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升级   78% TA的每日心情 | 开心 2016-10-15 15:49 |
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签到天数: 13 天 [LV.3]偶尔看看II
- 自我介绍
- 本人较内向,但却有浓厚的趣味和好奇心.再之本人叫诚恳和朴实.缺点就是不多愿与他人交流.谢谢!
 群组: 江苏建模 群组: Coldplayers 群组: Matlab讨论组 群组: 南京邮电大学数模协会 群组: 西南大学建模组 |
C语言设计谭浩强第三版的课后习题答案
. M8 V5 x5 d6 U. S$ S1.5请参照本章例题,编写一个C程序,输出以下信息:8 h+ ]* j0 o$ t' q1 @" y
main()+ v: l9 }) Y$ J2 y
{
3 c: v) n4 O: H) s" ] M7 eprintf(" ************ \n");' b ^$ R# s; J1 q9 W! t+ U
printf("\n");
; C) E; V+ ^' [ nprintf(" Very Good! \n");' i, B( I' V( Y3 i6 L% V
printf("\n");* B/ R8 H: o; F. x
printf(" ************\n");+ m; x0 a% r7 }
}( f/ r* A3 I9 d1 n0 k* \
1.6编写一个程序,输入a b c三个值,输出其中最大者。
, M% u z2 o: U" Y解:main()$ J" y' J2 P, }2 E. z
{int a,b,c,max;9 [6 n/ K/ y- a
printf("请输入三个数a,b,c:\n");
5 y# a4 K( `9 m+ t n$ p: J% e8 M; dscanf("%d,%d,%d",&a,&b,&c);
: c. W2 _. x1 q" [2 g+ F! @6 qmax=a;% s1 p1 g5 z+ Z: C1 M4 c" M
if(maxmax=b;/ P* d1 y6 _. A. M& I
if(maxmax=c;- Z; [: R$ u2 Y: _* v4 e2 `
printf("最大数为:%d",max);6 {) ]) S4 ~: b* g M0 S$ N1 w
}
' E% T3 R2 v; h9 M. `) x: \第三章' w! E2 L# w+ k9 y0 z: ~5 G' g
3.3 请将下面各数用八进制数和十六进制数表示:
& y& {. B& S2 _# Y9 {(1)10 (2)32 (3)75 (4)-617) s6 d6 r7 a0 t* \; b
(5)-111 (6)2483 (7)-28654 (8)210038 w. _7 p! H0 ?& n) B. j
解:十 八 十六
( F9 K4 W) q. f (10)=(12)=(a)& v. w2 p( z/ X6 ]
(32)=(40)=20
2 }" X( ]# ], t" ]; j (75)=(113)=4b
$ L, I/ B% b9 E) O (-617)=(176627)=fd97. p. q0 P! C6 Q( t
-111=177621=ff91
" ^$ I/ U& X+ n& p 2483=4663=963; V9 F( p; T, ?" x' n
-28654=110022=9012
$ q' z6 a5 p) E) E 21003=51013=520b" ?8 A; n( @, k8 o1 W9 D& e
3.5字符常量与字符串常量有什么区别?! U$ L5 k+ A- z( q4 b! O, g, w
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9 V. J) a H* C
而成,用双引号把它们括起来,存储时自动在字符串最后加一个结束符号'\0'.7 o2 Q+ g' R* Y" M: C4 }
3.6写出以下程序的运行结果:# d) P- B' I2 G, ^; m& ^# J5 O
#include
' d2 d. ]6 I4 pvoid main()5 a' z$ i( M( d H. R* W
{
8 [: B i. u" ochar c1='a',c2='b',c3='c',c4='\101',c5='\116';2 |/ U; \! X4 V) ~ N( |
printf("a%c b%c\tc%c\tabc\n",c1,c2,c3);: g& v8 p+ D3 b: R5 K8 h& i8 _ W+ o% @7 Z
printf("\t\b%c %c\n",c4,c5);
* Q- S' p* f( u+ ~ [解:程序的运行结果为:
0 {. x6 x; E# ?* G; Maabb cc abc
6 @+ n' E# s/ l% _% d# R A N
8 G4 V; n% ~/ E1 S* a: D3.7将"China"译成密码.密码规律:用原来的字母后面第4个字母代替原来的字母,9 K/ e3 T$ a( q. v3 ?# E+ W i
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".
4 d# |/ D" `0 O1 n/ @7 U! v% r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% `# G( e3 A+ H" l( J8 t# X* ?/ t
输出.; ~* d5 i+ h" P, y2 ?- K; }9 W
main()
6 Z8 ^! U* }3 W; V6 L h1 W$ l{char c1="C",c2="h",c3="i",c4='n',c5='a';
4 {1 B: k. \9 h# R5 Fc1+=4;) `2 K6 G& M ~
c2+=4;
" h. ]& A- ?. w; s n( gc3+=4;
/ [' [" n7 U, s w8 F7 Q' w* {c4+=4;
7 K% C" }9 d+ i& b- tc5+=4;
$ B7 E( P! u! \& E* u2 f3 zprintf("密码是%c%c%c%c%c\n",c1,c2,c3,c4,c5);* K, f# J& S) Q, v: h* c4 y
}
+ `' \# N s/ a. [3.8例3.6能否改成如下:. Q8 F# }$ |$ q p' ^& c
#include+ y" D- S* Q$ K
void main()6 d4 k4 W4 u% n h4 T
{
- v4 z3 i7 p+ Z* |int c1,c2;(原为 char c1,c2)
! p3 z- J9 A" _ R: R8 Xc1=97;
+ a3 r4 u& Y+ D- xc2=98;
& ]- g: ]& |( ~- gprintf("%c%c\n",c1,c2);
, _& l7 R' B/ Vprintf("%d%d\n",c1,c2);. H" s/ R: i8 a0 b" [
}" b& R6 M I. p% {; N9 }
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.; ?; k+ }# L- E; M; U! @
3.9求下面算术表达式的值.6 p2 @9 s L1 S# h. [
(1)x+a%3*(int)(x+y)%2/4=2.5(x=2.5,a=7,y=4.7)
: T& ^4 V3 d$ u! `+ X* V( k, Z- e, O(2)(float)(a+b)/2+(int)x%(int)y=3.5(设a=2,b=3,x=3.5,y=2.5)# A7 p# W2 |0 h! w
3.10写出下面程序的运行结果:+ M! A6 ~2 f, i D7 S. E4 [: g
#include
1 x) L# U: }, T- E( b/ j0 X" Hvoid main()
% }* A% h9 E, }: T/ E{
' A/ H. N8 f" X2 J4 U8 z( _int i,j,m,n;
1 o! S" {+ B; g# @i=8;1 q( B/ t$ R! X0 ~3 j' |- c# e& `$ H
j=10;* I1 @; C8 d4 w9 w$ {; N' s
m=++i;3 [; i- b0 X) G/ ~: s. n) }
n=j++;
& E) e, n% A, P' U+ Sprintf("%d,%d,%d,%d\n",i,j,m,n);/ v' a! L" l' t4 Y; q* k4 p
}" L: B4 Y0 A6 H
解:结果: 9,11,9,10
! U2 J; Y& x3 r6 f% U: n第4章
5 S& a5 n) {+ S; b9 E& K) Z7 y4.4.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
- R: H3 C4 r: o到以下的输出格式和结果,请写出程序要求输出的结果如下: F; T2 D$ G, C# I d
a= 3 b= 4 c= 5
+ D7 \+ [& Z. ?6 s$ bx=1.200000,y=2.400000,z=-3.600000
' E+ \8 H3 e1 _) c8 ax+y= 3.60 y+z=-1.20 z+x=-2.40* C2 P3 e. f- X2 Q# c8 s6 t8 S
u= 51274 n= 128765/ T% l1 C' l. G- n. T: x
c1='a' or 97(ASCII)
% g5 k3 j, ]9 o0 r9 p5 Dc2='B' or 98(ASCII)
$ s, E" j+ j( a. Y解:
! P c% M1 |& P; G* @main()0 z' }/ Q' c6 A0 ~8 t1 L
{
( x) m5 `( \: i1 dint a,b,c;
+ i. p) M, {4 \) D0 y; Clong int u,n;/ x! y/ w! {5 X- c: ~7 [' k
float x,y,z;' t* o/ l( i3 j, b
char c1,c2;! M+ M; W4 d6 O
a=3;b=4;c=5;
7 I |- E7 U* N! P5 I# ex=1.2;y=2.4;z=-3.6;! G4 n4 u$ Y) w1 B6 p
u=51274;n=128765;7 t- e" R; F% I2 [/ k
c1='a';c2='b';
3 f; y8 K) R7 @- m9 Jprintf("\n");
+ i% l0 c4 V$ w0 H. c2 Tprintf("a=%2d b=%2d c=%2d\n",a,b,c);
# W' B- o$ F8 \# X! \6 }* lprintf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);
c6 l% [ \- ~$ ?5 Kprintf("x+y=%5.2f y=z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x); \/ L4 L% Q% D6 K5 |/ c6 J5 n
printf("u=%6ld n=%9ld\n",u,n);
0 L5 S2 \! Q* F" s3 p4 Wprintf("c1='%c' or %d(ASCII)\n",c1,c2);
/ O4 w' w' K$ w* U9 Q1 Uprintf("c2='%c' or %d(ASCII)\n",c2,c2);) q2 F0 P' Z* J
}6 L( x: j3 q& D9 l% Y
4.5请写出下面程序的输出结果.
/ ~3 V' a! u! d$ l5 Z结果:1 i2 q" o& Y& |
57
" \5 \0 j$ P7 y 5 75 j$ o* n _+ h1 ^1 I$ y4 W3 ^2 z
67.856400,-789.123962" X1 E. {3 F) j' h
67.856400 ,-789.1239622 b) o* v4 [6 y$ V5 o. }
67.86,-789.12,67.856400,-789.123962,67.856400,-789.1239622 M2 Y; z. R6 P& [0 V4 f- G# S
6.785640e+001,-7.89e+0023 V+ W# w! _" m6 D8 H, H
A,65,101,41
+ ^4 e! J4 G ^; q U0 m; K1234567,4553207,d687
; T5 H3 _& L5 I5 f9 R) v) }65535,17777,ffff,-1
$ S. d7 q0 B2 iCOMPUTER, COM" S7 X# B" D; g `6 Q# j
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',. `% T' Q7 |: Q7 b' G+ X P% L
问在键盘上如何输入?
: K$ |; Z) y# N9 j) Dmain()! I$ ?- |# a5 j+ G% }) w
{
, i1 Z9 _2 b* c$ iint a,b;% e& k9 V2 I0 g9 X2 y& Z' P
float x,y;5 w4 f0 F5 l/ x* a* \$ v" N
char c1,c2;
6 z% f: N9 ~' P! t. x) gscanf("a=%d b=%d,&a,&b);2 Q3 d* q" ]2 F( {" n
scanf(" x=%f y=%e",&x,&y);" S' v7 d: M C3 E0 p
scanf(" c1=%c c2=%c",&c1,&c2);
% M. b8 i7 C/ V% C}# u& u7 s1 R$ H* G }6 X- B6 P) K( }
解:可按如下方式在键盘上输入:
- `6 g- E: B) g# `8 Z0 x1 G6 Q& D& Va=3 b=7
# t* P* [3 ?& T" C9 d, U2 P4 Ox=8.5 y=71.827 f) c, ^6 E3 v- Z5 }, c
c1=A c2=a
1 b# V; C( E5 U& j2 q说明:在边疆使用一个或多个scnaf函数时,第一个输入行末尾输入的"回车"被第二
1 ?, v v+ l! I2 X' W. R- S个scanf函数吸收,因此在第二\三个scanf函数的双引号后设一个空格以抵消上行 g7 [5 r8 e' p8 |
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% p' B/ r+ n" o
初步概念即可,以后再进一步深入理解.
; }, z# y3 n- ?, ~) {4.7用下面的scanf函数输入数据使a=10,b=20,c1='A',c2='a',x=1.5,y=-
$ J+ c) r# }: F- K" L# ^3 F3.75,z=57.8,请问
6 ^6 W1 o) b2 d3 w在键盘上如何输入数据?$ }; X! H) G: n& o& B4 @ f, |9 |
scanf("%5d%5d%c%c%f%f%*f %f",&a,&b,&c1,&c2,&y,&z);( d% l7 w3 {5 q* i7 f
解:
# k, |+ _* h) t( c% {; x% |+ b0 |main()! g& {& w( l$ T" @( ^; n3 C
{
* |9 Y+ r$ x" Q" {int a,b;- }1 \/ s1 K% F" G8 h* Y
float x,y,z;( Z1 k9 j& ]1 i6 w
char c1,c2; S1 t) ]& T2 O/ ~
scanf("%5d%5d%c%c%f%f",&a,&b,&c1,&c2,&x,&y,&z);4 t8 s% n" y2 i* O5 s# Z! x
}
4 n/ m& ]- }, X. t* ]9 e运行时输入:! W' n# @0 {, q' b4 G+ B
10 20Aa1.5 -3.75 +1.5,67.8
1 I- o) a! o: y: Q0 ~$ D: H: a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
- h5 l# h+ F* z9 v r& T' A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
6 L6 c7 S' R6 Y" q6 n会赋给任何变量。 f; C: {+ z7 p& F3 I
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,* N! r$ O$ K% ~( G5 h
用scanf输入数据,输出计算结果,输出时要求有文字说明,取小数点后两位数字.请编' p! x2 I b H ]
程.8 n4 ~$ v1 T& k5 a7 Y
解:main()
9 D. W$ V1 W3 j% B/ q, e# `& j4 a, r* e{
5 ]" x* K$ d. N4 Ofloat pi,h,r,l,s,sq,vq,vz;( r" P# ~& {/ K7 ?$ u9 A. o9 I
pi=3.1415926;
& Q @9 Q& |# Oprintf("请输入圆半径r圆柱高h:\n");
n1 ?/ D$ W( oscanf("%f,%f",&r,&h);9 e5 V7 i( ^' {/ s4 L" E0 W3 T
l=2*pi*r;
+ h: u, V8 r1 D# c8 Qs=r*r*pi;# N3 o4 v9 k/ B2 S4 x+ R0 a0 Z
sq=4*pi*r*r;) C! J8 c3 m9 ?( J( p
vq=4.0/3.0*pi*r*r*r;& T, W, j* _5 c P# c
vz=pi*r*r*h;
7 z& V, Z7 z' ?. w# n% G. K( gprintf("圆周长为: =%6.2f\n",l);! X0 m- U3 c& `; M1 X( n
printf("圆面积为: =%6.2f\n",s);1 _* j) Q0 G& E m7 K8 W
printf("圆球表面积为: =%6.2f\n",sq);
' ?* ]4 `" n; Qprintf("圆球体积为: =%6.2f\n",vz);. f# m1 B% h% c
}. T% L0 c- x3 o/ H
4.9输入一个华氏温度,要求输出摄氏温度,公式为C=5/9(F-32),输出要有文字说明,
: C# _7 @0 W4 O" m取两位小数.+ @% T# a6 [: m4 ]8 T4 b9 @: ^
解: main()8 \7 U. T7 Y* X4 @* h
{; Q' X- {) ^- k0 ], Z
float c,f;# k$ S% ~- S1 m1 Y" Y- V
printf("请输入一个华氏温度:\n");
- Y9 R U* L7 Q/ @3 }, T0 }+ T7 f# O; Oscanf("%f",&f);
/ v+ z3 U' z* Z. o0 d9 lc=(5.0/9.0)*(f-32);- v! r, x [, v, n3 Y
printf("摄氏温度为:%5.2f\n",c);+ c% @8 g* G: e! _' H& W" I
}! } ~8 [, B! v! C4 [9 j
第五章 逻辑运算和判断选取结构5 ^/ M6 M6 j3 U w
5.4有三个整数a,b,c,由键盘输入,输出其中最大的数.
