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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
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签到天数: 13 天 [LV.3]偶尔看看II
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- 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& I3 s9 w% z" Z, V
1.5请参照本章例题,编写一个C程序,输出以下信息:
9 ?& g, P( @* i) N$ u7 ~6 @main()
- E C: w% a$ G6 _1 Q, N9 `{
% t. x: V6 a r6 t7 A5 Eprintf(" ************ \n");
; U; q5 Q* @9 ^7 [2 lprintf("\n");$ q1 G. h' L" f, X# a; n1 }
printf(" Very Good! \n");
0 X& l8 c+ ~3 ^; m; u7 W* ^printf("\n");
+ \4 A: ^, k) m( E* u5 hprintf(" ************\n");8 k! i+ D/ y$ K) N8 u
}
8 _/ X! E7 d5 \1.6编写一个程序,输入a b c三个值,输出其中最大者。2 r& O$ e8 m5 S3 J8 v! W
解:main()- p7 z8 V# s. o' ?/ E! @
{int a,b,c,max;) G _- N2 M0 o! A7 I* e
printf("请输入三个数a,b,c:\n");% C0 Z6 {* U- F( `3 O) a- c+ L
scanf("%d,%d,%d",&a,&b,&c);
! [& e* b' k) V; g0 h0 dmax=a;
2 q- ]* d& A7 p, r i9 f! E% mif(maxmax=b;
- C$ T1 s# Q1 r! x8 r" G$ d, ~if(maxmax=c;
+ O+ s" y. X O: Q2 jprintf("最大数为:%d",max);
# D: Q/ O& \" D% R% ^5 O9 A}- I' V4 p0 p! I+ D+ v" M0 x
第三章' l) [* k! }1 P1 m* z
3.3 请将下面各数用八进制数和十六进制数表示:
9 e$ i. n: l; t: `& x+ Y u, l(1)10 (2)32 (3)75 (4)-6171 S: k3 W/ O; P9 g V5 P( D. B
(5)-111 (6)2483 (7)-28654 (8)21003
8 n7 r p4 c; w& o' a. S# k7 j& `解:十 八 十六$ M( J2 w5 t8 ? g" C7 x/ w+ J1 F
(10)=(12)=(a)
$ ^: g0 S. r3 @9 m2 t (32)=(40)=20
* S: h0 _" q+ I6 g: }7 r (75)=(113)=4b
# w/ o9 K5 C+ t# E! W- o4 k* \ (-617)=(176627)=fd972 [* E+ R. p2 V9 @5 V# Z4 F. m
-111=177621=ff91
: N# g6 G ~, x# V7 P 2483=4663=9630 u9 t$ o5 H, Z1 O
-28654=110022=90127 A" [+ H% g1 i% S
21003=51013=520b
! J4 R" S! i6 s9 t0 K: E: Q; P2 W: L3.5字符常量与字符串常量有什么区别?
5 N$ ]; K; Y) x3 L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
6 l& ^9 f6 m1 c* B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'.- a' ]% d5 h V" x- a
3.6写出以下程序的运行结果:
) N& W* r( g& X& h7 @#include) q8 \/ ^: k0 Z" X* x/ H2 p
void main()
+ Y1 O! t. F8 n$ c/ `3 `{# s+ i+ |$ W4 x. `3 F* X0 Q
char c1='a',c2='b',c3='c',c4='\101',c5='\116';
1 ~- S$ ^5 ^" b3 j3 |1 cprintf("a%c b%c\tc%c\tabc\n",c1,c2,c3);$ I9 G2 Z/ a& `
printf("\t\b%c %c\n",c4,c5); X' h" \! K' i7 a U
解:程序的运行结果为:; n0 p! T) n0 y% |
aabb cc abc
m' I5 }' Z; w A N
- W" p% ]) u2 l! i# y3.7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,% W3 {. D. u6 c3 E7 d5 I! ^+ s6 J% {
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".
" |: K5 L* F! |+ a+ D; h* ^8 h请编一程序,用赋初值的议程使c1,c2,c3,c4,c5分别变成'G','1','m','r','e',并+ ?: Z; S0 Q( d& V, k
输出.
. L9 v9 }0 y; H0 E) F$ N- {main()
5 R, ^# s3 a" r% N{char c1="C",c2="h",c3="i",c4='n',c5='a';5 M; [6 L* B- ]( G. K8 n
c1+=4;5 Z4 ~+ a% k( n/ t
c2+=4;, d, B5 ~/ Q0 X8 _0 S& G
c3+=4;0 l' G" N3 N G1 _- X
c4+=4;- X0 E$ [6 v \7 \) q! n9 G& m
c5+=4;
E9 e, x- Z; }" s# \3 z2 uprintf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);
2 _, Q u1 ]( q* P% R# D6 u}
) a- F3 d( O% c# M9 ~7 F( ^3.8例3.6能否改成如下:
, ~* w, H" p5 k; l" g8 f t0 W1 l#include
4 v5 ?/ \6 r. B' w, M, ]void main()
" J$ k& ], {1 U2 B' \0 A{
/ j9 k' Y6 C5 g. ^% n. h; Jint c1,c2;(原为 char c1,c2)( u1 l- u. m3 t% c5 y% t O
c1=97;
* i. l; @9 x4 Y0 o6 Pc2=98;3 t+ h2 }' ^, k$ l g) c
printf("%c%c\n",c1,c2);* e$ ?# x- y* g, a
printf("%d%d\n",c1,c2);
# z( a2 Z9 e( V) B}
1 D$ d6 G# S6 @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.
1 n n6 ^3 M V& W n, O7 `0 ]( k3.9求下面算术表达式的值.
: e6 a1 s3 H; C& \(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)+ a9 [) c# U: l |
(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)8 A9 k8 e- u8 r. Z; Y; N
3.10写出下面程序的运行结果:1 [, t) Z/ ?: Z
#include% F4 \4 d) m* N6 z) }- ]. w, F
void main()4 v8 M, m. y$ |! p- [" A
{
( U0 N, `- ?& Hint i,j,m,n;
8 M6 m3 V- y) bi=8;
6 d# o T9 K/ ?8 S) s oj=10;. m+ i' k+ a: t2 h" x& I3 ?& V! \
m=++i;, s& @1 `: |5 t4 E( F3 l# u
n=j++;' J$ \0 j8 A9 d, G/ U
printf("%d,%d,%d,%d\n",i,j,m,n);
3 }$ a5 }* y/ [: ?}: R& d9 |# A8 y& r' I
解:结果: 9,11,9,10% @6 d6 {! ^- L# G- x1 w) f
第4章
5 ]5 k+ [" P% L4.4.a=3,b=4,c=5,x=1.2,y=2.4,z=-3.6,u=51274,n=128765,c1='a',c2='b'.想得
8 |6 P! z1 z. @* J! S, B p- }) ?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:
$ H0 N4 j1 r8 ]; U* Va= 3 b= 4 c= 5
* Z) N* s% z2 q. O `/ \x=1.200000,y=2.400000,z=-3.600000
5 M5 q9 K3 H- N' L4 `" \ Px+y= 3.60 y+z=-1.20 z+x=-2.40 [8 z: Q1 [5 R6 H+ B
u= 51274 n= 128765
/ N9 _. z8 s# S; O/ g3 z" qc1='a' or 97(ASCII)
# I( T) J4 X6 r, P9 t) zc2='B' or 98(ASCII)
0 u+ x2 Q) t4 `6 {0 N' D, R解:5 z% W* R6 J: ]9 u( t3 B% g
main()
0 U9 p3 ] n5 X$ r- e) |8 R{
0 G2 Q, G) v# l( sint a,b,c;0 u! d$ s$ v0 E' `" X
long int u,n;
% ]. P W. d% H. C2 M* e# Gfloat x,y,z;& C5 s8 X0 I0 c& C V. D
char c1,c2;
# Q, n8 l2 l2 k. {- S, u' h1 Z Ua=3;b=4;c=5;# D1 J3 D: H) G: U* v$ n: P9 W
x=1.2;y=2.4;z=-3.6;9 C! A6 t# n+ p
u=51274;n=128765;3 }0 d; S: x4 D+ @% f! ~7 x; o: L
c1='a';c2='b';
7 y9 C/ W1 X& i. h& ?printf("\n");
* K2 W! e/ k, u# l$ w Z/ xprintf("a=%2d b=%2d c=%2d\n",a,b,c);
0 o, Q; D* [* } Kprintf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
! Y l2 V3 G4 x& ~0 V# K% b9 `1 @5 jprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);% i5 D6 I* u& o i- Y
printf("u=%6ld n=%9ld\n",u,n); o% A. I1 r/ M; O
printf("c1='%c' or %d(ASCII)\n",c1,c2);7 F% v" R( \6 Q
printf("c2='%c' or %d(ASCII)\n",c2,c2);
/ V) C+ H' ~1 f6 g}
2 F$ T7 R! j0 [4 ]% f* \- d) R4.5请写出下面程序的输出结果.3 r' G4 l& |4 Z* `3 ~6 W/ I
结果:" \& w6 _% o: q* _
57& ?& X2 P; u' Z1 u1 f g& j
5 7
1 @$ _% ^$ L# c6 {* ]' C. @67.856400,-789.123962
1 g3 g: u8 Q j: {0 p67.856400 ,-789.1239623 b* w! ?# h6 A6 G7 F- W
67.86,-789.12,67.856400,-789.123962,67.856400,-789.123962& J- N2 [8 V# @! U8 d; K
6.785640e+001,-7.89e+0028 k& ~0 ~2 e7 u0 u% `3 a" I7 G
A,65,101,413 {1 W" a* s1 {% H' R* [# I
1234567,4553207,d687
/ a8 [0 a4 ~0 `" g65535,17777,ffff,-1% G0 u" k+ g! s7 n
COMPUTER, COM. h5 P. m+ x# \4 f. h: ^4 I
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',
; O. `/ S4 q/ V& s& U2 U% H问在键盘上如何输入?5 b i- p. t! ^
main()
1 { \/ W3 C7 s' c" @7 q{6 |/ r0 Q3 H6 m
int a,b;8 [* G7 O5 M8 Z
float x,y;
4 [7 v; U3 C% J. k% B5 K1 W' {3 Rchar c1,c2;( |1 ^4 `+ x% L0 h+ h; p! g1 _
scanf("a=%d b=%d,&a,&b);" p5 \3 y% J! E; q3 D5 ~ Y- e
scanf(" x=%f y=%e",&x,&y);
3 { u6 ~3 p6 [( g1 ], S' G: z. jscanf(" c1=%c c2=%c",&c1,&c2);* t9 \3 R1 }$ k
}+ o7 c) ^( J7 [9 `' ?
解:可按如下方式在键盘上输入:
. K* X% b$ n1 I/ ga=3 b=7
1 o" K5 g" k5 vx=8.5 y=71.82
8 t6 W& u$ L ^( g" Lc1=A c2=a
: h* c B+ @/ }$ Z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% U4 p, c8 d& ^0 P6 K6 p
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: ~ ]& \ s3 M: V+ t/ A
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
% D) E; }0 a8 j初步概念即可,以后再进一步深入理解.
0 e+ ]* [" u6 b4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-
3 _( n( W& d' E& }$ G' B' q3.75,z=57.8,请问
3 D9 ]6 k) \" z& e在键盘上如何输入数据?
& T; f0 T: C" j; E/ k, Cscanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);
1 `7 l/ @ \& m# T解:/ b- ?" q# b/ k2 o8 l
main()
% S6 I6 v) \9 U2 N) G G{
8 {, I" Q" t8 e' G4 @/ V1 H9 ^int a,b;, M Z# ?& v* {1 w$ M' r; h, g
float x,y,z;/ b$ S& d) n% }6 r" I, V# l4 `
char c1,c2;
4 q4 J" N; p" Q$ U l4 P) P) Tscanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);
/ z f7 a2 i5 N+ B: @}" p. s& F; U8 H# h
运行时输入:- H$ C$ x+ s2 ~8 S
10 20Aa1.5 -3.75 +1.5,67.80 y- a! x8 m0 D& ~3 y9 m' D7 n
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$ z5 p. I/ L2 E+ N, D
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! a9 U w0 E2 X w2 S2 A w/ Z5 H
会赋给任何变量。4 s; L7 d! S ^5 B
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,$ O6 h+ }# O* T% ^/ r1 r. I
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 w, r3 U- I4 H
程.! M, w- n; ]$ C8 o$ t1 i$ _
解:main()
. f* N4 F) t# d$ |. [{
: d2 D( l5 m7 c% Qfloat pi,h,r,l,s,sq,vq,vz;
" [# k+ M3 S. ]) z) ] d3 Spi=3.1415926;/ U. V6 }2 j4 C1 T. l) y
printf("请输入圆半径r圆柱高h:\n");
* v. b5 g2 Z$ y! C7 Mscanf("%f,%f",&r,&h);. b7 K. M: k- p/ r- G% E
l=2*pi*r;1 t' k$ V! i/ h6 [" W
s=r*r*pi;
4 I1 X) ?% o- x, [6 c8 ? y. Nsq=4*pi*r*r;
. F. m+ x, ?# ]$ |- ^4 Ovq=4.0/3.0*pi*r*r*r;
f9 T# f; r+ Lvz=pi*r*r*h;* F, y9 |* o4 @
printf("圆周长为: =%6.2f\n",l);2 p$ G; c. k$ D
printf("圆面积为: =%6.2f\n",s);
4 X @! w+ [# p8 U7 x' y, }1 Kprintf("圆球表面积为: =%6.2f\n",sq);- v5 }. Y0 n; X( H$ f# I1 P/ V0 _
printf("圆球体积为: =%6.2f\n",vz);5 b. D7 I, M- C# P) s
}
& h# _0 K- ^4 A+ l4.9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, X- t" a& O! P4 Z% a" }0 n
取两位小数.
1 g# [8 ?7 D: d/ _0 P解: main()
+ Y/ k \# J( n0 S5 b9 j7 p{
}, M" r: w8 `4 R7 }float c,f;8 h3 x# [ D+ v! q2 g2 r
printf("请输入一个华氏温度:\n");- S! i' |3 w" Y6 s' I. @* b
scanf("%f",&f);
[5 A0 Z7 ]3 S* d6 l. Yc=(5.0/9.0)*(f-32);
! }, R$ P) @, B9 pprintf("摄氏温度为:%5.2f\n",c);
' j: u9 t0 s7 b% s& l+ k. @}
7 L3 p; p4 _. O9 [* D第五章 逻辑运算和判断选取结构6 a, W! }* N4 x! d% d, b/ O
5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
0 @1 v. u; Y- y) s" j; B6 G5 C1 emain()
" b, w7 N6 G' O{4 ^% C2 _+ p- }9 V: ]$ \2 ?: g1 C
int a,b,c;
$ E6 S! A3 R# ^) c8 n' \; Mprintf("请输入三个数:");: B3 l8 y: ]) B. P8 @+ w
scanf("%d,%d,%d",&a,&b,&c);& ^" \+ i, F* Z. | j$ D s
if(a if(b printf("max=%d\n",c);
O: T7 X' N0 W& T5 o else6 Z( m, q" [6 X. n
printf("max=%d\n",b);
2 U3 g1 J! d5 Helse if(a printf("max=%d\n",c);6 e* g& N2 M% V; r- V
else
1 C$ B. [ y7 d! `) V1 M printf("max-%d\n",a);( d% M; |: f+ D9 ^& C5 p
}, x3 e8 Q# G& w( C6 \3 g
方法2:使用条件表达式.0 X9 }- R# \; \% t
main()& a' [: l9 ?8 j2 A& @! ^1 @
{int a,b,c,termp,max;
+ E5 I# e/ {# b7 [- e: o z! E, v printf(" 请输入 A,B,C: ");
+ \/ Z- i9 x* n- Q! T scanf("%d,%d,%d",&a,&b,&c);% |. Z5 K% r; f7 Q
printf("A=%d,B=%d,C=%d\n",a,b,c);
* g% w& \$ ^7 A' X! N6 y1 u3 E6 R temp=(a>b)?a:b;5 [7 b& X: I9 H+ B
max=(temp>c)? temp:c;! I& h- e1 M2 G8 G
printf(" A,B,C中最大数是%d,",max);' A- v( h0 O& L# \( g
}
5 Z+ v- P8 w4 k5.5 main()- u6 [) Y. }" m* w
{int x,y;4 W. c6 J) O6 L/ k& \& P
printf("输入x:");+ b( D- A4 |2 g1 H
scanf("%d",&x);( c% a P4 `7 v! |+ J, |
if(x<1)2 o* j O P9 W6 \4 A( @9 {+ @; H8 {' L
{y=x;' F4 ~" U$ n+ c8 u6 e, t
printf("X-%d,Y=X=%d \n",x,y);, V- ?* |! b" N; c5 C
}1 {; N7 x2 [: ^
else if(x<10)$ B! m% t# i# Y8 R: @
{y=2*x-1;
5 b5 R, \5 ^" O, O; ?9 c+ U printf(" X=%d, Y=2*X-1=%d\n",x,y);8 E/ e: @$ ^0 ]/ B7 F
}1 ^4 o8 J, T& H; i! n1 @: u0 `) U
else( Z9 S$ c" {/ ~) A. P* E# X6 m
{y=3*x-11;
& ^6 }; N. ~4 W2 ]/ X printf("X=5d, Y=3*x-11=%d \n",x,y);; S6 L6 W3 d2 k9 o
}
6 ^' ^+ Q' L0 B! P; B}
4 s3 G1 f( L- ?(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! 7 F# Q9 G& _$ ?# Y) F
void main()
8 Q, ?" S9 L& [+ D5 f. K{, K) `( s8 o5 E# d
float s,i;
; d2 |* j4 `7 x4 f2 e& @$ Q5 r2 {char a;
( w' ]5 C4 @! Y% T5 x# xscanf("%f",&s);& i0 G0 S6 |3 D% S+ p8 g0 V6 |7 ~
while(s>100||s<0)
3 |* V. N# B5 }3 y% i{
+ x+ Y$ |( G* Bprintf("输入错误!error!");( {% V! p) b5 X1 ]2 A4 L5 R
scanf("%f",&s);$ n& }0 I F$ ]6 U B2 y
}6 W* ^' ?# Q* F- `6 C
i=s/10;
2 X! @4 n/ f" Oswitch((int)i)
0 } Y! b/ C- F3 @{7 A3 b2 L4 Q6 y* e# ?7 _ ~# D
case 10:
7 y) J s' E8 o3 {8 h; l% Vcase 9: a='A';break;# s6 P$ U- g: x9 B* @5 {& v$ F" h- Y8 ]
case 8: a='B';break;
9 m; W Z, n8 L! Ucase 7: a='C';break;. b' E' F' b' A
case 6: a='D';break;0 z' k+ T( T; J/ T8 T
case 5:
0 ^4 O) W6 M e; M. h! _' A" {case 4:& j2 \' A/ ~ l( h
case 2:2 f+ b) @6 f5 A+ ~4 Y
case 1:+ s0 ]2 y* x" ^, e, w! p7 l
case 0: a='E';
- Z4 M7 w1 t, ~/ s}% k) i1 L) Z3 n# |* U0 B8 [
printf("%c",a);
3 ~5 l; e( M- {3 g}$ s0 ?9 e3 J' w
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.
