,s; R6 k' a1 ^8 V4 ^( C+ V) m
for(j=2;j<=1000;j++)
) u( ~. y) z- Q/ G3 L8 b{
4 w9 C. ~1 n1 g+ a$ o: k9 l- r- T2 Gn=-1;
( n0 |" C( a5 R+ e- \ X" bs=j;
4 J0 e* _ p* ]: Q1 q" y. q$ ?for(i=1;i{
& ^0 f8 T5 }+ T: y( C1 j N: Zif((j%i)==0)6 C2 O0 H- [7 _) C
{4 m3 _+ \* l& @: ^5 j w
n++;
7 Y1 B' `$ u- r j6 zs=s-i;8 d8 [( H) ~0 M. y7 q
k[n]=i;) R2 B3 G' ]4 q: g7 D8 I
}* N, I+ @! p: `4 [0 J! F3 Y
}' {, Q7 @" H2 ?( O
if(s==0)
5 a- ~8 [1 v1 [{
0 N( L3 H% i0 H2 @! f F8 Gprintf("%d是一个完数,它的因子是:",j);
! V) h, }+ Z& ~% l3 @$ d/ }for(i=0;iprintf("%d,",k[i]);
. _: j5 y/ n( Rprintf("%d\n",k[n]);
6 O) w6 E+ Q8 k, ^" t}, m7 W7 B9 q7 r' Z2 j
}3 S C: P2 @( P, F
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.
7 b6 I; B/ G* Y9 Y# d# _0 h解: main()3 b% P& U* @$ V( r
{0 Z U4 T$ ?; f5 Q
int n,t,number=20;+ b5 X7 l' H1 }: j3 p
float a=2,b=1,s=0;; A$ s% v' K: S6 @* |5 R9 ^
for(n=1;n<=number;n++)
' ]* @' j; _: B6 K# I) z, y{9 o1 `- V. b9 [9 F! W8 N8 X
s=s+a/b;
& u9 ?9 s3 ]+ k: C p. Mt=a,a=a+b,b=t;
4 x+ \2 l3 Q) G+ |1 C}# i0 G5 s" R! c
printf("总和=%9.6f\n",s);! A. H" E5 W3 c
}! ^; N3 M$ D7 Q* P i
6.9球反弹问题7 u( R; E+ E' v9 [
main()% Y4 L; o+ x0 E. v$ B# s6 i
{
! E9 K' u( r6 ^. D6 r1 Ffloat sn=100.0,hn=sn/2;
( Y( `, o+ G, L: C' Eint n;4 Y' M" K. n/ ?, f. m. d
for(n=2;n<=10;n++)8 {$ d; Z) u o. ]2 U2 o8 |. m$ Y
{
5 w. F5 z% U6 f5 d/ |sn=sn+2*hn;" _; q: D$ P# Z+ k
hn=hn/2;2 I# Y! I3 _% C. d4 N
}
$ u- o! q$ u& f- O( [: x, e# Xprintf("第10次落地时共经过%f米 \n",sn);
+ q5 M4 b: i! Z2 b# I9 Kprintf("第10次反弹%f米.\n",hn);, Y+ [# ~' h7 E" b' k3 x/ L
}0 t+ b" U; k7 `, s! T; j
6.10猴子吃桃$ z* P# Q& ]; f7 [1 z0 I
main()- g5 T, v0 v3 K2 @
{; Y" o% m$ i) f* U2 C8 R
int day,x1,x2;$ D% a9 U2 Z8 M, b/ s. r7 u
day=9;; b" \( J& @+ y c+ X$ \
x2=1;
4 j5 c: Y& \) Xwhile(day>0)
0 h+ V9 [9 I% y g9 a' [$ G{
( x( l* x, o0 X( t2 ]" s1 Jx1=(x2+1)*2;
7 o2 F6 {/ Z/ x7 J0 |x2=x1;0 n, T9 O2 ?' h$ ?, q' R3 V
day--;; W, `: g; L4 e4 [$ w0 p- a
}6 V/ q1 t8 `: N) \& x& D9 ?1 ~
printf("桃子总数=%d\n",x1);3 @$ P( `4 |6 z: T* g
}1 |- Q# h' [- j X
& r* l! G) W- v2 o) f- a6 A9 i8 R6.12
$ B+ j7 o+ z2 C- } d8 p' m#include"math.h"
1 G; V$ m! W0 ?$ Fmain(). b8 h4 O" q$ X' q* r3 c
{float x,x0,f,f1;
% l: Z* R) x, G7 M; M% u x=1.5;
5 V( Z9 e: `' H6 A/ g- f do
9 y1 ^7 J x6 y, } {x0=x;9 U0 Z+ M8 t# b2 ]2 h2 M; Y
f=((2*x0-4)*x0+3)*x0-6;( v! R# F: n0 p" H
f1=(6*x0-8)*x0+3;1 q+ |; [( r! p3 q
x=x0-f/f1;
% F. ?/ _$ h) T+ f3 c2 x( T: c }
7 J7 T% L) w4 I. G4 [* ^) j; A while(fabs(x-x0)>=1e-5);. x$ c- ]+ @" x8 h
printf("x=%6.2f\n",x);
; T( W$ B- C% a$ Y) A R}
- [+ `0 k# I# m: J" Q& T& M$ W3 A! `+ `/ ~
) p2 o8 g3 M. U9 v& _+ w% \6.13
8 z7 M N( Y8 y' S& |#include"math.h"
" f# R# y( P* K* ?% Gmain(), E: C$ k$ {; |6 i, I8 N8 P
{float x0,x1,x2,fx0,fx1,fx2;- L% T! e: s" G8 u% h5 e
do
_5 g7 z8 P- `. \8 [& Q {scanf("%f,%f",&x1,&x2);
" d& `% i' [9 X) G# R- {$ h fx1=x1*((2*x1-4)*x1+3)-6;" M$ }2 _4 M. ~' L+ Q
fx2=x2*((2*x2-4)*x2+3)-6;
) Y, x1 ?% A* S }
7 H5 L* f4 W# S1 q& [ while(fx1*fx2>0);/ Y% p" ^: ^' R( e
do
! x; Y3 P+ e* ]; V2 g4 ~7 z* O {x0=(x1+x2)/2;
% u1 i4 h. `9 X- o6 ` fx0=x0*((2*x0-4)*x0+3)-6;
' l" v6 @/ M. P6 {' R4 w" a# k if((fx0*fx1)<0)1 h* Q: T3 {, `- s0 r& J
{x2=x0;
7 u* V- \# E/ M5 c( K fx2=fx0;
1 m' m6 w' T/ N; I: X2 [% C }
8 ^. h7 k/ p! o else
- s- `' g5 B! s7 v2 W! N {x1=x0;) `9 m; L) s' p% q6 F+ S- \
fx1=fx0;; z( C+ k- Q- ]' T: A
}7 Q7 E+ m# `+ H( c
}
4 \" O8 {, i/ ]( x& L while(fabs(fx0)>=1e-5);& P! l1 W7 y! c
printf("x0=%6.2f\n",x0);
4 e1 Y( Z% o3 z O+ L; M* ^2 Q}$ ?9 K E2 \ k0 ^+ h0 e8 E
6.14打印图案$ Z. i# [( k w0 f- x" m
main()+ Q2 Y- ~& u/ w4 t, d
{int i,j,k;$ m$ h# g) {' ?/ |) R! v
for(i=0;i<=3;i++)
/ h E/ x1 K/ C# s; z {for(j=0;j<=2-i;j++)
- |; s1 v8 }( w5 Z. y# Y7 k printf(" ");
3 [$ N# M6 U6 m for(k=0;k<=2*i;k++)
( I$ R" M5 P( T; E printf("*");9 L& `) i) s( N' ^) A7 Q: t' F
printf("\n");
) g' `6 y1 g/ t9 H2 x- R: E$ }1 j }
$ ?2 O) @2 E0 o5 t6 y for(i=0;i<=2;i++)
J3 |7 S" Y* f7 B5 l! M. a W1 D {for(j=0;j<=i;j++)
7 D( B! @5 s/ H! v3 M. D, A printf(" ");8 R! Z E# l U, h* [
for(k=0;k<=4-2*i;k++)
2 y& V4 w. v6 X6 T0 c- k9 P& b printf("*");; u# [" g( h5 T5 C
printf("\n");
/ N0 A' B3 Q: r4 Q' X9 T }
H& U3 |4 F/ Y) E, ^) i}
. w% N) Y+ n+ Y/ L2 F6.15乒乓比赛
& f3 S3 R/ p0 \7 `2 `6 k+ umain()
: y7 T& k2 f# d/ R+ j- Z1 |4 J" X{' |2 g. ^% @- I2 ?: \. |
char i,j,k;
4 F4 a7 y+ |, A$ W5 Bfor(i='x';i<='z';i++)
. v \/ o1 J2 F" N0 d% L. Sfor(j='x';j<='z';j++)% u! v: N1 p# y( {
{$ i; ~% V5 p$ x0 _0 N7 {
if(i!=j)2 j, u2 y+ G! F5 o0 `. O
for(k='x';k<='z';k++)
+ e- W$ z& {$ B$ R# w {; M6 C% T' i* J; Q: |+ j
if(i!=k&&j!=k)
$ X" G3 Q4 e( F0 I9 d {if(i!='x' && k!='x' && k! ='z')
6 z) T3 T C; g% Y* `. ]8 v5 M7 q7 H Eprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);) F' a8 j% R, M. K6 j" O, k
}' ~+ u# h* {% Y1 _3 W9 P
}
/ e$ f2 k! {1 E1 ]* e- S }
. |1 }# p& G# ?* U- \}
( A! }" s w( X7 s- W$ i& b, yC语言设计谭浩强第三版的课后习题答案
d: y' ]. b# k7.1用筛选法求100之内的素数.
5 F5 F1 s0 d1 ?) l @. _6 w#include) f4 p$ @) x( P
#define N 101
% w5 Y/ p$ o# X% v9 i! V. Tmain()
7 `* p+ L+ m5 A& ]8 v, @4 ]2 B6 o{int i,j,line,a[N];
. C! y4 P6 D! n: rfor(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)( `0 a6 \- J7 B8 d: m
if(a[j]%a[i]==0)1 {: u: g) j! E( g: c* f, f, {
a[j]=0;
$ U1 I# y6 D1 R1 n4 q6 lprintf("\n");6 k& e( Z! f( m2 Y+ _
for(i=2,line=0;i{ if(a[i]!=0)9 [8 j6 [7 w( s$ _+ `* q) X( L
{printf("%5d",a[i]);" b9 R+ o4 T6 S4 o# n" M
line++;8 G3 V. n& j" `( U. l' v& p
if(line==10)# b& a% }9 ^2 K
{printf("\n");
, ~+ C- @+ \* h5 @/ U3 ~: K3 l: { line=0;}4 ]# c- r1 d+ n7 J1 V5 k( }9 v
}& F; r! {0 E0 S$ ]) P2 x, T9 L
}
% D8 u. X3 S, L( d7.2用选择法对10个数排序.
! g% W1 n/ [9 j! z5 y0 i#define N 10) W+ n+ x6 l- a0 H
main()
+ l, i7 G8 @: C# m* U{ int i,j,min,temp,a[N];
% ~6 t# W) h! r( O" K; @printf("请输入十个数:\n");' |8 J+ T1 C" B; z- y
for (i=0;i{ printf("a[%d]=",i);5 F5 j3 G; I! \# A8 q
scanf("%d",&a[i]);
/ a8 k3 v9 u- \8 ^# _6 h) B}+ W* w1 v* F+ D- k+ b
printf("\n");
5 s* O( y. @6 X; Nfor(i=0;i printf("%5d",a[i]);+ ?9 ]8 _' k# y* s$ T
printf("\n");; V; s) T! B: p
for (i=0;i{ min=i;
) w6 i& |/ F- c6 Z$ Y for(j=i+1;j if(a[min]>a[j]) min=j;9 j; c6 } j$ j( F
temp=a[i];
# n8 E' O B5 y5 j' ?- l2 g% r a[i]=a[min];
' w7 S8 ^& T5 O1 {2 W, ? a[min]=temp;% ]: Y8 I5 H- K* @7 o |& o; K
}/ m1 V) v& c/ @5 l. n
printf("\n排序结果如下:\n");
4 p h* q! f% j3 Y6 w) ]5 f( _7 |7 Bfor(i=0;iprintf("%5d",a[i]);; A# B; l" y" E# F7 i( W- S
}# @7 |# y: b+ }
7.3对角线和:
8 f6 d2 ]! b( ^5 i. n$ bmain()
& [# \& `3 f8 O& e{
' R9 F7 f( B; w9 ^4 R" L. [* pfloat a[3][3],sum=0;0 e# i/ t) U0 K
int i,j;0 L3 a$ ?, Z; z$ y$ t6 d# @' i5 Z7 A
printf("请输入矩阵元素:\n");% J+ k/ b+ _7 U
for(i=0;i<3;i++)/ v/ v4 _4 t7 E
for(j=0;j<3;j++): [ a" ~( ?; J& w# _) T% Q
scanf("%f",&a[i][j]);6 a t3 N) X, A v/ G- ?- u' C) N
for(i=0;i<3;i++)8 X( e, c7 D4 T
sum=sum+a[i][i];
7 B* C" L6 K- d. V4 A printf("对角元素之和=6.2f",sum);2 G5 @9 S$ o# f! |: s" W* G! H a" Y
}1 i3 e J$ f2 X0 X1 z
7.4插入数据到数组
, T* U2 Y( o r8 D) P' L3 s$ w. C: umain()
" L' O( x# o. F8 E{int a[11]={1,4,6,9,13,16,19,28,40,100};) I- D0 I o: }9 ?
int temp1,temp2,number,end,i,j;
; B% w) c0 d, Z8 }* cprintf("初始数组如下:");" y# d- z8 h Y/ U$ M5 Z) Z$ [7 |
for (i=0;i<10;i++)
& ^; \4 ] Z6 y1 ^8 Hprintf("%5d",a[i]);; e( a! H. B) i. u
printf("\n");
8 k& j/ V9 U( l, e. E8 M; W* d% h9 [printf("输入插入数据:");
3 V0 h u" u7 j% x+ h' |scanf("%d",&number);
; G9 s g: m7 }0 q8 w b5 v1 @7 ?- ^end=a[9];
: X `2 |$ x/ J9 q: z. |if(number>end)
- ~$ e5 b: B; S, w1 Ea[10]=number;
# |, c7 x2 M& G) ^$ U1 Yelse0 E1 ^& E; ]4 l3 U
{for(i=0;i<10;i++)5 Y" p! d4 d# j- D8 l- j5 j4 o
{ if(a[i]>number)) r! n G [/ q
{temp1=a[i];
4 b A2 n; n4 G" z# H" c a[i]=number;3 y0 ~6 [6 m- ~* ]! B8 M
for(j=i+1;j<11;j++)) n8 w5 L' f8 v! E7 O6 r" L
{temp2=a[j];
* | |1 z$ W/ r4 y1 [ a[j]=temp1;
' Q7 _: g; S7 t7 S% E7 @" ]( ` temp1=temp2;8 I) f2 | l/ F' H/ ]6 Y+ ?