?; N8 `: W) A1 H! c3 Rmain()
# f" i( X% Q* C{
2 ^4 c( H1 B7 V2 sint a,b,c;) U3 i5 a' K) {
printf("请输入三个数:");: r2 @7 ^( Y. w2 t+ Q1 C8 j5 e
scanf("%d,%d,%d",&a,&b,&c);
( o& H9 ~+ D' u5 e% Iif(a if(b printf("max=%d\n",c);
! H6 q. l" ^0 r+ k5 c% j! o* J else
" T( Q% y9 T, @0 T- D( i printf("max=%d\n",b);
6 v. T5 l8 S. x) N0 Belse if(a printf("max=%d\n",c);
6 x0 G9 t) x) c: q0 x/ |4 o else
) j- H+ F* |( ^! e printf("max-%d\n",a);
0 ?6 a9 v5 c. Q}( V/ _: E. q, @" C6 o8 K) P7 j! [' t
方法2:使用条件表达式.) b. C3 b$ r; H/ N) R" v0 F' y2 |/ q
main()
& a4 o! Z. v& I{int a,b,c,termp,max;
) {9 v4 ?# ~6 e( t( f printf(" 请输入 A,B,C: ");4 I$ ?* v! A$ a- g/ f$ ` d
scanf("%d,%d,%d",&a,&b,&c);
: K- c5 M( b; e5 D9 o% G) H printf("A=%d,B=%d,C=%d\n",a,b,c);/ }% i' S8 w& y4 H, Z5 V4 @ m
temp=(a>b)?a:b;+ f# a6 R! }3 n# N5 o
max=(temp>c)? temp:c;0 @1 ^- f7 W2 Q' [; H$ u
printf(" A,B,C中最大数是%d,",max);2 a7 y! B3 q% V/ h+ G: I
}& R( ~1 u2 g r+ K' L" c+ k
5.5 main(): l2 v: L' N/ a4 x7 ]
{int x,y;% _5 G5 V: v7 [ Y5 R6 h1 H
printf("输入x:");
2 X# V$ J/ [& q5 M. ~1 j! D) zscanf("%d",&x);
3 ^5 ?, h) b# r( b: \/ \0 R& Tif(x<1)8 W$ V2 [6 i4 E; @% z
{y=x;
0 f% {; _7 b# H* k+ c printf("X-%d,Y=X=%d \n",x,y);
0 v" N& ^$ \9 x2 Z# i }( ^0 H$ m# H; {6 u I
else if(x<10)
7 `( A$ D q+ ~1 D) w5 H2 i7 ` {y=2*x-1;
7 h5 O, P% q$ C- x" c% q printf(" X=%d, Y=2*X-1=%d\n",x,y);
% A( Z `9 u2 S }, F2 l4 r! b1 I2 c5 d
else) n7 ^+ c: v" x- q! ~, }8 l1 [
{y=3*x-11;, o2 {6 V( y& ?% m4 d
printf("X=5d, Y=3*x-11=%d \n",x,y);8 m9 f ?& K9 @$ x D( {9 S
}
. J R! S+ k( `5 M7 `* F8 z3 x5 Y}" M4 g. m I' j5 Z
(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!
* o6 C; k$ X) wvoid main()
+ _2 U6 d& x4 C, e4 |{( C6 W5 f! N+ P( L! I
float s,i;6 d5 _4 k% ?1 i* W" X% n
char a; F- F' X# O8 n
scanf("%f",&s);
0 D- _! m! S( N2 z3 h" J. Jwhile(s>100||s<0)
* Q# ]. b5 x, m4 T5 Q7 s" x{3 X3 Q' F w3 I
printf("输入错误!error!");
1 Q' A2 r8 v2 E( Yscanf("%f",&s);2 N J3 S* u- Z8 j( F6 e6 i6 |% D
}
; M9 j* E: F( U. W- |i=s/10;+ e+ n. q: `5 a6 u9 F. G1 u: K
switch((int)i)% p' L- v! y7 y( G. T
{
% n' x$ p% b4 |8 d$ r9 |6 g9 lcase 10:
! v& {2 {. D2 @3 c; b7 dcase 9: a='A';break;* r: j8 Z, x9 O& V6 L
case 8: a='B';break;6 t& m7 w% R/ S3 ]3 y3 v% _
case 7: a='C';break;' T2 v. |, a* j9 g
case 6: a='D';break;
& h5 g- g* k2 r7 w1 \case 5:" O! x: V8 V8 |; j
case 4:, p5 O4 S5 O0 s; u$ n5 R* J* o5 T
case 2:
4 T: b9 k$ K9 o$ `) q- k3 ocase 1:; ]- ~" S% ~9 ?& [6 J, L1 B
case 0: a='E';
" ?6 y6 R* ^9 h! K5 v}
' i+ T2 j1 j. s. g2 z4 hprintf("%c",a);
- z- r. s* w+ H; c}
& U7 s" U* p8 X7 v5 A6 e# K' J5.7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.
# i. m( I. ?4 c1 ?; S4 j4 X1 y按逆序打印出各位数字.例如原数为321,应输出123.
$ N1 J$ u% y1 x0 B7 \main()! {2 ?; X3 q1 X: s: h
{
2 v$ B$ K3 m5 k8 p Q1 V# P. B long int num;3 \& l; l1 q. R
int indiv,ten,hundred,housand,tenthousand,place;
2 h4 e6 f9 z$ Y5 H. z: u3 x; i& H# G printf("请输入一个整数(0-99999):");$ h9 K; a1 b; u& w9 a b
scanf("%ld",&num);
" e+ @6 p" J/ G2 S/ E0 t if(num>9999)
! D& @: n9 M$ z# L( H+ ]* o place=5;
4 V& H: S, `# j$ V$ w/ w( Velse if(num>999)
6 i0 H4 m) `# _' N, o% m: L place=4;
( u" W1 ?+ l4 R9 T4 K* {else if(num>99)
# z1 e# O! [: l4 s* s4 C' k% ~ place=3;/ N3 Y9 l' C' \& P! s
else if(num>9)
$ h- u$ _* Q4 j place=2;
5 b+ t9 b/ o/ y2 K2 i% Selse place=1;# L6 b1 U, J0 S
printf("place=%d\n",place);& H7 d8 E7 R2 y
printf("每位数字为:");
9 P* q; I; |$ d) sten_thousand=num/10000;
+ ~9 Z! [7 O7 `; }1 T/ X" E: rthousand=(num-tenthousand*10000)/1000;. k* f# _' ~) \7 Q0 W; a
hundred=(num-tenthousand*10000-thousand*1000)/100; P7 u6 A7 n' I
ten=(num-tenthousand*10000-thousand*1000-hundred*100)/10;
7 X! D u- n1 O J# F. G! nindiv=num-tenthousand*10000-thousand*1000-hundred*100-ten*10;2 N7 |- j( B2 a$ a# N0 W6 F' Y* f
switch(place)
- Y1 l. h& D2 }& g% T* R6 {{case 5:printf("%d,%d,%d,%d,%d",tenthousand,thousand,hundred,ten,indiv);
, p; n# w! I# ?3 O' s printf("\n反序数字为:");8 }4 X" o. E1 [* w+ Z
printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,tenthousand);: W x8 N) V4 w a- |+ `% S
break;
2 W* \3 T( E- \6 U6 Zcase 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
5 o# Y: l8 H$ _' C% M2 e printf("\n反序数字为:");
, R( Z, s% F. M/ C3 R printf("%d%d%d%d\n",indiv,ten,hundred,thousand);! v/ L$ K* [( ^6 x, W% U* I5 _
break;( t8 E3 |" ?, S# K7 z3 d
case 3:printf("%d,%d,%d\n",hundred,ten,indiv);
4 O3 C. z9 [, x. U printf("\n反序数字为:");
, y9 }6 {$ ~ M% _+ s% B' v5 M printf("%d%d%d\n",indiv,ten,hundred);
0 a6 L* I, f# @- ?case 2:printf("%d,%d\n",ten,indiv);
: N5 @9 K$ S4 R* }. M, I7 D printf("\n反序数字为:");
; k6 }9 y- F( @( `4 P1 c$ Y m printf("%d%d\n",indiv,ten);
6 T- U- J0 b" z' icase 1:printf("%d\n",indiv);
1 G! ]" ^0 ^$ ^" c @7 \# \9 F printf("\n反序数字为:");+ x; L) t" r' {) t; N) K
printf("%d\n",indiv);
7 w- z- ?% o8 T5 X5 p }
R$ m; J. t2 @, E5 {$ P}
6 V$ ]. }' |+ }. l- B7 c5.8
7 }: M( ~" D. o% A+ p# Y! F1.if语句
% \* o# O% C mmain()
$ p' K: d- E% e/ t{long i;6 z* |/ E. y# E: }& Y! l
float bonus,bon1,bon2,bon4,bon6,bon10;. P" ]) L7 X& B$ }
bon1=100000*0.1;
$ z, N5 t( p' b3 f7 x bon2=bon1+100000*0.075;5 j* [$ P, Z+ }% M
bon4=bon2+200000*0.05;& G8 n: M) {$ n3 Z& A4 Y
bon6=bon4+200000*0.03;
9 l M: V4 r- M bon10=bon6+400000*0.015;
! t% P' F' B% b0 ^. l3 [ scanf("%ld",&i);+ q% D8 q* ~' s e) q* N
if(i<=1e5)bonus=i*0.1;( y+ ?( A+ W$ \: ]9 a, U2 u* C0 E
else if(i<=2e5)bonus=bon1+(i-100000)*0.075;
" j) q1 N" F1 o else if(i<=4e5)bonus=bon2+(i-200000)*0.05;
' }) d& v+ \. t0 n else if(i<=6e5)bonus=bon4+(i-400000)*0.03;- N5 s( D7 z! t- Y8 p
else if(i<=1e6)bonus=bon6+(i-600000)*0.015;
) P! s8 ~5 u* E2 H else bonus=bon10+(i-1000000)*0.01;
" x0 K/ i3 E6 Z/ N8 W6 Q printf("bonus=%10.2f",bonus);3 q$ g$ e- Y' k/ P% E7 B& F
}1 @: B: u) Q3 W4 P' [1 t5 S5 Q: x' @
用switch语句编程序7 Y! c! S& {/ G% g( O( j8 q
main(). [" E! o3 S2 t' l9 N! p5 t
{long i;, ?* n1 q5 K+ t3 ^) t0 b
float bonus,bon1,bon2,bon4,bon6,bon10;
/ Z" X6 ]6 K4 K, z2 H9 L int branch;7 U7 `) n! H2 _
bon1=100000*0.1;0 t S1 m! Q6 Y& f
bon2=bon1+100000*0.075;4 F% y7 D$ \5 H" \8 x# R4 m, u6 G
bon4=bon2+200000*0.05;
' ], }; e% |, q( P8 d2 z# @. z2 \ bon6=bon4+200000*0.03;
& O2 U" m6 K. i3 f1 H bon10=bon6+400000*0.015;
; M0 K) z- ?1 `) g8 n9 _ scanf("%ld",&i);
" D( n# m9 h6 c9 t+ E branch=i/100000;
$ ]. {0 \8 B* W1 ]: T( I if(branch>10)branch=10;% _7 m0 s* b, \( D' W
switch(branch)3 C! r. |- G# H% W; ]
{case 0:bonus=i*0.1;break;
: H9 F$ u; I# W5 I7 r case 1:bonus=bon1+(i-100000)*0.075;break;. b) O: ]7 j+ c/ g# o0 S. C
case 2:
! K( T M, I5 { case 3:bonus=bon2+(i-200000)*0.05;break;
% k6 O( Z1 C* Y f3 D: [0 e& I/ ~ case 4:6 [2 V4 U% ^" U) m% ^1 P/ D; S
case 5:bonus=bon4+(i-400000)*0.03;break;# e$ M* Z5 p8 n' U5 }1 d8 B
case 6:
" f: D: D1 n4 x/ _ case 7
5 ~2 L+ A3 r2 E! L' g4 w; B case 8:! g0 A' r, V5 k& C+ L& S
case 9:bonus=bon6+(i-600000)*0.015;break;: ]) w: w7 G9 z
case 10:bonus=bon10+(i-1000000)*0.01;* r4 P( n7 P. k7 R
}, L% H" w) O4 \, `0 [* O) L/ o* w
printf("bonus=%10.2f",bonus);$ j8 @9 ]$ h: v0 } p& E" W$ c
} 1 r" z; D& x1 r: \0 k3 r
5.9 输入四个整数,按大小顺序输出. A4 B0 P& [/ V# @) O
main()( K& |3 F/ e1 N, a( s/ Y
{int t,a,b,c,d;
- J$ _% Y+ ~. Q& q printf("请输入四个数:");
* C5 c% y* i3 X( V' A/ Y+ J5 D+ E scanf("%d,%d,%d,%d",&a,&b,&c,&d);2 U' [4 |$ V3 U5 f) h5 a
printf("\n\n a=%d,b=%d,c=%d,d=%d \n",a,b,c,d);
6 D( D% }6 Z* \# N9 d if(a>b)
0 I$ ]/ e- T! T4 b5 p$ _ {t=a;a=b;b=t;}) n. M/ u2 i, B, C1 X) ?0 F8 U
if(a>c)
: c6 a1 M Q$ m, S( K {t=a;a=c;c=t;}, R7 A1 }1 D1 @0 {* ~$ l5 U
if(a>d)
+ }/ ~* o% \ F7 v {t=a;a=d;d=t;}
! x5 F0 L+ y$ ^, j! l if(b>c)2 t; \! u# `/ U+ k8 R: _$ _# M& i0 G
{t=b;b=c;c=t;}
1 m1 ]5 w! s8 X# B) @. I/ N if(b>d)) B5 j; t, ]) ~ d: i
{t=b;b=d;d=t;}4 U2 U. v% P" ^
if(c>d)6 W S! L0 V. I; i1 y3 S% q4 h
{t=c;c=d;d=t;}" ~+ M6 {% r( U0 Y/ F
printf("\n 排序结果如下: \n");, b$ M$ V+ ?9 w# H
printf(" %d %d %d %d \n",a,b,c,d);
& X6 C' q- N; P8 P; g( Z}
& i7 _- } n% w4 s v1 N5.10塔& j0 e* C9 P& L1 z% \5 j/ [/ a- h( ^
main()
6 F( H) L' `! G6 E' l7 A{
9 Q+ X- |& k) K3 n" E& r: lint h=10;) M; U$ _7 h& d& u, h; G4 g
float x,y,x0=2,y0=2,d1,d2,d3,d4;+ n; o2 D" O5 F& Q
printf("请输入一个点(x,y):");2 N& O# [9 {% D
scanf("%f,%f",&x,&y);7 P3 F% |* T( D" A6 Q7 |
d1=(x-x0)*(x-x0)+(y-y0)(y-y0);
2 V6 e2 l( L& F3 z* Xd2=(x-x0)*(x-x0)+(y+y0)(y+y0);
$ Q4 ^! G$ S$ e' w. p7 J# xd3=(x+x0)*(x+x0)+(y-y0)*(y-y0);
% F3 ]( j; ]2 i2 u# v: l" Rd4=(x+x0)*(x+x0)+(y+y0)*(y+y0);# b; ] C, `+ c8 k( d2 I
if(d1>1 && d2>1 && d3>1 && d4>1)! L9 C+ Y& M) k
h=0;
) t1 v8 D1 ~, Nprintf("该点高度为%d",h);
7 N: P* S) | {* h9 O3 Q9 \}1 Y( @9 W8 r- a% P1 ?/ u5 m g
第六章 循环语句1 p* L& Y8 ` j _0 o$ W0 M! X1 j, x
6.1输入两个正数,求最大公约数最小公倍数.