7 V3 [- y2 U* Y1 s: b7 y按逆序打印出各位数字.例如原数为321,应输出123.
, H, X# |8 N3 u/ _2 t7 Tmain() t$ v/ [5 B7 H2 h s
{7 R& s" ^4 E6 C
long int num;
5 D& x! x7 V3 i4 C w- x* J: I int indiv,ten,hundred,housand,tenthousand,place;- K8 b. r) n2 S
printf("请输入一个整数(0-99999):");" h5 Y$ G8 D' S% d
scanf("%ld",&num);1 |% ^( G% i9 k3 o: N. c
if(num>9999)
! i& ^" X% X( i6 G# h place=5;
3 N& d, o6 P$ I% E8 welse if(num>999)
1 K t! Y8 ^2 x# S/ R( h place=4;: @! s4 X- A5 h
else if(num>99)! g8 Z3 J7 v; i3 \& O, F1 o
place=3;( Q6 i+ f( W& n" s
else if(num>9)6 f. w' I w: ^9 j- `
place=2;& n3 E4 D/ t) l2 c! L- J1 e9 M# O s
else place=1;
& ]2 ~) [) x; F) S L' @ }printf("place=%d\n",place);
! w; |6 G- r( _. {# ~' Mprintf("每位数字为:");
/ e( p& X* A% t$ d) L& v! P9 y8 M) qten_thousand=num/10000;
+ `( O- z0 k# \, _/ mthousand=(num-tenthousand*10000)/1000;. ]0 v& _( u, B, E0 n; {
hundred=(num-tenthousand*10000-thousand*1000)/100;" e5 l. ~9 \/ P6 U7 m6 P
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;. U! d8 ^2 n! m& u+ h9 z7 Y5 s
indiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;. b6 |9 X: J9 y% ?5 p
switch(place)
$ Q7 X! w( e' O4 G2 ?{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
0 w+ m* |. Z+ H8 K- s v printf("\n反序数字为:");
8 E2 K1 O0 |$ } printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);
4 w0 @) ~% y8 e2 t9 H8 l break;, \7 L# a0 Q+ c! r' g
case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
4 k/ d6 A% U5 {7 R printf("\n反序数字为:");% I" H6 Y# I* n, c
printf("%d%d%d%d\n",indiv,ten,hundred,thousand);5 o% J7 l' g) ~8 Y% q; V
break;, d* u7 W- {; W& L: m5 K% o# t
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);2 D5 v7 Q" |/ f% U) U* @
printf("\n反序数字为:");) Y1 @0 C5 Q1 T& z% H3 k
printf("%d%d%d\n",indiv,ten,hundred);
' N9 b3 H9 [! Q6 ]# _0 D( T/ u Mcase 2:printf("%d,%d\n",ten,indiv);
# p6 t- f4 z5 A- J6 f& ~ printf("\n反序数字为:");
. a1 ~6 ^+ S6 ^- _# W, O printf("%d%d\n",indiv,ten);) Q- X$ d3 U& O9 I8 ^. @0 X
case 1:printf("%d\n",indiv);8 j: }% V1 w5 R; m$ O' [1 C5 g2 u
printf("\n反序数字为:");
6 u) J' d2 } p/ X2 J$ h printf("%d\n",indiv);
% r& _, U1 {; O8 E }
, X$ d( c& u6 x, z5 ~} i. ^% s9 w8 V* D2 Z. [1 W
5.8
7 r* ^9 _1 ~; {6 ~+ C, G9 }1.if语句
* z& V6 M# D: Z/ ^7 ~/ Omain()
: A: o d& Y: d' b5 _2 T% g2 I/ }{long i;; K% a6 r% V2 m4 k! K
float bonus,bon1,bon2,bon4,bon6,bon10;
; Q: z# h4 [- Q& O- ]2 \2 U bon1=100000*0.1;
& ?1 k; E! S5 q: c* m bon2=bon1+100000*0.075;
2 c" [4 J( d, T' Z. ~" G6 L bon4=bon2+200000*0.05;
% ]/ Z8 |' s* b) Y6 j5 I/ b/ Q d bon6=bon4+200000*0.03;
0 [. I+ g9 W3 G4 B# v( Y bon10=bon6+400000*0.015;; A+ _: Q# q3 C% @! V9 j3 h& s
scanf("%ld",&i);) i, l1 ^+ r2 D* @1 B* Y
if(i<=1e5)bonus=i*0.1;
( ?$ V" K; m& [% z6 v, @* G else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
; q) H3 }) g1 u/ z else if(i<=4e5)bonus=bon2+(i-200000)*0.05; O- v* N( K" _! B
else if(i<=6e5)bonus=bon4+(i-400000)*0.03;6 r( G' ?% |. E, A* M0 [% x; E
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
5 h- R2 c' p7 L; Y else bonus=bon10+(i-1000000)*0.01;
! _( S/ j5 x5 W1 N" x printf("bonus=%10.2f",bonus);" F; a. \! y' m4 l0 {& Q& Y' J$ o( o
}
8 G$ {/ C4 M n9 A( k2 @8 E3 e4 P用switch语句编程序
+ R- X' a, G3 a, E7 mmain(); N6 Z3 l+ w5 j* M
{long i;
4 d) D' K1 r: o& G9 [- ` float bonus,bon1,bon2,bon4,bon6,bon10;6 s8 _7 s; g V: W' ?
int branch;) Y; P' I/ ?& x5 @% z) E6 P5 G1 q
bon1=100000*0.1;. d. d s: @ T) ~
bon2=bon1+100000*0.075;
2 z; d8 {8 ^1 q+ L/ B* c bon4=bon2+200000*0.05;
- }) j- a: M- { T5 F: J2 } bon6=bon4+200000*0.03;
- A _7 ^# ^; M! B; H5 j7 _' K+ F bon10=bon6+400000*0.015;9 G$ V3 o" J+ y3 O) a+ y3 J
scanf("%ld",&i);% p, A' C. K" B m( s/ H& m
branch=i/100000;
* l. u8 O$ _& q" R0 v5 V0 @ if(branch>10)branch=10;
% p# t* K# j, X& k! I1 L& | switch(branch)
$ A3 U( S, F9 F0 ` {case 0:bonus=i*0.1;break;( u+ K+ W4 G. \# `
case 1:bonus=bon1+(i-100000)*0.075;break;* p: x6 f+ ^) Y1 E3 m9 C9 I- d, G
case 2:- S3 U9 I# i& a/ Z
case 3:bonus=bon2+(i-200000)*0.05;break;; x- P! m/ D7 u. m7 l
case 4:
/ C- o. `8 H! _ case 5:bonus=bon4+(i-400000)*0.03;break;! u x) `4 N c; w; K5 z
case 6:( e; p$ f- `1 i$ l% Y! U4 h; B
case 7
- e9 ]3 O, }1 w" v* C$ c case 8:
, y2 p! g( v4 D5 i! j" { case 9:bonus=bon6+(i-600000)*0.015;break;1 @- l# m- Z2 {1 W* @
case 10:bonus=bon10+(i-1000000)*0.01;
. i! q1 g0 h& }% }& L, R9 t) N3 I }7 f: q, u2 m' I% s8 g! u
printf("bonus=%10.2f",bonus);
! D( O5 u) z' D/ U. [7 l: D}
% r# v6 ~% W+ F# G% B- o4 m8 Z5.9 输入四个整数,按大小顺序输出.
5 p% P( X4 F6 w( G1 Rmain()
6 ]+ [# _% A6 ^2 c0 O! K {int t,a,b,c,d;$ v. M( M2 q3 e( J: a
printf("请输入四个数:");: m1 C+ L' a0 a- W
scanf("%d,%d,%d,%d",&a,&b,&c,&d); y: `9 y" Z* v
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
+ t. Y! E' q# x% Q4 M if(a>b)0 ^2 T9 O2 H9 a- X0 o
{t=a;a=b;b=t;}* X" r% F8 W' N3 N
if(a>c)
6 X: C! x* }; A- s: H P P {t=a;a=c;c=t;}
7 J! Z' b1 [+ g# p if(a>d)# b9 t8 v5 }/ `# [
{t=a;a=d;d=t;}
0 |+ M, Z) I" T# [' B4 r if(b>c)
5 v& ~' c9 {* w3 e6 V {t=b;b=c;c=t;} }/ w5 q7 k" B X/ k! `: ?6 F
if(b>d)# s- g5 B+ t( f3 Z/ @4 }3 s
{t=b;b=d;d=t;} W- ^" d! P- q3 o/ k6 u2 S
if(c>d)
5 d) H% Q2 H s1 G {t=c;c=d;d=t;}6 j! C( F' B2 G; z' p$ Q; p
printf("\n 排序结果如下: \n");
( ^( ]- y+ |8 ~5 ?9 rprintf(" %d %d %d %d \n",a,b,c,d);) v* A9 F6 ]8 t( [( O" R
}8 x% b) G1 C5 Q; k" ]" o7 i
5.10塔) G4 j ^/ L1 A. z
main()
$ q W. B5 R, L& c{
" t" E3 K- S7 |1 p$ p- gint h=10;1 m7 x( m2 d I
float x,y,x0=2,y0=2,d1,d2,d3,d4;
6 W z4 B5 J8 u4 a tprintf("请输入一个点(x,y):");
" Y3 g: J7 ^: V M, j4 m# Vscanf("%f,%f",&x,&y);
& s0 Q- o# t6 @3 ~9 `d1=(x-x0)*(x-x0)+(y-y0)(y-y0);) R4 T2 @! m4 P( N
d2=(x-x0)*(x-x0)+(y+y0)(y+y0);8 P. s' M( l; P3 T7 ^
d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
9 t5 P* W* T6 Y2 i% j9 Rd4=(x+x0)*(x+x0)+(y+y0)*(y+y0);- j* x/ \5 R& r6 f' s% w# M
if(d1>1 && d2>1 && d3>1 && d4>1)9 S$ E3 E9 K$ J* f, k
h=0;- G7 m3 U% ?# ^
printf("该点高度为%d",h);
/ c; f7 W; P( m6 _; J/ d9 H}; B: M \- I8 v
第六章 循环语句, L5 C# e: Z9 I4 ^0 X# R/ ]
6.1输入两个正数,求最大公约数最小公倍数.5 T1 w& q9 j" Y; f0 r' H, K
main()
6 ^1 e5 } {' o7 T" [! `: o2 q{( Q9 Y. _, m$ u, ?% h
int a,b,num1,num2,temp;3 ]. @2 y& ^# Q
printf("请输入两个正整数:\n");+ C& Q, I* h0 _+ k
scanf("%d,%d",&num1,&num2);& C6 ]% s, }/ \& I
if(num1{
0 v+ V6 b+ E& p' ptemp=num1;: y4 c. n( w& G3 Z+ K; `
num1=num2;
" B- G$ K6 h0 ~2 r/ W5 m; Anum2=temp;% Y9 X# H2 t- k4 r! n: m
}* C. ~6 _ J2 V8 W7 t' @( l5 R7 `
a=num1,b=num2;
0 Z6 y# |* a# f- u% G7 Ewhile(b!=0)5 E( e; _& D" w0 F2 X; E: O4 Z+ j" k
{3 U3 n0 N) ~, n1 g8 F* t
temp=a%b;
" M, d, h) J- p4 F# j$ M/ s a=b;) R4 }. e4 a. R' J' k
b=temp;
/ K% S5 I& t0 Q }& x; _6 I" d, P5 H
printf("它们的最大公约数为:%d\n",a);
" f; v3 [2 d/ H. Z2 L) Y, m* Dprintf("它们的最小公倍数为:%d\n",num1*num2/2);
' x" a6 P/ C8 \5 s}
# f# }+ R4 r8 M- U# M6.2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.# s7 `3 W3 g8 D2 a' O5 n0 ^& N
解:
- P1 O# ^: [% S( k3 t' e) k- K#include < >+ r2 |9 S; L: n* E9 i. V+ d1 u
main()
1 ]! @' Y |, K/ q! q, b2 [{
! h+ S5 B3 y* Zchar c;
) Q# U% e8 z+ q0 xint letters=0,space=0,degit=0,other=0;3 E& w) y$ b' Q G) ~9 S
printf("请输入一行字符:\n");% R, j0 {+ M5 O* C( ?: R
scanf("%c",&c);; J0 [1 e2 W4 B& `0 O7 O
while((c=getchar())!='\n')
' M( c0 O6 i% G5 {{! s& b/ `3 f k* H' R3 v% ~
if(c>='a'&&c<='z'||c>'A'&&c<='Z')$ a; w, R- h6 T0 m! ~( m
letters++;
, K% n1 U; V) o5 welse if(c==' ')
, T5 p9 u. [0 `2 `: wspace++;
# m9 d: X8 e# H+ Eelse if(c>='0'&&c<='9')
. i6 A8 x) {' P x. ydigit++;
' h2 e9 D# N8 }3 _ _& G h: [7 Pelse" v! d( i6 B8 t7 |8 U% d; d
other++;) o/ h* Y7 b {1 g
}6 r) x+ ~+ f) c" l. e t4 w: E
printf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%
4 q# F( V4 K/ P9 g; `* H$ Yd\n",letters,space,3 z8 n) `# r5 w' b, y- J# q. Z
digit,other);, n% L$ B0 _1 ^( n& y
}
" m+ O1 C* H8 ^1 @. n" l6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.
* ~" ?$ c" D# j. b( H0 ]% b# ~ Y解:
" W" `# Q3 Q2 k4 W" d# J+ \main()
2 A; U% E0 E: s8 \# s$ _0 k{+ B$ Z9 B2 ]2 m4 ]6 ?4 a1 |
int a,n,count=1,sn=0,tn=0;
: _8 z" [! U( r8 \1 sprintf("请输入a和n的值:\n");* f' U* c; z* p J5 B* @- P" |( \
scanf("%d,%d",&a,&n);
! K; C! ~' g: k6 C Oprintf("a=%d n=%d \n",a,n);" E9 s5 O1 L j) W% D
while(count<=n)3 o% v, o2 V* K
{* c" j7 N) b# i2 N" s$ @
tn=tn+a;
( q2 I4 V+ v( {' ssn=sn+tn;
7 e& B5 g( c* p+ e6 c; `a=a*10;# v% Z$ O, k1 g' i' }. @
++count;
% v; I0 Q% `& v: O9 I}, H3 B1 w, h+ G9 Z' l% r$ b9 ^
printf("a+aa+aaa+…=%d\n",sn);
% }" D( ?; n( C, m' H9 L) _}% a) V, Z, J2 Z7 |5 G
6.4 求1+2!+3!+4!+…+20!." S6 N7 B I# i1 _, K5 |0 i2 X) x
main()
/ a3 d: Q+ @: J# s" ]' s9 v0 ?7 n{
7 {6 B+ P! x; G8 C2 Hfloat n,s=0,t=1;
: b. t# }9 }! Q5 K5 ofor(n=1;n<=20;n++)8 W' t- P" j y4 B P# d
{- w0 V" L; y- W& l: B2 K
t=t*n;
& U# @. s, I6 k! ]s=s+t;4 v: h3 A1 p# r4 I4 f# S
}3 f2 T4 q N7 O1 {# z
printf("1!+2!+…+20!=%e\n",s);. ~( n j" R+ ]2 s- N( `4 s
}
3 X6 z) h8 ]7 n5 }6.5 main()# O1 F9 t ]& Z, q" s
{
3 p1 h$ q8 f9 l: \- Oint N1=100,N2=50,N3=10;9 A% s6 U v5 X% s( m7 T3 x L4 w
float k;2 w7 @! D2 M" L
float s1=0,s2=0,s3=0;8 Q: v. z, c* N* Q) h0 ~- V) {
for(k=1;k<=N1;k++)+ U2 h% V1 U4 e$ B8 A
{
) y( C. O1 g2 `4 C$ }3 y0 us1=s1+k;
& c- m2 s# r6 [+ @8 {3 A. O! I; g, ]}
$ l. w) S u* b% c( B9 ]for(k=1;k<=N2;k++)
! ~0 V$ @* S% P) w. h{) B' U% ]' f9 J$ w- \
s2=s2+k*k;
3 O. ~2 i& F; o! R}
' u1 [* `6 B) j# ]. Cfor(k=1;k<=N3;k++)
/ x- }: l* O" B* }{
* b3 M' `" ~9 i9 Ws3=s3+1/k;
" H# h7 x2 C: c" C$ l}
+ N* B$ d% b8 j! c2 o: D% Zprintf("总和=%8.2f\n",s1+s2+s3);; G4 A+ F) n7 I6 M. d$ r: [. i4 X5 m
}
; t% f7 U9 D6 r$ E% f- {6.6水仙开花
5 q2 ~/ a' Y: \- ?) S7 l: G% j6 umain(). Z8 `5 L& h; T9 u3 `$ P
{
7 u0 x/ l3 G2 F1 K0 s3 Uint i,j,k,n;1 K! l0 p& C. Q( @
printf(" '水仙花'数是:");
4 k7 f# a$ m; i! \2 Rfor(n=100;n<1000;n++)
: V6 p9 r. x2 r* R6 f{8 `+ _/ N7 R( r$ I) B0 b) b1 t7 M
i=n/100;, U8 K o" E! ~& Y* K& B
j=n/10-i*10;( ~9 t$ J3 a) V7 o* [
k=n%10;
, S7 f. Q, t) n' s& f' Bif(i*100+j*10+k==i*i*i+j*j*j+k*k*k)0 M. T$ ^# G8 e1 F4 J/ [& \9 Z
{& I! Y1 k% a5 m0 b" V* }+ q
printf("%d",n);
3 e8 y0 B, r5 ]2 C# |) \3 S. Q}
7 x' `1 J- r9 c; _$ ]5 z; A: z$ M* D}* T6 f. y3 M+ x5 v/ x* L
printf("\n");
F) |0 w) B" H& J; O, g+ C}& [! Z T! B) ^5 f
6.7完数
0 d, `4 P, W+ U, Z$ qmain()
) C1 u' Z& c3 d# k, _* J#include M 1000
- r5 N2 N$ I; d( ymain()
( w. A& M% S2 v7 @' f$ e* @: o{
- ?5 U3 ^( b+ J. }# I" uint k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;8 W2 C2 n+ c4 J4 o, g+ p5 C
int i,j,n,s;
# I+ `4 u. w6 Nfor(j=2;j<=M;j++)
& C$ ?, r4 \% y# k& k. a2 C{
" b. i2 e7 C! Rn=0;0 h6 M% ]! T: ^ Z
s=j;
; B/ ]' q4 i/ mfor(i=1;i {
' f2 Y p" V _" c0 Wif((j%i)==0)* e2 p8 E9 y4 R6 k3 W
{$ Q, M) ]8 s7 D3 `! s" j: D! J5 J0 ~
if((j%i)==0)
" j7 z$ k0 n, l' q. E {
# c$ E) Z9 I8 Z0 c Q n++;
: D) Q- @0 y9 `/ Y( \' u- Z s=s-i;- I7 D5 m+ e. X
switch(n)
* _" ]: q: A2 p: p6 V4 h. G, E {. g7 t5 O+ X F P
case 1:
* w5 n5 |/ F7 U' P% n k0=i;
6 n$ }$ d8 T4 S9 }9 W% Z) v break;/ T t) F4 R9 y% X) U' H' i
case 2: M* m1 [- j; L) u' H0 I1 g1 h
k1=i;
5 P7 ^/ z' a$ [2 M9 C/ v# d7 ]. j break;
) U8 Y* y5 C/ l$ X# }% w3 H case 3:
7 z8 v1 y1 y& D+ U! B- _% i k2=i;0 M5 w! A" k8 |! ^& Y
break;
6 m6 z; ^; v, J9 r7 G: y case 4:4 o4 w& `1 ]/ ?6 @' ?
k3=i;9 a0 d; ~5 f7 k! v( b, \% i
break;* f2 \4 \6 [ Y0 {* F; F0 P
case 5:. g' g6 J9 N* k$ n3 n7 l
k4=i;: {1 O5 l' e2 |0 g7 [ P
break;
: |5 B/ g+ S' v3 F/ v) A+ W3 G case 6:3 c+ D, p# d9 N; U* h7 E
k5=i;1 O# U/ ~$ a' C6 }& a$ J
break;0 r: {! C+ O/ K7 A0 S N% O
case 7: B- X# i- H' w( i' C& q9 v9 J
k6=i; I' Q$ t$ u& K/ `6 n% I
break;: P6 p+ l" Z2 I' `' r/ S8 w/ |3 c4 e8 K
case 8:
& T7 M( _( n# k) l7 A k7=i;
8 J2 w; c4 ^! v% n( R break;$ F) M7 i! ^3 q1 A
case 9:
$ R% M) \4 ~1 N7 M& f k8=i;
+ f1 {! w/ j1 K3 c break;3 t- U: i9 Y V5 x" c8 U
case 10:$ q/ o! c8 L& ]% I$ y
k9=i;, N; ?+ O! d3 O0 z( A
break;6 R h# g% R5 f! G) ]( S! _ C
}# m6 d0 C* J5 l* s- b" {
}, E" R# P! f$ A
}
: _% R' s2 i! O ~/ Z- W( B7 Jif(s==0)3 U: j& w+ w. Y9 B0 O# @) N
{$ r% V+ x: x) r& u. S9 ~
printf("%d是一个‘完数’,它的因子是",j);
" r3 q8 ?7 n" s jif(n>1)
+ G0 {: M. b5 H$ w! q: r2 r3 a( A. Q printf("%d,%d",k0,k1);8 K, I' l6 T" @7 B* Y ?
if(n>2)4 C, R7 T! d" w5 `9 V. i
printf(",%d",k2);
' V/ L# p, b) ]' P3 M% s/ Vif(n>3) i# Z) U# x( j6 x& T( i
printf(",%d",k3);
! S v4 V+ ~4 j4 eif(n>4)
4 J9 j# A% u4 G, h printf(",%d",k4);* m+ b& N. e- }$ V/ s: e
if(n>5), B% v$ Z. t8 F" f4 \. i: M3 I
printf(",%d",k5);6 F! c/ A% }+ `0 S t8 l( X
if(n>6)
6 W: N7 k) Y% ~4 ?! W, z; Y printf(",%d",k6);
$ V6 r# f+ u R8 S3 Oif(n>7)2 V# G6 a; c4 A6 r' N+ B$ s
printf(",%d",k7);
, L) q9 G$ C' L5 J. h9 \if(n>8), u' ?3 P1 z: i" c( j
printf(",%d",k8);
! u) F. x' H* d, M3 U" {; W& rif(n>9)9 ?, V" k7 Y7 @9 S9 x' j5 m" |9 \6 I
printf(",%d",k9);5 V- x/ n2 [ d9 f K
printf("\n");
- u1 s+ b, a3 ?7 a6 f/ a/ N) V; b }& @; A2 p: x A9 C
}" i) H1 u! G; _( {& ^5 o* `: h8 o
方法二:此题用数组方法更为简单.
l1 i7 g5 I a0 `1 imain()
5 f+ G! o) V9 O- F5 d; X{
0 V4 E3 k% h5 v! \" Astatic int k[10];/ r2 ~$ k# C4 `6 Y, J) T0 M
int i,j,n,s;
. I) K" L1 E* |- S& sfor(j=2;j<=1000;j++)
/ T( _5 r \; f3 k{3 n1 N V# u |# ~# x
n=-1;. W0 w5 x1 S& p$ S9 M) N
s=j;7 I2 E- U* G- Z2 t- M
for(i=1;i{
: S: k/ R4 n+ W1 k0 zif((j%i)==0)- r. [4 P" h: R7 y
{. w, `: D/ F+ [, _) a+ B A
n++;0 r- g2 s0 d2 Q
s=s-i;
, w" `7 R; G! h2 A/ ik[n]=i;
+ M/ p: ^4 D& u7 U% V }
/ l j7 U0 {+ P; k( j }
# x3 F8 n% p) E; E$ T% ~if(s==0)# C* ]# \; G' p% P4 J
{0 Y( g+ I# }0 z: c# _
printf("%d是一个完数,它的因子是:",j);
) _' \* I# x9 h$ \for(i=0;iprintf("%d,",k[i]);
% l7 N' w* z6 Z& |2 Lprintf("%d\n",k[n]);4 e3 c; x8 G. Q1 k) F0 L; y2 F
}
6 Z! t1 V$ c$ r- k; e( T9 r}
6 n# F4 t0 E4 N/ r# p2 p6.8 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.