}: x- ?& M; A* H/ G; y; r8 H2 U
break;. j/ L, ^& z1 |
}( |. P6 i i) `
}5 o' k( H3 t! `+ l
}
1 F2 s2 g8 T9 ?7 E* o+ D for(i=0;j<11;i++)' ~: |+ v7 S& k6 Y
printf("a%6d",a[i]);
, u( N4 {$ l l4 v}2 z( N% F% z* i
7.5将一个数组逆序存放。
9 s' U0 G+ Z ^#define N 58 q0 p& J, W" J; ?
main()
8 s0 P G$ N5 B, `+ s5 ]9 Y3 f{ int a[N]={8,6,5,4,1},i,temp;" O* n+ `! a, n1 j
printf("\n 初始数组:\n");
1 O+ u1 `& V8 Jfor(i=0;iprintf("%4d",a[i]);3 l2 G' i9 Q* m4 w; u a% P
for(i=0;i{ temp=a[i];) E' Z# e/ A' o6 L# S6 V; }
a[i]=a[N-i-1];+ X& ^2 {% F' Q2 q* \
a[N-i-1]=temp;
6 g5 I: e W0 {}& K4 C( @" @& \" X/ d
printf("\n 交换后的数组:\n");
# Z5 }1 i' M+ [3 m0 [for(i=0;i printf("%4d",a[i]);2 ?9 l. M" e/ K. e: X0 U8 u
}
7 B0 q9 L, E! _7.6杨辉三角5 z6 H {" b9 c$ B) Q! i
#define N 11# k4 \% `7 D% k( {- l2 n
main(), T4 k/ v! Y' S9 o
{ int i,j,a[N][N];
# I. C/ x+ J- n" | for(i=1;i {a[i][i]=1;
& l7 e- @8 p, \1 q3 Y7 R a[i][1]=1;
+ e& V) k; |& h3 Q% B }
4 B* ^9 o/ Z: t1 `% }8 L0 \ for(i=3;i for(j=2;j<=i-1;j++)
* @3 r# B% D+ ]6 c" ^" l a[i][j]=a[i01][j-1]+a[i-1][j];$ S) o( c$ ]# Z2 B: w3 D. Z
for(i=1;i { for(j=1;j<=i;j++)' `3 j2 A' F3 S& ?
printf("%6d",a[i][j];
% }- m& j _& `: I$ Q printf("\n");9 S, B* |9 J' T# G- Z( i7 x+ W
}, s6 u% L7 V& [+ w0 {% L
printf("\n");
, [& U, [7 N U+ E}
% x: T' `- D! A* S7.8鞍点+ o# u$ L4 y ]$ J0 `. r
#define N 10* E* H8 X) z! `; V( v6 |
#define M 10% Z/ U- E' g0 b) ?5 Z# r& }* {
main()% Y! F, ]1 J/ h3 h/ T0 l" G# U
{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;) g2 u) v1 X8 V: w. F6 `; @
printf("\n输入行数n:");
' b( P& z7 o" e. j1 { scanf("%d",&n);1 T; v. |2 V5 j J! e3 l
printf("\n输入列数m:");- x/ r8 D+ ~# n" z d+ U
scanf("%d",&m);3 L9 N0 v, C) ^9 l" g
, q% x8 }7 y( D" X4 k' q6 n+ U0 [ for(i=0;i { printf("第%d行?\n",i);* g9 @" @- t6 y& N; E6 W# ~
for(j=0;j scanf("%d",&a[i][j];; c3 ]) E) R& d0 n8 S
}! i& x) p8 g! n4 I# l7 u
for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
8 b8 G z5 A7 s7 x! w- X7 C pritf("\n");
2 m; E' T0 e4 p8 L2 B+ \ }
6 n2 e) r2 |1 J! A flag2=0;4 k- X% {; l$ y0 a
for(i=0;i { max=a[i][0];' W \( R9 r" b' O
for(j=0;j if(a[i][j]>max)
9 P2 x2 V" p2 G B { max=a[i][j];; h+ \ q4 X3 O
maxj=j;& K: F+ w n. }3 V
}
1 f4 v7 i8 [6 q for (k=0,flag1=1;k if(max>a[k][max])
1 E8 e! ?/ Y1 I0 M flag1=0;
+ r v Y6 K, L' B if(flag1)
0 u& |) S' t- e# Q: U { printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);
: d4 F, t4 k7 v, ~8 N flag2=1;
% @* |9 w7 l: [0 ~8 j, j. v }
/ ~$ @% K) T0 c" t) D}& N$ N7 f: M" |2 s% A# e) `. s) I
if(!flag2). Q e. e. x! n- P1 I7 [; p- M
printf("\n 矩阵中无鞍点! \n");
: U( o/ {% }% \ Z) M}
# C! U% h1 w# f8 \! x$ T4 A
; E- ]4 \5 T6 {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.
( D$ m: O d8 {; T#include9 G. H& [) n! p* K
#define N 159 n ?& o$ T( e
main()* d. F8 ~# j# e. F V R
{ int i,j,number,top,bott,min,loca,a[N],flag;
, U2 S! |( `% p m char c;
" F$ o7 J/ _ S7 h printf("输入15个数(a[i]>[i-1])\n);
& \2 w% E4 b1 R9 k6 I scanf("%d",&a[0]);
- Z1 c8 {4 H( O# N9 e ] i=1;4 T+ t8 l8 E6 C% P' F/ V
while(i { scanf("%d",&a[i]);# C+ [) _, [ d- e0 F
if(a[i]>=a[i-1])
, Q; ?. _% C! X4 t3 A' W i++;
* a8 i7 z; D% C1 [) o8 R* T; r esle: g, g2 B- w) O" p
{printf("请重输入a[i]");
1 {% F, A' H/ u9 J# p, c printf("必须大于%d\n",a[i-1]);" k+ \6 h: c% Y! o# [& A
}
9 f5 q. j5 q: h- R8 d$ v5 |. N }) F0 N& V4 Y. O9 S0 z2 V/ [$ _
printf("\n");# k, @) X' E f. ~8 U
for(i=0;i printf("%4d",a[i]);
* c- u( u! k; Y( [, [( k. s. j printf("\n");- v5 W V0 @! V! P
2 \7 ? _" s# ^6 X4 C flag=1;
# _- B: S5 u$ w9 v+ O while(flag)
) v$ W E5 h( z9 u. d9 n3 }& ?0 o" B {& p' G3 |2 O0 c0 D
printf("请输入查找数据:");! U7 Y2 E% ^5 t4 J
scanf("%d",&number);
* B& V! V0 ~1 S$ {! y9 K7 f loca=0;
& K6 z- p' ^6 c4 D top=0;, R% i0 i9 k2 B* \: }+ f/ k9 x9 ]
bott=N-1;4 o- c; H7 y: |8 r o
if((numbera[N-1]))
: }6 O4 y: j- A4 x loca=-1;
3 n. G/ E( R r while((loca==0)&&(top<=bott))1 Y; y& C( Z1 a3 V& f8 ]
{ min=(bott+top)/2;
# R4 h# A6 l* z6 j; F5 X$ c2 K if(number==a[min])5 Y3 q" x1 ^9 {8 {7 X4 w
{ loca=min;/ m4 |3 @* q8 b/ Q9 n
printf("%d位于表中第%d个数\n",number,loca+1);
! X2 \6 G5 q4 x% [ c' p- N9 l }- l3 ^9 S) N& k' M. \" l2 m
else if(number bott=min-1;4 J9 T z, U# K0 l
else
, w+ s8 \7 i/ G4 B top=min+1;/ h2 C( O) _ K' }7 Z
}# U5 S: x$ }# [! W; {( p3 ?( x
if(loca==0||loca==-1)
8 ? M$ |( Q& p8 C/ u' a printf("%d不在表中\n",number);
7 j/ D% y2 l! H2 @1 D1 E4 @+ c P printf("是否继续查找?Y/N!\n");
- V9 _: ^3 I3 f% F$ c c=getchar();- A2 O6 y; ?6 a6 L
if(c=='N'||c=='n'), ~& B" B$ F( _; T( l0 n6 J* \6 ~
flag=0;& S: t; m- x" ^7 R' ^' w' e& t% D
}( C3 M* [: d# d- f2 z) ~9 h
}; W% f% O! X; x( c
" y; v6 m/ |/ w3 [
7.10
+ _ O; B; y bmain()$ ^+ G3 E7 d8 g( X5 B
{ int i,j,uppn,lown,dign,span,othn;
9 G9 t9 Q* `8 j! W' A6 v char text[3][80];
7 p Q; y: r+ n7 Q1 F uppn=lown=dign=span=othn=0;
]7 { X9 P7 h: q for(i=0;i<3;i++)
2 W8 F( B, c' q: _' Y& ]) q- Z { printf("\n请输入第%d行:\n",i);+ a$ P$ P7 E* w! l0 e
gets(text[i]);, c9 F3 o w9 ]
for(j=0;j<80 && text[i][j]!='\0';j++)
* p- T) \. m! c. t! n* ? {if(text[i][j]>='A' && text[i][j]<='Z')
2 U2 Z2 m- q+ D, `" M uppn+=1;
- W4 T( ^7 t6 T& `# L0 }8 J& l) [ else if(text[i][j]>='a' && text[i][j]<='z'); Y( ~; l6 F: ^7 x( g; G
lown+=1;
5 n+ r' }0 N3 J* j$ r% L else if(text[i][j]>='1' && text[i][j]<='9')
' q+ O9 [$ B: H h2 w dign+=1;
; Y9 E# |. R) D! D* \- X, y else if(text[i][j]=' ')
% }6 p5 r. s' ~+ c span+=1;' h( n0 Q7 O: Y ]# V# P8 k
else
( B' ^ }0 u* L6 Y T5 M2 I* Q& ~ othn+=1;9 V, H+ `# _2 V$ o
}
$ x# k* E& y: O J' n, y& i }
! D; x# |' [# } {" f for(i=0;i<3;i++): q' L* o. \" V2 B
printf("%s=n",text[i]);
$ N! d# x( K3 S# Z7 L/ @ printf("大写字母数:%d\n",uppn);
8 d5 g f% u0 X2 [, u printf("小写字母数:%d\n",lown);
/ Z; q7 k" R6 F4 i6 S: U$ I printf("数字个数:%d\n",dign);: v6 D y0 t( {- ^1 b
printf("空格个数:%d\n",span);
5 O1 M+ Q! z6 F6 V$ s printf("其它字符:%d\n",othn);, m1 ?/ s5 Z& y7 y) A* e& x0 |
}
! y( G+ j+ A: l
# \8 F: _4 j6 A( Z" s6 X% d' {0 K- D2 c, k: v9 B7 a( t
7.11
& v$ K. ] s, J* O% G+ t7 _main()
i7 P& L0 D# r' z {static char a[5]={'*','*','*','*','*'};9 w8 f, j5 Q- P6 ^. Y
int i,j,k;
4 D4 G! ]: w4 B/ f char space=' ';- B% y5 r& R# I. E: k* m8 K' n& v4 b
for(i=0;i<=5;i++)
& t- a' `5 b: m( c! ^ {printf("\n");+ z. U _# s( R) M S# T1 h- m
for(j=1;j<=3*i;j++)
9 @* @! {( e" U2 M printf("%lc",space);
) \: j4 R% `: x& p) i for(k=0;k<=5;k++)3 d+ h, ~' y8 A; Y$ p9 c% [
printf("%3c",a[k];$ C+ ~+ q& J4 z/ m; W4 @4 w
}
& U6 ?: c' ^6 Q1 y8 H}3 v- U- }8 {, d# a, l
7.12
0 N/ U' G3 e/ U6 e h; j#include
4 k( b: p" C3 {6 V [/ omain()7 p) G E+ v/ k% C
{int i,n;, h9 Q' I" a* x' C1 ]( `) k
char ch[80],tran[80];
# D4 W: J9 x- o$ [) `7 g printf("请输入字符:");
( R$ Z* E& x% ~0 g/ u2 _ gets(ch);* Q: B' D# p+ A/ k/ ]1 e' o+ b
printf("\n密码是%c",ch);
5 p$ E+ P: P1 ri=0;
" h6 X. B+ }, q5 m- I$ d5 v. Ywhile(ch[i]!='\0')1 M L. ?2 d* J: E* c3 @
{if((ch[i]>='A')&&(ch[i]<='Z'))
( L; j& T0 {" t6 G* q/ Q tran[i]=26+64-ch[i]+1+64;
3 E/ S p c7 pelse if((ch[i]>='a')&&(ch[i]<='z'))
/ f3 f, M: T# p0 E( U$ l tran[i]=26+96-ch[i]+1+96;
) V) ?/ G4 l r! Q+ E% b! J8 N% Ielse+ ^1 J$ G8 F' N T0 G, f/ g
tran[i]=ch[i];
+ P+ u. a( x- a7 E! f p) d i++;; W+ s- ~' B' I3 `# r+ w- C' z
}
; ]6 l8 ~8 \7 b4 L1 e" ^3 Dn=i;
+ }7 c' S) _) l$ P+ `7 uprintf("\n原文是:");. T9 L# j* b3 ~; B5 Q: h) ?3 z
for(i=0;iputchar(tran[i]);
# G( ?8 x0 {! [" ]}( s9 X& I- `; M; @
7.13
" _( v* l1 V7 l& M3 K4 J omain()
) Z. ?* {( J/ x6 q" v# c9 |+ { {8 C7 _$ I3 D: |* V% I. _+ o4 |
char s1[80],s2[40];
V) j. C- |) }$ E int i=0,j=0;
2 F. d9 q# X3 U% @1 _ printf("\n请输入字符串1:");
' Y0 x: o( ]& B1 v scanf("%s",s1);
# S1 k$ F( w3 ?2 a5 B8 i printf("\n请输入字符串2:");* ? n5 T6 A' P3 N6 w
scanf("%s",s2);
0 H5 g# L- E/ u8 |* b while(s1[i]!='\0')9 W$ k+ p, K0 D w
i++;. T* g! Y5 `" B+ O8 J
while(s2[j]!='\0')
! i. k* U9 R0 H, i s1[i++]=s2[j++];
+ b! I& E2 Y" p7 j c: l5 ss1[i]='\0';
# x7 N; s+ _- [7 n; l) k) }printf("\n连接后字符串为:%s",s1);
2 u5 c# h4 e. m) W6 b3 C* Z }
w/ _3 s e6 o* W: u
; I0 D) E/ Y% L7 v& @8 f
# z( P/ c1 z6 t8 I: ]7.14$ b1 u7 | N1 R6 o/ \
#include
! x+ t3 S+ H0 \" O- \main()
+ `, ?" }2 h @; t{int i,resu; h, D% H* r- x: A( L i
char s1[100],s2[100];/ J# K+ k# g& G0 l4 U+ ]
printf("请输入字符串1:\n");
- N$ N, A0 c/ a) G. c9 A gets(s1);
. q1 I7 S7 Q1 P. C9 J printf("\n 请输入字符串2:\n");
* d W4 E( X }: A l gets(s2);
& R$ q( `+ Z: I5 l3 |* j9 E i=0;
5 }: @, L+ q9 V" p2 L; m$ l while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;; R4 n3 V3 {3 B. [7 J) S
if(s1[i]=='\0' && s2[i]=='\0')resu=0;
% m6 d2 n y- c else
+ U3 Z# o$ B- O5 x resu=s1[i]-s2[i];5 T: s9 B% Y! B. y
printf(" %s与%s比较结果是%d",s1,s2,resu);
+ U' W- z% v9 V' z( s}
, Y+ R, D' e4 r; M9 j7.15
. l9 } ]: Y2 F0 j4 ?/ p& j/ w9 Y( ]#include- F& U3 _, h0 N
main()0 f! d) u) U, d) }2 R
{
1 p: O( Y/ ]7 O, a5 N2 } char from[80],to[80];
8 {* O/ _$ B! W/ X5 z/ ?1 h7 }: |0 } int i;
8 e4 M! A: E5 V4 _; X) k) p2 o# H printf("请输入字符串");7 y) K+ D$ J6 E ]
scanf("%s",from);
4 l( c% ~( p2 r/ G8 Z for(i=0;i<=strlen(from);i++)
/ k8 @( `# A5 R: }# { to[i]=from[i];" w$ z l& `$ ~, s: M
printf("复制字符串为:%s\n",to);