- g$ x( Z& M9 z* Fmain(). t" v4 \! f: j5 i9 K: x- ?$ d
{
4 V9 S" f( d6 R0 G9 v) a3 Q- d' Uint a,b,num1,num2,temp;
$ D. R1 [. ~; Q" [# T3 j! Sprintf("请输入两个正整数:\n");
" w$ [/ N" C2 U$ N3 @. yscanf("%d,%d",&num1,&num2);, V s1 u+ |! S9 O* M7 N2 B
if(num1{! }$ @+ T2 s1 U9 N1 U
temp=num1;
' ]+ m# {# J! O" T4 y. Anum1=num2;
1 D# ~. j, l5 h, F8 | Vnum2=temp;
: w2 B+ t' T3 I8 A}2 r( n( F1 R9 D$ p# d" a |& w9 s$ p' Y
a=num1,b=num2;3 G/ u& i, u1 t; k8 z2 l2 d
while(b!=0)
+ R" k3 t+ w! d* v3 X% R; `- V1 \ {) T; }* U, {, M5 s5 x: x' o
temp=a%b;
4 V! j: `1 W9 e( L3 D3 e$ M2 ]: u a=b;
! s6 M) S; O) R H b=temp;$ j+ H' e/ T: w+ C5 [4 j
}* F) e& |$ [) a( M# L
printf("它们的最大公约数为:%d\n",a);1 J: j1 S; t1 ?& r
printf("它们的最小公倍数为:%d\n",num1*num2/2);6 X& a' K6 E6 S# E) C
}
2 ~7 r6 N# i* b) [ K6.2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.( A* V, N8 T$ P
解:6 X& H" m6 m$ ^" J! y0 d& V6 w: |# q7 W
#include < >, m+ o5 ]& Z$ O+ |. G- I, X, U% ?
main()" W; g/ W: Z1 j n% G3 S' C
{
: P" K! p( ]( k9 m' H6 O$ J+ ichar c;& e: W" ^" l* C& ~0 }7 d$ z
int letters=0,space=0,degit=0,other=0;
/ q& X( t L- V" g2 J+ c) |3 tprintf("请输入一行字符:\n");
* R, S! i2 |- u4 u& O% escanf("%c",&c);3 k" h3 N1 {9 a
while((c=getchar())!='\n'); X1 m z3 e) P( M2 J
{
$ b$ G2 x3 W$ a. F% M- z m5 Hif(c>='a'&&c<='z'||c>'A'&&c<='Z'), m; J7 k- N. n8 L4 T; E3 Z
letters++;
9 h: q3 H) b( P8 @else if(c==' ')
T7 Q4 \9 t G) y2 s% Fspace++;# E1 u9 p4 n/ O: x- o" `& v8 s
else if(c>='0'&&c<='9')
( U9 O1 Y1 X# ^ t0 ^3 u& Hdigit++;
& K: {. ^- ~1 @( k( s7 q1 \* Pelse4 ]6 |3 t5 T5 ?9 Y) ^
other++;
/ s& W9 o8 `+ `# v7 c$ f/ V3 B% X}
( p4 y* ` ?- sprintf("其中:字母数=%d 空格数=%d 数字数=%d 其它字符数=%! }' w# V. _; i
d\n",letters,space,! F, R b& _0 |. C- G
digit,other);* `, J; ^7 i2 a3 t b4 s
}
* y- C7 k+ i3 k/ a* C# w; ], V6.3求s(n)=a+aa+aaa+…+aa…a之值,其中工是一个数字.7 V: q9 ^1 ^; ?; ]! s! i* z
解:
M7 ^! }2 r4 W( jmain()
% a4 Z) p/ ?' m{
( } l2 z% K: c) N# Z) [' G" {int a,n,count=1,sn=0,tn=0;
: |3 C) y. ?) jprintf("请输入a和n的值:\n");
7 g2 D E, s# a% ~8 l6 m+ r- Rscanf("%d,%d",&a,&n);
% E( d! ?1 S' ^' O/ h7 y# ~$ cprintf("a=%d n=%d \n",a,n);
# ~% }1 h; Y X! E! Vwhile(count<=n)
. Q) U# }6 `: f5 j/ Q0 R{
8 O& @6 @, ~6 j2 f" E8 I/ ftn=tn+a;
3 p: K$ p1 c$ i5 h+ h$ \! Esn=sn+tn;
7 r- }) K; V0 c x6 p8 La=a*10;
- Y. ~) n1 s4 }++count;
0 d; d5 ?& r9 B1 Q}5 X% t* n' Y0 r
printf("a+aa+aaa+…=%d\n",sn);
) {3 Q4 ?% S' b8 s# X}) e0 d- v( i% o; ~$ W2 E& S
6.4 求1+2!+3!+4!+…+20!." |# O( q2 c e9 b' O% d
main()# g6 D; |; Z. C
{
: J$ f. @) c" q; t% c- @9 ^/ ufloat n,s=0,t=1;
$ p- D% }) f" ?for(n=1;n<=20;n++) J# k' x1 l' O- l* V
{
( d9 v% n% K+ z( rt=t*n;
) p9 y# E2 M s) ]! r. T0 us=s+t;
0 A! Q& K, f2 d* u" [}' M9 [6 _: [7 J" ~& @- [+ o% I# b! c
printf("1!+2!+…+20!=%e\n",s);% e1 N- ?0 O6 D: i
}
8 X: F& R$ ~7 `) [) U- \% n- c6.5 main()+ V+ w9 j! k5 q6 T% n8 }2 y
{$ X; E: _3 Y% f5 M3 P. K) w# r+ C
int N1=100,N2=50,N3=10;. @. D0 y9 O7 g% V; b! V- m" ^: K" v
float k;$ X9 q! p- _6 Z
float s1=0,s2=0,s3=0;
) T6 H" }0 W6 D9 g, }& lfor(k=1;k<=N1;k++)5 j9 ^/ T# c! p9 l
{4 t5 ?/ y: O& X6 k
s1=s1+k;
, w: T/ F( H3 ]8 m}5 P$ T" R& U, J5 ]( R
for(k=1;k<=N2;k++)3 o/ w$ k1 ^5 @
{
+ O' g4 \2 ^, c% Rs2=s2+k*k;
- Z( W3 X) _' U/ j$ ~$ S}
0 `' X1 ]# u0 o7 sfor(k=1;k<=N3;k++)
- J$ d$ Y1 _& T/ B- j{7 o$ U& ] ^/ B+ _! t
s3=s3+1/k;5 o! |8 u7 _5 J7 o- o, b
}. f" q9 f5 c7 Z$ B' d
printf("总和=%8.2f\n",s1+s2+s3);& j# ]) S& h) Z) V; |9 L& @
}
/ o, b0 U5 r/ Y4 n( d6.6水仙开花 C" [0 J$ E5 g# Z/ m
main()0 O8 B3 M- e7 m' M! D: U/ b/ ]$ \( Y
{& l8 y4 ~- M+ c+ v
int i,j,k,n;
3 E" X2 I/ R- c cprintf(" '水仙花'数是:");
3 m1 u" ?* ~( K- ^- P( x( a! T3 Qfor(n=100;n<1000;n++). d; e" `; |8 ~0 J: g6 g% ?
{2 r) x- O9 N; J2 r
i=n/100;
* p, i) C* b' \ Xj=n/10-i*10;
8 a& p. S$ u2 x1 l, b% wk=n%10;
' v. W3 C+ r( N7 o" Sif(i*100+j*10+k==i*i*i+j*j*j+k*k*k). d5 T! g y0 M
{: N' J8 F5 S4 F1 E
printf("%d",n);
5 w$ M- N% V2 ^6 ?3 v" g. }8 Z$ Z2 d}. o- d; \4 d g& E, |$ @# h# |+ J
}
! U' V1 k1 Q( _2 f3 J: [printf("\n");
7 @9 a0 Z) o3 A' c. _4 n}
! D2 q, V+ ~! ]' d5 y, J3 u6.7完数3 j" w, r! x" [7 N
main()# x2 N( _% J, k1 F8 e, R
#include M 10000 _* M% E V& ]6 ]0 A7 q ?
main()! ]4 a0 s/ M* m' b
{ J5 l) Y0 H. v1 ]5 R* M
int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9;3 ~" f. U* x+ r& q o0 |6 f
int i,j,n,s;
# h/ j5 m7 u; B3 R' f# }4 |for(j=2;j<=M;j++)
2 I( J6 i; Z% W! P7 h{
* k* [5 \, ^2 E9 a6 W6 |n=0;! I4 A' Q7 O; M6 J6 ~9 S
s=j;
& T: B- n) E1 K7 V1 t1 qfor(i=1;i {
8 Q! s5 a( ] b5 E5 C, [$ ?+ Fif((j%i)==0)
, `$ N; @8 j$ i {
2 X2 r3 Y, w' ~! T if((j%i)==0)
( n- z: x, _5 S' v- [6 e. e {
8 ]1 X/ }' J" w2 O- D n++;6 b/ N6 r/ t+ |/ U' u
s=s-i;( m/ e* O; ]/ G x, J
switch(n)
% [8 d. N* }( W1 c9 L2 T' z {
/ d! j m, s8 r; X' i% G case 1:) w+ N: {; r h6 e. `6 Q+ C
k0=i;
+ X$ R; h/ \+ A break;
R3 V, i: y# B+ V+ U2 x7 _ Y case 2:
3 B1 S' n3 C- c% R k1=i;
' ]- [* _, Z) ?3 u break;0 G8 J, O9 a$ k: k' U& |9 }) G( N
case 3:. {% r/ r: n# p9 v1 b
k2=i;- z9 L# x* q1 ~" h/ T) i
break;1 P7 P! h& ]' ~8 y- H) g2 J
case 4:
$ t% ?0 c `9 p) Y C6 B k3=i;& k- ]! Z, r# u/ p
break;3 l C1 M" `4 w2 ?- Q3 {
case 5:5 B; ~8 o: D5 I) M0 j- z
k4=i;2 f2 s! {, n- u* a9 j% H) b8 \
break;
0 d' j; e a% r$ C! r case 6:
8 r$ s' F8 [2 E k5=i;
+ Y7 q, v9 f) q ?2 ~. p+ F5 W break;& j; ~2 K3 K% |- P6 W
case 7:! R# @$ N* |2 A* T2 ?6 E
k6=i;
! d. P0 U& t! k. o1 s, ?/ Z break;2 L+ ^& O1 j) E, o3 c
case 8:
; C" D4 e, ^) ]1 O& ?' h* I6 n k7=i;3 |6 `0 R- N. M9 _ H- p
break;
* b2 m' q0 n$ b4 t& ? h9 q case 9:* l* s' _9 n2 {; N
k8=i;" Q, U) t+ [8 g: [; p
break;1 |: o0 G6 U2 Z, m) p7 y, \2 K
case 10:
6 r8 A3 B* L& N" k6 O6 t5 e+ s k9=i;
4 M) j4 ^$ u) C" |7 [) ^ break;$ k/ m' [! F4 A c5 n8 X( E
}
3 J m n4 c6 J9 v4 j6 ~ }
. _, G7 u5 Q# K! B/ r9 o1 n3 J- g3 a T5 @ }. U+ B( b9 G; i, n
if(s==0)! X! D( N8 K& M6 ?. \' u
{! V& M; a) [2 |9 r- v' \
printf("%d是一个‘完数’,它的因子是",j); _1 Y6 v7 X1 y( o
if(n>1)
7 b, F8 m8 @) [1 ~ printf("%d,%d",k0,k1);9 |/ D% U8 T: q6 u, o* l" y
if(n>2)' b9 \% z" K8 E' D5 Z& V( A
printf(",%d",k2);: c' E! B q8 A% r0 V0 m$ j
if(n>3)
# W I- W2 ~1 T* Q1 `2 J printf(",%d",k3);
7 o% h9 L& I1 T$ x/ w* U" j- Mif(n>4)4 X) w1 q5 ?+ E1 ?/ ~5 Z k
printf(",%d",k4);! Z. `% m# S7 |/ `- o$ I7 d1 U
if(n>5)
7 M4 T! W6 O' C0 f) T' @ printf(",%d",k5);0 w8 Y- i+ l2 |. X4 w1 G
if(n>6) b2 d& E. g$ x6 C* ]
printf(",%d",k6);
6 P |/ N0 p, d: Bif(n>7)
# a5 ^, M' [; S+ a7 |8 A: r printf(",%d",k7);' L) p$ R$ x+ \3 V
if(n>8)
1 h4 O, T2 V: q7 C% ~ printf(",%d",k8);
6 y& t2 r6 k$ w$ `if(n>9)- \5 O" k. e5 H. k
printf(",%d",k9); h7 u. O: U. F/ R$ a, S4 ^
printf("\n");. G# @ x1 S6 ?% G$ u3 b
}: f* f+ N3 x- E2 M& i6 v& _, H, Q( T5 n
}
+ j9 I6 b( E( p方法二:此题用数组方法更为简单.
: s, b; ]+ @* G% } X4 w6 Z5 \main()
1 b( b. `& `: s, ~/ n, K+ x6 ?: W{
5 ]! _4 j7 g) ?+ T9 U6 \3 ?1 O% lstatic int k[10]; t( |* f7 G0 J+ i3 C
int i,j,n,s;& c$ R8 F, V( \9 } ]# j3 n5 U4 ?
for(j=2;j<=1000;j++)
& h5 _6 J7 x, M& u+ t; |7 {( l{
4 G$ J1 G& R7 I% tn=-1;
2 |7 g ^3 k* M. i1 ]# j$ A: {s=j;
- ~) N( v0 e' C4 r$ Jfor(i=1;i{" A0 n; o4 j. u- r- s z
if((j%i)==0)2 \# l, v" M% l1 y* w
{ K0 _# k) O5 w U3 X
n++;
2 K* }1 G4 N. g) \; ks=s-i;
% E4 i0 a; m. |& S+ sk[n]=i;% c8 b8 B: z, G7 T8 J# T* i6 l c
}0 l$ F L) `/ p1 _
} C( J; r. r' Z) V" b, E
if(s==0)
) c8 o; D; y; l' L{
4 p4 I: F) r+ l1 Lprintf("%d是一个完数,它的因子是:",j);
7 \1 @/ N$ [5 z' }for(i=0;iprintf("%d,",k[i]);$ \( {( B6 ?; m) O
printf("%d\n",k[n]);
) T1 j0 C7 _% u2 |}
( p1 R2 c$ b: g9 |6 f}
) `' Z0 I: ^) ]6.8 有一个分数序列:2/1,3/2,5/3,8/5……求出这个数列的前20项之和.: Y9 G4 e+ C" {3 ?
解: main()1 a# Q5 i8 I3 ]
{+ o# ]- r. e5 P2 X! ?& T4 S' t- {0 _
int n,t,number=20;
' Y5 C' b* p0 ~$ u. W0 r% Wfloat a=2,b=1,s=0;
! k3 A! v' z- afor(n=1;n<=number;n++)
& d8 f9 V1 E# `9 G4 t7 J6 h{ u+ I2 w( h4 l. h6 o8 E
s=s+a/b;
- \/ |% s2 J' |% m( @% Y8 kt=a,a=a+b,b=t;
) i) T5 _! s4 n7 T% G' w# q}
0 H- m8 ?, J! o$ A9 B' h, u V0 nprintf("总和=%9.6f\n",s);0 @' `, a! Q8 _1 b+ z+ o( _
}" k9 N4 p) l4 s# M+ ?8 p+ R% f; o1 X
6.9球反弹问题
. j. X q$ A7 ^: \; i' A! u$ Fmain()
! z5 p$ x2 r. `{, M* T1 l. _6 K% D
float sn=100.0,hn=sn/2;
/ Y0 r* K4 w1 v4 J/ f2 R& iint n;1 U' {; r5 ~" \5 L( Q2 P
for(n=2;n<=10;n++)
# w& R! y* @! x* ~, E+ E$ l% l/ j{
& b5 L4 ^- a1 A, ?" f% i G6 [2 \sn=sn+2*hn;
$ P+ m- h$ H3 T7 B0 Y$ whn=hn/2;
: w8 c3 N9 l7 \}0 e g9 y) h- h; z0 V1 C
printf("第10次落地时共经过%f米 \n",sn);' `1 w! N5 ]# j; P
printf("第10次反弹%f米.\n",hn);- t: O% s7 |2 M
}
; f9 p7 A: W7 L! X6.10猴子吃桃7 Q) l' o* }9 M4 e% C+ E
main()
7 i0 H8 b+ { e! V1 a; a: N6 t{. R$ a5 D4 I/ _4 [8 T0 Y
int day,x1,x2;
3 d9 f- S# U, X# x, l) Aday=9; j/ s4 p* x9 ~3 Z* Q
x2=1;
6 G) V, a8 a! v1 P! t. {% [while(day>0)% I0 {0 Z0 ~% {, C( j& P M
{6 P& @; L) x8 V% `
x1=(x2+1)*2;2 ~4 Z/ V, S; h+ z
x2=x1;
' v( y0 f C% d2 Gday--;
, X8 A4 ^5 ]; L; B- ]2 U}
4 b8 @( C& P+ q7 Aprintf("桃子总数=%d\n",x1);! F+ _6 e A: T; {" M
}3 F/ N' ^. r2 u9 p9 I( p, q
% L7 F" P, p$ k) _* D
6.12* g; N4 r& t" J0 |& E
#include"math.h"/ g9 f) Y+ \8 s& K
main()5 Q4 H3 T3 p. ^2 Q
{float x,x0,f,f1;
8 P& u1 m5 S: J0 f; {6 L x=1.5;% I; j, y j. W
do0 @- k# d, b0 Q$ _
{x0=x; ?% b" C# i( U3 R" X, _
f=((2*x0-4)*x0+3)*x0-6;& O& B) v0 x i, Y! C! J8 P
f1=(6*x0-8)*x0+3;
w0 P( ]" j4 s' S) Y x=x0-f/f1;, ^5 K; S- R. w) z
}
# G: E' G1 W! R6 {; p" y. H while(fabs(x-x0)>=1e-5);1 k" o, d+ o$ j! [, F
printf("x=%6.2f\n",x);1 F. |/ _" _/ A7 f
}9 ]4 ^. R. Z2 y! u6 I! Q
) Q( h& _, M- [; w
6.13) |: e0 p2 r) ?! ]. o
#include"math.h"
6 h0 X5 q2 m, Q$ h- \# J- hmain()
$ l1 c3 k+ ]; r8 |# z9 q{float x0,x1,x2,fx0,fx1,fx2;
7 f0 {; C! z4 |, O2 D# y8 | do- O* H3 r% d, W5 w4 C5 z5 Z4 X
{scanf("%f,%f",&x1,&x2); n. S5 R- k) I2 e+ v
fx1=x1*((2*x1-4)*x1+3)-6;/ F8 [7 w+ x- v# [: W) a
fx2=x2*((2*x2-4)*x2+3)-6;
2 Z$ a8 r" X4 A, @" Q/ u+ v4 F }
8 E! ^" O7 k7 C# K" T. f1 {$ W# m while(fx1*fx2>0);
" U# ^0 T3 s" L+ i do2 M- Z3 w/ y6 t6 s1 A
{x0=(x1+x2)/2;1 f( n, b" ]3 X2 g) m, U3 s
fx0=x0*((2*x0-4)*x0+3)-6;1 }2 [) {, b, j) _0 L, m$ y' o
if((fx0*fx1)<0)
! v( g" m9 X1 k; z% u8 P {x2=x0;
( u2 ?% F0 [8 e( d. { fx2=fx0;& \! e5 }8 J8 F' D$ A+ c, {8 K
}+ D0 ^+ Y/ d) |7 e+ {0 e( e) d
else* ? S! ^" _+ V4 `
{x1=x0;
. \6 u( [1 A6 C( f# n$ w# F: h/ X fx1=fx0;
u! k* F8 ?0 M% @8 P) q }, W4 Y5 Y2 H `" g) P4 S* d5 ~# N7 d) u
}
- t, X- K6 @0 a9 t" L7 [ while(fabs(fx0)>=1e-5);; e. t* H6 t8 p# G6 W6 K
printf("x0=%6.2f\n",x0);: i' E& _5 c1 W6 w6 s+ Y
}
H9 W8 `, p5 H! J) d ^/ P2 a s8 p1 ~: u6.14打印图案
L' f4 }% w) k g2 @main()
4 G: ]& t; |) J ?, |{int i,j,k;5 @6 f/ s) D( u; `
for(i=0;i<=3;i++)
% S2 `$ u" F: C; ]# N {for(j=0;j<=2-i;j++)* @9 e* S2 a$ N# v$ J
printf(" ");
l9 t3 y7 {. ?8 u for(k=0;k<=2*i;k++)
$ o, f0 L$ I R3 C/ ^% s! p printf("*");( W& g2 ? {4 w+ e
printf("\n");- D: q$ N$ B9 w9 G3 B a
}1 u" C" _1 n0 B/ t: R
for(i=0;i<=2;i++)% I- Q3 |0 X' f0 n3 a
{for(j=0;j<=i;j++)6 @' b9 g8 F3 [/ S# Z
printf(" ");" B8 ]: ^7 |6 Y! ^. b
for(k=0;k<=4-2*i;k++)' N& @' N0 F; f
printf("*");
5 a' s9 r/ H( `/ I% k( ~) Q printf("\n");
7 X k$ C/ h6 h* `% b1 ] }
- I( g9 m" X+ A, o}2 g9 r$ ?; T. D% L, P; }
6.15乒乓比赛
! C( p+ |# |3 J/ a9 |main(), {" @4 b4 t1 X
{" i5 T# Y1 {/ k1 Q, i, ^5 u- O* ~
char i,j,k;
# s4 T9 O2 t# T7 U. \0 P- afor(i='x';i<='z';i++)
, c( f: ^) P8 N4 \! Rfor(j='x';j<='z';j++)8 Y: c' |* _! n# A& @
{) a/ |1 S, t9 n2 e4 \/ b, t
if(i!=j)
5 K* }/ V3 H, v- ]for(k='x';k<='z';k++)
; K& ^- T) ^; Y& |. i {! q5 K6 t+ v# }0 y$ A
if(i!=k&&j!=k)
4 l3 i: X" k6 ^" i# W9 M {if(i!='x' && k!='x' && k! ='z')1 G+ e8 K3 F4 E9 X0 K+ Z
printf("顺序为:\na-%c\tb--%c\tc--%c\n",i,j,k);
Y- D, P7 W: U) U }
2 g* ^8 M; K" E8 e5 u }! m% q2 Y/ S, W& N
}
! K$ i" ?6 s, Q4 a}
/ A" s, T, ^0 b1 w) k7 }C语言设计谭浩强第三版的课后习题答案9 \& k' p4 ~, |/ I3 n
7.1用筛选法求100之内的素数.