1 P0 j5 W* _! X5 H解: main()
' p1 {/ r' i8 H$ z* q: |{
1 U1 Y3 p( c8 o! ^' ^! Aint n,t,number=20;& R3 p( c- X" ^- C' }
float a=2,b=1,s=0;* |1 e( [: e+ d- O! c6 F
for(n=1;n<=number;n++)* D0 m+ y9 c8 A+ l2 W3 _4 {0 T
{; U" B `4 o- h5 F% K9 ^ `# T
s=s+a/b;
1 q! o; B1 e9 n! ~! x! x8 {- s+ a, ct=a,a=a+b,b=t;, e, \* B/ r! o4 {0 u+ @
}
2 f: j" M* h4 K6 q8 M. mprintf("总和=%9.6f\n",s);! ^8 v; D$ \9 ]. D2 z s+ N
}4 X' _1 N: u7 v: W, l- H6 U$ s
6.9球反弹问题. q& h/ e5 L0 T2 Z' ?! g' \
main()
* |% L+ O5 G6 n{
3 X4 l+ k6 W$ Q" D5 Afloat sn=100.0,hn=sn/2;. H- H3 ^! V" r% _
int n;' g% v' i& f3 F2 o- w5 _8 O
for(n=2;n<=10;n++)+ U6 n+ v3 a9 _3 i `
{
& C+ e4 W; o. v$ Y" }/ |sn=sn+2*hn;% h( n5 g: o3 s% a* ^, I
hn=hn/2;
* k! b3 _" b4 a4 j% z9 f}
, F/ R7 O r1 k: v# U# vprintf("第10次落地时共经过%f米 \n",sn);
4 a! \* x' t# ~2 W" oprintf("第10次反弹%f米.\n",hn);, v8 \; m1 L7 z. E2 O$ P- B: [
}
/ `/ X& `+ p" U. q1 ]6.10猴子吃桃' H& m( W; |' F& m
main()( |/ Z0 f" N5 x
{
3 f! x9 ]# n# [$ Z# f8 [* D, Tint day,x1,x2;
7 K" L0 ]$ b! F; y8 Y tday=9;
! W; m8 m( T B7 Cx2=1;: D0 L2 |+ p8 M0 s
while(day>0)7 D9 b% w, H+ {, D" v2 U
{! Z2 n4 B0 `0 p
x1=(x2+1)*2;
8 r+ X) {+ {( X M; F Kx2=x1;! C9 U$ ^/ v" W- L* z4 z9 e A7 i4 ~
day--;4 O3 \" K, A' q# w) c+ v
}
, i) N E! K2 m2 nprintf("桃子总数=%d\n",x1);( v( p* v* M; c: C7 }' v
}4 y; F% V5 B) j5 u; E" I& D) X
7 Z, V: Y, J% R7 Z( @, I
6.12
) T3 J: {( ` d+ c" F5 r) a- h#include"math.h"+ Z5 r$ v( c( b7 x7 L0 n6 ]# g* U
main()3 D. I/ }8 z0 l* s5 z
{float x,x0,f,f1;5 U( F9 a2 O% f7 q4 ]
x=1.5;. w$ |, u) P$ A# L
do& b, E7 a' a! G9 m2 O
{x0=x;& t: b! W6 t6 W# C9 q
f=((2*x0-4)*x0+3)*x0-6;; P" Q' f/ J- V& N2 `
f1=(6*x0-8)*x0+3;) v: E! K9 ]% i
x=x0-f/f1;
3 Y, k$ p) ?) Z# x; t8 n }/ }; T* T; j- r) e. L$ j" O2 P- h
while(fabs(x-x0)>=1e-5);5 f& b8 j3 t, R; D0 ^
printf("x=%6.2f\n",x);
" W2 p+ D( t& P6 W) d& _7 t}
% l( {4 P6 I) ~6 K/ A0 a+ R8 K$ i: p% ]
6.138 B0 d/ X! C7 g* Y2 ^6 W
#include"math.h", B$ s% O3 e! _0 d" P
main()
! m0 z+ m; T9 @" t# w7 [0 h! J{float x0,x1,x2,fx0,fx1,fx2;
# N! c, c8 Q3 _. s0 f0 T do! ^% R5 ?/ H: g: V1 {
{scanf("%f,%f",&x1,&x2);3 ]7 p: q2 R( N+ L8 j
fx1=x1*((2*x1-4)*x1+3)-6;' i# j9 i% D) `8 |9 K- }" y0 v
fx2=x2*((2*x2-4)*x2+3)-6;
! d0 U: I& s! ^, ] }* A7 L% k, m4 p* q. K1 w9 y9 I
while(fx1*fx2>0);6 D7 s* A% B. T" o" V
do' @" o/ @9 V) d4 t) a' B8 `
{x0=(x1+x2)/2;9 P/ m& o/ K& \; Q# z
fx0=x0*((2*x0-4)*x0+3)-6;8 y4 W( C4 y4 g4 I- h/ O8 v
if((fx0*fx1)<0)1 w' L0 S1 U' y: _1 {
{x2=x0;1 P) s0 z- W/ d' X# \% ]
fx2=fx0;6 S% y! |4 ^8 m9 l
}
, d9 |( ^2 v' K' n6 ~& y else- U D& M0 k; y7 W
{x1=x0;1 P0 P! u- t1 ]' m% W, N) _, B
fx1=fx0;
* q. f) v6 R: y5 z* F5 j7 ]5 { } [/ f3 v- d$ Q7 F
}
% m( D$ T" f4 W& w9 x9 P& x while(fabs(fx0)>=1e-5);
* V% ?* x( S; u, j printf("x0=%6.2f\n",x0);7 C5 h, e+ w& u. S* _
}* g0 ?* Y/ ?* r% S; ] I
6.14打印图案8 s7 S% C) W; c+ A9 `* Z2 b' ]! k8 Q
main()4 d7 P3 o2 P3 j. E( d) x3 B1 @$ i* c) W
{int i,j,k;, I- r3 R0 m6 u* F3 l \8 H
for(i=0;i<=3;i++)
, |3 B% w' O- U) e: L; W {for(j=0;j<=2-i;j++)
* D* H3 I/ V$ o, x G, \2 N, n printf(" ");) o# ^8 E5 N$ y% j
for(k=0;k<=2*i;k++)
- Q0 p3 f4 J) {. O0 r! Y' } printf("*");$ b! X. V1 m0 `$ d( ]8 s% s
printf("\n");3 @2 _7 Q& {, x) a; O" l u/ g9 @
}+ Y- y( X4 Y4 z$ _/ L8 V# u* M( i6 m3 B
for(i=0;i<=2;i++)- X- C' e% N( b. W& ]% g, n8 R3 K
{for(j=0;j<=i;j++), w! S( v6 Z+ d
printf(" ");
# }; C" ~1 h* p3 U/ L' }2 E3 f4 O6 L for(k=0;k<=4-2*i;k++)
4 R& b* i, ^' i6 p/ G printf("*");# ]% {) F# J% q$ f/ D
printf("\n");
3 R! V- |4 a! j" T, M: E3 W7 b# V }, s; `9 T4 c7 M
}5 R& A2 p: Y- e( U, ^1 _7 K) Q
6.15乒乓比赛
& h0 ^! a# {7 V1 Gmain()# O s$ y ~( l
{
/ R3 ]. h$ I2 h$ Nchar i,j,k;
8 Y/ P: F0 D, Ofor(i='x';i<='z';i++); H2 Y1 Y6 ~8 {: m' x, W% E
for(j='x';j<='z';j++)( Q+ R1 X; J5 G
{
& p! G" {. M* a9 F7 c% uif(i!=j)% J* Y* ~& P5 T0 D$ e- v$ ^
for(k='x';k<='z';k++)0 u8 e2 v3 G" G$ }( [( c# y
{+ H$ A5 z( v- f6 @( S# ?
if(i!=k&&j!=k)! i7 [. \7 s; F O* J5 l
{if(i!='x' && k!='x' && k! ='z')
w( s# L9 C# }7 qprintf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);6 e+ ~' p! K) }8 v6 Y/ ]2 z* \
}
3 c) {+ f( y4 K( y0 o6 B8 a }
" Z! Q8 m8 B7 {/ u# b: H }( l E/ D* `) @% I
}
$ `' u9 u$ u( o: G. V* xC语言设计谭浩强第三版的课后习题答案; L/ u2 S/ j# P( I2 N. y: Z
7.1用筛选法求100之内的素数.% |. O3 X- l: m; W1 G9 A1 ~$ S/ m. ^
#include
- l, Q% w' h/ Y' B$ @: k+ k#define N 101
+ B! j* t3 I) hmain()
: ~3 X8 @* ]; F. w+ k{int i,j,line,a[N];
/ V+ R9 a+ y: F+ M3 v$ O8 Mfor(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
0 S+ ^" N+ b a* F; Z+ ` if(a[j]%a[i]==0)1 d! e* _. o" v- [& K' X
a[j]=0;
: w- t1 m {. k; Lprintf("\n");
- N8 M7 x4 z: T/ f7 R6 M" Bfor(i=2,line=0;i{ if(a[i]!=0) ]1 N) J' y5 g( W, x. j/ O. t! ^5 y4 ?! U
{printf("%5d",a[i]);
$ }) w: f1 X/ T1 I5 b line++;
+ p( q w6 N+ z8 b if(line==10)6 [0 Z4 {) \* b1 H
{printf("\n");
5 ]9 R+ k. _3 D0 s7 b3 v% a line=0;}
! j* M. t* {) Z# V" G) W6 o }
6 o E' K& j$ ` y3 R. k}
3 {9 |+ N/ e- @8 g7.2用选择法对10个数排序.
$ A. z7 Y. F2 {& R* L% Z0 M0 [0 g#define N 10 _4 ]& d4 c* m9 P
main()
' U% m0 Y, c" n; H# p$ p4 ]( n{ int i,j,min,temp,a[N];
+ m. W/ _2 L) ^ o0 l! U5 _/ R4 a. Zprintf("请输入十个数:\n");2 [4 h% a/ K. [. R1 U' [0 P6 g( r/ l
for (i=0;i{ printf("a[%d]=",i);
^8 W# ]7 P7 q m8 W scanf("%d",&a[i]);
+ J& p2 [7 | K3 S x, J) m}" q& @' K+ G: m. O
printf("\n");
2 _* J! f: _% j: u, k7 ~& Mfor(i=0;i printf("%5d",a[i]);
& x' K/ ?3 d" H+ O' d$ kprintf("\n");! d/ W) U3 E1 L, D% P6 d
for (i=0;i{ min=i;4 Z4 X! f4 Y( e
for(j=i+1;j if(a[min]>a[j]) min=j;4 s7 s: G# u0 N: X
temp=a[i];
0 V6 }; N0 c# ^) h3 @4 ? a[i]=a[min];) E/ s0 V; c4 O# O, w
a[min]=temp;6 C1 W( C/ {8 z; a+ c( R4 w' j$ k
}
( l; |2 Y8 C5 z( _. [8 c1 V. ]printf("\n排序结果如下:\n");
+ O0 K& ^+ M( x, y$ D& B" q' Qfor(i=0;iprintf("%5d",a[i]);& r4 E/ p1 l. Z5 P
}0 s& q' ~; s7 P5 s ?
7.3对角线和:3 j4 u$ q3 V7 B6 l
main()) p: E; E. X* R( b7 m$ r6 Y
{
& t4 b4 ]" V! z7 S: n* P% }; qfloat a[3][3],sum=0;! }, o% t2 h) d2 _ k/ L$ }, k
int i,j;
+ W1 a% V: n+ B# sprintf("请输入矩阵元素:\n");& G( C: n+ h3 F: e9 ?+ Z
for(i=0;i<3;i++)
+ g' W7 H: l _# L8 B for(j=0;j<3;j++)
& b2 v% W9 {: J! g1 C scanf("%f",&a[i][j]);: J3 F# ?* p& c
for(i=0;i<3;i++)
' g+ w" U& W8 G; | sum=sum+a[i][i];
& b- c5 `2 }: W5 R9 [8 K printf("对角元素之和=6.2f",sum);
& L9 N6 W3 e# y; f* X4 j% b}
* Y( g" ]5 X' ~# U0 R; g7.4插入数据到数组
, D# Y3 [0 D: c8 h+ p' o+ kmain()( `9 r. j0 |6 l m* p) i5 i0 p4 h
{int a[11]={1,4,6,9,13,16,19,28,40,100};
* p# g, z2 C+ H/ G: oint temp1,temp2,number,end,i,j; N- Z2 P" d0 a8 C6 y ]
printf("初始数组如下:");/ t. k( |' {( h; ?. t
for (i=0;i<10;i++)" P" O7 l4 Y1 S$ S! S
printf("%5d",a[i]);
# x% m2 k6 R2 f9 y Wprintf("\n");
/ z9 c. p+ [+ W. |printf("输入插入数据:");4 t; \# ]0 |. i) f: o
scanf("%d",&number);
4 B9 r$ r1 g6 K$ c$ t0 pend=a[9];
' b! x' w' B/ w; tif(number>end)
! i5 f( C& z U. Ea[10]=number;" q. F* m) H# Y7 ^
else3 m$ w8 R) Q9 [2 N2 v! `* s
{for(i=0;i<10;i++)
* B* D3 _8 P2 @- J0 F1 S9 x* l { if(a[i]>number)
% M1 g7 |( Y4 e7 R t {temp1=a[i];% p; S4 `; ]& ]4 N1 y
a[i]=number;3 o% H4 c/ D6 k% }
for(j=i+1;j<11;j++), Q) D8 G% M* F! g! r% @
{temp2=a[j];/ W* t; L* w& I1 Y% y: N4 M
a[j]=temp1;
6 q4 R, t+ {3 H" _3 d temp1=temp2;! k9 ^. u% w5 \
}* R2 \' v* g% S5 K# Y
break;$ Q7 ~& X$ U& B. N$ q2 r
}- v8 [$ c" r" C# ^7 B
}
8 R+ `0 p! V& z" P y# u) k$ C }
/ `$ J" R7 N2 V, A7 X for(i=0;j<11;i++)
! F* j& V: K9 I1 `3 B2 D6 b( ~/ t j printf("a%6d",a[i]);1 V, c' j" f/ H7 E4 B& H
}; \/ o/ f% G" Y+ K" k/ m3 k
7.5将一个数组逆序存放。2 O3 r4 Y2 B/ j# m- r
#define N 5/ a- F! W% m. H" A; `) |/ Q
main()& ?8 x% g2 S7 F% O% E8 U
{ int a[N]={8,6,5,4,1},i,temp;
( d( e8 L3 m% n( @0 O" Qprintf("\n 初始数组:\n");
' {% o6 ]. g0 Ifor(i=0;iprintf("%4d",a[i]);
I5 W5 V& s- @2 ~0 N" E5 V. O6 ]for(i=0;i{ temp=a[i];/ m5 n- i- B. Q
a[i]=a[N-i-1];
9 V* l9 Y! n6 Z; I3 G a[N-i-1]=temp;
% h& D9 |* o. ^8 |/ w L) Z* W# }}1 M3 L _! g6 G: Y# N
printf("\n 交换后的数组:\n");. B( a X F4 L, |# H
for(i=0;i printf("%4d",a[i]);
1 y* a6 k3 A# T4 X}$ b1 m! x- ]9 f. g, f
7.6杨辉三角; N/ f8 m; V% x
#define N 11
2 m' t+ z7 V, Y4 h6 y: ~/ H I9 Jmain()4 c, r/ X- t3 B3 h) V* Q5 Q, p
{ int i,j,a[N][N];
9 K4 R! h4 g% D$ G' l% U/ j for(i=1;i {a[i][i]=1;
* T3 m/ w8 `; ?4 f) ^5 a a[i][1]=1;
: W2 l5 r* w+ U. u- s# W( z }
; b( l3 b( S- V' n T9 A1 a9 [, D for(i=3;i for(j=2;j<=i-1;j++)! d7 X6 S5 `+ f' F: R7 l
a[i][j]=a[i01][j-1]+a[i-1][j];
' a d% x7 V0 g) @1 y0 ]7 f' n J# v, ^ for(i=1;i { for(j=1;j<=i;j++)4 T) {8 Y u5 _+ ]% A( H6 p% h
printf("%6d",a[i][j];, Z8 {5 F* }+ K" J
printf("\n");3 N' j" e; ~% [; s, b5 z
}' z6 U$ x4 L2 G' I) V$ K
printf("\n");' U& X T( F$ z' g) X( K& @
}* T! A# v9 | m
7.8鞍点; |$ D- V# Y4 p
#define N 10- w# y! C5 [: t# [
#define M 10/ Y4 D- Y) H" n5 ^% d) _
main()
! h( ^ }7 B l3 q+ k$ O8 d. x4 ~{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;6 x6 T6 Z7 w: G4 O
printf("\n输入行数n:");) W7 a3 c9 A" G$ D9 m
scanf("%d",&n);" |2 i& e" H# |
printf("\n输入列数m:");2 z% ` T: m; P$ F; K- r' D
scanf("%d",&m);) S; E+ {; {: [1 U/ V
8 o4 ~" S, u7 B& G1 \# `. o+ u for(i=0;i { printf("第%d行?\n",i);
$ }$ y: m% H. D8 y9 f for(j=0;j scanf("%d",&a[i][j];
" U4 W! e/ i' e3 P# H+ v# S }' X/ U- E3 b+ W/ a. h5 b& N
for(i=0;i { for(j=0;j printf("%5d",a[i][j]);5 H2 B# U( i* Z ?- S
pritf("\n");
4 k1 w" b( R5 Z1 q5 ? }
& N, q/ V3 l0 Z. ^* h5 a flag2=0;6 z S, B7 L% z
for(i=0;i { max=a[i][0];
b8 z. F1 y8 c3 b for(j=0;j if(a[i][j]>max)
: z: N/ t8 N; t$ ^ { max=a[i][j];
% [6 J/ _! I7 G. m maxj=j;
/ C _! x$ s" C1 k }
& S$ F4 v' D/ T' M1 Z for (k=0,flag1=1;k if(max>a[k][max])
8 B% [- M. D' g' k0 o$ w6 M- T flag1=0;% U+ f4 ]5 x z9 D: D+ d
if(flag1)' m8 D4 h/ Y- d. {9 b- E7 V! p7 r) b
{ printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);. S! m& d5 W+ Y6 m% A+ [4 E. a
flag2=1;# Z( r- c5 S2 Y) z/ P+ y
}) s, N+ ~" |* x, X" d; J5 f- D, V
}
! s$ c# J2 L& c/ Eif(!flag2)! J* r4 L' ~2 @% |2 |- F @
printf("\n 矩阵中无鞍点! \n");
7 e- d# o; Y5 Q+ T}
9 Y [' b6 i1 J5 M7 _
+ Q& } |$ U1 a/ }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.