2 t) r* g4 q0 }* ~' m* Q4 G5 W8 k }% p. @' d' X6 t0 S
; M7 R, X1 g; x6 v% u; m; i" F! h$ W" k, Z- a1 `
第八章 函数
3 M G3 ]" V o# B0 G7 Z+ e5 q8.1(最小公倍数=u*v/最大公约数.)
; H- |& l+ V. y" C/ E& Lhcf(u,v)
+ v S9 `7 Y/ x2 z* uint u,v;1 u* Q, |2 q6 D O3 z
(int a,b,t,r;5 D1 D( ^0 m! @( T" @( m; ?6 Z
if(u>v)
2 J! c4 ^. e; e7 r+ l# [1 Z {t=u;u=v;v=t;}
( y. n; K4 @% j2 [/ R( y a=u;b=v;9 E8 |4 g7 ?6 ^3 Z% ~' H, v, J/ E
while((r=b%a)!=0)
# P3 C( A7 U# u* x9 e! n- b+ | {b=a;a=r;}
5 K- x, y1 Q& H return(a);, |# q4 F) l+ x% R, a$ P
}
% ~2 T! ?2 [$ z* K: H! J lcd(u,v,h)$ h7 v+ ?; m6 _* D+ a- K
int u,v,h;; r7 B. N1 A& x+ l; |
{int u,v,h,l;
, [! G7 N9 J( s$ c scanf("%d,%d",&u,&v);
0 {$ h$ n9 b$ n$ b- r: K, z h=hcf(u,v);& j" n; _2 I9 B4 |
printf("H.C.F=%d\n",h);
5 B1 F1 ]& |8 j1 d: ]8 I( ]* y l=lcd(u,v,h);. ?% |- K. a o0 u& S* A( C
printf("L.C.d=%d\n",l);
) N9 a0 C5 O" k, [ }+ t) l: H ?1 E* E# ^
{return(u*v/h);}1 Z' d$ ?& f) y* {5 [9 r
main(); M3 a7 q M; s7 K
{int u,v,h,l;
1 k, d* n9 G. [( s/ w: R scanf("%d,%d",&u,&v);4 N$ A. N& L8 S5 {( k* t6 ?, H8 A
h=hcf(u,v);
* X8 i7 y3 c8 h7 q printf("H.C.F=%d\n",h);
8 I( U+ J5 @6 B9 ]$ `0 _9 E5 y7 J l=lcd(u,v,h);
% o3 O# ?& k# K printf("L.C.D=%d\n",l);
$ s' k Y: g6 p1 o$ k& R }* ^1 e e/ S' D5 E5 L
+ Q9 o- B Z: j {9 |6 K8 J* z' L$ f: N5 k: D* L
; ]6 @5 x$ N+ j. W6 c/ ]
8.2求方程根( j) j6 |( D( b0 ^' e* w. C
#include
& r# A, V% p0 ^float x1,x2,disc,p,q;, Z" y' l# S4 T- i M- }
greater_than_zero(a,b)! T( g# H" h' {" B
float a,b;
& I3 v- r) M+ H' x+ H{1 W, D. @4 ^ A& t- u" ?* h. q
x1=(-b+sqrt(disc))/(2*a);
& ^, ~3 T7 Q6 ?x2=(-b-sqrt(disc))/(2*a);
, _6 r8 r+ a0 B- t9 F/ ]}
+ W1 P4 G9 ?9 Sequal_to_zero(a,b): x+ B1 `* m1 m: |. Q
float a,b;* | D* z8 z( W0 B7 f' B
{x1=x2=(-b)/(2*a);}
$ z! X& a0 q/ Q1 y! G2 hsmaller_than_zero(a,b)4 K. Q4 \, G* i: J) u4 Y+ q: E: F
float a,b;1 W0 z6 |5 M( M( D+ R" a! P
{p=-b/(2*a); B1 X* \6 \; G1 w9 O
q=sqrt(disc)/(2*a);
1 \/ j+ ?4 M6 E- N}
3 c# l! F% B; O( g) N+ A- Omain()
, g, `1 @6 V6 y# G{9 u- ?8 y/ D' I$ ]0 |* r
float a,b,c;9 A0 v# m; G9 [! o4 p+ B
printf("\n输入方程的系数a,b,c:\n");
, S; D [! O/ v6 G* ascanf("%f,%f,%f",&a,&b,&c);
r9 V. O( t: w; z- \4 eprintf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);% {) L3 [2 x: N0 |5 S
disc=b*b-4*a*c;
, F& O9 P: }5 _2 @# Lprintf("方程的解是:\n");
3 V/ V: P0 S3 w5 dif(disc>0)& `: S8 I2 {% l' r7 O/ t6 |
{great_than_zero(a,b);" { r2 x4 ?6 E5 ]4 z/ H2 U
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);- v' g3 ?# D( p8 r8 O0 l; M, z9 j
}
3 _6 e' C& }8 f+ m: A C# X3 jelse if(disc==0)
8 L# s* _5 m2 Y' D! N- v9 Z4 V {
! ?& Q9 O# d% T3 K" Gzero(a,b);' V+ V# _. K1 k9 j; m
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
0 _8 T# D, ]" s _ K }
* b5 A$ t, v5 p( j1 V1 y3 jelse) j1 F$ h% D' ^! R( ? }
{
0 k. a5 r) { s; m( f small_than_zero(a,b,c);: m/ p# h2 M' ]
printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);
3 W) Z# n+ X3 W6 Z7 H }
! M' H3 \' i0 `* K/ ?3 Z V0 }}* i4 {( ?% f8 g( r, s7 u
8.3素数
* k; F- Y- ~ ?) D0 B: n#include"math.h"1 `( y- P0 t a+ o3 Y) L% k2 C
main()& e0 C4 X+ H" P5 N7 z
{int number;
}6 v6 ?$ d! v* T: a scanf("%d",&number);* I$ [8 ~9 a- n
if(prime(number))
- ?" D4 l" B" B, U) P4 Z printf("yes");
7 [* u; G0 A J else$ U# x& ]& A& ]2 I+ ?; |8 ^
printf("no");/ A/ [6 c3 t, ~3 t# U1 S
}) l6 h% F- R [& i
int prime(number)3 s9 K5 h) ^4 D( b. @3 F) t+ ^
int number;$ \2 n4 J- k9 i' L+ `0 V
{int flag=1,n;1 `; q( S$ Z# t
for(n=2;n if(number%n==0)
; c5 f. m6 n" A flag=0;
# |, e6 U* z! t! t2 c$ M% d return(flag);
" \! o' N" _. q0 i% a4 @6 _: Z. t}3 P2 H; S/ E0 ]
# L; W/ r+ K" [
2 [8 C; P# M3 E7 `0 m' l9 Y+ \. ^3 R2 t- {- Q2 F7 o1 G4 f# P
8.40 N: E% R' O- a G8 [6 D' l
#define N 3
. e! k, A8 M3 ^int array[N][N];
" X& Q N9 l. h6 v9 q8 qconvert(array)
, Y# E/ } \; V" xint array[3][3];
' P6 ]9 [4 |% o5 C7 E { int i,j,t;
, q, K3 s% m9 {! u$ V for(i=0;i for(j=i+1;j { t=array[i][j];! Y) J7 a* P$ q! u& D
array[i][j]=array[j][i];
& @. I. j f3 A% K array[j][i]=t;$ A/ ?' C# u; Y# W
}
2 g) X4 @- E" ]; p2 t; p }0 k" f) e$ B* |# Z2 r" e( u5 i
main()0 V5 T4 A7 m9 W6 Y# }
{
- I5 x( o& `8 E5 t% b2 s int i,j;0 c! w& d) W: i% f
printf("输入数组元素:\n");
# Y# n/ T0 J/ x6 } for(i=0;i for(j=0;j scanf("%d",&array[i][j];3 E' `5 L0 d- C$ K b# }- P
printf("\n数组是:\n");7 i- \4 P$ T. x& x: M' E
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
$ _9 f8 C2 R( F q# K1 E4 ] printf("\n");/ M4 B! z; K6 p, Y; s \
}
! y# g+ |' G }# Q0 X convert(array);1 Y& M ]7 ^( ~0 v# p4 C8 J
printf("转置数组是:\n");
8 t; Q" g6 v8 D' g7 N3 l9 l% Q; `3 k for(i=0;i { for(j=0;j printf("%5d",array[i][j]);! G5 _7 A) B$ x4 J4 S) L
printf("\n");
9 x5 R8 P8 G" W2 V- P }! f6 f* V4 c7 ^: H% N+ {: Q
}( m$ Q2 `* l" D$ i' ^
, H5 D+ d* k/ v
9 g! C& C4 _$ ^$ v6 F( [
/ [" D+ G/ _' J. U2 g8.5
7 L' O- J' U7 c. ^main(): b% `5 Q$ x2 d/ h. D) J1 s0 c' J
{' a5 `- N6 O1 f, B
char str[100];
: ^& H9 g" ?1 t$ f% X$ S3 X1 B2 B printf("输入字符串:\n");
8 i* T: v$ v# W1 o" C scanf("%s",str);
3 [* o6 {5 {: C' G$ f inverse(str);
3 v. S& x1 d1 G printf("转换后的字符串是: %s\n",str);
' W0 `' c& o5 N3 v4 M' ~; [}
, F$ Y: v4 J- d! F' uinverse(str)
1 m" U0 |6 S" qchar str[];, ^& e4 A- Y. p
{/ \5 d; a% A7 R
char t;
- ^) U% @! [- W* M% X- C int i,j;! F5 y( v- k* v; e! c/ o
for(i=0,j=strlen(str);i {& }+ J" N! e& y6 J% }& s7 P- J
t=str[i];9 x. z9 }' F% v) R' v
str[i]=str[i-1];
9 H7 Y `4 O, R9 y str[i-1]=t;
% A. a! l2 |/ Q w0 s: E }
/ @) [) a% s% k/ T% x1 q. ~}6 y! @* m0 w! M, }- A3 U+ N
3 s3 `: G( [* ^
. N3 n2 l. j3 W
: q# A8 `% N2 r) @% D
8.6
0 h0 m/ q" ?9 v" Jchar concatenate(string1,string2,string);
u9 }& G$ {3 o4 d- g0 c( ~- Hchar string1[],string2[],string[];: F. u- e4 {+ f' E$ }& S
{- n9 |) C; {" M8 t( q
int i,j;
' U! z# I" w/ G* o% ~for(i=0;string1[i]!='\0';i++)5 x9 V, H8 h% x& o; z$ ~% Y/ b
string[i]=string1[i];
: x4 f6 U5 l: \9 c4 b& ~/ ]/ ]* ?! _for(j=0;string2[j]!='\0';j++)+ J9 f( s4 |; [9 o: S, a
string[i+j]=string2[j];
. {- F# q4 Y$ a* } {, e string[i+j]='\0';% c" S6 }1 c f1 L8 j) P1 X$ N6 ], ~% V
}
+ S& H3 c) k; M# R' P; v7 [& {2 Pmain()
" l- p2 }4 _# f& U3 S' V{7 s' Y& e- ^2 {, i
char s1[100],s2[100],s[100];
9 U, P( e) P5 x: W) ^- R$ _ h printf("\n输入字符串1:\n");9 T" c* u4 r" A
scanf("%s",s1);
6 k6 P- U4 ^& ?% w: s printf("输入字符串2:\n");0 N3 n% V; Y/ k' J
scanf("%s",s2);
, P$ R9 k, K) h) P concatenate(s1,s2,s);" z9 E, K" n# P: Y
printf("连接后的字符串:%s\n",s);
; r3 e/ T# l! O( g' S# M8 I}6 F# T. u2 f; x I$ ]
" [6 h4 C' C- S2 h. H }7 S$ a1 h5 _6 L8 }* I
8.8
; L, k, f% Y1 Z. E9 ?6 cmain()' n4 C- P7 J# P4 E# ^( K
{8 `; z( C; P t* Z
char str[80];
$ |+ Z% j! \7 s printf("请输入含有四个数字的字符串:\n");5 {' r7 L& D. P3 n* S, A9 j
scanf("%s",str);4 a/ u3 K1 a- R" t
insert(str);; k- [7 I7 a$ i' `; n+ x
}4 |$ m6 z: c* R* a8 n. p
insert(str)
, I3 Q4 O: W: R/ X: S( B char str[];" a% y' |. A$ ?) z
{
S; y0 n9 r, k% x int i;' G8 \! _: V: t7 X. t) @) L
for(i=strlen(str);i>0;i--)
, W( x9 c6 u2 U6 m3 M { str[2*i]=str[i];
0 x1 ?1 G" \. b; _0 M+ ~ str[2*i-1]=' ';
4 ], g. m! p- t3 S' V/ p6 ` }
, u* `: ~- M, n, _9 ^% }9 ] printf("\n 结果是:\n %s",str);$ B4 O/ O6 m8 t0 K+ c9 _+ u4 I
}1 h8 l6 |! R0 m+ m7 I, X
" U. t3 `1 z3 Q _! R6 W/ B4 l
$ E4 u6 J4 l3 H$ P) m
0 |4 u p) m2 j, H( q+ u8.9
& Q5 v8 Q8 p' s, {+ a) A* E#include"math.h"; a; a+ U7 k8 [- R9 v
int alph,digit,space,others;
1 h( E' f! z4 e G. Mmain()! U0 k. H. p1 y" ]4 v' `
{char text[80];
, w9 k* V4 ^0 s4 S5 Y: p gets(text);3 b, |( Q f5 Q5 F3 p+ C
alph=0,digit=0,space=0,others=0;- \- t% Z7 A/ q1 [$ ?4 G- T, W2 B" w
count(text);
- P/ E, L: e1 Z. o2 h* a/ w3 | printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);, d3 Q# T6 N3 A; ?4 `
}
. @% \& j# R+ h+ H; ^' {count(str)# c; c, a, d! e
char str[];0 Y% y4 e# v! m# r: H/ W
{int i; t! T" ]; [7 U7 C+ H$ ?