+ E$ }' r W. K#include7 `/ q& W# N) l- M( I
#define N 101: n+ w: R7 Q+ g+ g
main()( i) m& K8 R! g0 X
{int i,j,line,a[N];% L" n/ y( l& M8 `4 [1 w& k1 K
for(i=2;ifor(i=2;ifor(j=i+1;j {if(a[i]!=0 && a[j]!=0)
. D5 @1 R! f: K( {: V) K# S$ [ if(a[j]%a[i]==0)0 v! Q$ k( `* B0 N4 y3 `+ v! ~
a[j]=0;
/ |0 l \; w2 W6 [ _# h/ fprintf("\n");0 M3 L6 b$ t* {
for(i=2,line=0;i{ if(a[i]!=0)
" k; T9 H7 R, W% m. n {printf("%5d",a[i]);
` A" w, @ }7 c/ U line++;
5 a+ y! f# _$ Q/ R% i( M5 ? if(line==10)
) ]( @" j6 }7 o+ y' S! G) @" ] {printf("\n");! e* f7 K6 m0 E1 i$ R
line=0;}
R; A* ~6 d" d3 ~. P. Q, Z }/ U! B$ u& l9 W* I. i @
}
# i3 R1 m& a' @7.2用选择法对10个数排序.
7 S& S+ m4 H9 Y+ U' G. D#define N 10
2 i" Z' |- X6 @main()$ s2 Y9 _: G+ V9 A' K8 W. I
{ int i,j,min,temp,a[N];
1 G0 L. q( L+ C; Vprintf("请输入十个数:\n");& A6 T6 s+ T6 b
for (i=0;i{ printf("a[%d]=",i);
1 p- v5 _' ~$ @. D+ N+ R. F0 o scanf("%d",&a[i]);
2 {) p" K5 C$ w/ i5 ~# o}/ R* h% t& Y; @% Z( ^
printf("\n");: }+ n6 y: e/ |3 l
for(i=0;i printf("%5d",a[i]);
$ I0 m8 |) G; A7 a% bprintf("\n");& y) e- V" O( N0 p+ @
for (i=0;i{ min=i;
. R$ j( F2 i% g! m- F0 |0 |# R: g) Y, i for(j=i+1;j if(a[min]>a[j]) min=j;* g! z$ S% p) T* T6 ^
temp=a[i];9 [8 A' A+ y) M9 f
a[i]=a[min];6 N: L" f3 X; f I, p
a[min]=temp;% ]2 o* o, H) D
}1 u8 M6 n7 L/ i% y5 n
printf("\n排序结果如下:\n");! v: m8 G, |% A: N7 A2 @) U9 D; _
for(i=0;iprintf("%5d",a[i]);
7 G5 E& N1 C* U+ B! T& [! M3 x0 s}4 M! S% B Y3 `" o' I' @4 t
7.3对角线和:/ K) g0 B& ]* ~7 `
main()- `. ?0 }- Z* B2 G8 P
{/ [# C) `+ ~5 N2 E! } F
float a[3][3],sum=0;
/ l* \6 Q6 X* a6 O5 R4 l4 e. hint i,j;
) d6 P3 M! }4 ~- X8 M9 D8 _. o# vprintf("请输入矩阵元素:\n");
% a; ~ `. i* D4 y! Sfor(i=0;i<3;i++)
- R9 L+ S3 c. d. H% ] for(j=0;j<3;j++)
+ }8 k! G% b2 t% j scanf("%f",&a[i][j]);/ @# F* k. M* h
for(i=0;i<3;i++)
% L$ E" I) X7 A+ U sum=sum+a[i][i];
$ i/ R z$ J. V2 G printf("对角元素之和=6.2f",sum);
- t/ B. t9 t+ G! j0 C, ^}
6 E- O0 c/ e1 O K C+ C( A2 ?7.4插入数据到数组. j' p5 [5 f. a" J# g. \- }0 N- p
main()
! y+ }( a. g. o& T. w! _. A5 G5 D{int a[11]={1,4,6,9,13,16,19,28,40,100};
9 t! f: y& Q+ k' d7 l/ r* n0 gint temp1,temp2,number,end,i,j;
& Y" ?+ J* d! D& Q& X8 Tprintf("初始数组如下:");! O* \9 z7 M6 h6 ^, w5 ]! Y
for (i=0;i<10;i++)' X$ c1 w+ X" d
printf("%5d",a[i]);
. @& R' a$ M/ n! x9 f5 p4 Y2 s# _: hprintf("\n");
5 o7 b/ i" E7 E, r) H! z. }printf("输入插入数据:");
6 k* x; a" E, |7 O% F2 mscanf("%d",&number);
) X+ v. v7 i" \" M! F/ ?- Dend=a[9];2 n5 h' z# [: O2 X, Q
if(number>end)
, L9 j) d$ J* l3 r& a( P6 na[10]=number;
! e1 E, U" @7 K; L8 b( Qelse2 B [- o# F! X
{for(i=0;i<10;i++)
, C$ ~: d6 I, U7 L { if(a[i]>number)
5 D* ^' {" `% i# H {temp1=a[i];
8 i/ a: e0 i" r) B) Z) _ a[i]=number;% o- d$ f" v/ z6 |: |5 e
for(j=i+1;j<11;j++)7 L; ?8 o' `7 C( _. p* {
{temp2=a[j];
2 r7 R/ V, }1 ?6 m a[j]=temp1;
& z5 `# d+ ~4 ^+ H temp1=temp2;" O1 N8 H1 f k0 t- y
}0 H2 r: E; I' B$ i
break;2 W5 n t, S# ]
}: J! x3 w" {* `' O
}5 Q3 _3 k$ A# m
}
# w. M. b5 W( h$ N; e7 l for(i=0;j<11;i++) W# @% W, A2 v. e; h
printf("a%6d",a[i]);5 Y+ C, X" [) C2 F2 Q% w
}+ K: V( K4 Q0 |6 Y# M/ H& l1 Z% p
7.5将一个数组逆序存放。
, P( _( u# C0 N9 J#define N 5- `% q* N' F' G) ]* l: L
main()
6 Z7 ^5 w$ H3 j' i& a2 o6 z3 r{ int a[N]={8,6,5,4,1},i,temp;1 n2 @) t% r$ B. A( h! {- p) N4 d5 j
printf("\n 初始数组:\n");2 k: e& S+ w/ \7 m v& u) m* ]' S
for(i=0;iprintf("%4d",a[i]);, U/ ]% S; ^% j. e6 W F
for(i=0;i{ temp=a[i];
' }0 ?/ C/ l% R# N5 I O a[i]=a[N-i-1];' G: T# e |! g) x
a[N-i-1]=temp;: U( E5 j f6 z4 S- `% i
}
: `& O% k1 i s7 c0 wprintf("\n 交换后的数组:\n");* P3 r) g1 Q4 ^ ?# k
for(i=0;i printf("%4d",a[i]);
U! \/ {- R: E}
% { l: ^2 J# `" w2 J7.6杨辉三角
* d8 X5 G/ X8 R- L) p8 T4 q#define N 110 [1 c- ?, A' t% C7 z
main(); Z, h% U9 T/ _3 `6 M3 l- P
{ int i,j,a[N][N];+ l' A% a) p9 \5 G
for(i=1;i {a[i][i]=1;% m4 J3 o8 F1 p
a[i][1]=1;* @- T8 ]& B% J7 ` o% P6 ]& x
}& \! @+ v$ h; t/ A
for(i=3;i for(j=2;j<=i-1;j++)3 J3 b- S. h6 k- C1 {
a[i][j]=a[i01][j-1]+a[i-1][j];% p) }5 W4 |, p, u
for(i=1;i { for(j=1;j<=i;j++)
9 u* g. l* }" e4 K `' K printf("%6d",a[i][j];5 |6 G3 e9 j' b" B* t5 F
printf("\n");
- }9 i$ ?6 c) q' ] }! T7 X& `3 p! o
printf("\n");+ `7 F2 W3 S7 y9 A
} ]9 x# B0 W$ q o% b
7.8鞍点
, k( i4 M3 m3 E; y3 N q7 ?) v#define N 10
7 }+ q7 N! ~. q% E#define M 10/ {! l9 ]5 c/ b' Y( G5 A" k
main()
& \& C( Q- ^% Z5 l" E{ int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj;$ p7 p" m" S0 K( L3 v" B
printf("\n输入行数n:");
. l& W& f- s* n. D/ O3 @# j3 c scanf("%d",&n);
. b' b3 {+ L8 Y8 h3 e printf("\n输入列数m:");3 L+ C8 ?' D. X9 v* f6 D% x9 X
scanf("%d",&m);+ Q4 A8 [% r4 ~
+ f# l! b4 X' ?, r3 K; M
for(i=0;i { printf("第%d行?\n",i);
% c: ~) ` o% _4 i& D9 _ for(j=0;j scanf("%d",&a[i][j];
& A! z; D' f) ~ }
- L+ B6 H: O5 u! c% l$ u& X; d for(i=0;i { for(j=0;j printf("%5d",a[i][j]);
9 n6 k H1 Z4 h7 ^! k' m pritf("\n");" q, f; N: G1 n4 A
}) F2 i7 C) [% D
flag2=0;7 p5 M# T; V4 u$ ~
for(i=0;i { max=a[i][0];
4 a/ v8 y( V8 l, i8 ^& P3 ] for(j=0;j if(a[i][j]>max)7 T; p1 e$ ^& q; W5 \6 I
{ max=a[i][j];* h3 A: |* e% `/ i
maxj=j;2 V4 p$ X4 {- l5 q
}% f3 y2 K% _9 n N( z6 ^6 P |
for (k=0,flag1=1;k if(max>a[k][max])8 |# v5 q, A. T
flag1=0;
" `( Z6 L9 o# e$ M3 T0 L if(flag1)& t- ]! r0 s/ u3 e$ `
{ printf("\n第%d行,第%d列的%d是鞍点\n",i,maxj,max);( L1 ~/ s! z. @- [4 H7 g1 |
flag2=1;
7 d# _4 I2 Z: p: e: ?% e; G% K }& }6 |7 i8 c$ u
}
2 P+ u. g; P" c3 Pif(!flag2)# i( Z: Y4 {$ v3 ^/ @6 f# `: J
printf("\n 矩阵中无鞍点! \n");2 p8 y1 u7 `8 Z$ v* U
}
1 j* M3 |, k( T( i% D/ q L0 F4 D3 I, k0 G* @: g
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.1 L/ }4 G6 m7 _) ]5 k( O" |
#include3 R. f* F( f7 A0 v
#define N 15
: k* }$ \& w+ {6 Umain()
9 b& I) g* H3 @{ int i,j,number,top,bott,min,loca,a[N],flag;4 e$ X: z/ u; n) z5 O# P0 Q
char c;
9 j4 I4 U. |; k9 d7 m printf("输入15个数(a[i]>[i-1])\n); N8 W+ P) O- Y4 F5 F$ ^
scanf("%d",&a[0]);6 }) {) v' @/ N' `6 [3 ~
i=1;. b7 w" A: `. q2 Q7 G( c3 @
while(i { scanf("%d",&a[i]);
$ Y4 a$ G# \3 r: c3 X" u if(a[i]>=a[i-1])
, a, g; {! z0 z+ g. y0 y i++;
4 o$ n N0 p9 k4 y. C esle
4 t) @9 D8 g1 k5 u, A* m( M8 V6 U {printf("请重输入a[i]");$ e0 i0 K9 n' h' y6 F( q" E
printf("必须大于%d\n",a[i-1]);
" H2 z, n2 ~. g' A% N }, X+ }# {3 d; T6 F/ V
}
7 |/ z) F; ]& Q4 K2 ^/ ]- G printf("\n");* I9 w6 ]% k/ f
for(i=0;i printf("%4d",a[i]);+ h: k( K/ E$ o: F4 B- U
printf("\n");; h/ |1 z! D7 b# T
$ T& V0 ^. `% |: h
flag=1;
- |+ V( F3 X* [8 s) [0 A5 P6 Y, t( N while(flag)
$ ]2 P( j4 H$ B. D {
# r* N! r: u+ k printf("请输入查找数据:");
1 n+ _: ]5 K0 f8 v) } scanf("%d",&number);
! f+ s) Y9 a1 L: g# q: \ loca=0;
* g! r3 l& |& W% g3 ^ top=0;
' [2 F Y1 L# a, c) n1 x bott=N-1;1 \0 ]4 T. Z; O6 \
if((numbera[N-1]))# L5 @" H4 d% J7 n+ S
loca=-1;, `1 E1 U* e% S S9 Y/ d8 G" ~
while((loca==0)&&(top<=bott))) E& u5 k0 `- j7 p4 A2 d
{ min=(bott+top)/2;4 u8 K; s9 W9 {3 M+ ?