- A3 S; }% f* [& F9 {#include+ u* J2 s2 \ F, S/ _
#define N 15" [* C0 w& I) g
main()- \: Z6 {& D/ h5 }9 w: u
{ int i,j,number,top,bott,min,loca,a[N],flag;- r2 T. F4 k: q. u
char c;0 ~$ ?5 ^. P* Q* h$ a
printf("输入15个数(a[i]>[i-1])\n);
, T; e5 h( }3 ]3 L/ X scanf("%d",&a[0]);* X( s: I4 ~$ c9 Q
i=1;+ G0 q# @9 l. [: w/ z
while(i { scanf("%d",&a[i]);
+ H+ h3 P& r" \& { if(a[i]>=a[i-1])
2 V$ c% d/ n2 H8 i1 n6 e( `8 ^ i++;8 c: g7 K3 P" W. N. A& o
esle5 X; n: M- d' S7 x$ J y
{printf("请重输入a[i]");
' P9 S" q# R" k& Z# y. ]0 J printf("必须大于%d\n",a[i-1]);
% _) f* h; A, @1 L1 [1 Q/ \ }
4 {1 [- U5 Q3 p }
3 G; ^' F0 h& `! s printf("\n");
, }1 q, K4 q/ { b for(i=0;i printf("%4d",a[i]);% w5 }* m# e# l4 a- t9 ~5 G/ l
printf("\n");
: s7 k: {* l3 x6 y s
3 l5 Z! V6 t! u# r: M+ g flag=1;
/ n8 u& |9 b& [- ^7 l+ h- ^6 }2 v, i while(flag)
5 j& A0 \7 k) k; | {. M6 W! X( g" P8 S) K( a
printf("请输入查找数据:");4 I) ^1 O! A2 b' _
scanf("%d",&number);0 P5 g q1 {$ c6 q: ~
loca=0;
4 ^+ O5 V# ^$ u$ v$ d1 g top=0;5 L: H+ o- K, \) O) H3 @
bott=N-1;6 J4 @( N6 G( t, G+ l
if((numbera[N-1]))0 i" p; N# h! h+ B
loca=-1;
# N* P4 k8 q+ R( U while((loca==0)&&(top<=bott))* R1 f" f: @" V+ G
{ min=(bott+top)/2;9 K5 k# B$ n3 Z7 x2 n' I2 @
if(number==a[min])# A: M6 D" W: o. U w7 h6 s
{ loca=min;4 y4 J& X3 L% H, u" B! M
printf("%d位于表中第%d个数\n",number,loca+1);
. B5 X( v. [5 E; C& @1 Z }# k Z7 U: M3 w# J* c: ]
else if(number bott=min-1;
7 ^' g4 u0 @6 }5 D5 l* {! l else
* x% Z- o$ N+ I top=min+1;/ D: w0 `, J; Z1 w4 o
}! O' ?. ~; K* N9 }1 y: s Z
if(loca==0||loca==-1)
2 X4 l, M% R2 t8 E& S* n printf("%d不在表中\n",number);: C6 H$ ?$ J' c9 r5 `
printf("是否继续查找?Y/N!\n");; ]: Z+ G% ? k- A8 I1 M1 { J5 p
c=getchar();; z8 L: _$ X! _& a% d. G$ ~
if(c=='N'||c=='n')
/ j! o) T! H7 _+ {: N) ^6 u: v g0 ?1 g flag=0;
) g+ }$ G2 Z1 }7 \" h }" Y& ]) k3 J1 O
}
: e3 Y- Y( d! o0 S- [% |- {9 i0 k
+ E, b# h, C6 d) I0 a8 j( P7.10
* O2 c) ? N8 R/ R+ |1 e" X# f) {main()
4 g9 n8 t @* w& W, @4 x7 V! q{ int i,j,uppn,lown,dign,span,othn;5 U k+ E* m A, a3 ~% z1 h3 K; M
char text[3][80];
+ j! k7 E+ k0 H3 O0 H" g2 R% B* _ uppn=lown=dign=span=othn=0;8 v6 Y' v5 s% m, }/ }( [
for(i=0;i<3;i++)
. O4 O, N2 S8 t- U( w { printf("\n请输入第%d行:\n",i);9 W4 z* X% s$ m
gets(text[i]);/ G; @" E7 M, l9 p7 B) H( W U
for(j=0;j<80 && text[i][j]!='\0';j++)
8 g& [$ v* ~6 U5 s& V: I- H {if(text[i][j]>='A' && text[i][j]<='Z')7 f& Z& E! ^2 m; j' B5 K
uppn+=1;
, E" f" z/ H5 h0 y! V, w" K else if(text[i][j]>='a' && text[i][j]<='z')
/ m! ]" D$ U( t3 D lown+=1;
) B- P' u- i! m5 d- X7 ]' n! U else if(text[i][j]>='1' && text[i][j]<='9')
+ j) N6 S5 J% g- R' T dign+=1;
; k0 T/ C* v r( e! |* T% d else if(text[i][j]=' ')) J' s7 C) u0 x+ {
span+=1;1 Z3 D# |/ p. w" p6 w
else
5 T0 \- x/ u9 }: |3 k othn+=1;) w8 X' K+ ?9 r2 y! g1 G
}
2 j9 t3 I* w2 K }" h7 }7 v; I& t% ]
for(i=0;i<3;i++)% L1 T8 Y; i- h ^3 A: F. F
printf("%s=n",text[i]);# e4 i+ Q3 q# r. ~/ C t: i! \! U
printf("大写字母数:%d\n",uppn);
! N6 _& H. H' R( ?% K0 l printf("小写字母数:%d\n",lown);
- y! _5 k4 @ U6 F! b K printf("数字个数:%d\n",dign);
0 y. ?6 Y2 N1 |4 { printf("空格个数:%d\n",span);
3 e; \: b% ?5 F5 }# N printf("其它字符:%d\n",othn);
! ?: |* {" Q% m0 h% x, d: s}
7 i0 i$ K+ u9 c5 {; o+ o: c* o6 K& v8 v4 b; H1 J+ E- S5 L
: G; C1 @3 Z3 k9 P% w
7.11
, z0 B1 ?" f. `. p+ i! r8 e2 H2 H$ W! d' ~main(), @6 i; h6 {- g* o0 J
{static char a[5]={'*','*','*','*','*'};
8 t' ?) @0 m. w$ R c int i,j,k;
: f) _6 u9 z( O9 m* _ char space=' ';/ d) {7 w* Q& i4 j g: X* @
for(i=0;i<=5;i++)
2 z# m1 b+ h0 X: L6 }' r3 Y0 e" q. Y {printf("\n");
3 a+ U6 d ]* ] for(j=1;j<=3*i;j++)
, i3 W W% c# m" ^6 }: n9 m printf("%lc",space);
; j G( w) l( O+ N o; }. L) B) S% x for(k=0;k<=5;k++)6 [& j6 N4 y8 u5 ` f8 S- n
printf("%3c",a[k];* j: z" m" z6 A0 o X0 }. }
}+ }( w$ v/ N! @* V, M! `4 K
}9 m3 d' Y1 y3 K2 X# C
7.12
) ~) O9 @, p$ A9 E1 }#include
7 ?: _9 C5 h4 Vmain()
?6 T% H: }2 y{int i,n;( [8 l% o# i: n; k
char ch[80],tran[80];
5 T" l8 B. J' v( m; B. K' { printf("请输入字符:");
. s- |$ h! n2 b4 K# ^2 n' u gets(ch);7 L5 s0 ~3 v+ k
printf("\n密码是%c",ch);0 R. E* z9 j ~1 k0 Z
i=0;
! S& Q- J. {2 ~% s) G2 ywhile(ch[i]!='\0')% T: D/ W. `. s# ?% g) R+ a4 L
{if((ch[i]>='A')&&(ch[i]<='Z'))
* C, r b {+ y* K tran[i]=26+64-ch[i]+1+64;
% ]3 C5 r7 q$ }5 s3 melse if((ch[i]>='a')&&(ch[i]<='z'))! c# Q/ J/ n# {. o
tran[i]=26+96-ch[i]+1+96;
0 E) W' {, j0 K4 p# relse
! d8 g8 N( H. s+ J! j) o" @ tran[i]=ch[i];
8 b9 t2 n8 a! O7 Z8 o/ G i++;
: T H1 y9 O0 e4 O}
) i' W' d1 F2 j% cn=i;
8 E5 Y: N6 i. F7 D& ]% o; U+ Rprintf("\n原文是:");
, N0 }' a! o7 |/ cfor(i=0;iputchar(tran[i]);
, B% G/ J8 u1 ]$ U0 T$ \}
0 ^: K# }/ [: k5 Q$ _7.13
9 x7 l, n1 E; lmain()
; ~5 ^% F% Z4 Y1 t3 c* a {
9 l, y0 {# T/ d0 l7 ? char s1[80],s2[40];
# d' V/ D% ~8 t- w& E int i=0,j=0;
6 H: E8 J& ]% R- \' P2 S0 J* Q y printf("\n请输入字符串1:");
. C# \7 ]% V4 g5 a" p9 W scanf("%s",s1);
, {& [8 Z$ E% I7 c printf("\n请输入字符串2:");
( i$ Q1 \+ |! \ N scanf("%s",s2);6 c- i% K: N1 B# s/ Y6 Y
while(s1[i]!='\0')
0 j) Z$ c) R( H0 S' U i++;
" m# I2 e) D' F* Uwhile(s2[j]!='\0')1 }% ?! t( s3 S2 j8 Y
s1[i++]=s2[j++];
k3 Q5 U% e' J6 J/ V* }* bs1[i]='\0';
" f" G: N, i( R0 Q, }printf("\n连接后字符串为:%s",s1);1 E4 }! \2 w* ~, S+ S
}& g% L4 Z K: h- k2 N4 D0 j5 ^" W
" _; N+ `5 K! z& X: \
7 r; c+ Q; G9 b) P
7.14
% u" S4 R& @; M, k. N* [) Y#include
% I& K( ?0 ]1 Wmain()
$ f0 K( z0 N4 x( H{int i,resu;6 r, M* H- \1 r% {
char s1[100],s2[100];
0 X6 q" k U0 y' w0 D printf("请输入字符串1:\n");( ^& j( \' r, Q5 Y% b
gets(s1);* k8 x' {" D# @4 u/ V
printf("\n 请输入字符串2:\n");
. z6 ]: ?# n: b* d0 { gets(s2);
. m6 h" e8 p& f5 w5 D3 i$ e i=0;
' b$ c2 y/ z) [ while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;& u$ X, S' ?& }4 h2 X8 A/ ]0 _
if(s1[i]=='\0' && s2[i]=='\0')resu=0;
3 @, H: g* R A4 q else
0 [# z+ ^ e5 j6 L* X6 H resu=s1[i]-s2[i];; B6 ]' z6 ]; t9 b }. M
printf(" %s与%s比较结果是%d",s1,s2,resu);* Q# S" z$ E/ t' W) f* ]; D
}# z1 g8 j5 R+ p4 z( f9 ?0 ~7 b
7.15
+ q) r' x2 u0 L#include, P# Y5 T w/ @+ c' A- `) x, z% K( k
main()* X, t! a/ `4 _: \1 c
{1 C4 r9 u" L& Z/ u8 r- V ^
char from[80],to[80];
8 k9 b' g; g% ~$ Q int i;9 D3 ` v8 H0 O; P
printf("请输入字符串");( [: w2 t3 F, k$ {
scanf("%s",from);
G* z3 d, `4 }1 Q+ x0 e& J for(i=0;i<=strlen(from);i++)8 Y9 {( n. H! d* b2 q' X! X! f
to[i]=from[i];
( P2 x' O3 u2 A8 X+ k2 y printf("复制字符串为:%s\n",to);
+ f& L! P6 h: [3 v: P1 g. l }
& O' f, P$ X1 c& ~! Q! t
" H2 v a5 I( C7 ^& z5 }0 y# K" Q7 Y3 ?) M4 e) w
第八章 函数. X) O- L- |. j
8.1(最小公倍数=u*v/最大公约数.)6 G$ i* x) a- v- J' V3 ]
hcf(u,v)6 _4 m, w5 z$ C1 H& X
int u,v;! H' S; W! F3 Y; A5 m: r+ c% r
(int a,b,t,r;
4 Z j2 d. H3 p% ? if(u>v)
, Q* b8 J0 s8 j' H. R3 ~2 k* ^ {t=u;u=v;v=t;}
6 ^5 y, k5 S8 i a=u;b=v;* w' w+ }. J; i' y- T* w; ?1 C
while((r=b%a)!=0)3 L, G2 P& S- H, ?
{b=a;a=r;}
* G( N Y- p$ | return(a);9 ^# @) V C# A( e9 ^: \( O" E6 p& ?8 b- T5 F
}7 _8 m) ?/ T$ @3 |
lcd(u,v,h)
0 {9 `& Q+ T7 w: o( ]0 U int u,v,h;9 p' u C) m$ z- j& h( l3 x' }
{int u,v,h,l;& ?' L. H+ A( M* Y0 H. Z
scanf("%d,%d",&u,&v);) F! |* i: _; r: u9 U
h=hcf(u,v);* L: j! z$ p; Z& Y" k( M. Y
printf("H.C.F=%d\n",h);7 }( C" x, D7 v: x" Q* D
l=lcd(u,v,h);8 ?" S% w3 O3 w+ D
printf("L.C.d=%d\n",l);( g" W, m0 u' G( A( M$ u
}
! O$ U1 L0 `! h; e- F# R {return(u*v/h);}2 f9 E# b7 m. E' z# y0 `
main(), H8 g4 f9 W4 S/ O& c( D
{int u,v,h,l;
* G3 f; ^" w4 k" ~% a1 a scanf("%d,%d",&u,&v);
4 T% }7 z, k6 w0 S4 \ h=hcf(u,v);
9 c8 w* W# a, S# s" g# J/ ~0 ? printf("H.C.F=%d\n",h);
0 L$ }' I' N; W/ C l=lcd(u,v,h);
; y" { P; Z9 G% `1 \ printf("L.C.D=%d\n",l);
2 D& ~% W* c2 B }. e8 x6 ]. G" [5 `- E+ x
% y9 t& ?0 g8 W4 n3 {
; |' L# l9 r+ `$ j+ S6 a8 B5 a
$ t; x6 P I) ?* E1 B# o: Z" G8.2求方程根
8 G$ a0 H" ]5 z G1 I% h) ?3 {#include
7 m5 E1 U8 a6 N: j1 Efloat x1,x2,disc,p,q;
$ v+ L5 V& @0 \2 H0 Z0 K4 jgreater_than_zero(a,b); {' ~# T& h) n; P4 y; q/ ^
float a,b;, N: a1 Q3 |1 ?' \7 j% a
{
: I) j: v* q8 y0 r, I4 C2 Z! hx1=(-b+sqrt(disc))/(2*a);
* l# b$ m) y }; r V0 kx2=(-b-sqrt(disc))/(2*a);
5 E. i3 ^! k! f$ A}
8 L) s e2 Z$ w7 i' tequal_to_zero(a,b)
8 }& y, c" a+ H7 R$ j. |$ Sfloat a,b;$ s$ n( g9 {/ ^! A$ W7 n
{x1=x2=(-b)/(2*a);}
8 m5 y# n% g+ F+ zsmaller_than_zero(a,b)
7 f! R/ U+ a( p. j1 Pfloat a,b;
( h, `5 C% y9 n/ E5 I{p=-b/(2*a);# H0 e7 E' |# S2 F2 e" D
q=sqrt(disc)/(2*a);
; F4 R* w' Z8 X" V1 }}
; x8 C) y2 M8 wmain()
/ Z& G9 s, M( t' n! k8 g{
8 x; t9 _9 s9 |: X8 x2 E: ~float a,b,c; e- a0 K" l$ c/ Z8 A
printf("\n输入方程的系数a,b,c:\n");5 z. J7 [) M3 _: |+ W- r
scanf("%f,%f,%f",&a,&b,&c);7 T# D7 g4 n1 n+ S
printf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);
$ l% }" s; G1 K% D7 adisc=b*b-4*a*c;
5 B+ g3 @! p: sprintf("方程的解是:\n");% G% h# Z6 i/ G0 f: e$ D+ K
if(disc>0)
- ~4 p" m! ]& V2 ^; R{great_than_zero(a,b);8 ? w5 l9 D2 y2 j) R; [
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
' Z) ?& M, Q9 C9 e% e} s9 c. ~% m9 G$ X) [& H
else if(disc==0)9 U2 Q$ d" G! i- v
{" u9 K% y7 ^8 Z; d ^
zero(a,b);, v* m2 u2 m2 Y" {% i' j
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);7 Z# h Q0 P8 g
}) S! p( Y. u5 d5 A f8 E
else; i4 `2 z1 Q( R; n
{& B9 l h' h' f" T
small_than_zero(a,b,c);
3 d* e Y. s2 J+ t2 r printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);
$ E5 @" J: ?& o d }# x0 ]( [. p1 M2 D
}
4 L6 s9 K" v" J& |! U% ^3 D8.3素数
& Y1 [. G4 e& j3 g#include"math.h") F- g) z6 w# r8 Z! I
main()
$ _6 \. v) R: \% ]. Z) N) b{int number;( e6 K' K" h$ ` ~
scanf("%d",&number);" @% u7 @) k& k3 h6 h
if(prime(number))& r9 E, J- X" l4 ^0 e
printf("yes");
4 W1 `' {0 C: l1 l* Q; |. O else
- ^0 s. o9 l0 u' P& V2 }! s* L printf("no");
N, P5 P8 h+ N. U}
1 X! R, J0 O7 p5 I' t sint prime(number)
1 y5 h2 I: f9 N" eint number;& ?7 L2 F, M( d9 Z- ^+ F
{int flag=1,n;
2 M& f) r' S, { for(n=2;n if(number%n==0)0 V1 c L$ K( f& K9 A/ i" _
flag=0;. W( P! n$ Z2 w8 B0 ]
return(flag);
/ c( z: ?1 a* u: {' {}
* E4 L! C8 {9 X, f; a6 I8 C+ b' I- F( Y# G
1 k5 }; q: B% ^" E( B E& Y# M% K. f2 U9 Z0 S$ Q. e1 [1 x
8.4
& Q4 o* O. g6 o+ U; S#define N 3
" X) N" D" G0 o l6 t W7 ^# Gint array[N][N];
$ Z ~8 c. j/ Gconvert(array)
& x# X y \! C( H$ l! Q9 Dint array[3][3];
1 M& K; q2 ` v0 G { int i,j,t;! k- P+ F1 L6 n+ Z9 ?
for(i=0;i for(j=i+1;j { t=array[i][j];
- S& j; e$ [3 U2 } array[i][j]=array[j][i];) O. w3 `9 Y5 r$ M; O' B3 F$ R; r
array[j][i]=t;+ ?! V/ ]. r$ ?2 ?
}7 b7 D" N2 v& }: _$ W0 x
}
" y/ x6 ^5 n2 E: o" n) Qmain()7 k8 W' G+ d9 a) i4 v0 w
{$ v! X$ I" v, r$ l
int i,j;
! ] v3 c2 n# c* W# R6 N printf("输入数组元素:\n");( M4 s2 f8 z3 _1 }) |
for(i=0;i for(j=0;j scanf("%d",&array[i][j];
# {5 \) ~. N; f; y2 l, ^ printf("\n数组是:\n");9 q5 k8 Q6 a9 W/ m
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);. {) l+ d1 D/ _, f
printf("\n");
1 l. N- R3 _1 K9 p }
& g9 V) D9 o& V. O convert(array);
# ~: r; {3 @9 T3 p& I printf("转置数组是:\n");
$ }0 Z, j) b' a" |, ^3 P for(i=0;i { for(j=0;j printf("%5d",array[i][j]);, u4 B; `# }8 U3 R
printf("\n");
9 Y9 a) A! g/ ~- Y; U( H }
@6 I. ]9 m( H}5 n( W0 X' S5 T7 V* ^8 r
' c) \3 B3 e0 C3 d3 n" d+ e& |4 m: C8 L; x: [1 D; y' f
9 ]+ U. ~0 r" h o9 _8.5' ^# X8 e' m' X( ]9 Y
main()
{# c. }; |1 ]* w5 P& H{
# q0 U4 V* J3 T4 D- uchar str[100];; j4 D& O7 n! D$ Y/ Y8 M6 \4 w, q
printf("输入字符串:\n");
0 @: ?: W, u+ K, R: G scanf("%s",str);: L/ f. {+ C% Y$ h* N/ k( G
inverse(str);
' e: I: Q% b# i+ U printf("转换后的字符串是: %s\n",str);
, P7 U2 h) D2 q' l# ?/ n. T7 A}' R6 j7 Z d) T5 V4 j
inverse(str)
& w+ [' m: a# @; n; p1 Xchar str[];
9 X4 M/ V# K8 |! C* s{8 ~* x, H4 g) i2 C, Z* t
char t;. f# P" y& ]! u i _" u+ Q
int i,j;
8 F. } L/ h+ f, [" ?" n for(i=0,j=strlen(str);i {
1 F5 ]' J/ E6 t T2 Z t=str[i];
) m S' {( i: A y str[i]=str[i-1];5 m/ {- j& T- y6 ^" Q9 R1 L: I
str[i-1]=t;9 I) ^ N1 d5 h; j% P2 X" u
}! ]6 @9 h) O& k
}
/ s6 z( j5 y6 ~
$ g, U/ g1 A% d
3 |4 i" r$ W! ~5 K5 V/ m% W' s; D/ K: e* r, ]
8.6, \/ a( q% L; P5 d
char concatenate(string1,string2,string);$ t! t w0 Q( q; T$ n8 ]1 l
char string1[],string2[],string[];7 b% a& F( e d/ a( X, p; ^- w
{
, `- x; o! ~$ }5 A! Uint i,j;# ?7 X# ^: @0 N0 K; n# K0 i8 N
for(i=0;string1[i]!='\0';i++)* o3 z6 x; t+ i, J0 F1 w
string[i]=string1[i];
& K% W! I B# J& J; Lfor(j=0;string2[j]!='\0';j++) r" f' A& H" e- N+ q- i) Y o
string[i+j]=string2[j];
! L- o/ s' c/ _: L! S string[i+j]='\0';: s4 S$ Z3 L( Y2 @
}
( H9 f% j( K# c1 S/ Jmain()
; P. a: W$ Q, P" \* K- R6 [{
- G6 d9 @; U g- d W; N char s1[100],s2[100],s[100];+ p0 l4 T; F9 q) x5 k1 N7 O, }
printf("\n输入字符串1:\n");- R0 d3 g9 H" [$ B& U9 b8 u
scanf("%s",s1);
5 W1 H$ p; {% H1 j$ |+ m printf("输入字符串2:\n");' b6 f+ ~+ B9 Z! [
scanf("%s",s2);4 y6 O5 ~" @% }( g
concatenate(s1,s2,s);
# M" l3 ^/ Q2 a! @& k5 k1 _. Z0 h printf("连接后的字符串:%s\n",s);
/ j0 I2 K4 Q8 m}9 P- [9 O* p A
0 \- L4 d* C' U2 k O: C' k* B. l& l
8.82 O( R2 B2 X7 |; G$ M
main()
( [# e& O& h. Y, h$ ]( a/ v7 U{/ h; y/ c% ]7 y/ O+ [( c
char str[80];, y; C: i* Z# u; i) X
printf("请输入含有四个数字的字符串:\n");: |' a9 x% @6 [: R6 e
scanf("%s",str);
; {, t% ^- x" ], h5 x% B insert(str);
' W) h0 l: E! ?: t2 v1 x}; Q; G8 J" m* g- |6 e
insert(str), R8 I& Z0 N; l7 N% p$ P
char str[];
/ I; K& I. |$ d{1 m! M' Z4 Y& [2 G4 R
int i;. I( V! _1 `- _8 K' {
for(i=strlen(str);i>0;i--)
! [ A- R" [ u2 p3 O2 i8 M { str[2*i]=str[i];1 h, k" S) `6 ]% c/ `! u
str[2*i-1]=' ';
) L* l8 t* X2 m S( w a5 l* z' r }8 V1 ~8 A8 T$ @: l
printf("\n 结果是:\n %s",str);
* N: f; f- T- N7 [$ ^$ e }
4 s/ D2 t& k) v; |: n+ t
5 l4 D) [+ |7 @2 \
) i: b- e3 O7 v$ U. {" I
: C, L! W$ d( c3 m/ ~8.98 m% E {' h) x6 I. h g# C
#include"math.h"* H5 K5 ^. l8 ?( Q9 c
int alph,digit,space,others;3 g4 Q6 V; ]% Q; `
main()3 ~8 G. s( ~" P" k8 K" @
{char text[80];
2 c! V0 A$ J0 Y4 g' B* @ gets(text);# M, ]2 y, \$ B# T
alph=0,digit=0,space=0,others=0;& g( x! t: x& X! F1 d
count(text);! d: E+ _; |7 Y
printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);* \! F) b7 x2 B! r4 P' c) c2 `
}
- X9 s, T' f- D5 b1 zcount(str)
* ~$ x1 C4 Z0 E0 j4 o4 R' Z9 @1 Tchar str[];* D7 z- j' t# b& P* u8 l; v4 P
{int i;" a u$ W4 c$ l! |
for(i=0;str[i]!='\0';i++)5 j, ^. C* H" x" r$ A5 `
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))4 Q0 L/ M6 |2 D5 \+ P/ Y
alph++;
9 M3 ?4 i0 } k4 J* q else if(str[i]>='0'&&str[i]<='9')1 [4 c5 T: j8 S: [) O/ H9 W
digit++;
) e/ M" b9 y2 Z! I3 B3 \" w1 ^ else if(strcmp(str[i],' ')==0)
& A* @% o& E0 p8 l- z) @ space++;1 `# r* `; m0 X2 c& ?