for(i=0;str[i]!='\0';i++)
' g+ w% L! K( y3 q if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))* v' Z/ J) `9 ?" k
alph++;
1 Y7 D0 _" R9 \6 Q else if(str[i]>='0'&&str[i]<='9')
. a* j4 v! ?5 d" Q- G, \ digit++;$ _+ o+ \+ G" e
else if(strcmp(str[i],' ')==0). a; {% v3 ]" ]3 V) Q' L' ^8 a4 x
space++;
( ^' {. X# Z$ v else5 S/ o. {7 \. I- g" l+ q( v3 I
others++;2 H, W0 W7 [# f& t1 J/ {9 F; a7 x
}3 Y6 p7 J" Q# {7 {' `
' S7 Y8 w6 K: E1 F$ y* X) D
% c6 m1 T- @ f8.10
6 ]# P* [ t+ _% i( R' {# k! @( W K) Wint alphabetic(c);* E. I( r) n3 ~
char c;
1 f X% u( L" D{
: P( D* H' l7 r: n v! Z if((c>='a' && c<='z'||(c>='A' && c<='Z'))
! h! w4 i! {7 o2 \0 W return(1);$ ~4 P, a& E. j% c8 @; `" }2 l2 O
else+ I [- f- I4 f& P9 E1 t" {- |: \
return(0);1 P0 y3 p7 B0 Y, {+ n
}& [! H- k K7 x1 ^
/ H4 {( l% s+ {4 q: j' ^. [: H1 r0 q- Aint longest (string)
4 ]0 \; R0 g3 d# X9 ]- `char string[];) e% z- R: d2 `$ L# N% {
{
4 }" D( {4 ] a) |- r- ~ int len=0,i,length=0,flag=1,place,point;+ w& H( a# i$ Y2 A9 k2 U2 {$ L
for(i=0;i<=strlen(string);i++)( t) a g) x# k3 R7 v$ a
if(alphabctic(string[i]))
( ~8 a7 T) r- Q- ~; q4 ] if(flag)
7 i/ t/ Z0 Q% c {
, O8 o& o2 |" e6 x& W point=i;0 [( e' _3 \) a" G. S
flag=0;
- }" M4 v4 Z$ z% |+ U/ y }& Q7 u# D3 B7 F0 m V) K0 n; p8 m
else
; l* P: ? }+ P7 F& T len++;7 W2 d4 p" U) g, l0 b0 ~
else3 c- `% V. h& N, r7 v6 a
{ flag=1;
0 k3 Z5 l5 p5 Y! S if len>length)2 }* I- R" d" N6 g) I
{length=len;
" F W1 M4 t+ H! [ place=point;1 n3 c/ q" a* L. a, N7 L" P4 p
len=0;9 p! G1 u4 }" q; H
}
# |/ ]* l: G1 y1 K/ e, ?; z0 a4 y$ n }
. r$ ~7 D& G# H* y/ n5 x! U; g* ^ return(place);6 j* L/ O& G& c" Q& q
}4 D- O) Y) q$ J" C1 j3 o% D& l2 Y
main()+ ~4 M( m$ @* ~4 w! ~
{5 J, m2 b- D3 [. d% X7 [1 i% {
int i;1 R2 @5 j2 L8 y/ k
char line[100];
6 o5 w& k3 D) qprintf("输入一行文本\n");
4 j3 O1 B. {4 @gets(line);. f/ P9 V, U) X* Y$ o
printf("\n最长的单词是:");
L, a( Y4 d' K* L9 s) Wfor(i=longest(line);alphabctic(line[i]);i++)
3 T+ x9 U- @4 |5 g, N+ M$ j printf("%c",line[i];7 O8 J7 U6 b/ x
printf("\n");
Q3 R# H" L- C {& y9 Y' L}9 ^" E4 e3 I. ] B5 t
% c8 l6 X }4 K* n# b
; ~! b6 C) ]: o4 J0 A! {( L
2 ^8 x) z6 r5 ]% E% N8.11
0 A0 s: Z) [! c% q7 E( P#include
; A0 _6 a0 B/ w! K G E, r, I! J+ a8 [- Y- _
#define N 10
( p& F9 _3 ?5 J" s& [char str[N];
- \- [1 y& k) I7 Q" S# Nmain()# ~3 n0 [& A- M' m
{6 N/ L7 O! i& |: O9 U
int i,flag;
$ S7 Y- u& L" f3 zfor(flag=1;flag==1;)4 S) ~, ?/ y# g, M8 I9 T
{$ g! m7 w1 V! E+ ^
printf("\n输入字符串,长度为10:\n");) n3 S/ g" N: l2 O1 O* b) h
scanf("%s",&str);- @ @8 P5 I4 s$ \
if(strlen(str)>N)1 c' I2 L0 Z# I
printf("超过长度,请重输!");- L3 G4 f: X. H
else
/ T8 e7 C- |( U6 s! E' E) X8 v* C% \ flag=0;6 w0 G9 J5 N# \ ^4 C+ C2 E
}
6 P% }5 X) Q# n+ Psort(str);
J2 D* V' f$ L S5 Hprintf("\n 排序结果:");5 t9 x; d3 ` `% j& k1 v7 b9 h6 Y
for(i=0;i printf("%c",str[i]);
4 F6 [6 ?! h& ]6 z}2 ?0 g) [ f, M/ \
sort(str)6 ]" u3 S! c# D* n
char str[N];
' B! z- U5 U3 R$ P# E5 X{% f [' L# N: n( `
int i,j;
4 }! U& b* B5 ]6 y% hchar t;
) w- q, A' j4 o+ K; ?; Pfor(j=1;j for(i=0;(i if(str[i]>str[i+1])9 K1 G5 X# X! L
{ t=str[i];
/ K, n" P: l$ h1 N8 P5 \ str[i]=str[i+1];8 U, R7 `! y. U
str[i+1]=t;) n6 n7 y, h! t
}0 E0 j! a$ m+ O; j
}
' }/ o* M8 o- @! c8 J* K8.12, R: w/ X3 Q0 e
#include; r$ K) G2 J& N- p W
#include
+ G0 v* u4 q4 sfloat solut(a,b,c,d)2 `4 i9 M7 A! l; z1 E# e, B
float a,b,c,d;- a/ H% b: X& d2 F2 G' G' T1 K
{float x=1,x0,f,f1;
9 d3 ^, c( P9 C8 Y2 t- e' L do5 {! R: I+ h% c
{x0=x;
, U+ X' m; E; a0 u) w$ @ f=((a*x0+b)*x0+c)*x0+d;9 \$ k) } n. J \1 X% T
f1=(3*a*x0+2*b)*x0+c;- K: j8 O# D* A( ]7 {+ D
x=x0-f/f1;
' ?& x7 E5 V! a/ e1 W }
* B0 {5 P4 n/ r8 Z. M while(fabs(x-x0)>=1e-5);* A0 E. H1 Z. G% L
return(x);/ ~6 ?1 C# w+ j: ?. D" U* V, y9 g
}, j9 U# e! ~8 k/ a3 F5 t- O ~
main()
# J3 D$ ^% P) o{float a,b,c,d;, U# U$ a% l+ a( k% A
scanf("%f,%f,%f,%f",&a,&b,&c,&d);
1 X m" T7 X( x: T2 {8 b printf("x=%10.7f\n",solut(a,b,c,d));
1 R! `# m( z9 \8 O+ s7 j}
& `, K! k* G0 ^' b. S, N5 K8 H) k8.13! @% Z/ x- R2 O
#include
; K% @7 W& T( W h1 o' |main()
5 C: B! a1 m: N' D5 m{int x,n;% a* |1 E) h+ F
float p();
1 O {( G, k6 W6 b scanf("%d,%d",&n,&x);
0 |, `7 b; O' K: {6 F: u) r$ H( _/ y' _ printf("P%d(%d)=%10.2f\n",n,x,p(n,x));" {* C6 `7 P! I6 O5 _/ H7 M2 a: ~
}- V7 ^! t( y8 { m( E3 i
float p(tn,tx)( _% m- C0 U( _
int tn,tx;' M$ D% [- a4 f
{if(tn==0)! k+ j% N% k" [3 H
return(1);
% \( o5 y8 O5 Z8 F2 X: R% O else if(tn==1)
# @0 d* q0 e0 E2 E5 c2 V' O return(tx);" V& ]! z3 @( z% l) ^4 u, j: T
else
( ]3 r; I" |1 m/ x6 b$ {0 y return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn); g. u) s8 q+ q! `9 F6 i
}' j8 J2 r7 G& }( P
8.143 j3 t* p$ U# I
#include "stdio.h"6 W, ]2 b/ E) x8 R. r6 g
#define N 10, Z [/ H4 ]% W; q/ N! t
#define M 5& N1 F0 f+ X" G# L5 P# X; S0 z2 x
float score[N][M];) h; d( k) }" d* K; y) L
float a_stu[N],a_cor[M];
0 n0 u' M S& D$ j: j+ t5 T1 Omain()0 v) _; `4 z* b+ @% ~1 R
{int i,j,r,c;6 l6 d4 j+ n; a% r0 Y7 \+ @
float h;% M; c9 |6 q+ ?7 J( e1 ]+ b6 E
float s_diff();
2 [% g$ A4 c* h7 E. k4 x' g float highest();" ~$ s/ x7 K2 k0 ^ D6 \
r=0;/ `* Z, F* X a0 g" W
c=1;
- X$ K. O S- r% F! x1 i input_stu();5 f8 k, |/ x N; Q& x
avr_stu();& h1 h1 w$ I/ @
avr_cor();
9 z/ E/ b9 D9 H, V$ u printf("\n number class 1 2 3 4 5 avr");
; q. ]* i+ N; O4 N for(i=0;i {printf("\nNO%2d",i+1);
" v7 L9 o1 K {3 E. ` for(j=0;j printf("%8.2f",score[i][j]);; ]# T4 Y9 E( G" n2 `
printf("%8.2f",a_stu[i]);1 h* l8 O. s6 `1 [9 G2 I; w% q9 r
}. D; p( q2 ?+ ~( W6 N# m( H
printf("\nclassavr");
" G' z7 J9 ~5 p H7 y for(j=0;j printf("%8.2f",a_cor[j]);+ t& n9 V! ?2 ^8 H4 @0 \! Q# j
h=highest(&r,&c);
5 e5 U# v, W) _ printf("\n\n%8.2f %d %d\n",h,r,c);
9 q) S1 ~$ y( l0 `- g printf("\n %8.2f\n",s_diff());1 p/ W. O/ m6 A7 T$ g
}
9 z$ J9 m4 J6 F9 ~* N7 k' Y0 Hinput_stu()* ~0 H0 l1 b/ D6 E, y
{int i,j;7 o ]4 X% X* ~/ \1 v8 a+ l K" L7 S
float x;
7 \+ g% {6 t9 b5 p* S& B- o5 r for(i=0;i {for(j=0;j {scanf("%f",&x);4 i6 U9 ?& G- C6 i) S4 b0 r( U
score[i][j]=x;: R2 \3 Y- Y' m9 N1 ?" {: d8 o( w1 B
}
+ @6 E$ y7 n6 M/ O$ V6 M }# j% w0 I" v' ?
}: L1 X* K* J- M4 o
avr_stu()6 `" d9 d7 w! G* V6 j2 C1 Z
{int i,j;
2 M- x9 y1 y R# r' y7 O' s5 \. l6 } float s;
& H7 _4 v7 t7 o for(i=0;i {for(j=0,s=0;j s+=score[i][j];; h4 M* H$ ^- N- S
a_stu[i]=s/5.0;/ `5 D s$ K- H# o7 Y8 z9 }6 V
}
$ x+ f7 X0 F4 N/ g: Q% ?, H}3 h: a V2 i! A$ X0 y
avr_cor()
/ g( S$ D/ H* F6 w7 F/ i, ?" q{int i,j;
6 Z% {( D0 ~5 t: V5 X% n6 b$ q( e float s;
/ l. J( d- w" f3 Q0 D' E for(j=0;j {for(i=0,s=0;i s+=score[i][j];
+ g7 Q0 T5 `0 r4 L2 Y a_cor[j]=s/(float)N;$ M+ C! n2 U5 [4 z3 r$ C3 h
}0 @) p0 g# p9 F" S+ x
}
: l9 `3 C$ R: p; bfloat highest(r,c)) h/ ~: K5 T* X1 n
int *r,*c;
9 L# w* H5 @' ~ f* X e{float high;8 I) l, {/ N3 a0 m5 U; Z
int i,j;
- B+ R+ G) X4 Q: u6 M. B$ Z/ z) ~0 g: t high=score[0][0];9 P9 B+ t. h8 v7 j% M4 l
for(i=0;i for(j=0;j if(score[i][j]>high)
# r. V& S r1 i {high=score[i][j];) w6 B i* A: [% f/ P
*r=i+1;
1 F" w% ^( n3 C8 g+ {& w# K *c=j+1;
. j1 y1 S1 ?3 e }# N& W/ V* ~, [/ j
return(high);4 g8 S; r) w9 q
}% W k/ T; D" q. p2 p' C" _5 u: H
float s_diff()9 ]* q1 w- @ N" Y: c
{int i,j;
% q$ C) [- f: n$ X2 l7 c; P) Q float sumx=0.0,sumxn=0.0;1 F. h) t% s5 }$ }
for(i=0;i {sumx+=a_stu[i]*a_stu[i];
1 D7 D( O+ v$ H, \9 u- G sumxn+=a_stu[i];+ _2 m' m( o. ^+ G. n; c% h p
}
, `0 b( k+ G$ T/ _+ V return(sumx/N-(sumxn/N)*(sumxn/N));
% W, c7 {* ^& k d* D2 ?" ]3 l9 p}
/ U j- B1 \9 b9 {5 K* O8.15- s* P. j8 C. T$ Q) Z- W
#include* K( j+ Y- ]" {" N; } @1 I
#define N 10" x" D' V; q) J0 L. R' ]9 j8 ~
void input_e(num,name)4 _- j0 R! T5 Z3 t' P$ X( \
int num[];: T$ O, G8 Y) J" g/ D1 G5 G0 l/ |
char name[N][8];
7 C3 S5 I7 A: ?{int i;0 _2 N$ r* X# [; b7 T% A* F1 {
for(i=0;i {scanf("%d",&num[i]);
& c3 E' V$ v9 h# M% T+ p/ V gets(name[i]);
/ d4 t; h& B- ]8 ^ }3 v" V4 u* c! y' K$ }3 r6 g
}
/ p- n$ o* q: P5 R% c% \) Ovoid sort(num,name)
( T! s d7 v4 ~; }$ jint num[];
( x# s: b7 ^* ?- U$ jchar name[N][8];
0 z8 ^, q+ A8 Q" j _9 U5 A1 X3 u{int i,j,min,temp1;; Q; P7 z8 f& H3 ^; I. d2 c
char temp2[8];# w. @8 J/ O9 Y
for(i=0;i {min=i;
6 E7 R8 J- K% R9 h# g, b for(j=i;j if(num[min]>num[j])min=j;
$ k8 q/ c) c. U" A, E- [ temp1=num[i];
, b+ ?" d3 t9 k4 j, i! R num[i]=num[min];+ }; \4 v2 Y& m: C9 ^; `
num[min]=temp1;: U. A" J% J. Y: Y" i' Z: [7 D( k
strcpy(temp2,name[i]);
3 D$ A& [# w$ _: d/ h strcpy(name[i],name[min]);
) R" }$ k# G4 i* D strcpy(name[min],temp2);. g& \# u F. Y3 J( m1 f
}: L: ]: ^7 o9 r7 H
for(i=0;i printf("\n%5d%10s",num[i],name[i]);
3 g; ?/ Q$ }/ K- u/ k' @* X}& H _9 y6 i! `: G$ ~4 W3 x
void search(n,num,name)8 i; w: I e7 q$ ^5 u
int n,num[];/ [% E/ V, T# b/ U
char name[N][8];1 |2 t( o9 p5 r* c. Q r
{int top,bott,min,loca;
% f' Y/ s6 M# s! `2 e& Q loca=0;
8 B5 `! \9 m5 Q1 Y6 W$ l$ N. S1 h top=0;
( S L3 ~7 S: P" k bott=N-1;
V- H& U" @) G- o if((nnum[N-1]))
' @; k: D- ^5 L5 t loca=-1;
3 K. g: V6 K' L U while((loca==0)&&(top<=bott))
0 O) l, v5 `+ E" M' o( x, k# w3 k {min=(bott+top)/2;& U# Q+ |8 k/ \6 z, R$ v
if(n==num[min])
1 N# y5 V( y! _" O. ~6 @( k2 Q4 _ {loca=min;
9 d; {) e; v- h1 ~3 | printf("number=%d,name=%s\n",n,name[loca]);
: ?6 ?0 o9 n# j1 S0 A4 C% t3 m1 |5 H }( U$ a0 ?3 s* B1 | ? z* p
else if(n bott=min-1;
G1 l. J% ^' ?9 P else% V9 V* R: G/ Q4 C) E
top=min+1;