if(number==a[min])
- L# G3 |+ R# l' M! q8 Q { loca=min;+ J; N# b; a S$ R
printf("%d位于表中第%d个数\n",number,loca+1);
' \* R: N$ ~* D$ P6 K }
- O0 Y% `9 L! U4 ?2 m else if(number bott=min-1;
u4 I6 R0 R' s1 B5 E" r6 B: D1 w7 ^ else
# Q8 Q* e9 \# U# A' Z% C/ e top=min+1;+ v6 G% L% {; m( `( Z$ q
}
1 R& D1 q. _* ~# m if(loca==0||loca==-1)
, n, C; h j, o# V; g5 G6 t4 ^6 Z printf("%d不在表中\n",number);. c; f% ?) |7 ~8 R
printf("是否继续查找?Y/N!\n");
2 @" Z; n2 w; X' g o5 Y c=getchar();; g! w+ e& J9 C: ^. X
if(c=='N'||c=='n')& C4 Y) _2 T7 |' r+ k8 t; }2 U
flag=0;
8 m+ q% ^( }1 P) a) ~, _ }
1 R& ]- U8 c' N/ O# c) J% S5 a}
m) `6 F# T3 t& e4 _, h- X/ U
) j; o. i! F" V! ?7.10, ~% P2 {/ d2 a
main()" R: U% _1 ~" j) V7 k
{ int i,j,uppn,lown,dign,span,othn;) X, | l H( _0 y. p6 k2 T! h
char text[3][80];
9 |, S# k0 n( F& F5 M+ l7 z3 E& W+ ` uppn=lown=dign=span=othn=0;8 |! m% B9 t2 \) F! x3 p
for(i=0;i<3;i++)/ Q3 D+ ~9 A: r: u& l; w; ^
{ printf("\n请输入第%d行:\n",i);
* K& \( x3 ^0 g+ \" r gets(text[i]);
( d q/ l4 H! }7 K1 Q for(j=0;j<80 && text[i][j]!='\0';j++) Y- i$ M5 e$ o4 B0 J1 M' v |. @
{if(text[i][j]>='A' && text[i][j]<='Z')3 c( Q P0 \4 }! G3 T
uppn+=1;
& T' `( Q. n" z$ A else if(text[i][j]>='a' && text[i][j]<='z')1 \; i7 q5 E' s+ N
lown+=1;
( [8 {) h7 @5 e5 D; j4 M else if(text[i][j]>='1' && text[i][j]<='9'); f5 c$ J5 F" f/ a" N# g- d9 A
dign+=1;0 p. _0 D+ R R0 C! I
else if(text[i][j]=' ')
. `; L. {! z* A7 n# | span+=1;3 H4 q5 c) ?5 f- I8 c1 O! W
else; g2 {; t$ a( F9 Q
othn+=1;
' | _: X m. \ }3 ]3 k* L% c9 l' B* H2 Q/ C9 E7 C
}
1 f8 C U6 \; C/ K& _7 S8 ?3 Z E for(i=0;i<3;i++)
8 \8 G5 @* N6 c4 B printf("%s=n",text[i]);
' R; b1 d: I+ G! C printf("大写字母数:%d\n",uppn);
# [- p" o7 [ x- n8 n printf("小写字母数:%d\n",lown);
2 r8 f5 ]6 y6 S' N b1 v9 R4 z2 P* c printf("数字个数:%d\n",dign);3 C" j* ~; r# _$ o8 {0 _2 l
printf("空格个数:%d\n",span); {! O; z5 Y5 V0 t4 F! j
printf("其它字符:%d\n",othn);
8 I& J9 w, P2 N, Y& W5 U7 d$ Y}& \: G: o, @; l, {7 @
$ e2 d! B. F0 Q8 E+ d
* n& ]" p# O7 T8 l) ~
7.11
3 z! ^4 H& }, ]7 r2 ymain()
) o/ Q6 {8 g9 l+ }5 x {static char a[5]={'*','*','*','*','*'};
, q1 k6 y: {, f* c( l E, n int i,j,k;# e9 v0 {4 z- m* a# h8 V1 Z8 G
char space=' ';
7 l M, \1 ?; U6 W* ?$ c# n for(i=0;i<=5;i++)
' P0 i9 T% { [ {printf("\n");' p+ E1 w! m) [ f" m7 ?2 b
for(j=1;j<=3*i;j++)1 t$ t" N6 H+ w( K6 E
printf("%lc",space);3 K, ?3 [ X5 t" ^& R. b6 b- U
for(k=0;k<=5;k++)
3 A1 {$ T. e9 d. ~& Y printf("%3c",a[k];
3 M7 W& R3 F7 h6 w# E }
1 |9 f# [2 F; {& P}2 I( c- A* V2 T" Q
7.12
/ e9 [6 N: O$ `9 w$ L, m# d4 ]: l6 ^" W#include h/ y; q9 F$ V2 e
main(). e: D6 w: m( @ }; o$ y9 U
{int i,n;7 S- `+ @* L0 R
char ch[80],tran[80]; @7 o4 A0 f- y5 i/ ] |) Y) S
printf("请输入字符:");
4 L9 E* P9 T0 q gets(ch);6 t5 d! Q3 a, D1 n2 Y+ t
printf("\n密码是%c",ch);
+ W/ Z% y; [% [/ Y2 `9 ^i=0;$ x" W4 j0 i l; u1 t
while(ch[i]!='\0'). M& Q2 S+ o# ^1 ~
{if((ch[i]>='A')&&(ch[i]<='Z'))
2 Z' c; W S7 T5 c( `) x8 L1 N tran[i]=26+64-ch[i]+1+64;- S$ N, Q2 i, \4 j3 n+ J; D0 M* R
else if((ch[i]>='a')&&(ch[i]<='z'))
, f# L" _, e. b% D1 A4 m, m tran[i]=26+96-ch[i]+1+96;& B) M1 a! O) z
else
( _+ n/ X( d* L tran[i]=ch[i];
! l0 E7 M/ I- x$ ^ i++;
% ?5 L% _7 D0 F+ d, F- m! q* V}
- a0 \1 _% w, _9 c+ W" h9 z4 R3 Un=i;; v8 a4 E9 ^) Z0 n4 T- V
printf("\n原文是:");* ~% D4 S, {. }% T4 D [1 o7 E
for(i=0;iputchar(tran[i]);
, [$ D( S' b% P& H5 S9 j}; S! i+ F$ ^% ^/ }
7.13
' }8 p: ~' I: U' v; Mmain()
/ `) W( ^! u" c9 V7 u5 Q {
$ a: O. A& W: o! r% [ char s1[80],s2[40];
. H# g( B+ s5 K3 c8 g& w: T int i=0,j=0;
# s+ |6 t% C- ]- h+ y printf("\n请输入字符串1:");! ]& i, L+ } J5 Z
scanf("%s",s1);
9 y9 K+ s/ f2 P# ?8 W# | printf("\n请输入字符串2:");: V% F. C- u) Q
scanf("%s",s2);3 V% {7 t8 i) k. t/ z
while(s1[i]!='\0')7 ?- U- q1 V; W& \5 D0 C: }
i++;! j# d' c2 _6 I& \: I/ u5 K
while(s2[j]!='\0')$ e- Q' Z& I8 U3 G* ]% m% T! _
s1[i++]=s2[j++];: B) f) ]3 c. \0 Z: B) k" q1 W
s1[i]='\0';8 _' O- t+ F! q3 { S3 ^
printf("\n连接后字符串为:%s",s1);
+ j r$ {. F1 n! x }
) ^! U8 }' `4 O3 E$ |( ]+ K! H3 V3 p6 b- _7 N; C: s- ~- ~
# _- N+ L8 L( b. y2 r7.143 W- k1 i: p! v H
#include' x: M7 m; D" E6 g
main(): F; _6 \1 ]8 Y9 A# s
{int i,resu;
N4 ~( _+ j* ^& ]4 ~/ R char s1[100],s2[100];
8 \. j% K8 ^7 E) Z5 o: Q' q# v printf("请输入字符串1:\n");
5 z0 G- {" L, `( { gets(s1);/ W$ s" A8 @# d2 S) a) H" ~
printf("\n 请输入字符串2:\n");
' @1 o9 \; T3 W gets(s2);3 h" o# P8 Z0 ]. N5 K) U- H
i=0;4 D0 Q; G9 V: a! p$ \4 C( a% v
while((s1[i]==s2[i]) && (s1[i]!='\0'))i++;
9 U, F5 K7 l6 l. @2 r* v: u. o if(s1[i]=='\0' && s2[i]=='\0')resu=0;$ P; }7 p4 h1 N. F6 O8 v5 A
else) I7 X$ s) J8 k! P
resu=s1[i]-s2[i];
/ d: }1 c% t9 c' [ printf(" %s与%s比较结果是%d",s1,s2,resu);8 _! }! |* M' l5 h4 B
}
+ R f Z( S9 {2 L7.15' T% N" w0 N0 A7 B0 p
#include4 t5 T% Q7 J1 |6 D" ^6 P
main()
/ [) Z- f% r# d6 L" V. K* {3 s {
( r y" c" m) n" ~' i2 Y& |4 x' H! H char from[80],to[80];
/ X4 H; Z! [) k+ r. j int i;
& ~. ^& |4 L3 J' ]; J printf("请输入字符串");; C3 Z, A( H' ~% q: K1 \# F% A5 U
scanf("%s",from);
8 [: w, m0 ] i5 ? for(i=0;i<=strlen(from);i++)1 F5 L i4 L! ^) z. n
to[i]=from[i];
3 T4 `: u9 [5 e; t) j printf("复制字符串为:%s\n",to);
: U5 w' v6 N( N, Q: U l3 N9 [ }% F4 d# [8 h% i8 W# p
9 @$ w% x6 x+ c8 S( H$ B0 W
" J9 r8 s" ? B% b7 O( T" x第八章 函数
6 U6 S* J/ {0 V8.1(最小公倍数=u*v/最大公约数.)3 d& L1 H" A9 X6 c
hcf(u,v), @1 P$ ^$ @8 S- N( G! D
int u,v;
+ @' X8 w6 u1 b8 y: y5 \/ y(int a,b,t,r; P0 S. T0 k8 P+ k/ B
if(u>v)3 j4 Z; b4 l8 h$ S* R" L
{t=u;u=v;v=t;}2 A$ W1 ]8 g; `
a=u;b=v;5 y" \& L' j. M# i# ~$ C- q
while((r=b%a)!=0)" i. `$ B8 d7 `4 F9 J" w
{b=a;a=r;}
1 j# k0 _6 {9 C3 l* l8 H+ u return(a);
2 m- t4 c# C! M }
2 L8 h$ \, B* i- } lcd(u,v,h)
" r; E( ~9 U2 I+ q& Q! t) R int u,v,h;/ M5 ~0 H# X+ Z& u) H7 ]7 b! H
{int u,v,h,l;
: N2 E: u8 h7 h. J scanf("%d,%d",&u,&v);
5 \) \/ h5 e8 j2 d: ~ h=hcf(u,v);
! m" V1 j/ `' H printf("H.C.F=%d\n",h);+ ?8 ]* j5 g' v/ ~: m/ I; A5 K
l=lcd(u,v,h);
' K% A" o S; Z( s- _& V printf("L.C.d=%d\n",l);
' h5 Z3 {8 k9 e+ |& H# \ }. |+ o% Y/ Z- s1 U; i
{return(u*v/h);}% P- v& F8 n7 X }$ j) \
main()+ n& ~/ h2 x7 L5 }' m }
{int u,v,h,l;4 K6 @; M) @& w$ ~7 e( E/ J1 n
scanf("%d,%d",&u,&v);! I" t; [1 o/ V
h=hcf(u,v);
8 f! \9 b- q9 F9 l: a6 D printf("H.C.F=%d\n",h);5 y5 a! b' u# M2 `; ? _
l=lcd(u,v,h);1 E3 P9 Q% Q; F8 ]" r- W
printf("L.C.D=%d\n",l);
3 U6 w& `' q6 g( J% F3 [) T0 n }
! ~: b7 i2 Z( x1 U6 d: f7 t5 d0 m- ~0 x5 r( B, V5 G5 Q7 ?
1 @1 B. l% r7 a& h/ H! A! X5 U5 y
7 ]7 O: A4 }) J5 i5 W9 r8.2求方程根8 a' j0 ~0 ~/ l0 H7 _% g
#include
* H3 t9 O" F2 D$ H" Z. Jfloat x1,x2,disc,p,q;
& A; s" A( l0 X$ z) kgreater_than_zero(a,b)" S, H! _; O3 F. m; p9 N6 w7 a- Y
float a,b;% k: x/ T- \7 v% L7 T$ Y5 o
{, k0 X0 t+ m' O9 N5 l$ d$ W
x1=(-b+sqrt(disc))/(2*a);1 C* l9 I0 K) {; l7 E
x2=(-b-sqrt(disc))/(2*a);
7 q- _ [; S- y}
; Y( K. d# }" W( L% X* w8 [equal_to_zero(a,b): F& J* N# P* X
float a,b;