else
6 }! W7 U3 z7 X- ?% \ others++;) }5 c3 Q7 J( a; i
}
2 @% R" Q0 _1 E5 ~) i, w
+ M" e3 C+ L0 Z$ i( f% U K8 x7 ]
3 A+ S" i5 q* b/ j6 m/ m+ G8 f+ F. Q8.10
' C' r% c7 q5 P) v# G/ iint alphabetic(c);
$ u7 E6 E: A* q' k5 q, M6 q2 kchar c;
; [- w. o0 X; m _% E{+ X) d- n# U2 b* T+ l) D2 w
if((c>='a' && c<='z'||(c>='A' && c<='Z'))9 `; i# F" `- b6 N& B
return(1);
7 i0 Y, T/ W: D ]% u! S else5 P( Z0 C# Z4 I* E
return(0);
g1 s6 x B4 Y% \# G) c}
4 H& o2 w) e$ F9 h
3 V# ]' L" v" ~9 u' Aint longest (string)
- k6 l" A3 ?* H: i* u6 `8 s$ Nchar string[];
2 I" G2 \ O1 _ ]$ @+ a{
1 j/ `8 u3 G* Z, E( @ int len=0,i,length=0,flag=1,place,point;5 H. z4 Z; {! d& {# l8 h
for(i=0;i<=strlen(string);i++)
1 v: J$ D5 h2 g* s& p- u6 Z& i if(alphabctic(string[i]))5 d i4 T9 E/ b9 |
if(flag)" R" V- k' X! ^% Z4 |
{
7 U8 J+ R* p. y: [ point=i;
7 l4 H0 L9 ^3 R9 X( H0 [9 K! V7 s! p flag=0;
' V) G a, A+ O8 ]( v }
3 H% e. m6 c- \% `( |: R else
$ U5 x. R" c# Z len++;5 g" T( Z7 |9 K" c6 S" f
else
/ s8 e# I8 Q5 R/ b4 q6 j; d4 {) ^ { flag=1;; v+ j% t6 l( x1 A' E$ ^0 i* L
if len>length)* E( @* W4 e: P: a7 O
{length=len;
$ D) l+ b0 T* j2 A4 Q place=point;7 [: J; Q# e# A: K0 P' q
len=0;& d4 X4 @! k8 M j! _
}
% j* O' p/ |# e% N" ~ }
0 u6 C) n$ p [* y% O8 v( N) {8 J return(place);- e( I+ W7 a5 w- N- Q3 @* r% ^
}
) s5 e, A5 F0 T7 mmain()
- c6 Y1 w1 Z5 I$ F, l3 z# L8 i7 ?& L{
8 B8 I& s' ^9 `7 z- n# H, Dint i;
& X! ~" X' I `+ E6 q. E, hchar line[100];
5 t1 {. V9 E! K2 C0 j. qprintf("输入一行文本\n");) E# s$ v/ s$ u3 W; ^) u* g
gets(line);
f; j# P P- p" V5 M- Iprintf("\n最长的单词是:");8 q. I" z0 P/ g' v1 G# z1 C1 l
for(i=longest(line);alphabctic(line[i]);i++)
7 p) t4 S( i8 _5 l printf("%c",line[i];
) F: Z' v' P! p7 B# f7 ~! Rprintf("\n");2 c' u y, [7 N5 ?6 ] Z7 V
}' d( M% M, N) E$ a" c
3 F" S$ P& |% j
" s6 l/ H1 ^' B# E
3 i. G3 I+ L9 J+ V8.114 f q( y" x& A% u
#include+ ^" w* T7 Z6 _; S' r2 t* f6 S$ `
. V& ?/ a" X$ i1 ^+ s7 Z L: C
#define N 10
& p: b( ?8 I$ B) w- {7 c8 ^char str[N];
* g* i* [6 N% O& Vmain()' G4 C3 O3 g1 `. K1 a3 q
{
/ |; A0 p; S+ v% v6 z* Xint i,flag;
1 W( P4 u. S0 Q9 y+ Jfor(flag=1;flag==1;)
. Q, _; d/ I% @# b1 K' ?{( V! I% ~5 H4 W0 X" M: J" s
printf("\n输入字符串,长度为10:\n");
: Z5 R3 N1 {. T& |+ Z scanf("%s",&str);9 {. @+ R5 ~* N" Y
if(strlen(str)>N)# s& x9 W7 m; X7 h4 i4 B$ z
printf("超过长度,请重输!");
! K; }# V$ h: G3 o: b9 r0 b/ M else/ V2 R B0 p/ e2 [3 }: a
flag=0;+ a, e2 c f6 s$ ?. E# B8 a
}& `: `3 A" Z- s9 S0 Q
sort(str);, l( {0 v0 ^- H2 f S. D
printf("\n 排序结果:");( i9 d e5 m$ ?5 L" g3 Q
for(i=0;i printf("%c",str[i]);) @4 Z ^" t q! `2 W
}
2 j6 G% f' k9 Hsort(str)( v4 Y' L" D Z: i" E( n
char str[N];4 Y* ^% s6 F+ V; G
{( P1 A: Y! J$ H$ M" I. x
int i,j;
+ Q) H3 r4 Z8 b+ {char t;4 F l) R( q( C6 B3 }( [
for(j=1;j for(i=0;(i if(str[i]>str[i+1])% c/ G7 z: ~6 c
{ t=str[i];9 l f# T$ q; O
str[i]=str[i+1];/ }2 S3 E7 Y- V
str[i+1]=t;
* K4 }. ?- e/ B5 ^4 W: z0 r }9 H8 ^3 }! Y3 a) {0 _' {9 f! D
}9 v: Q% ]3 u$ v2 h7 G
8.12% y* ]$ v$ o" R- w& u
#include
7 O3 N- |2 n# r#include
! m H+ x! I. o" i! f" Afloat solut(a,b,c,d)/ O) e3 u8 z* Y1 f4 v$ H
float a,b,c,d;) s. |. m! B' p9 i
{float x=1,x0,f,f1;
5 @+ z. w' D2 S( V6 o5 r do
$ i% f& U% W3 m' a {x0=x;7 P2 G3 D; p* A, @( E6 n
f=((a*x0+b)*x0+c)*x0+d;) L* m+ R- ^2 v' s" b: N0 H2 y8 v
f1=(3*a*x0+2*b)*x0+c;
/ } P+ O2 q3 i& G( j x=x0-f/f1;
3 n9 x' {. _. b9 Q" Y4 s }3 V. k @1 `: z0 r2 d# s
while(fabs(x-x0)>=1e-5);( ]1 n0 Q5 @7 j& _& S
return(x);) p" D3 ^2 Z' O9 a6 c& M
}0 `' t4 {0 P, J3 I) B
main()
- y( r g3 P. ?9 Q; l{float a,b,c,d;: z) o) a+ A; r+ g$ z9 Q$ [8 I
scanf("%f,%f,%f,%f",&a,&b,&c,&d);9 M& |0 k. [' N2 u C2 v" A: V
printf("x=%10.7f\n",solut(a,b,c,d));. F8 Y0 H$ [1 z' M
}
1 L( u2 m. G& Z% M$ @1 }8.13
/ ]! z4 ^, S9 O2 a2 n; h: k#include3 c6 B# M6 l. o& \, Y! l
main()
5 E/ Y- R; g$ U; M9 B& K{int x,n;
6 u6 V, b" [; h+ U' s8 D float p();
3 H" s9 T n8 _5 T4 L scanf("%d,%d",&n,&x);
! L. D. c. K# k/ k- a/ a8 d printf("P%d(%d)=%10.2f\n",n,x,p(n,x));
( w8 ~7 p! d% @! u2 x( A}
0 `9 G. c! R6 O3 Mfloat p(tn,tx)
4 Y: v. A0 o r! a7 H- zint tn,tx;
( X* s) G" B- z) ~{if(tn==0)' X, V% q& O, \; ]
return(1);% u1 @# t9 x. W) g: ?- E! V6 K+ l
else if(tn==1)
$ R: x8 w1 l/ H5 _ return(tx);
5 ]; }7 K P, |/ \( W else
" c& Y( Y& L+ e7 O6 @: ? return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);. h; m/ w# `) C& ^& m3 o: q
}; {1 q ?, U* M! F2 G- |
8.145 @# q3 ^! s, Q o7 s: I+ Z& d
#include "stdio.h"
" ]# H6 N' ]0 y% c) b#define N 10& S: C5 e4 S! u5 u8 a3 C3 h. |" S
#define M 5
b/ Q( X% q# [1 y* S$ F# Cfloat score[N][M];
. i+ G; B9 I' y+ }6 ffloat a_stu[N],a_cor[M];
. a; D: K% m+ d( ^8 zmain()$ |- M9 t5 h8 S, g
{int i,j,r,c;
f/ \9 F0 x& D& L/ Q* q8 v float h;1 [# K' [! _4 |9 K- e" H+ V: ]/ s
float s_diff();
2 C- }) F; D; x4 L$ A float highest();* w( ]2 ?& r: ?
r=0;* ]* k1 l9 N2 y
c=1;
8 B# I5 h( ^9 X1 _* o$ s input_stu();, t3 z- ]) z8 f2 K$ F; U! d
avr_stu();: t& B& n" P2 O4 {: F5 O
avr_cor();
9 [# k3 _! N6 g: U9 Y8 V% N$ T printf("\n number class 1 2 3 4 5 avr");
8 ~/ ~: {& x. \, ]$ n$ y( w& { for(i=0;i {printf("\nNO%2d",i+1);- N" X! ?$ |# P; H$ y
for(j=0;j printf("%8.2f",score[i][j]);. m3 W0 W, l# j& l9 e2 `& s' @
printf("%8.2f",a_stu[i]);
, c) \5 l4 G6 b+ D8 l9 a }: P4 v5 H" ?# K% y
printf("\nclassavr");
/ [7 @ F$ Y9 t; V; m for(j=0;j printf("%8.2f",a_cor[j]);8 \7 t! I5 \/ ]
h=highest(&r,&c);
* p2 D. i q1 t4 d& v' Q printf("\n\n%8.2f %d %d\n",h,r,c);/ M" G* D, q# W
printf("\n %8.2f\n",s_diff());6 ]8 s# U, R1 c2 R4 P. T3 d6 c
}4 E( h7 w- g. ]1 y- \/ N
input_stu() B K6 S5 r+ c) ~/ f
{int i,j;
1 R8 h9 B4 h& A: i$ t float x;
" Y" x: H( i4 o3 @0 i) K7 ]# e for(i=0;i {for(j=0;j {scanf("%f",&x);
4 U4 g- l8 U0 x9 `- f' M score[i][j]=x;
* _" d, x9 n9 ^" c- {9 o }# n. q; W* \8 Y, l2 p) u# b/ B
}" R3 |2 D' \ W, s$ ~
}8 `( ~- x, L9 A3 _9 G: P
avr_stu()" J( x( N: C+ Z: K
{int i,j;
" b: S. L8 [' e: \7 \' n float s;
' r; k" Q" d9 I U for(i=0;i {for(j=0,s=0;j s+=score[i][j];
5 S5 g2 M. c; ?7 x/ @* L( u4 n a_stu[i]=s/5.0;
% m9 o! X4 a8 |) P I3 [; ]. d Y }
, i3 ^- a; b& l; Q$ ~}9 A. \8 j& v+ B+ |9 O2 b* b$ u
avr_cor()
- I Y1 c7 K* c) b{int i,j;
3 b- a; C! g% ?% r" T float s;
) n- e$ W5 B6 |& s for(j=0;j {for(i=0,s=0;i s+=score[i][j];4 i1 S% K0 k9 w2 r) P2 ? D+ n) W
a_cor[j]=s/(float)N;" E( D9 A+ g9 s. q
}
* m3 n. j- p7 h8 ~# z9 @}
q8 ~3 Z8 `7 W0 \' L7 vfloat highest(r,c)
" l4 p) N& Y/ T1 u7 }int *r,*c;1 I$ M, H; F8 X' Y, C
{float high;
2 w" L2 o2 V4 s. Q int i,j;
1 a( E; A* f5 I4 i$ ~6 J high=score[0][0];
. M9 R8 y* t$ Y7 L, O for(i=0;i for(j=0;j if(score[i][j]>high)
: ~* A0 O5 o% N( N; i {high=score[i][j];
* n" t& B: b6 R8 R7 { *r=i+1;- S6 Q6 u5 ~! B' E; d
*c=j+1;6 E/ B/ r2 |6 u; s7 [
}
+ f; G0 F- X2 K, K: J6 X return(high);
, R! u0 G! T6 r7 B) R2 G}
) S. D0 G6 l2 N6 v) _/ ofloat s_diff()
j' n2 r% V% I6 }{int i,j;- t0 {3 F% _# W) y5 S: j1 a
float sumx=0.0,sumxn=0.0;
4 D4 J) w* E# }* B0 J2 ? for(i=0;i {sumx+=a_stu[i]*a_stu[i];
9 W1 D8 h7 _8 B3 i sumxn+=a_stu[i];
0 c/ i! P+ X% ]( p }
% B8 p. |6 R% |6 b' T return(sumx/N-(sumxn/N)*(sumxn/N));' ]- K j2 H4 }7 I' ^
}" W3 G4 N) A5 s7 P. s
8.15
0 B3 _3 p' N F#include
! h" ]! q; z: d3 i#define N 10
* l% ~; @) n6 r6 E2 K" Nvoid input_e(num,name)5 C- E0 N( t! r1 l
int num[];
6 Q8 R& J; K' tchar name[N][8];
/ ~& L- _' [; A4 s{int i;; b& N: i) d( }
for(i=0;i {scanf("%d",&num[i]);
: \& a G* H9 a gets(name[i]);
1 r( @3 y" K% L8 U, { }
. S; b% P5 b+ x2 k5 M. O4 M}5 s) x) n7 V, [$ l, L% \4 K
void sort(num,name)
1 H% _5 g3 U0 g; eint num[];
* Q: q8 \ h- E0 S8 V* Y$ b' Mchar name[N][8];# D; p ^3 [/ g0 a$ r" i
{int i,j,min,temp1;( O+ L1 ~* I$ s) k, C
char temp2[8];/ x$ f+ i9 o& k8 b
for(i=0;i {min=i;
4 N0 j: E# b1 W7 e" T for(j=i;j if(num[min]>num[j])min=j;
/ q, P) d& v7 x* h7 V( t temp1=num[i];# ?+ h8 ?& s4 {
num[i]=num[min];
7 t; f5 V: n) y' | num[min]=temp1;0 O* @4 s; K/ S
strcpy(temp2,name[i]);" I7 x1 {( Q7 ]2 a9 l( l
strcpy(name[i],name[min]);
8 t7 ^) e2 m1 O8 m strcpy(name[min],temp2);
1 ?2 D; R D" R3 E0 _ }
3 b) y9 {' i6 @8 v3 ]+ d2 q for(i=0;i printf("\n%5d%10s",num[i],name[i]);
6 P2 {- }( ~5 R4 ] I3 y' d}/ Y. a. J8 }' W" l
void search(n,num,name)$ I, ]6 `/ Q. a, E% D" y5 l5 T& j, k
int n,num[];& H% s& q3 A/ a% h+ t% Y
char name[N][8];+ r' H3 ]0 t% Q! @
{int top,bott,min,loca;% V5 O2 P, q, i' L
loca=0;
- I, m: a, X1 c) |# n% X" h) z- k* {$ W top=0;
4 t) e# f _5 F bott=N-1;
. B4 r$ S9 c8 \% c1 a3 X if((nnum[N-1]))& Q/ |- H- u. }1 j6 ^
loca=-1;
4 Y8 { l. L! Q' p3 [/ B6 P while((loca==0)&&(top<=bott))# q) `0 e0 |3 U8 n* G, R$ B4 S
{min=(bott+top)/2;
. {( X/ |% |0 B" ~$ H. D9 F if(n==num[min])
$ @( T4 s% Z2 T {loca=min;
/ r" K; L4 c, ~* ]9 U printf("number=%d,name=%s\n",n,name[loca]);
% T, M" H( U" z3 ] }
$ G+ J! p5 w; K4 E% t* b: P else if(n bott=min-1;
; C! |( L, Z9 t0 x/ h else# j1 ^% l& l2 g3 V
top=min+1;8 l) D" _9 k: c. X
}; m9 N) f; [) F! h+ T& d) x/ E. g
if(loca==0||loca==-1)
' c: e. C& a( v printf("number=%d is not in table\n",n);) |2 I7 g$ ]* o) W7 e
}# m) B0 H. Y& C
main()
, t/ P; B" t* j/ L# r' u5 i{int num[N],number,flag,c,n;2 G4 ~4 X, H/ ^3 ^$ m2 q$ @
char name[N][8];
5 R' G; y) z; D B: T. P: s input_e(num,name);
& R f, P/ I5 [( u9 O sort(num,name);
, f' l- O7 B0 g. ~ for(flag=1;flag;)3 r4 x7 }2 l; h0 H. T* i8 T- b8 n
{scanf("%d",&number);
. r1 s2 n3 x0 F A! b& s; g7 ]8 ~ search(number,num,name);
2 ~& |- V7 R+ p. S printf("continue?Y/N!");
1 n6 \- s0 @+ m c=getchar();
3 s5 X8 Z8 A4 K& ^) A4 \/ u if(c=='N'||c=='n')6 |& l! |! Q0 n8 Q- R0 a- u
flag=0;
( P! P& I5 j8 i+ X" b }9 f5 D9 N- r* s7 i2 D
}
. ~+ l8 q0 i6 T; I3 |! I3 o( H
9 c# i# o3 a- c/ m$ t/ I; J8 Z4 l8.16; J/ Q: ^: B1 x0 {- f. V( C# n1 [
#include
* G& x7 c7 |" j/ _8 Q5 K! A8 j#define MAX 1000
9 W# q4 v) k% W( A; i/ @main()
: v( p1 N# y; |1 S{ int c,i,flag,flag1;
- u8 p. X1 P+ P9 T char t[MAX];% w7 m7 u. X% I7 |6 W
i=0;8 P4 t/ R$ p4 Q0 U) g0 s+ e
flag=0;8 I# b' ^7 n# a6 B, M
flag1=1;
* ^; p5 I" N% I9 K" X printf("\n输入十六进制数:");! w# w# G, `" N) o9 T$ U+ i# }5 [
while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
) P2 `, O5 W% G& a" E# r {flag=1;# d+ E' S$ r- I! p/ C0 B: p- I
t[i++]=c;
9 ?4 G6 M7 Z H V }
6 h' E& I) i9 g" N: Q else if(flag)
( d2 V) l* I& g" } {
9 T# n( G4 t% G. P6 W7 X t[i]='\0';, z% g" }0 G% y# ^& V5 a
printf("\n 十进制数%d\n",htoi(t));
, [9 t4 [, ?" {' D printf("继续吗?");6 v1 \1 n: M- t1 B
c=getchar();
6 L. p! [7 F1 u1 _ if(c=='N'||c=='n')
- v/ A" \2 k1 V- \ flag1=0;
9 x+ M5 |: U8 c- e" V7 I else# \8 G$ U# E* r% o
{flag=0;
2 K4 {: C+ ?' V* g i=0;, x/ A+ y: J2 @! n+ E0 G
printf("\n 输入十六进制数:");! p% d7 z2 o( \, J3 q
}% W5 \; d, [9 _+ Y5 g
}) H! d7 y& C3 V! ]$ X- Z9 s0 Q' P
}/ @; q6 B# }5 G: U# l
}$ t m$ I( l% j! ? E7 m
htoi(s)$ J3 T1 o1 H/ R1 o2 ]' L
char s[];1 Y% Q3 z( @8 O! W" b! ~9 c. \6 t
{ int i,n;! `1 @# R% W2 h8 B$ w9 |% c; K9 R
n=0;9 b: \+ S- b v2 Q+ F# h
for(i=0;s[i]!='\0';i++)
$ v" [1 n) h5 r# }9 x j {if(s[i]>='0'&&s[i]<='9')
/ M& }& M( x. h# Y. b n=n*16+s[i]-'0';
' m! O# c5 A- y' I; N if(s[i]>='a'&&s[i]<='f')
4 [3 i7 |# X3 A( F! A0 Q n=n*16+s[i]-'a'+10;8 f/ b* Z: U/ X' w
if(s[i]>='A'&&s[i]<='F')5 E: v2 E( T. i) Z6 E* T5 E1 N3 l$ V
n=n*16+s[i]-'A'+10;5 J1 n2 [3 @; ?% ^
}3 w' x5 B5 s+ w, n7 J0 v) e6 }
return(n);
3 O+ H9 _* d! a. \+ P} Y n! _, \, L& H3 |4 k- A" Y* v5 n
3 V ]. D8 }) C: d
$ [' L8 _0 C- L% q) M. F$ ]2 _5 Q, r/ E) c) x1 z# N- q. X
8.179 |. Y5 C: a1 ^, }0 R
#include$ v/ h- v1 H. x5 ]0 {) K4 w/ m
void counvert(n)
( \+ J- ?6 G M) j( e; k' c; Qint n;8 X8 n1 \# s2 H; s* G. ~
{ int i;; b3 A6 V5 r2 H
if((i=n/10)!=0); h/ c2 f$ m: K* ^: n" R
convert(i);" f/ n$ J; ~. o( N9 n% k
putchar(n%10+'0');, h; v' ^9 j5 X. j( Y/ G
}4 d* G) {4 I* E% M
main()6 _: }/ E3 |; o
{ int number;# q7 M8 x% h3 @8 G$ l
printf("\n 输入整数:");- E& q( t) N. m" u
scanf("%d",&number);- r$ D2 Z {& A [. N, ]' |
printf("\n 输出是: ");3 ] Q+ C$ J+ ]& ~) s# X' D
if(number<0)
+ p) q+ z$ z+ u" r7 y { putchar('-');
6 A P! c& T+ r! ^ number=-number;
& O3 }; K$ Z" B3 U. f }
5 b! v* c7 r$ rconvert(number);
) K$ _+ _. H$ r w% m, ?0 H}2 Z% i4 D/ i* j/ o1 w$ J
" v) n7 \: I7 h7 v2 Q
8 W: H3 q7 S1 W8 @( Z
( f2 j c- f( b, {$ o7 _6 M7 B8.181 h# [( }3 J" J# L4 W5 O& Y& S
main()
" v1 x9 p- q2 m; J8 B{
" S }; Y A( g( b int year,month,day;! ^/ o# W" b! d( V5 l
int days;0 O9 e% x7 n6 c4 L; \4 t4 }5 N; y
printf("\n 请输入日期(年,月,日)\n");4 \1 w7 z. V- V8 X5 g U D& z
scanf("%d,%d,%d",&year,&month,&day);& r) Y9 x9 ~9 z0 p: N; V9 M- R
printf("\n %d年%d月%d日",year,month,day);
8 C' i& t7 J) M. S2 B# @ days=sum_day(month,day);
H* ^, \! \1 M if(leap(year)&&month>=3)8 k4 i _2 r7 ?. s4 b9 F
days=days+1;
! M( @2 y6 p m( `" F2 Y( p* r" q printf("是该年的%d天.\n",days);
$ S7 F4 D5 m+ b$ L$ r" h }
2 _8 m7 D4 j' f static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
D+ @8 ]7 i2 ]: j3 J7 E int(sum_day(month,day)0 _3 [3 t3 P% O# H6 q5 v6 ^; @
int month,day;
) ~. z& L6 a8 C% C( k& j# o {
2 y: y8 O1 ?, g/ C int i;* i* B* j3 ~# G. l) V0 D- z
for(i=1;i day+=day_tab[i];4 y2 p. ~+ j$ i; A
return(day);
0 h$ R% N3 p2 ^% X, C. i+ U7 b2 y }9 e* I* v/ |& [' W$ i$ ^: S
int leap(year)
+ r7 F! m, V. |3 s: Z int year;
# F" Y" y6 n" y* B {
( M, g+ z3 @, m1 h; ? int leap;
0 K2 T; N" x/ b leap=year%4==0&&year%100!=0||year%400==0;9 b: n" S% T* w1 K3 o, F3 v& u
return(leap);
! R2 E$ l; ~) b$ j }
% [7 w1 K. L' K* M' q7 c第九章 编译预处理
9 ?6 T' B5 q: k9.16 o- e. v# N) b L* f
#define SWAP(a,b) t=b;b=a;a=t
0 J+ `. I$ Z c: d" x; R& s0 v5 mmain()
/ `3 V; g% F( E{$ ?5 M3 a3 {0 T9 p
int a,b,t;# R! I }1 I" s
printf("请输入两个整数 a,b:");
* T: l4 B" T' I; \scanf("%d,%d",&a,&b);# o; J4 H% c. @' d# \( X
SWAP(a,b);. f6 J3 B1 i7 Q' A
printf("交换结果为:a=%d,b=%d\n",a,b);2 L/ `6 V o9 L; w- ?