/ U3 q P! ^- ]$ L9 w& X }3 `9 `) i( T9 r% S# z. h% v
if(loca==0||loca==-1)
8 b/ z4 W1 N. `8 U4 B, d" w% e printf("number=%d is not in table\n",n);
) W6 M$ m% R3 f4 s& f S}
" Q: u9 y% d1 ^7 c; x6 v/ Kmain()
: H9 v' h- [3 r3 G{int num[N],number,flag,c,n;
S: _9 K. [3 r- o8 x char name[N][8];0 Q4 H$ `, L$ n$ I3 i0 h
input_e(num,name);
( X+ U9 ~' M2 ~" o& Y sort(num,name);9 J$ n/ Z6 D, Z( l: y! O4 F
for(flag=1;flag;)
4 z$ S e# l1 N" \& S6 x2 c; ]0 G {scanf("%d",&number);/ ~( @: p2 F& h- \2 m. u2 A' S
search(number,num,name);0 V9 v: o0 s/ S* ~
printf("continue?Y/N!");
8 r4 `% j' Y1 V$ u c=getchar();
) l0 }5 f- J. _ if(c=='N'||c=='n')
+ J) `/ U7 [# c) G0 { flag=0;+ C) l$ K+ [" L2 U6 `5 ^2 x
}2 f! I# G3 V5 O% C
}, O; n4 e* y! u# ^8 T
5 h" H$ G+ u- r( ?* e8.16( w6 ^. U0 c$ g
#include* v" G/ K5 r' j" M
#define MAX 1000" C5 I; r+ t3 P6 g
main()/ x' i9 z7 B9 E
{ int c,i,flag,flag1;5 f& I, h# ~3 b% B: P Y+ l
char t[MAX];! p' P2 k1 j6 Q7 y& D% m- r& z
i=0;
3 ~# O* A; n( f7 @2 ~! v+ o1 T flag=0;5 x+ N8 a! N2 \% W% g- n! T
flag1=1;, H; \# o' C& v% A* B% P$ ?4 D. d
printf("\n输入十六进制数:");4 o( i1 }+ C# }& D; {
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')0 K* @* z$ p% v; Q
{flag=1;. u7 X' f ~( C1 Q' T$ F, _
t[i++]=c;
. u u$ q7 P p ]& q J H/ P }/ k/ Z$ D* {8 l ]1 [2 U
else if(flag)6 z! e4 U9 M- i5 }
{8 H" w1 l i+ G4 k! f
t[i]='\0';
8 a1 y8 U6 y' U4 w3 ? printf("\n 十进制数%d\n",htoi(t));' Q7 b0 |0 t2 h, x1 Y
printf("继续吗?");
% B% d9 s8 @ k% c) v c=getchar();% o# ]- A( X, r
if(c=='N'||c=='n')
2 f( B5 p' d3 T/ b/ I9 p% J flag1=0;
! j% {% |/ t' Y3 @5 K% ? else! Q, U1 \2 @) u9 k1 k
{flag=0;+ q S0 K8 T! O. p: }
i=0;
& m+ y, J* A# E" V- ]8 \# O printf("\n 输入十六进制数:");
! J: j7 i$ ?6 N }/ x. ^% T" |: r7 q8 ]
}4 Z9 t+ I- R/ z2 M: l
}! k" Z7 @* i5 h
}
! k' q) P, M/ I m- j) dhtoi(s)! G6 Y6 z/ A$ l3 R' B
char s[];
2 e( B' F6 g9 N e6 A{ int i,n;, d: ^ c3 ?$ i4 g& w
n=0;' {4 n* T, S2 O y$ P
for(i=0;s[i]!='\0';i++)
4 b6 k& b9 X& a {if(s[i]>='0'&&s[i]<='9')% {% |& ?( y; {& T+ M5 P
n=n*16+s[i]-'0';, H6 c: `6 z6 ], n' i5 \
if(s[i]>='a'&&s[i]<='f')
" V# n& T/ B( Z n=n*16+s[i]-'a'+10;- t4 o8 f8 V# R2 q' d
if(s[i]>='A'&&s[i]<='F')
! O! Y- B; `6 C7 H9 u9 w! n1 i n=n*16+s[i]-'A'+10;' U' a0 A, q' Y$ Y: @ r
}
0 _4 M8 a4 W1 Z0 w( k return(n);
3 i# }: `9 H9 H}8 P* M3 E' `5 }, s) Y
N# z1 \. R8 l: d
0 n! M2 n9 T+ o/ [9 ^# b& g i
7 Z' C- Z+ \% C' \+ m5 K8.17
; u1 B/ c1 p; C" i9 `#include( l1 M6 k# i2 P/ u& A q0 J
void counvert(n)
4 G8 ` D, m/ H7 T- \4 Pint n;: Z! \. `6 P( n8 W+ u, q
{ int i;
* Q5 h4 p* Q# g0 `* W if((i=n/10)!=0)
5 s- e1 @1 ~: L5 v* b! D8 l convert(i);6 S' H- S4 N! b4 @( o/ H
putchar(n%10+'0');
. E& a/ j7 F0 g% W5 w; T; q7 I8 V}
: |$ I- \" e2 @main()
; d2 q9 N! \$ Q2 Q+ `$ q{ int number;
, |* z# I U7 ~* k) d X0 p printf("\n 输入整数:");
4 @* S/ d$ @; n0 ^ scanf("%d",&number);: v0 D# h D: y
printf("\n 输出是: ");
: a, e+ J: F4 f" e) j; X/ i& q if(number<0)
$ m" Z& m8 {/ k3 }! A' D { putchar('-');
7 j2 o0 P+ n4 C$ Q number=-number;
2 @+ h) y; c3 y* p, p6 V. [# M }/ I/ C( E$ P( |8 w
convert(number);
0 r# _ G( l w& Q7 o& y}
8 a* ?$ M ~9 B* e- ^; F8 g
- W+ v* \+ X- y/ a b |2 t' @# M7 p3 v9 ?+ M4 w) X* n3 J" K
. D8 t; M2 Q1 M: Z$ m1 y
8.18
8 b1 N0 `: L4 x# dmain()
+ q$ Q8 M$ R6 P; g( ?3 f; }{2 p: D4 E: B; |
int year,month,day;' ~9 P$ R+ R1 i* m
int days;
+ B5 Y, {7 b* R* Q8 w printf("\n 请输入日期(年,月,日)\n");
; E, n& k+ }: w2 J scanf("%d,%d,%d",&year,&month,&day);
* e- }6 N8 Z W" ^+ S8 ? K printf("\n %d年%d月%d日",year,month,day);
8 A. |' |/ L+ C days=sum_day(month,day);8 O; |% b8 b$ m7 X* j' y
if(leap(year)&&month>=3)5 S4 Z5 @2 r1 p- K' j+ E6 F' K2 w
days=days+1; G M4 @3 L$ A: g3 W! B7 X
printf("是该年的%d天.\n",days);
4 ~% G% H f4 D }
) B% e4 V u+ q" l* i static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}6 a; f3 ~. B; J2 p ^) T5 C
int(sum_day(month,day)3 u# l$ i G2 x3 A, m% v
int month,day;
4 f/ s9 O3 U$ v/ F( ` {
6 p! H2 ]% j) ^ int i;
7 B, |% h+ x6 z: f9 Q$ E: U for(i=1;i day+=day_tab[i];
' u* ]' o/ ~1 D return(day);
6 ~) T. ^- z4 P4 n& N: r6 u$ K }! b" q0 B+ j/ u) C4 L1 J# K |
int leap(year)
% A: i @+ G5 y9 M int year;
6 U1 Z& y6 p+ C- q# I, {: | {0 `$ t8 F6 H/ X& J5 J
int leap;
& y4 T7 d/ Y3 P* f0 i+ u7 u2 T- w leap=year%4==0&&year%100!=0||year%400==0;- o" I3 ^3 [9 m2 ~7 x- v
return(leap);2 C6 L* n8 a: m) T* y( Z$ g" |( d V
}
5 f1 e2 P" y) T3 ~. _' E第九章 编译预处理& m2 ~* l' M# T: [' W/ Y% _
9.1
- R8 z, {- Y3 k0 o5 l I$ [#define SWAP(a,b) t=b;b=a;a=t
1 {' C! _9 A4 c$ Bmain()0 a$ f/ v4 b7 r7 O- d) H1 d1 e7 ]! B
{
$ S( {; e" j: a% e' f$ }0 Mint a,b,t;+ r5 h8 a) a! ~4 f( _
printf("请输入两个整数 a,b:");
/ B$ T7 p5 B) bscanf("%d,%d",&a,&b);
+ p' f" y4 R" w; vSWAP(a,b);$ @6 z% X5 Z+ J) r2 U# G
printf("交换结果为:a=%d,b=%d\n",a,b);
( h$ s' D w* g; \3 z+ W}
1 e9 N" h; h" j# b( a4 V ' i9 L) W2 l7 w
4 N ^: z( v; u& }9.2 W1 N& ^' c; F4 P- H$ _/ n2 f
#define SURPLUS(a,b) ((a)%(b))
' O& F6 F8 _9 H% tmain()
* F3 F9 E4 A2 z7 J; G2 q$ P {
" e0 F5 i) J0 v4 _ int a,b;& T# G% Z/ s% X# i3 W
printf(" 请输入两个整数 a,b:");
) W, l6 k" \! D3 E: H scanf("%d,%d",&a,&b);3 k7 z0 `" g5 v6 D) p4 l7 {
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));
# p$ I3 y1 O& R1 n* A, [! H0 G0 t: V }
% W5 v3 Y/ _! r) i! N/ G& j4 [8 b: S. }* ~( h; T ]+ }. k5 i
, K7 U7 [1 \& ^% Z6 n# ^3 n9.3
0 k: L3 }* J* S, l#include
( g1 c( e5 @& z+ }#defin S(a,b,c) ((a+b+c)/2)0 x, f/ Y! O+ H4 K1 ]
#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 o2 D! h4 B5 C7 Z5 r
c)))
$ ^$ }8 W2 E) `: Zmain(): w6 E& ~! n" {1 o9 E
{
]* B! r" r% w, O- j d. T" F; @9 @ float a,b,c;
, w/ C9 Y2 @$ q printf("请输入三角形的三条边:");
6 p7 u% Z4 N9 I& f f' I6 c" Z2 r scanf("%f,%f,%f",&a,&b,&c);% o' U0 m. [0 ^8 }) g
if(a+b>c && a+c>b && b+c>a)
/ x! ^) k9 G2 D/ b! z printf("其面积为:%8.2f.\n",AREA(a,b,c));
1 i7 Z6 n2 ]$ v9 d# r' o7 k9 a else
0 O8 D7 m2 f' S$ I' d8 U" C* e printf("不能构成三角形!");4 M& I4 v+ _1 E% c' p1 }# N8 m
}" N0 O7 n, k# `0 l; I
; y6 c6 d0 ?/ c) w9 j9 E* g
; r o; }4 w7 g& v4 r0 O
. v; a, r" n$ b3 @9 A6 B9.4
# y0 \& [1 x* \* N+ W#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)
# }6 R, v" J2 amain()3 h" p9 I5 v, H. g0 O. b
{6 T, [% s4 t( R# F) V1 T ?
int year;
9 c2 [# P, X( G: f printf("\n请输入某一年:");
; A k- ?- l1 ^/ L scanf("%d",&year);
" Q; U" M! L& e2 ^+ m if(LEAP_YEAR(year))! w) T% r3 Z3 w- J; L; X3 `
printf("%d 是闰年.\n",year);" ?( F( i# T6 @, m4 X( p
else2 H+ ]8 T6 E! i+ p' C
printf("%d 不是闰年.\n",year);
( h/ b- z. X7 |) D }
3 f @" z+ ?9 q ^- n
5 L3 z. J9 ^4 [6 M9 V+ f
" l0 g9 w( X1 Q+ U: ?+ K
: E' z5 S+ h! K9.5解:展开后:1 N1 A7 o% d* i- W
printf("&#118alue=%format\t",x);
& k1 {6 y: l" p- p' vprintf("&#118alue=%format\t",x);putchar('\n');! B7 f5 n1 d% {* B( b& m
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');% i( T2 S0 c# F9 |
输出结果:
: _: l8 g9 p+ B0 u/ ?+ ?&#118alue=5.000000ormat &#118alue=5.000000ormat* u8 Y# S" @% `3 d& D$ ~; C; F2 ^
&#118alue=3.000000ormat &#118alue=8.000000ormat
8 `4 ?* b4 U" m/ C& V# G8 b4 C+ e7 \6 j j1 v9 S5 _
' O/ F8 H6 [& f: R
9.80 o9 C; Y$ V# \3 q, B* k5 }
main()0 B- H- L, p* m: d
{3 P) a( T2 s& f9 F8 O
int a,b,c;
: _7 W# P9 l, @& i6 G printf("请输入三个整数:");7 j9 m: Y/ A, y
scanf("%d,%d,%d",&a,&b,&c);
% J! b9 E8 Y2 Z9 O! p0 A N' s printf("三个之中最大值为:%d\n",max(a,b,c));7 O* X1 |! _/ t6 N! p
}7 Y' v! n1 e. i! d+ N& G. @
max(x,y,z)
6 x# O( B- `. E. F) Z( ~ int x,y,z;
. L1 q+ ], l) A5 S+ Q* H {
& j" v! t" h1 K+ Y( G8 R! U int t;3 F$ H2 u- m; C8 Z) }$ T
t=(x>y? x:y);$ F V( X! Z# K4 H& {
return(t>z? t:z);6 p3 t7 n" t+ \( M! e
}
2 d/ y( X4 n8 @7 o+ {# d9 }2 T! A1 w y b! c% D. e' c
9 q( o, {9 x& C, t+ {
/ S) c4 `, ?0 o9.10
% ~1 u3 [2 I, M8 N/ e9 l |" U0 K#include$ A' U4 e# ?: P+ X' Y
#define MAX 806 D3 N/ l5 s4 S8 z; \6 b' W0 P9 B. c
#define CHANGE 1
$ c3 I1 h" O1 R# B# wmain()- Z' O+ t7 B& C. f; A
{8 H n2 i9 J# a4 i" g" p
char str[MAX];* I$ q% l9 f0 S' x1 B6 u/ n; q
int i;
" R( ?) ?: K5 Z, k/ O printf("请输入文本行:\n");/ {6 A# d9 H; m
scanf("%s",str);7 D1 i7 F! F, M, L* F- h
#if(CHANGE)3 S" o! D) C- \5 Q$ D, n! E9 m
{
+ S% P% ]# a$ i3 _ for (i=0;i {
A5 U- N n7 V' T if(str[i]!='\0'
4 ?7 _! _' \8 w9 M* \5 w0 I if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')) Z7 s V, K+ P4 R- N$ T
str[i]+=1;. ^ q* J; V- {3 C. ~2 P2 X8 R
else if(str[i]=='z' || str[i]=='Z')0 _' P1 I0 V9 r+ F1 ]
str[i]-=25;5 `4 e4 `8 \1 a R6 S7 ]" l
}
6 i) k% f: `% o w) A}) [5 ?* _7 A0 @" V$ q6 [
#endif4 O, \. ]) o* I
printf("输出电码为:\n%s",str);
5 m3 B2 p* _- E6 _1 E}3 @4 n$ z3 k7 V s6 p# V! U
第十章 指针
0 Q+ R9 B# O& [0 e10.1
" c- W/ H( A/ X+ e2 Tmain()
: `. h9 W# H1 d( c/ ~$ V% S" Q{int n1,n2,n3;! a! d- w$ ^" L" z7 r
int *p1,*p2,*p3;* d3 l/ y b/ n! N* M' d
scanf("%d,%d,%d",&n1,&n2,&n3);
8 U# M' N+ n3 x# Q2 C0 q, o) F5 [ p1=&n1;
1 O$ n5 b; L* G6 l% p4 m p2=&n2;
2 P P7 G7 [! ^' c& i p3=&n3;* V: l3 |" W7 q- d9 A
if(n1>n2)swap(p1,p2);! v6 J6 U9 g8 S( Q. d8 D
if(n1>n3)swap(p1,p3);5 w: J. E5 `4 [& q4 h
if(n2>n3)swap(p2,p3);
2 n: M4 y, I! y6 u. G printf("%d,%d,%d\n",n1,n2,n3);
6 u: |- [; y3 H* u9 S6 K1 c}
/ D$ l8 Z0 @* B( N6 [- M' o$ j8 Xswap(p1,p2)
: U/ \ Q* R1 Q- t. K3 J3 I( Zint *p1,*p2;
. r7 z- S/ u. P1 w0 h/ `{int p;
: Q1 q- @& F1 ]& d p=*p1;*p1=*p2;*p2=p;4 Y5 g y% L* M
}
- O" X X: m5 Y10.2
0 k! I d* l7 p* e+ _# D: `main()! E0 o2 f+ X: w. O9 r! v1 J2 N