3 P3 D. ?- d$ X0 P6 }/ S( p; ]" O% k8 L{x1=x2=(-b)/(2*a);}$ Z: W5 N D* G; p: ?
smaller_than_zero(a,b)) ]4 w0 S1 p% O
float a,b;( E' \0 N, p; B
{p=-b/(2*a);
$ S, t5 o3 ]* v& uq=sqrt(disc)/(2*a);
, {% u2 X$ |7 d! t) s}
9 q# ]) ~% Q v {5 I7 [main()$ G c4 ?( ]6 t' J3 K `
{% M+ q7 T; ~( T6 N G
float a,b,c;
# {7 q& y5 P' d- k7 i `printf("\n输入方程的系数a,b,c:\n");
9 V' s6 @9 O9 _' ? ]1 {& s2 {scanf("%f,%f,%f",&a,&b,&c);
% X8 n' s' y# N: h' [$ z! ]printf("\n 方程是:%5.2f*x*x+%5.2f*x+%5.2f=0\n",a,b,c);, A8 g. A" {" `* `" H% ?! B# ^
disc=b*b-4*a*c;
2 i0 X6 M1 l- c+ Iprintf("方程的解是:\n");
- [4 R4 T4 ^6 v! O9 T+ Q& s; T* ?/ Rif(disc>0)
. y% V f" Z. f; ]{great_than_zero(a,b);! |! B+ `6 A3 G7 Z% C
printf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);
8 \& b/ I4 P$ W; F' x, X" U}8 q' p l4 r. ^; Y
else if(disc==0)* Y0 ~& d& D( M2 s$ H, G, x; y
{
- i/ U2 i7 K6 Q4 fzero(a,b);
2 ?0 O( v. _' F5 v% S8 p# e( qprintf("X1=%5.2f\tX2=%5.2f\n\n",x1,x2);7 {; L% k. R6 D! d8 \2 G/ r/ ~
}( Y; |# m% \$ g" o5 ~1 M0 U
else
7 M6 W1 W* h8 S. x7 X0 x {
4 M) `) G/ g x small_than_zero(a,b,c);
, u$ p4 N! P, @9 U7 Z: z' z9 m+ ?: | printf("X1=%5.2f+%5.2fi\tX2=%5.2f-%2.2fi\n",p,q,p,q);
) \+ y) K. s' L( A! M }( v+ S( w, Q4 j" L G$ }
}
/ t$ F) n; P, I4 R/ X* T8 p$ N8.3素数
3 D# V8 i1 n( e, q6 }- k% C# H#include"math.h"2 E. E& o3 W7 M9 h3 \$ N9 h
main()$ m; V7 N0 ?0 T' B
{int number;3 r+ ]6 x8 @% {! z Z- D) w! {& `
scanf("%d",&number);3 Y, p1 Q6 }" h# h- `8 V6 c+ w
if(prime(number))7 N5 ^( S8 Q" A" a! f9 d% z6 F
printf("yes");
# F; y! N' c* y1 v5 ]9 E else
/ S1 K& t; o) H8 m printf("no");# D8 ~- H! l. W8 i O) S* B, S
}
, v% X" ~( u% j6 v; [$ e6 _* Q' ~- ?int prime(number)" _# w% u% S$ y* Z
int number;
8 U" v- v8 T; Z, ]( c/ C0 S) ?7 w{int flag=1,n;5 k q$ t& B t$ s
for(n=2;n if(number%n==0)0 C* ?1 q0 L% g1 x8 v/ o
flag=0;
9 A( ^0 J: `+ F5 s8 T return(flag);
/ o4 `* @/ b7 y4 T1 u1 S}' d4 X- m! t0 o6 K( I# j( Q4 O/ u" G
7 a7 B# q/ W' l
+ |. [9 e/ e: ]' R7 D
' l& A! x) o4 i; T8.47 _ a% U3 N3 K( W0 a/ H, ^
#define N 32 l+ P- @! P3 ]9 h, B1 p) m! A( @
int array[N][N];
; R2 j' A% E( n5 t" c0 h5 f( yconvert(array)
* V8 \, e- E; J; v: Rint array[3][3];
; j% S! V! {. [- a { int i,j,t;
$ G# ^/ L. J1 |6 r' q for(i=0;i for(j=i+1;j { t=array[i][j];) ~8 H1 a% F3 I3 k# ]2 J/ Z
array[i][j]=array[j][i];
$ Q t$ Y' q |0 u8 h array[j][i]=t;
# x( V' i( a* g/ G( P }
' ^4 o# M7 m7 ? }
/ l- I9 Y. r% f0 X3 imain()
! k. C: o( f! x9 U# { ~5 E% I1 Y{6 a/ e: R5 C$ n4 H, D% l- M9 |0 u
int i,j;5 {+ v3 Y: m- l ]$ n* W+ ~. L
printf("输入数组元素:\n");& o; [: G3 t f- @# `
for(i=0;i for(j=0;j scanf("%d",&array[i][j];* S) q S( L: J' q/ [
printf("\n数组是:\n");
6 F/ k& u; F. {for(i=0;i { for(j=0;j printf("%5d",array[i][j]);
& d: _ P3 f) h. G. N8 U printf("\n");
! K) }% O+ f7 R7 D I L }
8 T) [9 d2 o# n Y8 S convert(array);
6 m* D, ~7 W W `2 ^3 i$ P9 K printf("转置数组是:\n");9 I& k/ |% Y$ x; E0 R
for(i=0;i { for(j=0;j printf("%5d",array[i][j]);' X. t/ Z" M5 `/ O* |6 [
printf("\n");* J4 l( _3 ?+ R4 s/ k# R
}! k/ u @$ n8 G; _" D
}
/ u3 H2 b& ^+ Y( Z; ^; E! E/ o3 E7 h: g* Q. o' M4 f
# x o, U6 Z' U
: i7 F8 s e# o- p. ~# c+ ~8.5( @; d: l2 A* N6 J: S
main()) X; x+ L9 `; C! T
{1 N3 G) Q5 X, }' T
char str[100];
$ O% F: H6 X' u printf("输入字符串:\n");
3 _- C% Z7 a: B# E scanf("%s",str);
+ w% P( V0 q4 ?8 ]0 s I* o inverse(str);" i G- s# N5 x' Y, ?9 d7 S
printf("转换后的字符串是: %s\n",str);
# G1 H. x# g0 ^- o}
7 V' M. H# F' m9 Uinverse(str)
& x2 R6 b2 g# R1 N& X Lchar str[]; j6 f" T3 J7 z" c2 ?1 y1 z1 |+ n
{
+ o& q! U3 u( w4 z& Q char t;
8 M' [ m: f9 G int i,j;/ X# e8 a; m1 L: E4 H' e/ }
for(i=0,j=strlen(str);i { I; U, K4 B. d+ P1 v
t=str[i];
9 d" S1 v( L' o) ^ str[i]=str[i-1];/ ]6 d$ I, G/ D2 E/ L0 j
str[i-1]=t;
, z* O n& o0 g5 b K0 b4 f }$ }) [( S, N$ b2 Z. o
}4 B" ?( U5 J8 @
" y4 T" c F; z7 H7 |# {
: [- ^0 h1 L+ P9 [/ n' a0 ^* M2 T/ f7 T
8.6/ y4 V( z, u- j4 \+ l7 y
char concatenate(string1,string2,string);
% k) ^! q s! B0 @" e- Echar string1[],string2[],string[];# Z1 d- v' Q) U: }' |- `- v4 ~* O
{+ S& s3 ?5 n3 L5 Z
int i,j;( w/ T8 L; \8 k5 a3 P( e
for(i=0;string1[i]!='\0';i++)
5 H7 u" c8 U/ T F% `3 K( J5 L string[i]=string1[i];
" j' x5 }1 z, _" P# k% `$ p; kfor(j=0;string2[j]!='\0';j++)% q) ?" _ {8 P; D3 @+ b5 R
string[i+j]=string2[j];% l9 }. j( o: O/ Z. k
string[i+j]='\0';
; u3 h9 Y) E8 u6 O' x2 n}2 p d. T% \7 h& L U( j! o
main()
/ U' w9 h# V1 k) I$ q1 u{$ f) ~# O1 B& s1 E! `
char s1[100],s2[100],s[100];
: e" N) _7 o9 v" ^2 o printf("\n输入字符串1:\n");: t" y% {2 k$ c3 r3 t& ]
scanf("%s",s1);
& W+ G( {0 d Q: E) B printf("输入字符串2:\n");" ]% [+ n5 Z1 @( E
scanf("%s",s2);& o+ K. G% m& m
concatenate(s1,s2,s);& u0 p! y+ R" A' q* Q
printf("连接后的字符串:%s\n",s); y. j$ x+ F, J6 k
}
( p) G3 K% [% a. l3 [, N. X" l
$ j" f! J# S: P4 h8.8
& N& ?5 ~; f! ?4 W/ omain()
& g' H, ^3 {. E- M4 H, k& Z4 o# J# _% i{1 z5 p" D2 ]5 f- k, Y+ O C/ h
char str[80];% c! o# _' T* S
printf("请输入含有四个数字的字符串:\n");
" y+ e0 B- Y7 l' V" L0 D scanf("%s",str);3 N5 N' E' M/ @+ ^
insert(str);
6 p2 l( S$ H# b. _& k0 n* v9 v. q}
+ z$ L$ Q- n! Ninsert(str)6 q3 C2 l& D0 H' P' n4 p
char str[];$ {7 ]" G2 x9 ^- ?4 N# V: a2 h
{
% `: E# ?, n' ~ x int i;3 }7 v2 T. [; |5 h9 L+ b' _
for(i=strlen(str);i>0;i--) }# b* s* \# m, h; y( \4 E+ w
{ str[2*i]=str[i];
1 ~" h' k* P# [' ~9 w str[2*i-1]=' ';
2 G1 o. w" G' v% K- m }
* ^* P) C! \% @* V* P printf("\n 结果是:\n %s",str);
, o6 g+ }/ v6 s! Y" z: h" x }
) e4 h' q) L1 n# s+ E
9 A1 P7 _: d4 E' }" U; M
) Q9 A& z* P; X; z8 n5 ?! |; y* ~7 h; u8 J- m, E( {- \
8.9$ J. }* N+ ~! }
#include"math.h"
2 g/ q. T; U% @ e- c% C& Oint alph,digit,space,others;# C" d, i8 r/ J( u
main()
$ o4 T- g A' \, b) E, n( K{char text[80];
# Y5 w1 A. ~. \, \ gets(text);
7 W' I: ^: O- D3 ]( x% P% f alph=0,digit=0,space=0,others=0;3 n( R J) x/ f+ D
count(text); V' g; D. A9 C) F
printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others);
* ?/ N8 g$ F( ^* B0 ^! b}# t$ `4 d: |3 O
count(str)
- K; _ K4 c3 k6 a, A, Qchar str[];% [& Q- F, k; n* N* P4 N+ F
{int i;; k3 z. P/ M- a: a' K
for(i=0;str[i]!='\0';i++)) f6 c* n9 h0 h$ o
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))! H) K1 g. c6 c! U
alph++;& H) R! v9 m* h7 m# J! X
else if(str[i]>='0'&&str[i]<='9')
# X |3 B; C- y8 w; o7 c9 C digit++;* k, X1 g- n S& T
else if(strcmp(str[i],' ')==0)
3 w. s* S8 R+ R space++;
5 l8 f6 f! ?" m- M else0 E, Z1 m! Y0 q+ W8 P
others++;
; h8 i4 V' T9 u% h4 s4 M}0 t# b2 T" s9 B
/ o5 S( G- S1 C( ?7 \
/ _7 j C" Q2 G8.10
, z1 D8 D, Z0 j) U. b+ Hint alphabetic(c);7 R* [& K5 S3 O( F3 M" K9 V7 h
char c;
, f X! I0 P/ C, e{. E; t8 k$ p# \; `% K1 [
if((c>='a' && c<='z'||(c>='A' && c<='Z')) n# T% Z4 c1 ~, B
return(1);8 j$ _, g' |2 |1 ]6 l, H
else
& y7 X! j: v1 L( w) j8 ^/ B. T& \ return(0);1 }6 o1 F n6 \: l, \
}/ s: s4 i }! z$ y1 d3 i
+ O) q. y5 x# U. F# \$ t: n! _int longest (string)
8 p, L- Y/ }1 W9 G+ Echar string[]; m4 U' @$ v5 p! c h# m
{
! e+ c ^6 U' [1 z3 r int len=0,i,length=0,flag=1,place,point;- D% V* e# {. Q- N
for(i=0;i<=strlen(string);i++)
' j; A( J9 G; Q+ ?- o) E/ m if(alphabctic(string[i]))
$ U$ R- ~3 m: ^+ t6 w if(flag)
9 ~( m9 x O2 `' N3 c: _9 ^5 k {0 J& L1 A" u F( r$ x+ ?- f4 ^
point=i;
( y& C1 T, y2 V1 N' ~ flag=0;) J* h4 b+ v+ O) z; R; k `
}/ A: t$ Y. j/ m; h) w2 ?
else# z/ x* V& B) K4 u+ H) x
len++;
# d6 P4 f' h# n0 [% N4 B else" b0 |7 Q; H* ]9 P, _6 }
{ flag=1;
' o: v. ?0 S1 {+ W2 v* o if len>length)
* |$ i4 _! V2 b; s {length=len;
7 R) v x2 j7 P) |9 `2 u place=point;& v5 L$ A+ P) {0 M0 S4 r ]* i
len=0;
9 [& e- l2 T/ k' A4 K }
, g) A6 b8 I! R; p }
5 u5 ]( E; s' o" K6 Y- O return(place);
. `1 Z& P* F2 k }3 c* R4 X9 Q7 C9 w
main()
# j- i+ T- U# @5 d, C{
% m9 m2 r0 P6 Z: B# m/ jint i;
, e% @- B; G, I3 w7 dchar line[100];; R6 ]+ K4 ^9 N# i
printf("输入一行文本\n");3 b& W- Z( T! E$ I% \% J
gets(line);3 n) G* e5 Q# a3 G8 O( i2 d0 Z2 h4 v
printf("\n最长的单词是:");, [& E& A1 B @2 ^, f0 d; U
for(i=longest(line);alphabctic(line[i]);i++)5 c4 o" s! G/ L n0 D
printf("%c",line[i];
! S! h: T8 o3 P8 F Gprintf("\n");
& K4 x8 R) j/ [; e& Z$ Z}
0 T! P% h3 P+ g, K6 v3 R( A) \/ W0 e( W C
& o H2 ?+ X! \
6 B& j) s- T0 M f1 F. Q9 n* k
8.11
0 J: j5 I6 z( s6 V6 |* N& p K#include
2 R" }9 k$ W \+ d4 l5 E
% E! S% A' G8 [+ y( ~/ a( s#define N 10
# ?/ ^% y6 e5 J. F! Hchar str[N];
) u/ ~8 j1 J5 ?. W7 Wmain()
1 z2 [( P+ e7 a" f/ x) |. o{
! e% }9 J. h# h! _1 Q/ Bint i,flag;
* d$ A0 N; [3 h; rfor(flag=1;flag==1;). @$ B8 K. p( [! S- ]
{
% R- D! b% b4 {% L. o printf("\n输入字符串,长度为10:\n");4 J) q4 ^* b/ c& M
scanf("%s",&str);
. F/ X5 p5 V- O/ S$ N if(strlen(str)>N)6 Q0 y$ Q7 x+ |9 U# {( F( l
printf("超过长度,请重输!");4 y9 x$ e W+ r
else! H, v6 n* I8 L
flag=0;/ ~3 ^1 O) R& Y+ y8 ^: `; r3 l
}$ [- t Q( N0 _, ]5 w) K
sort(str);! \& _' _) X3 W4 S
printf("\n 排序结果:");. c! u* `% ?# F) T+ u, u6 U' m
for(i=0;i printf("%c",str[i]);! D8 b/ U; L# v8 |& `$ ?
}9 p: j8 v( h! r6 H2 P6 d8 X
sort(str)
+ X* S8 U8 E# r# P) Nchar str[N];
! ~. u3 _8 h& g$ E: s# p! f0 |) f- ]{
7 R9 Z& U; T) b& _' k, h# |8 nint i,j;8 g8 v( u' P7 a x, M4 r0 J7 ^( s8 f
char t;* z5 m$ n0 {& F# m