} ) D/ v) o& U# l: g/ o0 C
. |+ B$ P. y( t/ m
s" B' F9 r$ B
9.2
- i8 `8 ?2 q7 L/ W#define SURPLUS(a,b) ((a)%(b))
# q" }, R6 m5 a( N! V/ ~6 jmain()
2 p j6 A7 a/ c$ l& r {3 W% i& p- O) N9 r8 _
int a,b;
8 F z# `+ [, K. g9 r" I printf(" 请输入两个整数 a,b:");( n& I a- B* d3 B* e H
scanf("%d,%d",&a,&b);3 n4 C7 I, i& D3 J4 c1 W- W) Z% h
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));9 d( ]$ Z; Z# I6 u6 z
}% x2 Y" _- A" g {
$ m7 k% u& k7 C1 J2 o
' s8 X% e5 M) k8 b
9.3; v% ~1 R0 y4 s& R
#include; j# y* z% n9 D, h3 U/ W
#defin S(a,b,c) ((a+b+c)/2)6 j8 F) x n+ T2 \3 k
#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-
( i; M: ?3 x1 c# o* \/ Lc)))5 n. Y6 {$ H$ d, ~* J. T9 Z
main()0 e4 ~$ l$ y) Q
{
6 u" C! u, ?% K. H/ A float a,b,c;
) V8 w( ?" _2 l6 {7 d: y4 I1 E- T a printf("请输入三角形的三条边:");2 | {2 \" y$ M+ [- S7 B+ a
scanf("%f,%f,%f",&a,&b,&c);
/ P6 i- X3 K5 _: d( V if(a+b>c && a+c>b && b+c>a)
* O# p P) [: j2 } printf("其面积为:%8.2f.\n",AREA(a,b,c));1 W, X6 ^, y4 V Z8 A9 j+ S5 w* d) J5 B
else
! N/ ?% w8 e* [- V8 r& g printf("不能构成三角形!");
/ ^8 p- Q1 c5 ~1 } G }
4 q, h: Y& k# W+ e1 }0 f* M; j
/ n3 a8 P1 X0 b+ o6 m8 U5 c5 p& Z1 S4 {- q- g' y$ k
+ ?( C( D+ P9 U! z- f }9.4
8 C0 |5 _& d' h) n8 m7 `/ m, z#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)0 W/ R) a! t$ j. A* F9 A& r1 o6 l
main()
. O& |4 }3 f# i7 w/ N; s {! k) [2 {& j0 @
int year;
! I5 I6 F2 F4 w0 V3 u printf("\n请输入某一年:");
; _' h x1 e/ b: j- D, E7 B. H scanf("%d",&year);
' A! [% J2 ?1 q( m if(LEAP_YEAR(year))
, H1 T3 o3 h6 j1 R( r/ l printf("%d 是闰年.\n",year);4 B/ q' ^& R2 \2 ~: I
else" N7 z4 ?1 K3 v9 [% B: b: }
printf("%d 不是闰年.\n",year);
6 R' C/ l2 f8 B7 y }
' F- Q9 L9 F. |* p7 [6 W$ M! x- x: p' k! `' t- ~; d( ~
' ]5 G+ F, u9 J
0 p7 d) A' l4 Y! ?1 u9.5解:展开后:
4 L1 h% P) G/ [ [) iprintf("&#118alue=%format\t",x);9 L) V1 v2 r3 \- P+ v; C2 v* {
printf("&#118alue=%format\t",x);putchar('\n');
4 _$ t) ]. E) H/ T7 @! @printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');, x/ | O4 O) w2 \6 D3 ]# W+ S
输出结果:
8 ]. k/ N- L0 }, z. i; C&#118alue=5.000000ormat &#118alue=5.000000ormat8 e1 ^1 i k8 F. L4 `
&#118alue=3.000000ormat &#118alue=8.000000ormat
( Z9 K3 ` t! q/ I$ p) j7 _- c! Q
( h+ G$ k7 Q) X5 {. V& ^9 h0 T/ }8 ^0 V, F+ ^5 ?. s' @3 l$ X& S
9.8/ x+ M" I0 u% X, U1 C: X+ |
main()
" O7 B; d* |. B0 Z! o' z. g% p {
' h& J1 `- `3 Q% `& k int a,b,c;* g# j5 F" s' i, W( j. s4 L
printf("请输入三个整数:");
3 S' |% @& ~/ q7 ~9 ]. U scanf("%d,%d,%d",&a,&b,&c);
# t) E0 w3 z) W6 W! r t printf("三个之中最大值为:%d\n",max(a,b,c));
" Y2 |& B+ @1 o# { }
3 k6 }7 f+ @4 g3 b max(x,y,z)4 z1 d4 @' E+ O1 c% |- a
int x,y,z;
* k/ B% ~! R# [ C$ c/ [2 j {* I. i, W0 R3 C. X, w/ R4 p
int t;
* F9 c, I& t: Q4 }& I! n* Y# `- H. ^ t=(x>y? x:y);
6 i9 \4 p) W* }& T# I return(t>z? t:z);
, Z' p/ g, o0 l }
% s; h: k$ O2 Y
5 E: f( s# ]3 p: c; G5 S# \ h) c$ K) U9 p5 V6 t6 Z
) W) ~' Y8 v) ]" G7 ^ L
9.10
3 g6 d$ q7 t; D" Q7 {#include
\2 W) b4 g* i+ H) x H8 _#define MAX 80
3 O/ C: {: W" l+ N3 @: m#define CHANGE 1
$ r7 m; o* f# p* z" Y& l% D5 P2 z; [main()5 [( r; h p; S7 W1 g1 [' o% h9 y+ K. i
{
9 Z* j1 l. ~' ? char str[MAX];
8 m4 Q* D R1 u0 \2 b* A" r int i;
6 U# U/ \$ F" `9 y printf("请输入文本行:\n");- N; S F) X. w a; r$ ?% j5 z' [
scanf("%s",str); ^8 B6 v* \- K! B9 j" \
#if(CHANGE)5 F6 c! W2 w# w: l, X
{9 h* l6 I4 ]5 r% C$ T5 O
for (i=0;i { u. T# i. G+ M. T
if(str[i]!='\0'+ U! E0 E6 Z, E: t
if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
1 {9 w% T, i- |9 c* |( m str[i]+=1;
- F9 A5 t r8 l' T8 y else if(str[i]=='z' || str[i]=='Z'), z' H* T' j6 L, p) J" _ l5 o
str[i]-=25;) u2 U c( |$ ]$ H2 Y/ r
}- a q6 o. R8 Z( E* R
}7 e" M2 @2 ~3 z
#endif
5 c. B9 y6 I3 \printf("输出电码为:\n%s",str);) l/ q! G( v( J# i( f6 j
}
$ ^' m3 ^- I$ D, D3 }第十章 指针
& H( h6 I' A9 e4 S10.1
9 B/ G8 ~8 _% h& `* i; Tmain()% o: v! B3 J8 T
{int n1,n2,n3;
. p5 H6 \" `+ \2 \ [$ ^" G int *p1,*p2,*p3;( Z5 d0 g+ e+ X" Y$ q
scanf("%d,%d,%d",&n1,&n2,&n3);
4 [+ ?& N, E1 k [% P p1=&n1;
* g' X( O7 h% p) e p2=&n2;8 _" c) w2 a- {! g. B2 @
p3=&n3;" N# W \ b- R( \. `) ^3 V+ r
if(n1>n2)swap(p1,p2);
* E/ [8 E, l) j' W- V( j# v/ a/ W if(n1>n3)swap(p1,p3);( [4 c( R: W' S9 C+ v: L
if(n2>n3)swap(p2,p3);
: K) Y7 D! K3 o3 a- J/ f* Y printf("%d,%d,%d\n",n1,n2,n3);
' v8 P& f6 N F5 N; `: u}
+ O% W" E8 F. t7 ~swap(p1,p2)
+ k, ?: Z0 V: @) }* Aint *p1,*p2;
1 {" S8 q) H% r: b# P{int p;
' t' b, h* e+ Z' g2 J* l8 a p=*p1;*p1=*p2;*p2=p;8 j. s, y$ }; i) g4 i
}: H1 x( d4 H; p% q a; u! U( s0 l
10.2
. x* x2 U Y( |# N& D/ ~main()
- O: e2 h) n5 t{char *str1[20],*str2[20],*str3[20];
- M& m! V0 j6 V" }0 O/ ?% \) E) x char swap();# B D% M8 S# m5 ^
scanf("%s",str1);
! p9 z. T+ e' Q, g: A scanf("%s",str2);$ W1 d1 `6 n$ V; _
scanf("%s",str3);
# x( C/ X. t" Y) A3 O. j if(strcmp(str1,str2)>0)swap(str1,str2);
6 k& @3 Q# D4 j* o: b if(strcmp(str1,str3)>0)swap(str1,str3);/ j* H" c; ^9 e4 T$ @
if(strcmp(str2,str3)>0)swap(str2,str3);/ J- b" H, R) ~1 x/ R, j! N
printf("%s\n%s\n%s\n",str1,str2,str3);% D. s* V- n5 \$ H( ?