{char *str1[20],*str2[20],*str3[20];$ V8 B$ K6 b5 w
char swap();. t' w% L+ z/ h/ z7 e
scanf("%s",str1);
( Z) Y% b; b5 y* H' B scanf("%s",str2);6 b3 y0 U+ u5 u# z4 N% x* C: C
scanf("%s",str3);6 H$ ^) I/ z$ U! V' X# }
if(strcmp(str1,str2)>0)swap(str1,str2);/ \% k2 }' k! P1 T( L
if(strcmp(str1,str3)>0)swap(str1,str3);$ o5 o) |1 \1 v" }1 Z+ _3 x
if(strcmp(str2,str3)>0)swap(str2,str3);
- t _8 p, C5 `: Y printf("%s\n%s\n%s\n",str1,str2,str3);
) T; g) X7 v% D3 k2 }' d0 v}
' x ?, W0 e+ S) @6 Dchar swap(p1,p2)
" ^. g& B: ~$ d; M9 Schar *p1,*p2;
$ g# B+ u- w) N+ `6 @# X, @$ _! P{char *p[20];
?+ L# j! o: q: @( M strcpy(p,p1);
1 R# x7 y$ I( g2 n6 H+ ^ strcpy(p1,p2);) E* X) R. d+ d7 h5 ?- H, n# H5 g
strcpy(p2,p);& R- e! ]; D4 O* d( s3 N$ {' l
}
' z3 g1 g( X' ~6 R, ^; E, N. H9 l" e10.3
! o+ H9 w: |+ x; Q" O% Y, Kmain()" J0 G! S: r9 ~9 M
{int number[10];
7 l7 c0 X9 h7 I( A& z input(number);( q7 `) o8 t8 g6 b) [' G6 _( L% g7 g4 _
max_min_&#118alue(number);
8 _% O7 S- b8 y2 G- s" t output(number);5 A1 j1 q, h' v
}6 t+ W5 }6 j- h5 e
input(number)
5 r7 z- w! R# o" A" x9 Wint number[10];
9 ?; e7 L0 @" d{int i;! z2 [% b/ N& w/ ~# u0 `0 X; r4 n
for(i=0;i<10;i++)/ z# G& c% b4 q% D
scanf("%d",&number[i]);4 p# F7 U1 i1 y' S0 V+ m
}
( j+ Y) ?7 r* Gmax_min_&#118alue(number)
4 N% b+ s9 x; {2 A% y+ ~! ?int number[10]; _4 O7 h$ B9 X0 ]
{int *max,*min;) g1 _/ |. Z: M( g4 U0 ] N5 V
int *p,*end;
! i/ N; |! S5 r/ R( i( D end=number+10;* o' e9 v" n- J3 ~9 A
max=min=number;
* X8 y6 _( e( A* I0 m8 z for(p=number+1;p if(*p>*max)max=p;
- [* C; W) ~' P4 M# ^8 c else if(*p<*min)min=p;2 Y7 P$ x5 k3 }: M" t8 v4 T1 X
*p=number[0];$ \1 N) e. z% f
number[0]=*min;
4 I8 X- |" C: ]0 U' u& S Y$ R *min=*p;( \; \" q4 Y$ z1 f
*p=number[9];
; {1 n0 C7 Y( w4 ?/ B: L number[9]=*max;
! x# D+ p% y9 `7 D$ m: B7 p6 { *max=*p;
* e G! F$ Q; r return;
0 F# W7 D; q) J; @' s}4 z: ~. Z) ^0 `. m
output(number)5 o( B9 K* ]9 ]$ \0 s
int number[10];
3 x/ V4 Z- s0 B. g% G# e J{int *p;
+ N7 M; d# \: q, b: H% X for(p=number;p printf("%d,",*p);
( d" x4 _. W) \1 x4 d printf("%d\n",*p);' j+ w6 w; f/ J: z
}# r) |" v# g/ }" M& b7 N0 A& v
10.4
3 h" |/ T( `( Qmain()8 K; Q: v/ W6 y
{int number[20],n,m,i;
+ P) o- @7 x2 v! d scanf("%d",&n);( C* F g$ B9 ?, O( W3 A3 w' B# x
scanf("%d",&m);/ J; s0 b8 M: c0 ^2 n* F: R
for(i=0;i scanf("%d",&number[i]);3 r* e# U' x6 f: U6 F! @6 P
move(number,n,m);( T: ]: L7 m, ~4 J
for(i=0;i printf("%8d",number[i]);' S; f' _% b8 v5 P4 B; x% `2 W5 z
}6 ^/ E/ O3 v5 P% i2 O7 P
move(array,n,m)# e: X" ^4 \/ e2 S J# J1 W
int array[20],n,m;
7 G; W w+ t) s. C{int *p,end;
: {; l2 {( v$ s( x+ _, d end=*(array+n-1);
8 N0 U8 e* I" o/ z for(p=array+n-1;p>array;p--)1 J' S" k/ e) S# T
*p=*(p-1);' q8 m p0 E( g* y( ]
*array=end;5 C: O5 M: U* c1 V1 w
m--;% A1 a3 ~5 R! N7 B' G4 V( C
if(m>0)move(array,n,m);
! S' f* U1 Z& v1 i; f}
7 c' S G, O0 j* {: F) C- D! X10.55 ^& @- n# h; w1 _) t" {, F. v
#define nmax 509 B: H( k$ \2 ]
main(): K- u( V' A4 ]; j7 r# \! @
{int i,k,m,n,num[nmax],*p;' f% z1 K! J% E6 T
scanf("%d",&n);$ }5 X0 k5 E/ V
p=num;
9 L( W/ O# h5 R( y k* ]& f for(i=0;i *(p+i)=i+1;" p, \! V G9 W: B9 q
i=k=m=0;# L' r1 u4 D, |7 N. W3 ?5 Y& w
while(m {if(*(p+i)!=0)k++;# Q, W2 B# x0 o3 F5 [2 y, I
if(k==3): v3 }& L! V; \* q
{*(p+i)=0;& n, D% u/ Y8 V; P
k=0;! m' ~( m* G( j/ r6 X0 }
m++;/ r7 y( U% z6 m8 T% H+ a$ Q! a* C" I
}% |" Q9 d' K' ?( }
i++;2 F4 | k: L2 |. o. J
if(i==n)i=0;, Q# L3 B } a* T6 m+ x' y
}
) U( `6 F0 c! }$ Q# W' L while(*p==0)p++;
7 T: M% g9 L9 `. W& ^ printf("%d",*p);
( s, G* m% {) C" U& u0 d7 _+ k}
9 t! c: T+ c# O# j. Q10.6# F. a5 ^2 c2 o" ?& Y! K
main()/ N5 ]) k5 P4 N! H& ?
{int len; a8 C ?4 q/ N/ t2 W6 c
char *str[20];
: a; P/ j3 }, P: F5 q scanf("%s",str);
( K' ~; ^ n+ Z/ K2 ` len=length(str);
9 Y& h7 B/ r7 N: f* p- R- D- ~ printf("\nlen=%d\n",len);
' ], ^$ {) ?& a; ]& u$ i}5 p8 x9 P2 ]0 v6 u6 T7 S
length(p) k2 @) Z( k' t) z
char *p;
! F& N7 \3 X& \' S! j{int n=0;6 C b4 [1 h6 I
while(*p!='\0')
. x: q) Q6 D' W# E+ B* O: o {n++;p++;}9 L Y9 ^& U: t0 Q" }/ T6 d
return(n);
( J; \$ ~8 ]# [9 ^ ~. H* o+ _* V}
% j7 w- [6 o5 h10.7- y) x& j# M C6 T+ G. j
main()' X9 J2 l0 S& f. N6 G5 {1 m! D6 l4 p
{int m;
" P1 S! R" V6 X8 v) T char *str1[20],*str2[20];
! I5 @) a1 v) ^5 S3 S1 v7 T1 ~0 d) \ scanf("%s",str1);8 A* Q( e) t/ U5 K7 q+ v
scanf("%d",&m);# K, Y1 t% C$ E5 k U( |7 e
if(strlen(str1) printf("error");: Q6 i- z8 E6 l P& a7 S5 Q
else
2 K. e, X) T+ w/ j, G7 V4 Y {copystr(str1,str2,m);
2 h) m* x \4 Z/ `. z printf("%s",str2);" \& _3 g4 S- d6 P# W% ]+ @. Z
}+ q3 x% W, Z3 T: I6 ~/ }
}
$ r0 X/ g' _; z' w' E, J9 \6 w* X- U3 fcopystr(p1,p2,m)9 ]; U+ w$ H: _# O+ ?# Y
char *p1,*p2;
0 i5 c! m6 u% pint m;
6 L& P1 [* Y& {# v& V: }2 `- U5 K5 I{int n=0;
2 k3 c. ^8 @1 A# e. `, H+ W& e( ^ while(n {n++;p1++;}! ^ l; G2 B2 G L; s
while(*p1!='\0')
/ g w# p. W+ O' i" ~/ \4 E# j' @ {*p2=*p1;" L5 @2 Z3 ~/ i4 T6 I/ f
p1++;& l7 M9 g! n& }# E) {; m2 G5 M
p2++;
) j6 p; i3 f( E* Z8 T& l }( h4 @7 X" }8 J3 R3 ~ {. u
*p2='\0';, a9 j1 B9 x8 f) w
}! j/ r, v* ~# r3 V- ~" Q& {
10.8
" B; C5 W# T2 m6 X#include"stdio.h"
6 r; Q* ?3 X4 Rmain()
W& p4 x* d* L: R# ^$ K4 z{int cle=0,sle=0,di=0,wsp=0,ot=0,i;& v4 z c& q0 L/ X1 X
char *p,s[20];! |: H. N* j" |9 o: K4 h0 s
for(i=0;i<20;i++)s[i]=0;
4 M2 [" s' y/ W4 N& P' m i=0;
! U& t% {- q" V6 M! v/ t6 c. J' s while((s[i]=getchar())!='\n')i++;. h% r4 h3 L! T" L; E; z
p=s;. h( n \1 H' d+ e2 l, d% s
while(*p!='\n')
! E$ C. H2 I2 j `6 E/ q# R9 a+ d4 o {if(*p>='a'&&*p<='z')3 R! T# o" }. q
++sle;4 y( Y' i, C7 F# U$ H( ^
else if(*p>='A'&&*p<='Z')# b ?) M: C: F; R
++cle;
; t& O) z+ J* L; \2 } else if(*p==' ')
Q! Z# n: O, R; G6 ^, ] ++wsp;% \: I/ I1 V' Y' W
else if(*p>='0'&&*p<='9')
4 [: \0 i3 s- ], e ++di;7 _: I4 a8 W @
else Z& l. C/ s- P6 e% z/ _$ g7 B+ O: Q
++ot;* i7 p# e1 \ ?, [" v" u+ O
p++;4 {7 d! C4 g! G. B
}* I* [0 Z; ~3 N Y
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);1 G2 y0 c+ ]' N6 ^, B
}( S, d) m- Y, b$ \' M( u/ ~2 f% V4 ]
10.9
# `. q" n G0 M% n; Y2 J' d9 {# fmain(). Z0 g$ S/ S% W( a. q0 Z1 q) b
{int a[3][3],*p,i;' B8 T. H' x1 D1 |9 i
for(i=0;i<3;i++)
! o/ o) h" t' |8 ?! ~* | scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);2 C) \$ }; u% u0 h, n; @! w S
p=a;
. o& l; n0 T0 S X) t9 F U move(p);) m1 x0 Y" l! w( x& a
for(i=0;i<3;i++)8 J s! W7 G. M: n9 D
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);6 P, }4 D5 {! C; i# o1 r4 S
}# d4 X+ r6 T' Z9 q' w
move(pointer)
: l2 S4 S$ t' T" h8 I4 `int *pointer;( H' N3 ]/ y7 T4 ]
{int i,j,t;
! v; O, u7 s% j. F for(i=0;i<2;i++)
4 p) _6 Q6 c1 @3 }+ K7 S, Z for(j=i+1;j<3;j++)! L, {4 I1 F3 C. y& g
{t=*(pointer+3*i+j);; \( c% ^# e+ g4 \9 z5 a) f
*(pointer+3*i+j)=*(pointer+3*j+i);- n. @: S1 j0 v8 Q4 H% j( w
*(pointer+3*j+i)=t;
! Z- [& e9 b* ~' w! ]( }3 p9 ~9 e }
9 X9 g5 i1 I+ S! A% S}
! K! d9 Y% U% A, @10.10& v. u" |; n$ i, C7 x
main()
3 e( g$ k6 b' t. ?2 G8 m1 ]9 L{int a[5][5],*p,i,j;
) d9 t0 Z. R& ~/ y. U0 U for(i=0;i<5;i++)
' v z1 v v5 `8 P6 X4 M4 \ for(j=0;j<5;j++)
/ X5 F$ ~$ e. }- l+ M4 _ scanf("%d",&a[i][j]);* _, U5 K4 d& O7 p6 \' q+ _0 C% Y
p=a;
/ x1 H" E+ ^! I) L+ Y6 Y5 E% \8 q change(p);
) m9 Z" d6 e: U( \7 n- y3 o for(i=0;i<5;i++)* H7 {; [' y+ X1 Q8 T" o* B- w- F
{printf("\n");
+ ] Z; I# P6 X) x5 s5 Q0 J) ~ for(j=0;j<5;j++)
4 t2 a: r# C( d* ? printf("%8d",a[i][j]);
& _9 M& }9 p# m$ u% F' X' j( A( x }
; A1 }( Q7 o6 {' w( R}# Y7 J" ]- G' U- F5 r
change(p)
, G! a; ]# W8 }/ Y# q' [int *p;
) Q8 Z/ ^# f0 f{int i,j,change;
% T Y5 ^) y/ j' { int *pmax,*pmin;# y4 ]5 U7 s. j* J
pmax=p;
. {& x Y7 _5 Z/ ?" F pmin=p;; I3 H, S0 G, |+ k
for(i=0;i<5;i++), y$ a" p2 Q0 ^- p5 O/ h8 W
for(j=0;j<5;j++)3 {$ y2 q q$ w
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;2 n& S5 x$ a2 l
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;: S ?2 \$ M' [; ?0 O) I
}2 f! t% v3 L$ T; R; W
change=*(p+12);
4 P9 @8 y$ c+ H7 [2 W2 Z9 O! m *(p+12)=*pmax;3 M% \/ B( p' o9 ^& M1 D
*pmax=change;% }. {9 D/ B5 g& Q& O
change=*p;
( \( g0 T5 }" N" _8 ^. O1 j *p=*pmin;
* a' h- X; I+ y R3 a0 s *pmin=change;
/ g& m3 ?$ ]4 b4 l5 l pmin=p+1;
1 t: b; u2 P/ q8 R* z for(i=0;i<5;i++)( Y& v6 g6 \5 y+ y z( z
for(j=0;j<5;j++)5 w( s" i( _' o' {6 u
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;7 M( m$ z( N2 R1 F+ S- `: J3 m' r
change=*(p+4);
4 ~2 d' Q- h- n/ \) w *(p+4)=*pmin;8 g g9 k0 s8 i3 B1 O6 m
*pmin=change;
/ k, F+ Q9 k7 i5 @ pmin=p+1;# O3 F0 q+ |+ p& _& \7 \7 n4 o
for(i=0;i<5;i++)$ O# k1 B- F1 H2 F: L, p; r
for(j=0;j<5;j++)9 v( V2 @: J' i- U1 H% `" F7 d0 j
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))# L5 Y7 k' F% p2 T+ n
pmin=p+5*i+j;
6 I3 ]- m# h3 ]( p change=*(p+20);# b1 Z$ u2 `9 J
*(p+20)=*pmin;& H9 L+ o3 D9 N/ h! |: _
*pmin=change;
1 }9 g) s; H- S! B' J+ U6 f pmin=p+1;
2 l) S6 t" I9 b, z/ v* b+ e8 m3 L for(i=0;i<5;i++)9 x3 @) b0 [ v
for(j=0;j<5;j++)/ c. \" e* Y1 h# O5 k. l$ I
if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))6 I: j2 F/ } U7 B' ]. d8 x1 ^
&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;. }7 V) y( N, B- N+ y% h' t
change=*(p+24);4 R5 a) a3 J9 j+ b8 g
*(p+24)=*pmin;' i: z3 S& }5 L+ G# S e `
*pmin=change;
' W0 n" n% `1 \# @5 O" I}8 Z$ y d7 d# S' V- u# Q B
10.11
1 {5 {" y0 U& J Qmain()1 j7 K5 { D) j
{int i;: ~5 o* }! ~1 o2 ]3 t