for(j=1;j for(i=0;(i if(str[i]>str[i+1])$ y2 O1 Y2 N; j& K
{ t=str[i];
8 g! s% b5 X" F0 t str[i]=str[i+1];9 r4 H6 B+ u' ^) `) L3 A
str[i+1]=t;
7 v% L: J3 M7 `/ @ }
2 B7 B$ s, L9 E3 F) q. g}1 [9 c( y6 q. v8 U O
8.126 \% @$ _7 x5 v
#include
: H, L, \1 Z5 }' B#include* o6 R1 K2 H" \' A
float solut(a,b,c,d)% |% U3 n' X& {2 k: g4 A5 d8 y& W
float a,b,c,d;
; z* T2 Y) D5 ?8 Y6 e/ A% G; ^# e{float x=1,x0,f,f1;
- {5 s+ z% [, e& i, Z do
1 o) p. C* T4 z8 o* { {x0=x;. l# P; G4 U. J$ N
f=((a*x0+b)*x0+c)*x0+d;
) U' t; \: _/ Z+ U. R$ E, E. ?) R f1=(3*a*x0+2*b)*x0+c;
. R: @" Q% V- m5 z% Z/ ` x=x0-f/f1;
% c" `- W! x+ ` }
* L4 B& `8 {6 i) {9 I6 r N while(fabs(x-x0)>=1e-5); b" H9 C9 M5 H9 b" k
return(x);* c" p% ]8 H4 R V8 t/ E
}
' g. _- M9 K0 s+ E& D0 imain()
! Y2 U- F0 m- ?{float a,b,c,d;# y+ u2 x' f8 F2 m! ^5 |, |
scanf("%f,%f,%f,%f",&a,&b,&c,&d);
: \/ [7 u/ s: |# H printf("x=%10.7f\n",solut(a,b,c,d));6 B$ Z. C' _- Q+ g
}
s, }# V' F8 T; q( \3 F" N8.139 n( Q3 P f( S+ _) O: @
#include
7 Z$ `5 u9 ]1 B0 C$ e0 v& Zmain()
; M2 S U5 |! r{int x,n;
( n0 q" b; s+ q0 w8 E; V& z$ B float p();
* I- F( D; }: g3 @: U+ z- p9 @1 O scanf("%d,%d",&n,&x);1 w! m& P2 w. B. `& M! Z
printf("P%d(%d)=%10.2f\n",n,x,p(n,x));% \! ?* w( ^+ v4 e
}
) W! u3 Q, X1 h) n1 g2 bfloat p(tn,tx)1 q0 K# b" B9 Y1 i! ^1 F5 G. d
int tn,tx;1 {+ ~% e) D# R6 p
{if(tn==0)
y3 @" G4 ]0 W return(1);& V# e1 _2 `& {+ }; \& e
else if(tn==1)
( t+ D6 e+ n/ C. R, q# Z9 R return(tx); a1 I& h# W! x' s b
else
# _) _+ E% o# i! S return(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn);8 d+ a! I# b4 I, k, O0 W
}
0 q. z& v6 k% I0 }: j' Y8.140 g7 K( M6 T: E8 |: X: s* }
#include "stdio.h"
6 `) O2 ~- t! U5 W#define N 105 v" Z7 A: u4 I/ _7 u
#define M 52 n. i1 ~. z9 ^$ |3 A0 n5 n f7 K
float score[N][M];1 [4 [+ G$ K/ f9 F3 h) i& i- o
float a_stu[N],a_cor[M];4 u; J/ A' @. _
main()
% A4 ]/ T1 |9 c% }{int i,j,r,c;% H! W) ?7 ]1 h% X
float h;
3 \$ V. b+ v& s7 M' b float s_diff();
' _; s$ D) y; k) C/ Y* R* q float highest();8 }2 U1 s1 _# Y5 T( t
r=0;
' g7 M7 C# C; A1 C: z, _ c=1;
$ Q. j" u. ^, b3 [ }5 t- G input_stu();, e/ t4 c- O; {% ~
avr_stu();
) X1 M" g! e. u; ^2 y5 ]0 Q9 p' u$ ` avr_cor();% \- M8 F+ C8 H' {5 R+ D
printf("\n number class 1 2 3 4 5 avr");% k/ e6 F/ T# B, i
for(i=0;i {printf("\nNO%2d",i+1);
& U, d* X% N4 W for(j=0;j printf("%8.2f",score[i][j]);6 j' T9 R0 w& x4 K
printf("%8.2f",a_stu[i]);3 V( n8 F0 X% c* ^& z3 v
}
0 p$ U) V( ]" b$ O1 T! {* c. W' t printf("\nclassavr");
1 r" s: J: q6 T. w2 B) V/ \ for(j=0;j printf("%8.2f",a_cor[j]);
$ L" G( Z$ X* O; V. n- M h=highest(&r,&c);& o/ y2 @" C1 a; [/ z. P2 v
printf("\n\n%8.2f %d %d\n",h,r,c);8 a* U# m. |* d/ S1 f7 ?' o
printf("\n %8.2f\n",s_diff());, F( F5 G& u/ F
}
4 O$ Z% X& K. {0 u& @- Tinput_stu()
7 \9 o1 J. S* C& ^{int i,j;
, W e' h6 q& i* q: G! B& I float x;
/ x$ f- l% r2 M& M' L for(i=0;i {for(j=0;j {scanf("%f",&x);
3 R; q* c# }: u( ?% m6 x1 Y) F4 n score[i][j]=x;
! j- U. L( B/ U& I }2 F3 Y8 M9 A6 h4 g( B% G7 k
}
$ B9 _$ G% N" U0 Z}
2 K. r% K3 e9 Q y: X) X$ }* {7 havr_stu()1 f% _0 ^5 Z& H8 w8 S9 p0 Z; u
{int i,j;
' G2 F8 W/ [, ] float s;; n9 J5 o& _7 v% C A3 u
for(i=0;i {for(j=0,s=0;j s+=score[i][j];8 u- O4 \8 I/ u7 B, Y$ ]* u; p
a_stu[i]=s/5.0;- i4 g* s- q/ [. u4 r8 O, _
}5 R/ o! T6 \( ]! x( y" b C% s4 G
}
# ]1 v1 @& [9 ] R! W* k0 \) v7 Zavr_cor()
/ i" W% Z' h' g7 l& D{int i,j;
( h$ F6 T3 v5 i( F float s;
. s' T, A0 W I/ F2 Y; F1 c { for(j=0;j {for(i=0,s=0;i s+=score[i][j];! ^2 O# L3 N5 T% x/ ?% ^0 f
a_cor[j]=s/(float)N;
. k" P, F0 G" W }4 }% ?: Z7 U O
}, `6 j# x$ c& T% i' `
float highest(r,c) Y+ a- H# X$ u9 i: q
int *r,*c;, \! I+ X B7 h3 c* Z* f/ j
{float high;4 `. h; \# E6 g
int i,j;
! b, z9 l' @" P' q$ D/ n8 K high=score[0][0];! H T( ^( r D) L1 L% V% l
for(i=0;i for(j=0;j if(score[i][j]>high)
: |& o' s6 v! J6 P# ^, y& M* e {high=score[i][j];- I, z4 }! |8 ]; U5 q2 C0 B
*r=i+1;
* Z- z: F& ]5 [' J *c=j+1;
Z6 g( o: L# q; f/ z) \ }5 V& Z6 I0 \" v1 U# c' C% N) s
return(high);- B# B$ I& w B) _' v
}
: N7 ^1 U" l* p# F& `float s_diff()
* |; _/ g' z% Y3 w0 o3 o/ O( V+ I{int i,j;
: H: L, ~9 Q R float sumx=0.0,sumxn=0.0;( B& L' K, G# }+ }7 W
for(i=0;i {sumx+=a_stu[i]*a_stu[i];: _0 s3 X$ i+ ?+ N$ E/ X/ D
sumxn+=a_stu[i];0 ~3 W1 G$ c3 u
}
5 I1 w' W; T) N return(sumx/N-(sumxn/N)*(sumxn/N));
1 y$ {% k" s, o+ s8 o6 _' j' [& C}
/ J; f% F, d; A, r3 C8.15: w3 i* v" }7 p" ~) S7 F H' m1 A
#include
: L# J7 Q, g+ t9 D Y0 F- e& }#define N 103 _9 X. [) D& W- g
void input_e(num,name)
5 s- v; o7 m+ x3 w7 f* Y3 Nint num[];2 o/ p* S s' c D/ u1 L
char name[N][8];! c; j4 n$ _6 n" S+ k$ Z; Y9 M
{int i;
% |7 J; P' i9 C6 D% V for(i=0;i {scanf("%d",&num[i]);; m8 F& N/ y5 o' X* [6 R
gets(name[i]);
u6 r/ X+ S1 ]- x }
% P- z8 B0 n! L1 [# s" j}
. g+ x; m2 ~4 u9 kvoid sort(num,name)
( D5 b5 y* a! \! s% k' }8 |, jint num[];8 U# z1 g8 C1 G; J) [8 O/ I1 ?: W4 Z
char name[N][8];
" o+ `1 @. g; J{int i,j,min,temp1;
& |9 a1 X* d6 n, O4 N D9 H0 g char temp2[8];* j6 f3 O# E( @* y
for(i=0;i {min=i;
' f/ e; B; `! K% e$ ~* c [( W for(j=i;j if(num[min]>num[j])min=j;- K$ f3 h5 S( J: o3 ]
temp1=num[i];4 Z+ K) O, ^4 y- B
num[i]=num[min];6 T/ R# p ?. ^0 F
num[min]=temp1;) W' g$ r9 p* s) B9 h7 e; `
strcpy(temp2,name[i]);
& a( \- E- v6 v! T strcpy(name[i],name[min]);5 E% l6 {' s: O% t4 A, E0 {! i! {
strcpy(name[min],temp2);6 V! h: ?4 j/ ^5 m
}9 O. p# q% s8 n9 M9 z( l( u- j0 W$ b
for(i=0;i printf("\n%5d%10s",num[i],name[i]);
% H" T0 j8 D, }( ]8 J; [! k$ J}, K& D& S) ?2 L( Y# w9 Y8 K- _
void search(n,num,name)
: m+ \5 F4 D0 T9 Pint n,num[];
% S1 o8 S7 K: w1 S6 A f. Wchar name[N][8];# ^0 j- B: L" t& @
{int top,bott,min,loca;& N) O g" }; F+ `5 e( y
loca=0;. V# F$ v; f3 Y. j! i
top=0;; P4 ]& t l0 I8 @
bott=N-1;
$ M( N6 U: R4 `; a5 w( x if((nnum[N-1]))3 m2 @* J. O. c' v! U* L5 H
loca=-1;
( w+ \; T+ \, F while((loca==0)&&(top<=bott))# ?& w2 t; l' y9 U
{min=(bott+top)/2;
6 f! y3 H J/ q3 u% @ if(n==num[min])
/ e1 ^" p' h) i5 q {loca=min;
5 b; L( ^9 K7 ]+ ]. i/ m( v printf("number=%d,name=%s\n",n,name[loca]); z! Z$ f% i4 C4 H& Y& p
}
4 e* {+ H" x) f, a6 R$ O else if(n bott=min-1;
/ r/ S7 ]) [9 m1 \/ R; q else! X( u* c) N3 [8 x
top=min+1;
- B& B) A* E% K7 V" ^5 j }: O5 m; w* B/ p: G& G
if(loca==0||loca==-1)* T/ ^$ d- C& T2 B
printf("number=%d is not in table\n",n);" p* p/ [5 F2 [6 ]$ `8 h
}+ x9 ^5 |" m, O1 J* k7 d6 z
main()
! K" S) t: v* D' I{int num[N],number,flag,c,n;
# ?$ l: x& u0 D, K char name[N][8];% H v9 {: d, L- R
input_e(num,name);2 E/ X( d$ K9 @0 w
sort(num,name);0 ~- z9 x1 `3 |( O8 N ~
for(flag=1;flag;)
% l/ _ ] F% ^" s {scanf("%d",&number);1 V. p+ X' J7 z( K' [; i! B) F- a9 R* K
search(number,num,name);
7 |+ h1 e V" |5 T0 Z; i* K printf("continue?Y/N!");
! @; Y* G: V; V. t c=getchar();
, h! b/ q1 O/ Y9 Y$ B% q if(c=='N'||c=='n')4 d( R" M2 a/ Z- Q
flag=0;
$ ]9 Z) e$ x' O6 s. W5 N. [% ^& R }
9 _, v y# \7 I# j+ H, J) s}
, W+ T0 `: a5 O2 x! s4 W
a4 i9 ]: A; u* s: A' {8.164 J* H0 l$ k7 c9 G# h
#include) q# \: J. {% j. ?
#define MAX 1000
3 }" ^0 O- A# k+ ymain()
: k+ Y1 q0 ]/ H{ int c,i,flag,flag1;
8 C& q; S+ W! D4 t5 P char t[MAX];
! l7 R/ Y! B( ~: [8 S1 @ i=0;! Q' n8 t- K; b7 x1 v
flag=0;# w" i- j- B5 g# j u$ X5 G
flag1=1;4 J: P: ^! S [' E) m
printf("\n输入十六进制数:");
$ o" j5 A4 o: j. T/ u while((c=getchar())!='\0'&&i { if c>='0' && c<='9'||c>='a'&&c<='f'||c>='A'&&c<='F')
9 R" q! j, ?. Q4 h/ q5 |8 Y {flag=1;5 s0 X% L# A: P! C! T% M6 I/ Q
t[i++]=c;
. R0 M+ e& S7 x% L: Q9 ~: Y1 K! C }# G) N% \: `. P h! V) s1 l
else if(flag)$ l+ [% @- z9 T) y% E7 I
{
( ]4 l P- M3 o t[i]='\0';+ r- h' A: o& [! k; j5 l+ J
printf("\n 十进制数%d\n",htoi(t));! p% V3 O9 Y# X) G& N! K) B# v% H
printf("继续吗?");
) I2 b$ [( J9 R! o ~2 a0 m c=getchar();
+ i; a6 V" _$ d2 S if(c=='N'||c=='n')! O4 j/ P! b: G
flag1=0;8 e4 C6 C k5 v8 ?- s1 @
else
3 l- X3 B6 d0 t$ H7 d {flag=0;
+ z8 u/ f5 |" E0 g3 f$ b4 J* T i=0;
; R+ r; G) k H5 u8 i7 Q9 M3 `% Q printf("\n 输入十六进制数:");
; D! B; m F O( q, s }6 v w) r( p4 u" `
}7 p N$ r; s. {/ h2 z0 L
}
' k" a, X+ |, j. y9 c}
, B) {8 ^) J1 k ?) O3 c: Dhtoi(s)
. Z9 M% h% u1 _+ q% @char s[];
# e+ u* y; n) a* h- o: l{ int i,n;
' c; V h/ x4 k& ~$ D ` n=0;8 G: [. J. {+ P l, v/ _* W0 F9 f
for(i=0;s[i]!='\0';i++)8 i' u* c& N% [
{if(s[i]>='0'&&s[i]<='9')
6 ?. ~8 o3 F3 h. }' S: P2 J: N+ E n=n*16+s[i]-'0';
3 o: Z8 k T2 j" `8 v% p/ f if(s[i]>='a'&&s[i]<='f')/ `7 h+ ]$ B1 x" ^+ x+ w
n=n*16+s[i]-'a'+10;; L7 V4 x* T. Z; d
if(s[i]>='A'&&s[i]<='F')1 o5 O* y' B4 \, y1 ~# w( x
n=n*16+s[i]-'A'+10;5 U7 `; J m, U2 r
}8 g g+ L% l8 t: a
return(n);
; }* d2 Z0 D, r}3 l8 b q8 k' @+ s% Q: h6 d, `
# a. i* C& ~# y2 ~% d( s
+ I/ c$ U" b `: f' q
% ^& Q# }8 n/ N8.17
# X3 Z/ [2 T3 ~3 w' k) E#include
& g) T9 [( P- C' `! i/ ovoid counvert(n)2 r, `$ ^2 J6 P% S
int n;
6 _1 s8 M' S6 X5 a" F9 ~{ int i;
* X7 P8 ]" h$ r6 ~( ]) Y$ b* u if((i=n/10)!=0)8 }! O3 F, r( f1 G2 H
convert(i);
2 |$ O& e$ n" d; D9 S) ` putchar(n%10+'0');% `' E, b5 H W |/ S8 ?; K
}
/ F+ g0 @8 C# S0 O, K7 }main()6 a2 N9 d7 y5 o0 e" f( Y
{ int number;
& C5 I8 @8 }4 n. ^* ^ printf("\n 输入整数:");2 M! u7 X/ D g G7 ]9 \- x+ ?