}
9 E: v& d6 p q9 v# ?% k9 y7 ychar swap(p1,p2)7 f3 W9 b) ^( y, k2 ^
char *p1,*p2;
, O Z4 L c- b+ e1 g1 A{char *p[20];! i, {4 _* Y8 n& b0 r$ J
strcpy(p,p1);/ t) V! d0 L2 E5 \8 {8 E
strcpy(p1,p2);9 _. V& r) [, `: b$ P
strcpy(p2,p);/ J( V. h4 v; \8 ?# ?2 x/ `) P
}( J( w! i, y% ]5 _% K- r
10.3; K3 C) l" b- @ B. {
main()
6 I) R0 o2 T# H. {{int number[10];
7 \6 s9 Q; v' u& ^6 v4 P input(number);
- w$ u% b+ ^% Y+ f% v/ ^( t max_min_&#118alue(number);
7 Q$ H4 \# n" z# ], n; x output(number);, W) a3 }1 m# y! M* v" w
}" ?" P$ u" K; n |: B
input(number)
* p* y' I0 ?+ n, u+ xint number[10];4 E9 g' K, o! H/ y
{int i;" m7 K2 }; i$ y. l7 {- r+ n
for(i=0;i<10;i++)
7 `5 j4 T6 s. I4 r# m scanf("%d",&number[i]);" ~: K% K0 f2 q( J* H! Z* i% D
}
# A j8 U( W$ hmax_min_&#118alue(number)$ V3 q6 V# M$ b6 ^5 J
int number[10];
" {" S4 x; g6 b2 o* \# X( t{int *max,*min;& q# G4 x, Q9 p; \3 ]: @3 d$ T* E
int *p,*end;& i8 @- U0 q* X8 Q& _6 Q
end=number+10;& G& k7 I u2 e: b% d" S. R
max=min=number;
0 z( k1 W/ y: ^ z# [' d0 i0 C for(p=number+1;p if(*p>*max)max=p;
- L6 I) a5 n$ O% g) c else if(*p<*min)min=p;
# d7 a: N6 L! D9 K *p=number[0];% @* O& h" y. Q6 w* L
number[0]=*min;
" w4 a+ @6 b \% u: c5 \) y *min=*p;
, v5 P6 B: f8 v" u *p=number[9];
0 L4 G1 F' B; X" c number[9]=*max;5 y' D% T: T) V1 o
*max=*p;* J6 ~) q3 j4 a, B# {+ R4 f
return; s0 B) S" Y3 W4 G4 H$ I
}" l: F% w" @) a. W
output(number)
6 @2 B; X6 }: |$ Cint number[10];
: y. k k. k6 a, F6 ?{int *p;
7 H5 `5 E" `" H9 h for(p=number;p printf("%d,",*p);! U& d0 T/ S i$ [0 X' s3 n
printf("%d\n",*p);
/ H; ~) V* ~7 w; y7 F, T5 w}
% }6 t7 @ \" j# b- \8 T4 g10.4; g3 S' u8 K9 n1 B1 ~ h% p
main()5 u0 c- X1 N" h6 q
{int number[20],n,m,i;
5 q9 { d1 [* n5 d0 U# \* v scanf("%d",&n);
3 }9 f! [( q7 _- J scanf("%d",&m);
A3 J5 q1 P1 ~4 ]& Q' I for(i=0;i scanf("%d",&number[i]);
, @, N5 {: y; X O. o# b0 P# ^ move(number,n,m);7 s( ?) }; C8 X# X
for(i=0;i printf("%8d",number[i]);7 {$ q8 R: [2 r9 J3 ~) P
}8 \# g5 Z/ O1 W: ^- ~8 N
move(array,n,m). X! t' e z) o! |8 r) g- |
int array[20],n,m;
0 m& z% _4 w5 ?{int *p,end;( b2 ~8 s9 R1 e1 u) k4 y+ T
end=*(array+n-1);2 U* a$ N4 ~$ L0 ?" Y
for(p=array+n-1;p>array;p--)# A- |/ B! `' X
*p=*(p-1);
) [1 }: K s" c: P' \ *array=end;7 F+ W8 R2 k1 j3 f+ Q
m--;
" ?3 B% G, T/ k V2 m* P' s* R( U if(m>0)move(array,n,m);9 y) W1 J; h; `0 w1 y: c
}- j9 y$ Y4 Z" l3 z1 \8 c) D
10.5% d0 |0 r9 K0 B9 G
#define nmax 50: u' U/ G3 G/ D4 e# L3 e
main()
0 C$ h' ]( y7 Q- {3 K2 i; F4 Q) M7 ~{int i,k,m,n,num[nmax],*p;
. w/ V9 P1 ?+ p% D6 E" @' | scanf("%d",&n);
7 f. c5 _8 Z/ o8 K1 k p=num;9 B7 c* I* v4 m s. y3 Y
for(i=0;i *(p+i)=i+1;6 f- x3 W* a5 D6 C6 ~* \3 [- H
i=k=m=0;: P: W. A# a' D/ H! q6 i) r
while(m {if(*(p+i)!=0)k++;9 [: h% Z0 |1 h @2 v$ W2 f1 h L
if(k==3)
7 o% o3 U1 |; Y5 j l* S {*(p+i)=0; L/ V. E" H# j, G: }3 @* |6 v1 ]
k=0;3 a; l* ~5 O& [7 |4 R' i
m++;7 `) x& `" p; | I C
}
# E5 u% W6 h, X O0 b* Y8 Y- z; L i++;
$ u% u, A0 b5 Y1 L if(i==n)i=0;2 x8 K, e0 ?0 q# c
}3 ~. K, |2 {, ~; Z, K3 \
while(*p==0)p++;9 e4 I. T7 G2 ~& _! T; V" _
printf("%d",*p);
) p7 h# ^: d( F( ?% d O* i}
9 A( I H' X' @9 ? N4 t! C10.67 a9 q3 d0 c; Q6 P+ G, O4 A8 z$ p
main()
l m& Q; f( m0 x3 N+ d9 h{int len;, ?. J; F5 j! h4 L, G
char *str[20];
/ G+ o: z7 ?! P; m/ f scanf("%s",str);' Q) G+ G6 t3 l8 B, Y+ |. E/ s; t
len=length(str);+ g) s$ s6 e1 W! N; |1 D" S5 b
printf("\nlen=%d\n",len);
& H3 {- p# m I9 `& G}
. @0 Q7 e+ p6 W4 `3 J! K8 y% L) H Flength(p)9 r) t- j4 v' @
char *p;
1 Q& d6 {2 C( c7 |5 Y q: D{int n=0;
" P+ z: b# L" P" `- V while(*p!='\0')
9 C9 x6 @0 F' R- }: Y% g a {n++;p++;}
$ C+ c, N0 C) O6 T* I' c7 s2 X return(n);5 [, d$ g9 \/ Z: Q( e. V. c, [. r, P, J
}/ Y% i1 Q# ^, k" ^4 d5 G R
10.7
' K+ _! P" ~4 _main()
^: a: D; i6 z- v{int m;
; h, s l3 @9 K char *str1[20],*str2[20];
1 f. K6 j% G3 B5 N: n( j: o1 B scanf("%s",str1);
9 P [( o+ _: k$ f0 n. l2 h scanf("%d",&m);
; v+ S* I( A" {& O( {1 P if(strlen(str1) printf("error");. j1 j ]; x! |9 B- j
else
o8 T+ B! Q; N3 S6 d {copystr(str1,str2,m);
$ q0 G; V5 p" V' b8 d0 M printf("%s",str2);3 I1 ?" i. F9 [* a# N6 `1 p) I
}
! ], u* Z/ b6 c}6 Q5 K" y- y$ }+ d, G* t; b
copystr(p1,p2,m)
. V) `9 ~$ D# ~2 Pchar *p1,*p2;
' N$ O4 w0 L# A: G' Aint m;
Q5 B7 p& B w6 b2 g4 h* i, p{int n=0;- h0 V/ ~$ H0 @9 T' v& B
while(n {n++;p1++;}1 g5 u( T: u& y5 C# h# t& N r
while(*p1!='\0')
T% O' w n; F6 x8 z: V/ N N {*p2=*p1;
. P9 w* y" k* l2 K- j) Q p1++;' \/ ^# z0 C8 k* B1 x$ O# O
p2++;" }) ] z) J" i& D
}
) ], Q* ~- r6 {& W *p2='\0';2 \; X) s! Z0 w4 g& Y. R
}
7 s( W' ^9 H1 b V10.8# `2 b- L. m0 N0 p0 x5 |9 Q
#include"stdio.h"
}- G; z5 y5 D% I5 i0 ~: N8 Dmain()
" i1 [6 l# r7 i8 K' J5 J( _ Q{int cle=0,sle=0,di=0,wsp=0,ot=0,i;( K5 I" A8 A: `' N) O t- I
char *p,s[20];
, Z8 T$ i! { ]$ n for(i=0;i<20;i++)s[i]=0;
1 y. t: f9 {( W& c, T+ Y0 [ i=0;
0 w4 Q1 o. g9 _ while((s[i]=getchar())!='\n')i++;
x7 D7 I) a% ~% x2 w. j$ h3 N p=s;
1 K: k. V z Z8 t; s while(*p!='\n'): A2 h$ Z" }6 a+ g' T4 ^
{if(*p>='a'&&*p<='z')
7 b1 r0 |# [; Y; W' Y ++sle;
5 j4 P8 T1 d% l7 } else if(*p>='A'&&*p<='Z')
! d- j1 l. a% W2 t ++cle;
4 h9 b. k8 ]$ r* f' ^ else if(*p==' ')
' |6 s6 L4 H0 u( J# i# B3 J8 E ++wsp;$ l) P% o, D4 Q* n. f( a
else if(*p>='0'&&*p<='9'): F1 x9 S$ x, \% Y
++di;
/ F& |+ i0 R" |. M5 }/ D else
! P: r( G& h$ a0 b* i1 O0 P ++ot;
; X. s1 u% d( H9 B/ m p++;
- ^) z5 o: y$ g( x" k* s; T }9 ^( [- Y, Z0 T
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
/ e2 ?4 Y9 |# W X6 l- s" v}
- d* b! o8 L6 b+ [- T10.9
' Y$ e3 V# l: Vmain()
5 f* V/ n5 Z& J* ~{int a[3][3],*p,i;( ^- B+ H7 X( w* v, Q/ R
for(i=0;i<3;i++)
- {$ w" o5 |2 |! w; m scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);
" B6 p* Q+ L* w- V( b5 L p=a;
4 k4 x8 v) M9 Q: C% b3 x move(p); T1 X u" U+ Q$ O* J
for(i=0;i<3;i++)1 Y$ \9 u/ i$ ^# t- |" t7 H
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);1 z- [8 r1 ~+ I) G; R9 `' u; b
}* V4 ^! m* k% f2 @6 g. |$ k8 [7 y% V
move(pointer)
$ m6 ?9 c$ h" qint *pointer;
9 L: u( X$ E5 R G! V{int i,j,t;
3 d: [, e0 D$ [, w+ w5 T for(i=0;i<2;i++)
/ ]* [$ d3 o0 P/ k% L for(j=i+1;j<3;j++). ^0 e$ U2 d, s3 r% O
{t=*(pointer+3*i+j);3 Y. H9 H3 W" K) X z2 t# }2 E
*(pointer+3*i+j)=*(pointer+3*j+i);
( K4 c1 R B) U" C z *(pointer+3*j+i)=t;, \3 V" h/ C. Y. E$ I( v8 S. r3 q
}3 X" N) c$ I9 X" U9 C
}1 E; g! l1 x+ o$ Z
10.10
" r2 r4 a6 R: d$ l" w a" Q8 i5 X; Smain()( q d+ N/ N- m- Y! S2 B
{int a[5][5],*p,i,j;
3 |6 ]/ p; N/ n. _/ t for(i=0;i<5;i++)
' }. D) J. T: } for(j=0;j<5;j++)
# [5 t) x" \) p2 Q2 k scanf("%d",&a[i][j]);; f! w, T$ R0 H: l! U
p=a;
( q5 s D" A8 j; l change(p);, x" _! s) y% q" }# B; E) A) `
for(i=0;i<5;i++)
( o* B9 H1 H e1 y+ L {printf("\n");
' c5 y& E/ i/ a+ Y8 ^& L/ Z for(j=0;j<5;j++)
`) g/ s8 m6 N printf("%8d",a[i][j]);
4 ^ I1 D( z' g$ d1 P9 P" }5 m1 a }
9 X# r E& }$ I9 A- A}0 ?& M7 ^ @0 i' U: j* E. p# E8 y
change(p)4 P' C- C4 ~+ U! o
int *p;
$ C3 }1 G, I, b) ^4 k/ q{int i,j,change;
: x# \/ P. [ Y* R( T int *pmax,*pmin; v! I8 ^' L% \' ^) G$ {
pmax=p;
+ c! ?" b# s: B1 ~ pmin=p;/ x/ e7 \/ y* `7 Q
for(i=0;i<5;i++) u& t- d8 k1 \6 y+ c
for(j=0;j<5;j++)
* L4 l H0 ?9 }9 D& a {if(*pmax<*(p+5*i+j))pmax=p+5*i+j;
5 ]5 Y: c7 O! ] if(*pmin>*(p+5*i+j))pmin=p+5*i+j;. k7 w5 z' V. B8 J! |
}
* G' J# e# ]9 N8 w9 v" t change=*(p+12);# Y8 n2 W S# e( f
*(p+12)=*pmax;
$ j2 p3 ] J4 e7 _1 d *pmax=change;
# `) Q+ X; x) p! k% J change=*p;1 U- B- P, q# Q( R7 V7 x H2 i
*p=*pmin;7 y7 g/ @, ^1 N6 j- M
*pmin=change;3 C8 i l+ ?- }9 j: }
pmin=p+1;9 e g3 J4 d+ _- G3 z' S
for(i=0;i<5;i++)6 u% K. H5 S: l4 V. @0 c
for(j=0;j<5;j++)( |, e( Y2 V/ X' I
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
2 s* W2 V% G U; ` change=*(p+4);- E0 Y3 R* [1 `! W6 Y4 F. s) N
*(p+4)=*pmin;
2 P0 G9 G: ]! H$ m *pmin=change;+ o! |- l* P% u; v; |# g" w
pmin=p+1;
. I0 [# N! A) K5 T# f: p for(i=0;i<5;i++)
; s V: E& l/ Z$ J for(j=0;j<5;j++)5 s2 g$ |% u1 n
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))
4 c- W1 d2 ^9 V pmin=p+5*i+j;: [( o% r% }- G" b& q8 U
change=*(p+20);
1 W) f7 d9 t9 x$ p, g1 P3 F# A *(p+20)=*pmin;
1 p! ], f/ `8 S *pmin=change;
1 J M- O- w9 Y2 c; L pmin=p+1;+ U7 K2 _- q1 r) ~
for(i=0;i<5;i++)
$ }# I3 I' r U' P0 Q for(j=0;j<5;j++)+ _2 J! U( Y6 X1 u! j3 U" j
if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20)); |/ G. \; e, f( ^+ B* @
&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
% [, g, B- K Z6 w8 S' x change=*(p+24);; i* ^ C$ t% w
*(p+24)=*pmin;' X% G1 a3 S' T9 B( B: \: f
*pmin=change;1 k* C( W: R4 a* A
}
, [' y( B" h T1 A$ F4 X10.11- W( D4 [, e" B% u1 N
main()4 g/ Y+ |5 R7 b, O+ e. C+ J
{int i;
8 i% t9 B; N l/ A char *p,str[10][10];
& i B% M5 @5 ] ?; E& ]9 O for(i=0;i<10;i++)+ [! m5 [( O! A! K* g2 N
scanf("%s",str[i]);
& @& N. f; M' f9 W- r( Y5 j p=str;# |! H/ I/ U+ q7 G$ a8 |+ v
sort(p);
1 i# O3 [* A" |0 n' N2 ] for(i=0;i<10;i++)
3 X4 n" l1 g* c$ `' o1 W: @! p" _ printf("%s\n",str[i]);( | K+ T* j" n# Z
}# K; c. @& x( Y0 i; U5 I
sort(p)3 d! V4 L% I, l3 Z
char *p;
& C+ b5 L( y B5 n- @3 C{int i,j;
4 ~( |5 F% J- X0 g; _6 B+ s char s[10],*smax,*smin;
* S) r6 N5 {* B9 y* P3 i1 l+ R for(i=0;i<10;i++)
& a* A% X3 I0 `1 k. B {smax=p+10*i;
, ]# P, N& @5 d8 r! k+ O7 i4 A for(j=i+1;j<10;j++)1 g3 ^1 X' M9 p8 o! y
{smin=p+10*j;
% m( v# h0 ?" n1 h if(strcmp(smax,smin)>0)7 b. P) E2 i* A) ` K N+ h
{strcpy(s,smin);
, X" }& s" A6 c1 [" w strcpy(smin,smax);6 _2 G3 ?1 |9 u) S9 Q
strcpy(smax,s);( z) } D1 M' e1 R. B2 u
}
! w1 j/ v4 _ Z. C E4 Q5 E' m }
3 K4 u2 `( `" f+ \" z) n }7 @7 @8 L: D( h$ D9 e' j0 U: R
}" w) x9 P4 Z$ D, E# n8 e4 v2 T3 ^
10.12+ Q2 e2 i6 i3 v2 N4 ]5 ]! [
#define MAX 20% S; t" V& L4 N6 c
main()
" W" R* g6 I; c5 }{int i;
/ _* x5 e* j d4 O2 f m7 b1 ? char *pstr[10],str[10][MAX];
+ i4 ~7 T" L2 I/ h( d) b5 S7 i for(i=0;i<10;i++)7 d% ~" t2 D) u ^7 `5 g8 {6 f6 E
pstr[i]=str[i];
# I9 [1 G a9 ]0 a- g for(i=0;i<10;i++)
- o# \+ H9 c6 V/ t) g5 F: k scanf("%s",pstr[i]);
: K9 o' ]! e1 O) Q8 k7 E sort(pstr);
2 j7 G- G. x) _% N) j. I- H3 z1 b for(i=0;i<10;i++)
8 Z0 m. m5 P' W- D0 ~* R6 R printf("%s\n",pstr[i]);
4 f' M0 g2 N* w9 i9 P; M2 u' V}- [6 U" h+ m; i
sort(pstr)" @# y6 G" j/ q4 V! g, e5 N! a
char *pstr[10];
+ {9 F/ |+ T" j1 ^" n/ }0 l2 s{int i,j;* |# u( I$ K7 S2 d
char *p;* [) X9 C; _1 f% U# O6 D
for(i=0;i<10;i++)
( B6 M6 f; t! g2 h {for(j=i+1;j<10;j++)
j4 p# J& ]1 g, f, Y2 w {if(strcmp(*(pstr+i),*(pstr+j))>0)" G; B$ |- A# t( q+ b( d
{p=*(pstr+i);
w3 x8 e5 }; }7 e0 v! u *(pstr+i)=*(pstr+j);
3 S3 S* O/ N% Y8 u5 Z *(pstr+j)=p;" w0 B* o) E4 {& P" P( h7 ~, u
}
: R$ w# Y0 K; g# b/ j }
9 T5 U' K: I7 Z0 @( L* x# ?, t }
% y' k( x: q# Q a+ @/ k& X; l} m! L+ E8 k. M
10.13
/ _! J* r a! K$ ^% d#include"math.h"/ d, h9 d& h. f
main()! H$ c: A R6 _7 K1 ~* x; I: @
{int n=20;
8 _9 v% b P j4 p) P+ a+ H float a,b,a1,b1,a2,b2,c,(*p)(),jiff();" H: R! b9 e& Y
scanf("%f,%f",&a,&b);+ U* p9 H1 S6 u, J( T# I
scanf("%f,%f",&a1,&b1);$ G0 J5 i) Y6 i4 c
scanf("%f,%f",&a2,&b2);! R8 j& ]: i* E1 V
p=sin;
# ]; _1 g* _- T p5 O1 |8 V c=jiff(a,b,n,p);8 h: j6 @! Z& k: j' K* s1 c* Q; }
printf("sin=%f\n",c);
4 ~8 M. e, V2 u4 D% X& n Z p=cos;
/ u6 k r$ l, M, _ c=jiff(a1,b1,n,p);3 f+ C/ e: n& _' I
printf("cos=%f\n",c);
$ s' A, G t/ o# B p=exp;
0 N; J0 K1 L% H- \' t0 E3 N6 k c=jiff(a2,b2,n,p);
/ I+ q% ]1 W: a1 v/ l printf("exp=%f\n",c);
. R1 \7 X0 x# D0 F}5 m# _( o: g D6 a4 o- Z
float jiff(a,b,n,p)
4 Q7 q- R: Q" W$ K1 R# h6 q/ \float a,b,(*p)();
* w4 t% I+ e; Z" g4 R9 J ]int n;- M0 T7 R7 Z5 k& e+ N
{int i; [# Q. m$ i! L g1 |5 X- N
float x,f,h,area;9 t6 [2 F+ f; e" T0 b
h=(b-a)/n;- N7 N/ t" ~1 Q
x=a;
7 h( C) g7 v1 z$ N area=0;+ P, a# Z- v0 M2 u9 R
for(i=1;i<=n;i++)
, g* x) Y- ?' }- q- g {x=x+h;" _# U; V& J( r, l
area=area+(*p)(x)*h;
1 |. F- U9 Y+ v5 z B- }9 {( l }
1 T6 h* }$ v; O; R. P+ `7 G' d return(area);
5 J/ A i2 y& j- C u% L# T' Z}7 S" e' h, ~. D" L5 _" q
10.147 {: v; x' n+ M( u( n
main()" l$ L7 |& v0 U: p3 @" ]* b
{int i,n,num[20];
0 {/ Y7 `3 Q G9 O) z1 U char *p;
. g8 l$ J& L3 F- L! T/ _ scanf("%d",&n);
m5 t3 F7 o9 I. L u M for(i=0;i scanf("%d",&num[i]);
; M! K" s# K0 D6 @ p=num;9 j$ l/ `/ s0 ^% `# X! j
sort(p,n);
+ Z1 G+ I' {0 w; ~ x3 U& I: t8 F7 c' h for(i=0;i printf("%8d",num[i]);
: R/ }* s/ Z6 w, n- c5 m3 f8 \0 r}
8 F, u- P9 g% I( Csort(p,m)! c# i" A# E( J* Y5 x. U
char *p;) `! _) }7 v, }7 K! t a
int m;
8 O& I8 b$ V6 y/ S{int i;( p+ ]" F% K, T8 [1 R
char change,*p1,*p2;
9 ^) Q" d2 y1 G9 q+ ?$ u for(i=0;i {p1=p+i;9 u' S) d2 { B( I/ h0 z- A% a# e
p2=p+(m-1-i);
2 W/ ^- N$ G0 I& D/ g, W change=*p1;5 U3 W" Y- q2 P8 |2 J' b* |+ }1 I
*p1=*p2;
1 x6 o' c+ Y1 r q& d3 S" g( U *p2=change;3 c7 n: V. ^: _! Y) d6 N# n& }
}
6 @, F7 T/ i" K) A}9 ?0 H8 d) J" x0 s2 J) j9 W
10.15
2 X6 x: O1 W% ~4 _: u cmain()
3 @2 T V" a# ^& P7 R{int i,j,*pnum,num[4];; V# J8 ^6 C8 {0 q( w
float score[4][5],aver[4],*psco,*pave;
! M! j' T4 Q7 `! i% l. N' h char course[5][10],*pcou;
1 f+ C) k0 [- { pcou=course[0];, Q' n$ [( X/ y( n# \4 p- A
for(i=0;i<5;i++)+ Y$ l1 }& e+ ~% ^
scanf("%s",pcou+10*i);
+ p+ X) T& N6 C+ |( P& B7 P- T printf("number");1 p& A0 p) |' o; p
for(i=0;i<5;i++)$ \7 O" g; z' I+ }
printf(",%s",pcou+10*i);% d& F/ q4 j: l8 U. [% ~( v
printf("\n");
6 Z1 B9 Y( O0 K7 l' j6 p: N1 u psco=score;+ l& @% Y' b2 k
pnum=num;
0 c/ h) S, f: X1 N0 B# d2 p3 G; i for(i=0;i<4;i++)' l' o% s. O, t' W, T; Z
{scanf("%d",pnum+i);8 A7 ^4 S* @ m8 [( N
for(j=0;j<5;j++)
1 ?9 o# H- F: R: k scanf(",%f",psco+5*i+j);
$ X M% G. t! ~4 b }) |* K0 X& @, ^' c x/ I+ o
pave=aver;
. K' x% H/ w! _2 N) x' j8 }+ A l printf("\n");
2 z& Z: A* y1 A, [' [9 O avsco(psco,pave);' X7 J, y9 h% r6 M
avcour1(pcou,psco);/ U' S0 Y) P' @" P) n- } e0 H
printf("\n");
$ ]+ x/ C, z, k- F! D fali2(pcou,pnum,psco,pave);4 g9 V9 w# I( _. c% I4 U
printf("\n");
# o1 ^$ b0 g+ `+ [0 O good(pcou,pnum,psco,pave);
3 M9 b. K3 b# {* V5 y3 o/ l}- ^' p4 u; @' F; Z' u) M
avsco(psco,pave)1 N" b5 a7 U. d5 H9 _. y& q
float *psco,*pave;
2 b% S7 ~' n7 j) C. g5 \7 M& e C{int i,j;
, H! p, i1 Z# f+ X* T# b/ n float sum,average;
9 ?" d( \2 B; @7 \/ j4 Q9 f' i) k for(i=0;i<4;i++)
. a9 I0 |. ^9 x m- H* ?1 q; j# p {sum=0;7 j0 I. \- o0 C+ o: C: b) t* P
for(j=0;j<5;j+)
4 G* _+ ~5 B9 Q! X H sum+=(*(psco+5*i+j));$ B& F+ q* C! u" U
average=sum/5;
$ n7 _2 ?2 f3 K) L8 g *(pave+i)=average;9 N4 b* E, C5 A8 P
}9 [8 [+ f' `* w. M! V4 m, }
}- G. w7 a7 Z0 v3 N9 j% \$ k
avcour1(pcou,psco)0 T4 V; n' O; F8 f- C
char *pcou;
* j4 w1 [! [; M' ~9 Z$ b7 j% Kfloat *psco;
' c% Q0 x$ I% j: K. j5 N{int i;3 g4 j- H4 e5 c0 Q
float sum,average1;
0 m/ _+ s; F4 o* F, b sum=0;
1 k9 e) |% S L) \ for(i=0;i<4;i++)' Y v- P5 `* l/ d7 n% Y5 r" e5 |3 n
sum+=(*(psco+5*i))5 r G7 o+ ~8 e2 S3 l- [
average1=sum/4;' [5 O7 V& d# S" H
printf("%s %5.2f\n",pcou,average1);; {8 D, K; s" R$ R0 _1 v
}, t/ i* _2 _ t5 f) n4 Q0 [
fali2(pcou,pnum,psco,pave)- |$ n4 b' q4 K; f) K* }
char *pcou;( u# G5 a: I: F
int *pnum;
* i7 a( b6 E8 nfloat *psco,*pave;
5 P$ B! S3 [3 w, H{int i,j,k,label;
) i( I- x1 T- o5 }$ h- r printf("\nnumber\n");3 \. d, a8 s* G& x4 @9 J% Q
for(i=0;i<5;i++)
* I- D f. I9 h# v4 Y# P" z printf("%-8s",pcou+10*i);
3 \7 N+ y" l! M5 N printf("\naverage\n");+ J- A% f& s% I+ \
for(i=0;i<4;i++)% w" l& j8 ^* V/ a( z# a
{label=0;
+ T, S: K: O1 X7 v1 X1 D8 k, P for(j=0;j<5;j++)
6 y) d7 D- ? V7 q1 } if(*(psco+5*i+j)<60.0)label++;
; P1 x7 ]0 _; d# A# e if(label>=2)
; M+ @( j6 U0 l2 \6 J: ` M" [ {printf("%-8d",*(pnum+i));/ ]8 ?4 T \* u5 V
for(k=0;k<5;k++)4 u1 }" c% h; J8 p- p
printf("%-8.2f",*(psco+5*i+k));' N9 b0 Z5 M3 j' N$ ~ i
printf("%-8.2f",*(pave+i));
9 P& p) a% P E/ G$ [, [ }
" Y c4 C8 h9 v4 f7 A% C! F }( ~7 S' ^- k- B% N. P
}& Y! a& \: K6 N; b, G
good(pcou,pnum,psco,pave)' z. N! p; Q: W, Y
char *pcou;5 n+ ^" Q4 |7 J. S2 j- Y" S- [
int *pnum;6 [6 A$ ~6 x" k6 w5 m
float *psco,*pave;% V# M* e( M9 e. I9 ^: T; |
{int i,j,k,label;! Y! ~" C7 ]1 w5 M$ Z$ f
printf("number");% P) a' |1 ] P4 K
for(i=0;i<5;i++)
( z$ C! J4 D4 r0 Z0 m7 h1 _- J8 ^ t printf("%-8s",pcou+10*i);
0 _. T9 a8 H/ k printf("average");/ R+ W/ K" X0 q! Z" s+ R2 o
for(i=0;i<4;i++)
- U1 H" k" k! U {label=0;
# y/ {. _2 D! v for(j=0;j<5;j++)
4 L1 h3 Z( h, X4 j6 v if(*(psco+5*i+j)>=85.0)label++;8 p/ b7 [3 }( B. C2 v- |1 G" i
if((label>=5)||(*(pave+i)>=90))# I( Q3 D% X( D7 Z* I6 O
{printf("%-8d",*(pnum+i));8 o5 {' { a; W) j& h
for(k=0;k<5;k++)
0 a7 b2 d* T; D7 e0 v. J printf("%-8.2f",*(psco+5*i+k));3 P3 L$ G! @5 ~ @; q
printf("%-8.2f",*(pave+i));: }6 I$ ^7 H7 F$ \) m
}2 d' q( [. `2 w0 l4 A: X
}# m( O* v8 L& h2 @# B9 @
}
" F* t$ u% D* z10.16
% `8 f# ]5 g# C( M#include"stdio.h"
7 S' Q% n* E* y0 J) P0 @ X1 imain()
# u. q+ u' x5 x, R# u{char str[50],*pstr;
( z- C. T$ g0 W2 C int i,j,k,m,e10,digit,ndigit,a[10],*pa;9 J5 Z. ?# y/ P0 U8 v+ V& G
gets(str);+ }6 a& _; E6 I- B/ h4 w/ {( A
pstr=str;
5 u4 i. |6 b1 B" B2 { pa=a;
( t: L+ G6 N2 s3 `5 b# U ndigit=0;
% P1 Y2 M$ c6 A# B4 y% W% l: I+ h i=j=0;
- f. ]; K" g4 x- v8 h' ~* z while(*(pstr+i)!='\0')
" V: b$ z/ [9 y$ j& i {if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))
6 |5 P( s4 {* m9 t, [$ M9 N: } j++;. Z( d/ O" Z2 z3 U# d* L5 D
else6 T' a8 j* I. e/ o7 o5 N! a! Z
{if(j>0)4 e! `. m5 h i
{digit=*(pstr+i-1)-48;2 l/ M2 H' m, m# e5 x: y
k=1; ]9 T9 _* K3 L: O6 `. v9 ]
while(k {e10=1;1 y' {1 c3 j x' S
for(m=1;m<=k;m++). F4 a0 ~ I0 J3 Z1 g9 d- z
e10=e10*10;
2 n7 \1 g6 I5 p1 W: Y digit+=(*(pstr+i-1-k)-48)*e10;9 O H4 ^& U3 K8 e
k++;
- s/ o! }: S7 {' {# [ }, \% n! e% x5 O
*pa=digit;
6 p) N! k& n- E2 d- b& b# U ndigit++;& x& v( o9 k% [# e7 W$ p% f
pa++;
( S3 ?6 }5 W4 s6 k' r. K# J& { j=0;" N# g; `" k9 Y5 j5 w1 z; w4 B( V
} B: C; W) M* T( |6 `4 x
}& r" d3 m/ |3 o$ ^
i++;
6 j L0 ?0 W# t3 w1 G/ I, p7 k }, Z/ Y; R7 O: f4 ~7 A: V
if(j>0)
$ X8 U! V/ v; B# a {digit=*(pstr+i-1)-48;0 R1 K' R7 q% E% o8 ?