char *p,str[10][10];
$ f8 s9 G0 d% S( X/ z' N8 } for(i=0;i<10;i++)( d% u! o+ ]# G) \
scanf("%s",str[i]);5 m% M: o8 x9 v* o% F, v( s+ j
p=str;+ U5 [: f: v# k7 L! V% c% z
sort(p);
* w" _/ M# p7 f0 y u E for(i=0;i<10;i++)
4 p8 W" c7 P( G1 I0 Z printf("%s\n",str[i]);" ?- `* [" f; d0 ~* i4 i
}# y/ v2 a3 P; T, m, h1 _1 e+ |
sort(p)
" g+ K u: `7 c/ D9 @char *p;
- Z# @5 I/ {+ l$ q; }{int i,j;
1 }8 n* w( @% Q char s[10],*smax,*smin;
; `4 J2 |# w+ D! l for(i=0;i<10;i++)
+ m# q6 h! N( h6 o f& C {smax=p+10*i;
4 a9 G& j9 A9 _ for(j=i+1;j<10;j++)
8 M) f+ a3 B6 l' Y( L {smin=p+10*j;- k# P) H" N3 _/ ]# j2 E. G
if(strcmp(smax,smin)>0)* g5 \% | A$ U' B
{strcpy(s,smin);
5 i( ~% e$ p5 ?' t* Y. G3 H' F strcpy(smin,smax);9 b& h8 u* D6 Q" d
strcpy(smax,s);" P' S( p7 a& t) Q% F" @( l* r
}
. H( w+ |& a, q5 U }- a q, y- j" H% i2 l0 V
}
4 S5 V1 x9 @3 m( h}
( X t {0 Y/ {' }3 h, P10.12, A) y% R3 `, f5 V3 d) W3 K' J
#define MAX 20
6 I$ @) V. o' U( Jmain()- `; G3 I" m! D! L8 b
{int i;/ E7 q* P+ e5 }1 G* I
char *pstr[10],str[10][MAX];
8 [& r- p1 M9 S0 M' K1 a for(i=0;i<10;i++)
& N: C. z1 K" I+ y H; t2 @ pstr[i]=str[i];
5 C4 d! {0 Y; J for(i=0;i<10;i++)& p- B4 W5 w0 }. b& F* |* d; w
scanf("%s",pstr[i]);# p3 l T$ S: H0 ]! d+ x) D
sort(pstr);2 O4 ]( u) w7 V
for(i=0;i<10;i++)
U6 u7 P1 f/ \; g* D printf("%s\n",pstr[i]);, u" d1 o3 j" h
}5 e6 s# t$ S6 h2 \3 m
sort(pstr)
" A5 _: P; D7 `9 T, v }char *pstr[10];9 U# J; Q w% v) {3 ?6 O
{int i,j;( _* C0 o% O) [* o3 d
char *p;
% r- q7 x0 _: v" A. @ for(i=0;i<10;i++)( E7 w& U; Q. p( ]
{for(j=i+1;j<10;j++)
: M$ M* y' l& q1 s0 |7 a0 T {if(strcmp(*(pstr+i),*(pstr+j))>0)
' V; ^2 @+ b5 r" a+ ]$ e$ j( }9 H {p=*(pstr+i);
' }4 N$ U# ^7 P *(pstr+i)=*(pstr+j);9 |! z' i8 f% a; d3 C. m9 N# F
*(pstr+j)=p;
- X9 R7 Z y: M4 m }) d+ I( X a! s5 D
}
; a* [; U7 D8 L/ L+ S% } }
! \0 Y7 l3 z/ Y' e. `. L9 a}
7 c) a9 J- J9 a( [10.13
# ~% l0 g4 E" i& Z. {7 o8 ^#include"math.h"$ u& Y# U: U6 u( z
main()
( Z! l1 {* q- ]- ^{int n=20;
5 s! R& t; Y6 ], r4 H# M" N( I' [ float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
6 s; e, q4 T3 o scanf("%f,%f",&a,&b);
7 f) |, f. U' c, c scanf("%f,%f",&a1,&b1);6 z- [) Q% O0 e
scanf("%f,%f",&a2,&b2);
" ^+ v3 `$ V, ?4 Y; F p=sin;% I6 A1 d! ^( K2 p+ x8 O2 d9 u
c=jiff(a,b,n,p);% P: { V' y/ Q+ b1 n
printf("sin=%f\n",c);) q& E$ ]5 O1 q5 U
p=cos;
) t2 i- Z8 p' g; i9 z c=jiff(a1,b1,n,p);
3 Z' x' ^7 ?' m+ B( q, ?" { printf("cos=%f\n",c);- u, C/ Z& U6 F# d2 U( X0 K
p=exp;
+ C d3 h- s2 w9 C" y) p/ U5 J c=jiff(a2,b2,n,p);
, W1 _# K& q# Y' c v+ ?# z$ L5 Q printf("exp=%f\n",c);
( |* p8 v, Y. d}
6 o& ^7 D, {- x" G- Ofloat jiff(a,b,n,p)
|, R$ v9 p- T3 O( lfloat a,b,(*p)();
2 ]9 v; T3 V/ z7 aint n;! O# o! @# a0 E
{int i;$ m) K6 s" R( H# ^! P7 W7 u
float x,f,h,area;$ v' p1 j2 D2 a4 P+ s$ l
h=(b-a)/n;
7 [3 r/ v/ [9 u0 f# @: N x=a;
+ @3 d% _6 }; _8 R6 P" K, A area=0;5 c: j/ O8 T: K2 J1 e/ h/ O
for(i=1;i<=n;i++). L9 n$ |: E0 P. y, v, V$ c1 G* M
{x=x+h;
9 D) ~: c2 H, r+ W1 b$ K area=area+(*p)(x)*h;; n3 m% ~- d# l; S
}; [. _, U9 `% _# n. R* U
return(area);
, V6 ?6 U, M- N7 R( M; ~# I}
! ~7 b; g8 S% T0 C; D: X1 j) v) ~10.142 x- f# Y2 ]) u% J# C/ }* @, A; i
main()& ?9 J q2 K+ T3 u$ ^
{int i,n,num[20];5 r% e, q# k. ^* V& B# i' t
char *p;7 b4 j. {7 H1 h; v6 c7 I3 }" \
scanf("%d",&n);
7 g1 E% o1 G) z6 U6 | for(i=0;i scanf("%d",&num[i]);
% g3 q$ _. l6 T" o& X( R p=num;4 c+ _. H: Q' Q( _1 m+ K- e
sort(p,n);6 a+ V1 `4 X& p5 x
for(i=0;i printf("%8d",num[i]);
* j* z4 a$ J2 k' B}
6 r& {$ F' p9 ]* z5 l% dsort(p,m)# {4 Y& d' ~# e" E
char *p;0 ^: f$ \3 ?+ A; R* [) d B
int m;- H0 U" }- H$ A' }
{int i;
0 E9 G# N; _1 ]& A char change,*p1,*p2;) U8 J. S# F, @- ]6 I
for(i=0;i {p1=p+i;" F6 e! p- Y' s% g
p2=p+(m-1-i); k! p( |9 J& _0 d! A
change=*p1;
! ]/ q0 M$ C, l! }8 w. X/ c6 u3 B *p1=*p2;
. {& j& N K& k/ h1 U2 h *p2=change;
( f$ ~ W$ s o7 \/ w% R }
. h3 `+ F q0 d) j+ M}/ ^3 l" b# S9 \" D. O) ^
10.15
; h1 W3 D- w0 E# g' ?1 s8 q* omain()
. V$ ^! n% `% G$ J{int i,j,*pnum,num[4];. X& L9 w1 Y+ t0 n
float score[4][5],aver[4],*psco,*pave;3 Z& w/ L$ C# c0 H- h' r0 w
char course[5][10],*pcou;
* n8 u4 Z5 ]7 h; e$ S6 l1 { pcou=course[0];
3 e% _4 c' X; u9 A5 ]; l) P* W! @ for(i=0;i<5;i++)
: i$ G/ D X2 r9 I P' V5 l( D scanf("%s",pcou+10*i);7 h3 |0 v9 B1 e" f- f
printf("number");
7 H# ?* n! A9 A4 y: }4 \ for(i=0;i<5;i++)# [5 `9 A6 P; c( c
printf(",%s",pcou+10*i);; @7 L) v) Y9 E8 \, p/ h% w
printf("\n");
& M, d* W( f; M) M6 O psco=score;, x6 ~. W) w* T4 X$ L- }5 x& F$ B
pnum=num;! ~- ?+ w6 u4 A- [; k9 e
for(i=0;i<4;i++)
# b6 d4 p5 i" \4 H/ \# S {scanf("%d",pnum+i);7 S9 d* y, P5 N$ j
for(j=0;j<5;j++)! K6 A, I4 _' u9 \. s
scanf(",%f",psco+5*i+j);
, k2 b8 g0 F; T: v2 m8 c' u! Z }6 w& Q9 q: K" V. H% h$ ~
pave=aver;
: k, c& p7 F0 }7 u' Z8 x; B7 v1 i printf("\n");( z: s* b9 O0 k3 X9 A
avsco(psco,pave);
9 R& o! y- _/ t) ^& x avcour1(pcou,psco);
7 n- F3 R; N6 C9 {( z, m7 `3 E' ? printf("\n");
, ?" N a4 _( r9 H* }% J fali2(pcou,pnum,psco,pave);6 v o+ C; e. L( P" C1 Q
printf("\n");7 ^; Z" q. c5 Z, C
good(pcou,pnum,psco,pave);
. I( M( `! L- T7 [/ ], ]}5 Z9 r5 |% u3 Y7 z' r, F3 F( { P: f- C
avsco(psco,pave)& t# v3 y O/ L' O4 |8 l$ c3 Y
float *psco,*pave;- c) c" e4 d; w0 `$ G% E
{int i,j;( ~9 j7 _0 ]- h0 Q/ b* L
float sum,average;5 ]% e( V5 e' g8 I
for(i=0;i<4;i++)
. Z$ Q: J& e0 r3 |) d! I b' P- b+ m {sum=0;# I* [7 q% U% T' l: c( b
for(j=0;j<5;j+)
" [0 J$ l$ S9 p; t. M* } sum+=(*(psco+5*i+j));
; \1 N: N+ d7 f6 f3 T0 q& d( X average=sum/5;
5 ] ^" a: O5 `; F. G( k *(pave+i)=average;/ L4 X- r+ I- R: g
}
* T8 V. H) r' y! C. n0 p/ S}
0 u. D7 o' U/ x a* zavcour1(pcou,psco) ` g( y: _* L* c8 Z, ~1 M5 M4 L
char *pcou;/ [! c* o9 P8 X7 Q: o9 F, [
float *psco;
" B6 ]" x Y; X: V: o" ~" D{int i;
! M- @1 S" J1 a float sum,average1;
9 e9 ^8 b2 C" |( ^5 N% ? sum=0;
1 q8 q( f+ p5 j$ _ for(i=0;i<4;i++)
0 M' Y6 `5 I# O; h, K) H ? sum+=(*(psco+5*i))4 \$ Q& u% W: H7 L% @8 y2 u
average1=sum/4;, @& b: s( @' x$ g! E
printf("%s %5.2f\n",pcou,average1);! P1 D( u& B0 N- ^6 R* {4 t1 f
}
4 q' w0 `- L+ g! K# J# T! Bfali2(pcou,pnum,psco,pave)1 t0 v* H, j* L8 ^+ `9 U; w
char *pcou;9 X6 N7 Z* A% G# x; N
int *pnum;6 h& a4 U0 l4 e Z7 ]: v
float *psco,*pave;) U. z" M" @# [$ I0 [5 b3 T- G
{int i,j,k,label;: N( P M Z# Y% ]5 ?3 P% k! f
printf("\nnumber\n");# \3 S7 y( A8 w' v* c: C
for(i=0;i<5;i++)
; d" {- R" P ^ printf("%-8s",pcou+10*i);
) Q7 q: t& ^" l8 ? printf("\naverage\n");* m7 g* v8 Y( |5 I0 j0 V$ \
for(i=0;i<4;i++)
8 |- g0 M- a7 c6 p {label=0;
! ]# h) m: M6 W" [; U( `; C/ S; g for(j=0;j<5;j++)
# N6 E6 O/ o) b- v8 c+ a3 T if(*(psco+5*i+j)<60.0)label++;
# a9 }+ T8 I; ~2 r. y/ F if(label>=2)6 }6 Z! a `/ G8 o
{printf("%-8d",*(pnum+i));
1 ?+ i3 T- ?. K for(k=0;k<5;k++)
9 r$ {0 y8 B4 D- k2 w) Q printf("%-8.2f",*(psco+5*i+k));
0 Z: o2 u& X- q) G printf("%-8.2f",*(pave+i));1 f. \1 S; F4 g- C Z5 r
}
4 I) M! ~1 g3 j% s, K/ C: X }$ T7 ^" z5 X- s4 K9 C7 L4 ~8 H) r
}' o& ]1 x, z* C; [
good(pcou,pnum,psco,pave)
+ L o* _. M6 @/ b: x8 d4 P$ Lchar *pcou;
Y% c; O+ [" U; [- U* m- bint *pnum;1 x8 o+ U6 C7 {" I) W- Z0 B7 q( U" T
float *psco,*pave;
( k) A" n4 j; J! @& m{int i,j,k,label;
: ^9 s% p) a! c+ E printf("number");
1 k* D& N* e0 B5 d S5 {0 j6 t for(i=0;i<5;i++)- I+ n" A. G8 D9 h2 A
printf("%-8s",pcou+10*i);. j( f# D+ V$ ~: L M) |" e
printf("average");
$ V; u5 U/ H9 B2 ^ for(i=0;i<4;i++)
$ y9 I* I- p* a( Z. c {label=0;$ _9 G& {. e2 i
for(j=0;j<5;j++)
: E7 V9 ^2 q- P if(*(psco+5*i+j)>=85.0)label++;
0 y/ S: V. d! q; H* E if((label>=5)||(*(pave+i)>=90))
H- V* u0 H! H- @$ e4 `, M2 p {printf("%-8d",*(pnum+i));3 p5 N2 G+ z6 ~- x* N
for(k=0;k<5;k++)1 c1 I1 @7 H K) Y
printf("%-8.2f",*(psco+5*i+k));
5 z/ v- i: P- m( c h% U1 |0 @ printf("%-8.2f",*(pave+i));! z; F' B( Z7 n) z( B
}
) {9 C% r5 [ @, S( D }
- `* s& ]! J, S7 y}
( F$ n( a: T# C( I& s: F10.16" U, c/ V2 O; s
#include"stdio.h") [0 w6 Q4 S7 }8 t: z9 E
main(), _9 \2 q9 _, X! z
{char str[50],*pstr;7 b% s4 [" W; k7 S2 n
int i,j,k,m,e10,digit,ndigit,a[10],*pa;8 E8 C& U8 O7 e
gets(str);3 e* H2 h4 T* t/ Y( N: Y
pstr=str;, N6 b Q) i4 h, \( _
pa=a;" s. F5 s' S; p' S4 B
ndigit=0;8 e6 @8 O# c; e; d" c
i=j=0;" B, H7 A6 m6 T
while(*(pstr+i)!='\0')
: b h4 C" H: \+ i+ R: ?* f4 P {if((*(pstr+i)>='0')&&(*(pstr+i)<='9')): P$ s' Z+ Z! |/ _% V: a1 n. s
j++;
+ C; I; S$ [/ e4 h) _ @ else
5 }( i/ v- D/ i1 s. D% D! ] {if(j>0)" j+ i/ h+ V3 E& F
{digit=*(pstr+i-1)-48;
- i: d& z& `1 s7 g k=1;
5 V! B+ O$ {9 E: y( U" y8 r while(k {e10=1;0 q7 U: n+ l# g
for(m=1;m<=k;m++)( L0 A1 d2 q* v. B+ }4 W' [
e10=e10*10;5 [5 s j2 R6 E+ G N3 i( |
digit+=(*(pstr+i-1-k)-48)*e10;; o* y' W( S; Q
k++;3 t( Q5 O3 w+ R, `4 k- S8 r
}& B4 {- Z Q# |7 p) `
*pa=digit;0 U4 ]& B5 J' [5 N
ndigit++;
1 p C9 K, l* @4 Q5 A' R pa++;' d; i: U+ _; G/ ~! t
j=0;* p2 T- n$ L2 t( S7 U, `, q
}
! T2 x4 U4 A. t) K* T }3 a) t ~5 ^. z4 [
i++;
3 P$ w$ x; m5 p% M5 C) A: P3 { }
3 C# @( Z% i T' p' { if(j>0)
9 s0 C, _! R/ A e, O! L% H {digit=*(pstr+i-1)-48;
2 U/ e! s! v- G8 V1 t5 A k=1;
/ l: _" X& J0 s+ x: z& W8 Q while(k {e10=1;
. u6 T9 z+ A& ]6 {) R" t0 c for(m=1;m<=k;m++)
& Q8 R K5 ?$ T* t- C' h$ F e10=e10*10;
# ~; \$ H. X* m digit+=(*(pstr+i-1-k)-48)*e10;
8 r; q: `" i. J6 [" d. { k++;
; J. l: \+ ?$ I0 t }
6 P6 \# T% Z7 y8 N0 r# \ *pa=digit;, Y( K Z6 C8 [