k=1;- d* Z$ C, G) T" q
while(k {e10=1;( B% N/ U- R$ U; z
for(m=1;m<=k;m++)3 R% J& B7 `* P9 @
e10=e10*10;4 g0 P4 J/ k) e' V0 q7 A C" F- O
digit+=(*(pstr+i-1-k)-48)*e10;+ H4 h# f7 C0 U% j) Q. V1 t
k++;# \5 S/ {* I- N- [6 ?, j
}
3 X! ^/ z1 Q# j ` y* o/ ^! R: Z *pa=digit;
! S' j( H9 {6 z, x3 E. w z ndigit++;* o1 p2 n$ L6 S P+ K8 j3 @% m3 Z
j=0;
% ]# d% g1 x# V1 x. n$ } }
s4 u' z. l0 e7 r% h( i M printf("ndigit=%d\n",ndigit);
/ a0 T- r0 i3 Z2 B j=0;' u0 f/ Z5 f* {5 }5 t7 ?
pa=a;
/ |" D% |# l. y9 H+ D9 z3 J3 J for(j=0;j printf("%d",*(pa+j));
, d7 ^& G& R! N) E( E! ~}
+ T. ^$ w& K2 d0 u10.17( @6 }+ p' A% y/ k7 R3 g- A
main()
. }, [+ o; {( o: H{int m;+ W* |! V5 r7 ] |0 o" f5 s
char str1[20],str2[20],*p1,*p2;7 e% Q: _' z* M) Z3 w. p; O
scanf("%s",str1);) N- I; S5 w* G! M C& C
scanf("%s",str2);+ f2 b6 J' n7 f1 h5 r" V2 z/ `
p1=str1;/ c V# ^* j$ p f5 [7 d
p2=str2;2 G! m6 k6 j0 L
m=strcmp(p1,p2);
+ F1 h; b- F# k6 w" ]. W% i+ C2 x printf("%d\n",m);
, ?. C& |1 A" H% D% {" r! C8 e}
^9 I; q: V" b5 C2 r+ U7 L1 t! Sstrcmp(p1,p2)" b, ?: {0 o. s8 i7 U' x7 g+ d0 c& I
char *p1,*p2;
' R; e( O6 @5 a t' u# G3 S: W; ?{int i=0;+ {3 ?% P+ e+ D
while(*(p1+i)==*(p2+i))
# ]8 @ e F: p. x* o8 E if(*(p+i++)=='\0')return(0);- \( a- z" x9 t- k. R. ]
return(*(p1+i)-*(p2+i));+ \/ e! T$ C; q/ w" d! O
}
5 d9 Z, t7 Y0 w3 o10.18
0 a9 ~7 E3 N* P$ hmain()
/ d: V+ _; w) H" j& A{static char *mname[13]={"illeagl","January","February","March",& i& r* J: r0 e8 ^% t* T
"April","May","June","July","August","September","October",
y5 o9 d0 J0 } C0 R9 N. p "November","December"};
- y$ S. q1 p2 P2 P: Q int n;
* y5 J, r; Q0 C9 B scanf("%d",&n);
5 A* m( A, d) S# B; a7 {$ l* @ if((n>=1)&&(n<=12))* J( j5 t9 o+ K2 ?# \ X9 ~$ u
printf("%s\n",*(mname+n));
# z; _( G5 o# w; g else7 |: q/ ]5 O1 C2 c# p# j& s; w. u
printf("error");
, ]4 }6 N9 A8 }. h7 E}$ A; _+ k% h( I; \" y
10.20
- p& o" i; @/ r% Q2 k* ~main(). E3 l6 B; b* O5 V7 j: w( l/ C
{int i;0 A. ]8 U& n' w
char **p,*pstr[5],str[5][10]; X/ _% A' d7 d2 r
for(i=0;i<5;i++)) F! a v0 Q* f$ |) d G
pstr[i]=str[i];: X3 Y$ M' u. S- P0 H+ A
for(i=0;i<5;i++)
* u+ T7 p! X: i' o scanf("%s",pstr[i]);% d- @0 j1 \3 H! R8 c! B
p=pstr;. |. i& u1 {' f/ s0 y0 r
sort(p);) @( M3 e+ G+ h
for(i=0;i<5;i++)
5 u. n4 v& T! {: o# d* j3 t printf("%s\n",pstr[i]);' N# X3 _' V2 N" }$ L7 k' Q
}* B5 j4 N9 Y9 B! P7 F& V
sort(p)
4 O! E+ z$ [( d) a1 Fchar **P;, y3 x* z# A, h& D
{int i,j;
% N! k( d! z' E! T) ]8 t* U- y char *pchange;
- @) e7 a5 o7 { ^ for(i=0;i<5;i++)7 H" ]! p9 }6 e: @6 Q7 [: g
{for(j=i+1;j<5;j++)* P" ^4 W0 o/ |) K( ^- e" ]: [; g
{if(strcmp(*(p+i),*(p+j))>0)* u8 _! {7 \( q- ], V
{pchange=*(p+i);
# @% ?8 Z: x! I/ `, o2 _2 l *(p+i)=*(p+j);# k/ q3 j6 d2 v! n4 a
*(p+j)=pchange;
' |6 v$ I6 y7 x8 [7 W3 r }- t$ @ t- `, ]# e5 ]) ?
}. Y, d' p, y8 M8 r6 M' [1 l
}* z4 l0 j) `) h5 P
}2 o* m9 E( O: i& k$ o
10.21
* D$ d, N9 a4 A; x d1 G; E8 Emain()4 Z, v# L d5 Q- h8 F' |
{int i,n,digit[20],**p,*pstr[20];
|( l& ]3 k( ^/ B8 y) ]3 @/ Q scanf("%d",&n);9 m) l( v# g6 E3 V) E w; K
for(i=0;i pstr[i]=&digit[i];& P/ j; K, z5 C) c: j1 u
for(i=0;i scanf("%d",pstr[i]);- @' R) i$ w# Z& `' {1 I
p=pstr;& o' ^8 [& u( H; \6 w) _- y+ O
sort(p,n);
) @ {6 Y8 s- d for(i=0;i printf("%d ",*pstr[i]);1 H& t) h, Q0 m. d9 b+ Y
}
. W0 ~$ ?: D# i. o/ T/ O! J+ nsort(p,n)- P% ?! G" `$ H. |$ z }8 K
int **p,n;
" h9 C" A$ [1 h; N$ o3 f{int i,j,*pchange;2 @# z9 ?7 c7 C% Z1 t2 A8 a5 A. A
for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j)), R- n' @3 Q% l9 w( W
{pchange=*(p+i);" X& b5 N: n" o4 d
*(p+i)=*(p+j);
) V7 ?" m3 a" u4 t/ |* s9 X+ V *(p+j)=pchange;
% G2 q1 g$ C! {; V2 I" b }; X$ C( y' U' v
}
9 u5 J; r$ L: u) E1 `" E }
& w9 K6 v! ^3 [* Q( P}
: N9 B- E2 f! M X) x第十一章 结构体与共用体
* R1 a$ [9 ?8 w: v# w1 S11.1
% O J5 D7 l- h4 P+ ^1 t0 `struct
- M s# ?, X1 i {int year;2 E3 i$ @* ~/ @# V+ i9 V5 x
int month;
1 J5 E, O+ S. Z9 `8 s. j" t4 N' T int day;/ r e i' k% P) L: G
}date;6 F7 L: S# ^3 p+ X7 x/ t4 k
main()6 w& j- c+ I8 H* ]; B5 a6 c# ?
{int days;
4 Z. v8 P( ~7 p! g( o, T scanf("%d,%d,%d",&date.year,&date.month,&date.day);0 ?: e5 ]7 ^" l! h
switch(date.month)
& ^: w& x8 S" r+ c6 W/ {' m {case 1:days=date.day;break;/ C- _- w7 @$ I* {# `
case 2:days=date.day+31;break;
. |& y; V/ d# u& R! s1 p) [- U case 3:days=date.day+59;break;
- `& g W0 t( y case 4:days=date.day+90;break;' q; ], |7 v! z- C# o2 G
case 5:days=date.day+120;break;% S% j4 u; I" ^+ i
case 6:days=date.day+151;break;
# _1 R. T* }* Q, l4 R$ @2 G case 7:days=date.day+181;break;/ P8 m$ e( X* F( A
case 8:days=date.day+212;break;
8 x9 C# b9 i+ b( k4 b4 _4 b case 9:days=date.day+243;break;
( k8 q- ]# c% i& g* t case 10:days=date.day+273;break;# c* E# r2 _+ P: T) i
case 11:days=date.day+304;break;
( J0 ]& r# N+ P1 K, s: ^6 t O6 b& @1 ? case 12:days=date.day+334;break;
6 U8 H0 U6 M* O [0 [8 l0 Q }# w0 m4 f2 B) a& Z5 v
if((date.year%4==0&&date.year%100!=0||date.year%400==0)0 I' D/ H6 \( H b T* l# t& M0 x
&&date.month>=3)
6 j( e) p# w. d! L2 M4 X days+=1;
$ {5 Z6 O. Q3 f' c7 F7 M2 e; \ printf("days=%d\n",days);
' m9 K0 D7 V, ]. _1 K2 U}
( t' U4 p+ }; k11.2
- c" }9 D+ ~% f; D: tstruct dt
* {& V, t2 K- _& r {int year;( E' W! ?* X- c0 o" ^/ r0 H
int month;
; v0 X/ O% d3 w0 y& K# q& O/ I int day;7 ]6 B% y& ?# r7 D3 L6 B
}date;; A6 M5 d' W$ T- C
main()
1 `0 P3 t T1 d8 B3 a- y- L& F: n{
6 A5 Y9 \1 p0 u9 q4 d scanf("%d,%d,%d",&date.year,&date.month,&date.day);8 ~6 Z2 Q7 ]& ~" O& H- z, P5 v
printf("\n%d\n",days(date.year,date.month,date.day));
* Q+ g0 l7 }2 ~: ~* F}6 S9 o6 X* d# w9 [
days(year,month,day)9 ~4 M! e4 b l* p- ]
int year,month,day;
0 d8 f: |9 K( i6 M6 w5 }7 B [{int daysum=0,i;; n& l2 R/ Z+ b+ G: `( p- O
static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
: I t! l9 S5 \5 c/ b w5 N for(i=1;i daysum+=daytab[i];- ?/ L, e1 q0 {6 O% P: p
daysum+=day;
* X( n( M" P& D% P( K' O7 c if((year%4==0&&year%100!=0||year%400==0)&&month>=3)
! A& o5 r. c' k2 v daysum+=1;3 T G5 m8 @7 s0 }; f* a0 T
return(daysum);" n/ ]" A, h' W! ~ N1 p
}
8 R3 \2 ~4 {" s9 s8 p; s+ n* _11.3
" ^% R: E- I. B0 U4 N, d+ }11.4
m+ A/ y: | ~1 }7 G) O# N#define N 5+ |! d! [9 \$ z/ h$ M. z
struct student. Q/ ?4 E( X) r! J. {0 T- C
{char num[6];
$ s& H( Y* }6 g3 u$ a- g char name[8];/ p" P+ u, S: }; c' }: u9 S) K
int score[4];
) Y# j8 a9 N ]$ J/ i }stu[N];
* i S, O& ]1 |: t' Mmain()7 S B% b; h3 |0 \2 k& O9 O
{) L ]: Z/ B4 o, x4 U
input(stu);
( v' ?" h5 @6 x+ d8 X8 V print(stu);
7 ~3 A/ U V1 w7 `}
* V* ]7 R, ` i' J! e1 B; X& sinput(stu)
1 j3 K8 F- w' A# W( F4 |struct student stu[];
6 W6 G: |* e8 |{int i,j;5 V/ ^) ~9 l. \% U# N
for(i=0;i {printf("number");5 x/ a3 H) b. M" C) b
scanf("%s",stu[i].num);
4 @! \. o% R8 N$ i6 E printf("name");* x. P8 e$ v* k4 C* ~+ [
scanf("%s",stu[i].name);
4 R8 {4 j" r1 B x0 q2 W for(j=0;j<3;j++); }2 ~" w! K# }3 _- u
{printf("\nscore\n");
" K% X+ @9 b+ A& @0 ~ scanf("%d",&stu[i].score[j]);6 Q+ a$ H) z9 h/ R4 l% ~' a/ K
}
# L6 l) D5 g( D* q5 q printf("\n");
+ ]: P7 p1 W; N1 F8 p$ r }
( X* M. h# M. V) L/ n, V9 J# g}$ f# V0 E% x8 `, N* p" S {* [2 F
print(stu)
2 g% \& r! E7 S9 t- W+ V& astruct student stu[];
8 W/ Y( T0 N5 l4 H; v) J{int i,j;# {* j2 r U# [5 Z: A
printf("\nnumber name score1 score2 score3 \n");. K/ e$ q' K; G$ q4 o
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);
6 A9 E$ n4 u* D: r for(j=0;j<3;j++)
: t+ w% ^" \* Z* z* h% l2 I printf("%7d",stu[i].score[j]);
8 V1 u' M' j% m# x: f$ n printf("\n");
/ w( `, H& O% W" c }
1 _( H+ r8 @9 y! F6 s( A5 N}/ K6 U d5 k9 M/ u0 i
11.5 j1 c6 g$ m1 `7 A& @" _0 x
struct student1 m( ^/ i& H* Q' c' G
{char num[6];
! ?. u; T* E0 X! S( [7 } O1 } char name[8];
: W9 e: e6 [ `, t; u5 a+ I int score[4];# b$ _4 F; A. T
float avr;
2 H/ U! m$ B4 x% |9 t1 n }stu[5];# f, ~+ G6 v: c+ A+ D, [* ?& K
main()
6 w& X& R G3 |& v' E- u% U{int i,j,max,maxi,sum;* H% k+ v" q7 T( u( D
float average;) w# x5 F' B5 ~" j/ ^
for(i=0;i<5;i++)
) a. w$ Y' T0 c6 T8 r3 `9 K" K. q {printf("number");+ m B3 `$ r) ]/ S6 ~$ U, r
scanf("%s",stu[i].num);% p+ [4 B, Y4 k# s
printf("name");; I% p4 [6 \2 q* }. b
scanf("%s",stu[i].name);
8 o( Y& E' E) A0 g& ] for(j=0;j<3;j++)7 g! }( g9 ?" ^* c8 A
{printf("\nscore\n"); c/ P! X: j- F9 b
scanf("%d",&stu[i].score[j]);9 B* \3 _9 ~. |- H
}
' W0 `5 ~% @4 b. l }: }( t% ^1 P2 d* j/ p6 w( q5 y
average=0;
" {' ~' P, M$ W max=0;
2 M; P( N0 L) k2 g) U5 K maxi=0;
0 P, n2 @: o6 M, s for(i=0;i<5;i++)1 W0 {; [* g" {- l5 o/ f8 l
{sum=0;( H# O7 M- s' f
for(j=0;j<3;j++)$ J: ?& m: [( N; F2 d
sum+=stu[i].score[j];+ {5 @7 [0 ?. ?
stu[i].avr=sum/3.0;, v' s3 U7 c- x
average+=stu[i].avr;
9 ?; x3 r Z9 O0 V# m if(sum>max)4 Z6 [! E# \- W- ^. v& J
{max=sum;8 W: J! v7 X5 s' \" D+ Q
maxi=i;" P, c) @' o& p/ `
}' Y2 z! [9 |' O* N" z0 N5 `* _2 i
}5 k; z" h& H3 V- u* f; T1 E( Q; K
average/=5;- y# f4 M. D3 |; g4 W* e
printf("number name score1 score2 score3 average\n");& K- p6 [+ x1 Z* {6 d
for(i=0;i<5;i++): k4 T$ Z3 v9 p8 B$ g
{printf("%8s%10s",stu[i].num,stu[i].name);
6 P4 j- a4 A8 p4 S for(j=0;j<3;j++)
0 b8 N1 N) F& M) p$ l: y' l1 _0 l$ K printf("%7d",stu[i].score[j]);
N( A( ~; _: I1 B* l W printf("%6.2f\n",stu[i].avr);
3 T2 X- F# S, C# q }! k1 e3 o2 ^, U( r2 U0 L U5 t9 y
printf("average=%5.2f\n",average);
% N$ c' P. ?: O3 `7 T printf("The best student is %s,sum=%d\n",stu[maxi].name,max);
# x" T" }4 r! ]5 ?) A}
2 i8 k- f4 d7 s* v, t+ s+ g. Y8 S8 ^# l# r! G2 r& e
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