ndigit++;3 N2 E# W7 a2 f5 |
j=0;
* X" g0 N7 N0 q/ U0 W }
" l. j0 L+ N6 D1 |" N0 T$ b printf("ndigit=%d\n",ndigit);4 v2 p6 `1 s0 D3 {. U1 m5 H1 K# Z0 k
j=0;
5 w c0 @2 v" q$ j2 A7 C" z: o0 j* Z pa=a;
' Z: N2 i+ o7 v" n# F' I for(j=0;j printf("%d",*(pa+j));. F+ |0 C/ u# t0 X1 r1 n$ C" M
}2 d# {# ]/ O# N2 Z, H# q$ s7 n* o3 h' p
10.17- [; C+ M! t4 r! g7 Z
main(): t6 l0 ?8 V5 j1 X6 I6 I6 B8 e+ \
{int m;
4 |9 O4 z& ~- q& m0 U0 X- T char str1[20],str2[20],*p1,*p2;. o7 N, M; e8 Y
scanf("%s",str1);
0 ?* `+ E/ u+ o: w scanf("%s",str2);6 j, i- [+ x% E+ \! L2 x( U! h
p1=str1;/ ?' I# q( b- Q! L# I
p2=str2;
! `$ n+ Q" l6 g- c0 I, a! d m=strcmp(p1,p2);
s) H& j4 r2 f4 P+ |+ C! M printf("%d\n",m);( l. Y; G3 L+ q% b
}# |. V- X U) n' o( e: W8 F" {
strcmp(p1,p2)' [3 z2 l% t4 f, u- V8 s
char *p1,*p2;
K- P- [- L) H% y8 w! D$ w{int i=0;$ t- @7 G* F9 K
while(*(p1+i)==*(p2+i))
1 }; r$ }4 t2 x" n- u" ]- B5 W if(*(p+i++)=='\0')return(0);
1 C# R$ a+ G2 O. M' Y$ L return(*(p1+i)-*(p2+i));
! ~5 B8 t$ g* r$ [. E}
0 J( q$ Z2 |' H9 ]! ?10.186 B7 b) K' x: k9 P" m
main(); Z% Y8 ^( P2 e5 D6 K6 H3 U4 b$ ^
{static char *mname[13]={"illeagl","January","February","March",
2 Z p6 \% }- R. A) d; o "April","May","June","July","August","September","October",
0 @- D6 a+ D. b! e4 ^, Z6 o* ] "November","December"};
2 ]) ]8 U' q, {* N3 t l int n;0 S. X) B/ a' ^+ Y" f2 e9 ^
scanf("%d",&n);
; p8 j0 a- Q! i' d% P1 z if((n>=1)&&(n<=12))+ `! ~7 Q) l+ o) B2 _+ N% o+ q
printf("%s\n",*(mname+n));8 w- r8 J6 O* s- Y8 t* I/ m! e
else
0 x' ?! G' E0 i2 R5 Q- l6 a printf("error");
: n1 l& a" b7 M5 ?% f5 i+ X: Q' W}
6 E8 g& j% W1 f5 q* U4 B10.20
( B4 l1 e B% b5 Q3 D) ?4 i5 Vmain()
0 P' K/ v! }! `3 s- }* F{int i;
( ~, p( X' O l char **p,*pstr[5],str[5][10];
6 t; \+ _$ A' p3 B I0 D for(i=0;i<5;i++)7 s& F/ m, d' t( Y: x- \" _
pstr[i]=str[i];
5 Y' A; i" A6 z7 u/ t+ S6 P' U for(i=0;i<5;i++)$ A0 ]. Y; E% ~9 y5 o
scanf("%s",pstr[i]);* `5 l2 ^5 X* G8 J
p=pstr;
' h! Y& Z# b! v4 n' q2 Y& S sort(p);% v1 r. i- ?( A& p+ l8 k
for(i=0;i<5;i++)$ i+ L: j6 l+ b/ K* D
printf("%s\n",pstr[i]);
6 B) M B8 d* I+ f}# U$ g8 q$ N$ H3 L6 \) |- r
sort(p)
# E1 s$ d4 n: e1 i5 Ichar **P;: n9 J% s3 i) i( h: ]
{int i,j;
~$ X+ N" G& B char *pchange;% ~5 i! K3 {: F. n) K! }' X! G
for(i=0;i<5;i++)
9 l' Z6 B3 H1 }: T( H" a {for(j=i+1;j<5;j++)
! Y9 r( v" D( } {if(strcmp(*(p+i),*(p+j))>0)
' ^' U% G* `5 i/ P$ I5 {. Z8 h: h {pchange=*(p+i);
. q% ]& O# p. F% m1 s. d *(p+i)=*(p+j);
# L7 u; C$ T4 @( X2 q2 P- t *(p+j)=pchange;% w! v1 F: x2 |6 U
}
2 ?9 q* V% x6 B% r& N6 [2 ^ }+ t" P, T* ?- F3 w T- ^
}0 I( C% h" S2 d7 g# r( f6 _. S, }& |
}
8 F5 y/ r! o9 b% Z10.21
; {" K& v% C: a( D! k4 Wmain() m7 K$ g" r" O0 `9 E
{int i,n,digit[20],**p,*pstr[20];% w5 G& k. m; A% u; {8 X1 ]" a
scanf("%d",&n);: c, D3 r& H' v' O9 ?
for(i=0;i pstr[i]=&digit[i];
' b/ M4 |5 ^' c2 m8 C! B for(i=0;i scanf("%d",pstr[i]);- v4 H. h* H2 T. O9 x9 B# M. L
p=pstr;+ X+ |/ ` W: D+ a/ W' f8 o
sort(p,n);
+ B% O% g$ w& \# _9 A- p6 j for(i=0;i printf("%d ",*pstr[i]);
8 i" _! D- Q4 R( r8 e}" b' F* T1 ?4 Y8 h$ [$ g
sort(p,n)
: S* ]5 y z# ?, J, K6 b; K0 Kint **p,n;# ~- g3 ?9 N2 \& v. _
{int i,j,*pchange;
$ c9 U" \% |1 d. _. v: }/ R for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
/ W( O* Q+ P: c {pchange=*(p+i);/ o [ \' }0 z- ]( e( \
*(p+i)=*(p+j);0 s& m* `) Z5 m j2 j. G: ^6 s
*(p+j)=pchange;& z$ P; S* t1 a2 H1 T9 z7 U/ P
}6 X( T" r$ o' u) T8 {
}9 C7 u8 b- n) H
}- E7 ^& {$ B6 G+ B
}. a* o! s7 _; Q9 M' c3 m
第十一章 结构体与共用体: P% i A- U3 x* Y8 p
11.1
" U9 p$ @5 P; O4 V$ u `5 Dstruct9 U& \8 f+ R3 V. L$ c
{int year;
# y; `9 S& |) i) @# P int month;( m) I) y1 P, E. `
int day;
1 U3 s+ E7 p- y }date;
3 z4 z; x+ T4 U7 x0 Nmain()
6 W _ s3 R, @{int days;
5 O, v1 }, C7 m3 H" g scanf("%d,%d,%d",&date.year,&date.month,&date.day);# S/ `% Y& P7 n! P3 N6 w/ D
switch(date.month)
+ G. d1 v- @6 Y* T6 U {case 1:days=date.day;break;. p' w7 m* q7 I# | J
case 2:days=date.day+31;break;! @2 Z3 b- f7 ^8 G9 F0 s% {
case 3:days=date.day+59;break;; a. [/ Q! m% Y7 T) O* b0 x
case 4:days=date.day+90;break;3 Q9 u/ i' Y# Z R B4 D# |5 [. V
case 5:days=date.day+120;break;5 g; U! Q6 u4 t! ?0 W& u
case 6:days=date.day+151;break;+ A# ~; t. [3 i" i- L- V
case 7:days=date.day+181;break;$ a6 V0 a" \! L
case 8:days=date.day+212;break;* u1 f' Z# s2 V1 f" V) Q$ [
case 9:days=date.day+243;break;" T, A' R% [6 \- ~1 [
case 10:days=date.day+273;break;/ _% d9 L9 v- E4 t
case 11:days=date.day+304;break;
" [, m( {0 f% r. u$ E& Z case 12:days=date.day+334;break;
( v9 o, U% V8 f9 \4 C' Q/ P }
- w5 `5 q8 }2 ?+ | c3 g2 u& T if((date.year%4==0&&date.year%100!=0||date.year%400==0)
/ z& `0 t/ q0 {3 u7 E" f5 V2 n) ] &&date.month>=3)2 Q+ I$ X- O7 {6 `& }8 B8 q6 W
days+=1;
7 e1 m$ t+ s) K$ [ b6 M$ C printf("days=%d\n",days);9 v! f, v* X& {7 m) n8 w2 E
}4 ^; i6 w' @7 W/ m( N
11.2
9 [" ^/ \0 J* K) U, \- p2 tstruct dt
( a1 ]' t. y* e/ w0 K6 q$ ?+ A {int year;
M! Y1 H% `/ k/ k3 _& z& `( b5 j* } int month;
* A6 Z7 [) s# ^: c: U int day;/ c6 [4 Y! l4 W F3 n1 l/ _5 x
}date;3 a, M+ A+ Z8 j' |9 n9 F
main()
s& I6 A8 B& N4 e6 f+ C( H2 ^{" w. O+ r+ X* K ^2 n: D3 z
scanf("%d,%d,%d",&date.year,&date.month,&date.day);* u( l- D8 r1 s9 _' _' ?2 d
printf("\n%d\n",days(date.year,date.month,date.day));
3 [3 o! ? C1 H0 m0 [$ l- P% l}
. S+ X8 K2 Q+ o6 x2 udays(year,month,day)
; k+ `2 i% s/ _' U3 ^1 ~int year,month,day;
* |4 p- _& T' |& ?{int daysum=0,i;
; G' A" n$ |0 S static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}" R. q. W0 Q* r
for(i=1;i daysum+=daytab[i];
9 P$ W; {7 e. _% y1 c* f. C daysum+=day;
9 b+ G8 V4 ^/ [& W8 R if((year%4==0&&year%100!=0||year%400==0)&&month>=3)
h. R0 {" S. q) U' s2 R9 y* n daysum+=1;! |. X8 Y+ s8 \1 N$ F
return(daysum);
! S/ P& k3 }, {9 M}
5 g- G9 f, l6 A: ? u; l/ x. v( z7 P3 f1 u11.3) @1 G4 @* T; |! a" l+ P0 i
11.49 o& N6 k1 E) I9 T2 P
#define N 5) r/ ~7 S; b/ Q/ X/ r% u" T$ m
struct student
# m! v2 H) B( ^; Y {char num[6];
$ D6 I+ j& e( u3 R( m: y& f; s! J7 G char name[8];* e2 @$ O% b5 \# n6 ]) a
int score[4];
3 W$ k# R; Y& F/ v' V, p) N }stu[N];; r$ s7 |- e' f! M
main()7 G! T* L$ d" p# ]9 w
{
4 a% x2 `$ L# b; |% h: z input(stu);9 j5 S) ?: q7 Q2 p( Q6 y
print(stu);! ?/ d3 k& A: x) f6 v# m
}" c4 _$ o: O: N" ^
input(stu)' ^6 U. A" Y, o% _6 f9 ~8 V
struct student stu[];
d0 A- L$ i, d# N0 v" S3 s{int i,j;
0 O4 t+ t& n A+ ~2 w for(i=0;i {printf("number");3 g/ e9 t7 B/ z4 A2 T
scanf("%s",stu[i].num);3 k5 V: q# \6 s2 l0 T: |! @+ ^
printf("name");# Y5 [4 N, M* N- ~5 |) Q) S/ g
scanf("%s",stu[i].name);5 f h8 O, A) H) J5 ~
for(j=0;j<3;j++)- g0 q- r4 |# P4 y: a9 m. Y
{printf("\nscore\n");( F! Q( ^2 p! q5 v5 j
scanf("%d",&stu[i].score[j]);# A% W* G, F4 w+ s
}8 Q( i: U% b9 \4 e+ S. ~
printf("\n");: \) p% n9 Q/ p$ m; p& B
}6 C6 d: ]$ u3 N9 h+ U& @( }8 i3 \
}+ ~" t7 g6 P( M8 p" X4 ^6 \
print(stu)
- d7 D+ j4 g) W& |* M& lstruct student stu[];
8 z+ _1 _% R7 ~{int i,j;
4 ~3 o# ^% I1 y! M7 @. B- l printf("\nnumber name score1 score2 score3 \n");! K0 ~! B5 n# l6 e0 N: Y g
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
, P3 `/ j6 i, y7 _( {8 d* O+ {& w for(j=0;j<3;j++)
. x. f& e3 w: e$ k) I printf("%7d",stu[i].score[j]);) Z$ M% W, _) Q) f/ Q* x
printf("\n");/ |' ]! ]1 L& ]( z) N4 S) e
}
- V. W) l/ a0 W- G) F}( \2 {2 z& K) A$ S2 G S
11.5/ P0 x5 q. k8 ?
struct student
) j: f1 k! `& x# r$ n1 |* d {char num[6];
& n9 J! r; J) z$ E$ M1 ~ char name[8];
2 S9 R5 g3 e: \1 p! T+ A7 @ int score[4]; b# R4 z& m$ N
float avr;
% T4 H+ W/ K8 x5 B }stu[5];
2 b# s1 @7 j: O* nmain()
2 W( r$ ] o) D: a+ t% {0 q' T" h- l{int i,j,max,maxi,sum;
3 h. S& s/ E( L, n5 d0 q float average;
% _0 h/ N/ e( Z0 F" E3 m; v for(i=0;i<5;i++)/ r% A1 r" ]9 h, J' N; C1 r
{printf("number");
0 u' Z9 ]5 |% a2 \* \/ E" o8 Z. x scanf("%s",stu[i].num);9 @- q) A9 r: @1 X D# n
printf("name");
) O w% f' h7 M" Y# [2 X scanf("%s",stu[i].name);
; P9 M% ]6 f/ r3 E. w for(j=0;j<3;j++)
4 q$ H( D4 |3 Y, k {printf("\nscore\n");
* `/ m* N' ^% ^, }! G1 ] scanf("%d",&stu[i].score[j]);' B! W1 _, B% S1 ?
}. P( j- _7 C# [3 J
}
1 v, T# b7 N6 N) h9 h average=0;9 c* j; |: Z7 G S
max=0;8 R# n' R+ R, d* m! s
maxi=0;. m4 R+ ~% s* h- U
for(i=0;i<5;i++)% q# L9 c7 P+ H; j4 u& G; x( K
{sum=0;
8 B5 \# W O, _0 { for(j=0;j<3;j++), x- p* |4 Y/ I1 _( x) r9 T
sum+=stu[i].score[j];/ d D; y, Y; h+ k5 o
stu[i].avr=sum/3.0;
& x7 L& y* l: c* S- W average+=stu[i].avr;
9 a) `) O; T: H. P if(sum>max)
- }! [ e; r T0 H5 @1 ^6 l7 T {max=sum;
& X9 ~% o$ t5 `0 g: P- \) U4 t maxi=i;
- V/ W! n. g; a. b2 G2 u9 p. ^2 \; n }
: Z. p O8 W9 ~- I( X) l, e- Y }0 a1 V7 d6 I S9 u- c; F f8 T
average/=5;
4 y U7 a4 ^- |+ v0 T; k& Z: } printf("number name score1 score2 score3 average\n");
i. v0 o* P. n, Q* e) n$ U for(i=0;i<5;i++)
2 j4 _8 z4 u" k8 K+ `- ] {printf("%8s%10s",stu[i].num,stu[i].name);
8 G6 `2 `( Q5 i- w- } for(j=0;j<3;j++)
6 \5 c( @2 n8 P5 j# |8 b+ e( v printf("%7d",stu[i].score[j]);6 O$ h4 ^( a) C0 Z e' G c
printf("%6.2f\n",stu[i].avr);7 V( Q1 y) t1 | H' h! f
}4 O7 S( i( `; q9 V4 @
printf("average=%5.2f\n",average);
3 I& W3 i# q9 o6 r printf("The best student is %s,sum=%d\n",stu[maxi].name,max);. g" P9 _1 O6 p
}
w; d N7 A5 v$ D: J F% K% D0 g" Q1 `3 d# p5 c S0 z: ~
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