scanf("%d",&number);
4 C9 j0 N+ n; V6 X! C% a; q1 O printf("\n 输出是: ");
3 s# y4 {9 O0 J6 g+ }$ @ if(number<0)
0 j8 t: O" s/ y( N# B$ [ { putchar('-');
4 J* P5 d/ C) J4 C number=-number;
' d6 L$ Q- ^5 W }3 R. @# ]& [* F# P4 f
convert(number);
2 l9 u% P+ l; Q3 H/ q! j: X}
5 C3 a5 r. |6 U
! p7 z& ], d7 ?4 T4 l0 l) O7 M) O# n9 R! \. y
. b# _; ?. ]: b% E/ O% V! x. E/ y- g
8.18
/ P+ V$ J* P l ?' Smain()1 _% {% C k6 _9 T& a/ A, N8 M0 e @
{
% z6 k% t- Q( h+ Q9 ^) h int year,month,day;1 {+ ?$ G8 I1 d! C2 w: j
int days;
D% o) q- L8 x/ Q2 a" [ printf("\n 请输入日期(年,月,日)\n");
, c4 J# q+ l: B* [' L, E scanf("%d,%d,%d",&year,&month,&day);
( q" x( Z* v4 y, _/ f printf("\n %d年%d月%d日",year,month,day);- s: [/ `) j5 C3 ~; c" O
days=sum_day(month,day);
- R) P+ ~, }5 S8 O9 k1 X; U if(leap(year)&&month>=3): @1 L4 J1 d/ |/ e$ ?! Y, h5 d) q
days=days+1;& H7 e8 v4 C/ T: a! L
printf("是该年的%d天.\n",days);& e$ d1 X# C, k1 V
}; H! Y4 c' j9 _0 x: x9 a- B
static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
' W( g' B- }' Q4 l1 f2 \. @ int(sum_day(month,day)( ?; I7 j# K+ f, }5 V
int month,day;/ L0 l; K; |* d+ Z3 H# p
{9 `, f7 z" |2 f
int i;( e6 q; ~9 Q. W \1 A
for(i=1;i day+=day_tab[i];; c4 E; y8 l$ i9 j) L# G4 I# w
return(day);3 E4 k, ~( R; Z5 n. w$ W
}
8 ^. L6 P: ^" }2 B5 A$ e int leap(year)
. J4 W( n* f: v9 q5 d' s0 h9 Q6 x int year;
+ U u) s% o5 N- v% G5 p {, B& F$ s6 w7 y. b8 c+ A9 W; \) _* \& z
int leap;9 r( j0 e2 y2 w
leap=year%4==0&&year%100!=0||year%400==0;1 ^0 I6 {8 O" c
return(leap);
& I) i& o6 L! D3 O, ~ }/ L0 p+ ]1 Q2 Q6 Z
第九章 编译预处理. x9 R F& V1 V% c
9.1
i! l! h, O' r/ M1 U( C: F' V#define SWAP(a,b) t=b;b=a;a=t
3 z+ H) B/ C8 H5 l- lmain()% \ y) U( u% k' E, u9 g
{4 D+ B% f0 L! I7 E( P2 W
int a,b,t;
4 T6 ? P2 k+ z' @: Mprintf("请输入两个整数 a,b:"); u8 F4 D) C9 ~& M, e0 `0 N: G! Z% W! u' r; _
scanf("%d,%d",&a,&b);( D' T6 J/ T( ?, @4 J& I
SWAP(a,b);$ X4 `( p. ?6 s/ S" I; N; J1 U
printf("交换结果为:a=%d,b=%d\n",a,b);
/ ?4 J1 w" n0 f! Q} # k8 G6 x$ N/ k
6 {1 z A) G$ H- h* P! u$ _. `9 b- q0 e% h! b" a8 I1 Q
9.2
1 ~. ^$ U# A8 X#define SURPLUS(a,b) ((a)%(b))
9 V9 @! v8 J# B. W% _3 jmain(), Y1 R% d# A& |; I* A2 e6 x7 g+ H
{
3 p& z' C; Y# K6 ^ int a,b;
( ?; l+ |' k; T6 g% c5 s% b printf(" 请输入两个整数 a,b:");
" C$ ]9 h4 e5 |( N* u" } scanf("%d,%d",&a,&b);8 M3 P I k' B! U+ [
printf("a,b相除的余数为:%d\n",SURPLUS(a,b));
( d! V, u7 [- L3 @# } }
* {# L5 j. x: B2 g" G3 M7 R9 F) E4 c
" D$ O9 p! X5 C# O4 e5 U
+ |% Q; m, O: ?+ {* I$ t9.3 v3 R$ S# n6 b8 R5 s
#include
* U: K! K9 F- ?5 |#defin S(a,b,c) ((a+b+c)/2)$ F3 a1 O2 ?9 {# _, Q) p
#define AREA(a,b,c) (sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(s(a,b,c)-
, K/ R3 N8 S' d) B6 tc)))/ S8 a- W2 `. u* p. g/ |! G
main()
. k/ ?% f/ }: F, u, ` {
% k% Y' C- P. ]1 W4 f. ^& [" X float a,b,c;; X1 W- @# S- g7 C# D
printf("请输入三角形的三条边:");6 f1 F( O O" A, u4 E
scanf("%f,%f,%f",&a,&b,&c);9 t( G6 S- X5 P
if(a+b>c && a+c>b && b+c>a)
% C: Q( U; }' i# e% a5 [9 N6 R printf("其面积为:%8.2f.\n",AREA(a,b,c));
3 B) r# v4 p9 o* o9 K$ L else
" Q1 v9 Z3 q6 R' D! Q+ B+ w+ O printf("不能构成三角形!");- ^ h; Z$ D# f( }0 g
}. }: [5 W, w) y- p, J1 s! n
- S1 @2 W7 A# Z4 X3 F: n4 @2 O0 o! Q8 E* q. ^8 D
2 j# X- j5 G4 D& a: \9.4
, U) h) y& W$ h5 b#define LEAP_YEAR(y) (y%4==0) && (y%100!=0)||(y%400==0)# ]: t6 f; |( \/ `
main()
5 u* A6 J; C) M! H6 I {
- w: A- N% a% t* B int year;
- a" {$ y- z% }0 \: B printf("\n请输入某一年:");
- {3 k/ \6 F& I' T# g scanf("%d",&year);2 _) o" N# n' o2 R$ M+ z8 ]
if(LEAP_YEAR(year))8 k/ g; H7 z9 Y: }7 Z
printf("%d 是闰年.\n",year);- C, ?/ }7 S1 l
else
2 M2 s; M: V( m9 h printf("%d 不是闰年.\n",year);
2 l+ K" K( o7 D( E% ? }
- y% L, [/ d2 b" ~+ b; v% x1 O2 k1 W
* W5 A/ y5 }7 a, V t
8 N# g7 ~9 z! E/ M' M0 a
1 K% h6 Q8 N9 Y4 p* i$ q& ?2 g9.5解:展开后:! F6 c: D/ E: ~# V1 Z( B! o! U6 Z4 L
printf("&#118alue=%format\t",x);
6 |) e6 P2 E0 _- v) {printf("&#118alue=%format\t",x);putchar('\n');0 r# t( S6 b9 G" o
printf("&#118alue=%format\t");printf("&#118alue=%format\t",x2);putchar('\n');
5 ~6 a! s0 C Y% f& ~0 m. ]' A输出结果:8 `1 ]7 T$ Z1 \5 |4 J
&#118alue=5.000000ormat &#118alue=5.000000ormat
A8 [2 y4 S: b&#118alue=3.000000ormat &#118alue=8.000000ormat6 O4 i6 n& L) b1 L# y
3 `- O/ q8 k3 k, N9 b. X! c
# g. P z. L- M0 s+ v. n9.8
4 l- B" y$ m! u) [" ]! hmain()
/ J. \3 A! S/ b' K3 s8 n4 Q5 B {) X7 \& v, C; M" D4 {
int a,b,c;& s. I$ ^7 B0 Y5 l
printf("请输入三个整数:");
9 {* M' P+ V( v% q7 N3 e scanf("%d,%d,%d",&a,&b,&c);/ _3 P3 j. [, ]$ q( X# W! _4 n: h
printf("三个之中最大值为:%d\n",max(a,b,c));# b4 |* P# k. j! A9 f0 B* e
}; v, K$ K4 O/ I: [
max(x,y,z)
" n& p* A) v; [, g( t D0 Q" { int x,y,z;/ C2 q3 o; ^1 Q& n2 P* L/ R/ k
{; m+ C- h8 M# K! s8 d+ Y9 y- B. T
int t;
: Z c- q7 I. U3 b- G, h& d) y t=(x>y? x:y);
7 d* N2 r9 v+ Z" H$ y' S" v e: d. M return(t>z? t:z);
0 M4 Y+ i" d3 q: Y }
+ N% N9 n2 a6 g9 ^8 l0 j6 |! e
3 l" C* J& ^7 p( {( Y( t0 o
% p$ F! T7 z0 a4 n) ? v- | h4 S; Y, c2 k% X3 u. g
9.10
+ U5 D8 E3 q# C; O- A- B#include0 ^/ {, F8 G+ K- p' q' N9 c
#define MAX 80: _# Y- M: z" a7 {2 Q) z5 |9 |9 T6 u/ o9 H
#define CHANGE 1
( Y. v7 I. K! n) ]# ]: ?( Z8 `/ _main()% s2 r& h9 }( h$ f- G# _
{2 C+ P( o- Q7 y7 m _0 I% n
char str[MAX];8 R- f! G8 D, B* T! c
int i;- y1 Y" _9 M a) w
printf("请输入文本行:\n");6 l+ v, e# K, O/ B/ b
scanf("%s",str);8 }! d" v* p v$ X* j
#if(CHANGE)
0 B1 \7 _4 } w5 I {; e; y5 z, `3 S6 L+ U
for (i=0;i {
* X, F7 v0 ]/ ~" M) u1 b- ]" F if(str[i]!='\0'
" G; d1 W; b) {8 P- e" ?! k if(str[i]>='a' && str[i]<'z' || str[i]>='A'&&str[i]<'Z')
* \6 a- {3 o. D( N9 k! X str[i]+=1;/ c# B" U( W: C4 _6 ?
else if(str[i]=='z' || str[i]=='Z')
) M, R; T. U9 a str[i]-=25;9 G" I# L1 |2 Y2 ^ w+ }& f- v c2 r% y
}
, w% l- ?: @5 k' M}
: h1 u& E( e9 t! [1 d8 l#endif. b9 ]# U1 P% w ]
printf("输出电码为:\n%s",str);
9 _( L1 F$ q' V7 i4 I1 J: j. |}1 e" D3 ~' E z. @" q9 h
第十章 指针
3 J/ F& I4 Q& L% Z9 W; k10.1; y# Y; S1 i; [# v: }
main()
4 s9 }* p$ I: j+ t; P- p$ c! [{int n1,n2,n3;! j/ X3 l* W- B' e; I3 r4 n
int *p1,*p2,*p3;
. d, O4 s! T, l: b, E! |2 q scanf("%d,%d,%d",&n1,&n2,&n3);& l- s: n; `+ U# C) L8 N. ` j
p1=&n1;
) {0 I* S# w! z- W O, I6 T: v% k, X p2=&n2;$ o9 B& n3 Y) |
p3=&n3;# [$ l" o. l1 Q8 i
if(n1>n2)swap(p1,p2);
& a# q: T- |( s/ t3 Y# G) @& d" j if(n1>n3)swap(p1,p3);( L) u1 A0 z+ a. d4 K, F
if(n2>n3)swap(p2,p3);. v p- F0 q) R3 y- |: b* }
printf("%d,%d,%d\n",n1,n2,n3);4 Q' P4 D! k* N4 q1 H: e- q3 J
}
; Y' d7 A! }3 B1 `7 m2 f3 u, |swap(p1,p2) S0 C! r( C$ s3 f4 z1 T2 B
int *p1,*p2;
1 `3 f9 P: }5 a5 ?6 C d{int p;
3 A, p9 c( h1 n2 V2 ^* z1 i& \# o; G p=*p1;*p1=*p2;*p2=p; e: d, m# Q0 C3 ]& U* S- D: i1 C
}
, B5 T8 I- N% g* D6 u d* q5 A10.2
" P- m8 _; [( q" c% wmain()
+ @9 ^, ]: n3 U: W a{char *str1[20],*str2[20],*str3[20];
o0 L" p& I9 g H char swap();9 u- b. d2 u ~" z# F4 x. l
scanf("%s",str1);/ z2 Q# {6 H u
scanf("%s",str2);
. q5 e/ U3 {" ?3 V scanf("%s",str3);
% F. R9 G2 A3 R1 s4 d if(strcmp(str1,str2)>0)swap(str1,str2);; ?1 k( e: |' f: @1 a' h# S
if(strcmp(str1,str3)>0)swap(str1,str3);9 m2 D5 c- D& C% x3 [
if(strcmp(str2,str3)>0)swap(str2,str3);
+ O+ ~6 ~: \ \/ k- J printf("%s\n%s\n%s\n",str1,str2,str3);" ^2 O- g0 l& w" J$ t
}
5 Z% Y+ Y% \ w4 C( o# _char swap(p1,p2)& p. P5 J0 m3 }, I
char *p1,*p2;
) ~% i! n0 e7 N, [7 t3 F, _4 v{char *p[20];
4 y% D2 A% o G: S strcpy(p,p1);2 K6 Y4 e- { t2 J
strcpy(p1,p2); M! C% x1 T# j5 u9 h4 X1 X
strcpy(p2,p);
# {" {1 ^7 u4 M4 ~8 b}" \) i' _1 m# Q/ x" c
10.3
( g9 Z/ h) Y* S- {. N$ N3 Y" P8 i# Nmain()6 P& |6 [3 @. J& U8 F
{int number[10];
3 }! ^9 R4 u1 D6 c- C" _ input(number);
9 G! O+ e9 _. H1 c5 s" c max_min_&#118alue(number);
& ]! w. j+ n0 f B" Z7 R- x0 x0 }& j output(number);
8 P9 z- ~2 ` j* @5 }4 `0 t7 M }* g) G* E3 \8 J3 Q3 h
input(number)
* [7 F# B# W1 m2 n% |, Bint number[10];2 I6 R/ c/ o% q Z/ }' L2 \, Q
{int i;7 G! G7 @3 ?; m4 ]3 K; D' Y, K
for(i=0;i<10;i++)0 G4 O3 N' g1 \& s X5 X
scanf("%d",&number[i]);
9 [$ V& _2 m* C) u}
& w" C: P& v% l0 @. b- Fmax_min_&#118alue(number)
: W& ]! x! s+ g% [3 [6 dint number[10];) M; Z! f t9 E6 P4 v1 Y% t( x
{int *max,*min;
: | o6 ~4 m1 z c int *p,*end;2 V% Z3 m4 D \2 f- R3 m7 Q1 P9 v, e
end=number+10;
1 `0 X! H# x. ^ W b# S, d7 F/ t: l max=min=number;
% w! f/ j9 X$ V3 H: p& O for(p=number+1;p if(*p>*max)max=p;
* \) Y- z6 h+ \: m* _) q7 ^ else if(*p<*min)min=p;
- T9 G% f5 W* f; v. l *p=number[0];
1 |! [+ A$ r+ ~% p3 J% F# c! ` number[0]=*min;0 W6 P. H! [, ^# Q1 Y
*min=*p;
/ S. h3 w9 S1 u: D: N, [' n- m# t *p=number[9];
7 W1 Y# S- N& M1 g5 A7 d number[9]=*max;
4 s, H' j4 C: x, ~0 |: W *max=*p;
! |0 c! v; R6 [1 b return;
5 c$ v F2 G5 Q2 A}
! M, s/ U! V; _output(number)
1 q v6 ~) }/ d3 W3 qint number[10];
1 ~% ?7 b+ v, R6 g0 z* l{int *p;* k* `: w# E" f1 u; k: p. T
for(p=number;p printf("%d,",*p);/ Y3 J/ U/ s3 {
printf("%d\n",*p);* f& s6 x8 z5 V; F" C1 L
}: m) T5 x5 ~! @1 z" E
10.4
3 b- T. ?2 T' e+ Cmain()0 ~; H6 v0 t7 @6 }9 T
{int number[20],n,m,i;
/ F( H! v! V$ F scanf("%d",&n);
$ q) @+ S; n/ M& `" b scanf("%d",&m);8 l" z3 M' }1 f2 V
for(i=0;i scanf("%d",&number[i]);
& J1 r5 k- i$ t C move(number,n,m);
) s8 ?, B! k3 F) B for(i=0;i printf("%8d",number[i]);
+ S% W6 L O7 F u; _0 I: E* u0 f}
8 U! v! i+ g( [* \move(array,n,m)
: l) X9 Q" y6 a8 L2 P6 sint array[20],n,m;
% E( ~9 A2 ?, w9 K! V{int *p,end;
4 o: k8 w2 X' K; b# @6 `6 A( @3 C% R end=*(array+n-1);, X5 `5 o/ G" `4 O% [+ Y
for(p=array+n-1;p>array;p--)3 W2 i, z S0 x `0 n; r; O. }
*p=*(p-1);- \: \. o. |; H; D# _
*array=end;% J) g d3 D1 R2 V* ^
m--;
9 H( l) ~' ~& c if(m>0)move(array,n,m);- Z: c( H9 E" e+ K# E' ?; q
}4 n% z' F, Q- d! O! t
10.5
: y! a2 B4 A+ u* e#define nmax 50
9 n* R7 F B, v- ^& W7 ~main()9 Q7 N3 [. p' i+ o4 d+ Q* t
{int i,k,m,n,num[nmax],*p;: W! c. b& b. v0 a: b \
scanf("%d",&n);% |1 J* V$ Q3 u, n, k! v% `' d+ R- V1 W
p=num;' x1 v9 k& K+ _+ f0 o/ e% [
for(i=0;i *(p+i)=i+1;
3 c! O8 o- Y& @0 } ?: R+ n/ e i=k=m=0;' ?2 f- `" g" q; {1 v& j
while(m {if(*(p+i)!=0)k++;- \, C6 x- u: S
if(k==3)
* Q6 o) p# V+ j* l. G$ n {*(p+i)=0;
% t4 P, _: r8 s- ]; M$ N k=0;
* I/ K) k) G7 U2 L1 P! s! [$ {- x m++;
7 A/ M5 V- S+ A9 f( r }
3 h& y! v, h2 x" z4 s- ] i++;) l+ l; m, i1 B( P
if(i==n)i=0;
2 R T0 L3 D3 t V }
/ z0 A- t5 C3 b1 T while(*p==0)p++;0 T' V1 {# _5 }* F
printf("%d",*p);, s# L# ?' q3 @2 C' x8 Z
}
5 N' c; z& H9 O9 |2 W2 o10.6
- C; A2 `1 R& d1 s. s5 P& tmain() j3 w& H, i* B) N. X
{int len;
" j0 P+ s5 y* f/ i char *str[20];
4 H1 D1 Q8 y: d+ b$ u( v scanf("%s",str);
/ p$ y! H" A V8 I3 _4 N, B1 P len=length(str);. Q# D/ F( u9 V9 [
printf("\nlen=%d\n",len);! } j, R" _8 t' o/ K D2 ^
}6 f) i* ~* G4 g S
length(p)7 t8 x) P% ~5 I4 H0 p
char *p;' z# {& Z7 A) w2 a3 ]! [, k
{int n=0;
$ G( G w4 x" `, J! C3 ?# S0 C. y while(*p!='\0')/ v1 g8 e# X' E0 h5 V: ?1 q* G
{n++;p++;}6 ~6 I% ~' F! w2 r
return(n);
. I& y) a% n; d, M6 k; [ _}
0 F: G! g& q$ ~, H& c10.7
, B0 b2 L. W- umain()
# B3 u# @% U/ f{int m;
/ s/ w) n5 S: S5 e char *str1[20],*str2[20];
7 J# W4 `$ h! m3 l4 x5 X" O scanf("%s",str1);' p8 A9 H! U# u8 [( B4 _
scanf("%d",&m);
& q8 K# l# V. L) V7 n2 [ if(strlen(str1) printf("error");
& {) G$ S; m6 F* m* z( Q else x9 i- Y2 g8 R; a/ E5 i
{copystr(str1,str2,m);
- @- m" {0 N# u) R( a- n, a printf("%s",str2);
7 J7 L# a, z! v* G) A1 E6 G/ @+ c }
! y1 y; X0 n6 m* c+ e- j}
: K8 Y9 Q% j0 C! b# `copystr(p1,p2,m). L* _: f9 k: S' N$ l0 B2 B
char *p1,*p2;
4 g/ F# z5 C( R$ ^6 [$ `$ h9 sint m;
2 W: e, b+ G4 [4 O2 g' V{int n=0;
$ M; E7 i: `* ? while(n {n++;p1++;}3 K6 [8 u$ U5 H) U8 U$ o$ r6 q
while(*p1!='\0')7 ?9 j* F- z% p7 w
{*p2=*p1;
5 [! A; d) r/ x: U p1++;
6 A: B8 M# d3 m p2++;
# u2 x+ R- S* S4 ? }4 b$ V- k6 ~) }
*p2='\0';3 E% d8 d' t2 r' Z" ]0 T
}) l* V* a/ W; i3 n# O
10.8; q! W+ M) F# p) h2 S
#include"stdio.h"' m# P! x) M) V; q; f6 [
main(), p* R& E: r+ e; x( `% f: U
{int cle=0,sle=0,di=0,wsp=0,ot=0,i;* S0 Q2 t% A; y- J2 u
char *p,s[20];
. k) N. G2 ^% l" ~" t( d% y for(i=0;i<20;i++)s[i]=0;% j; o6 b) z; K5 H3 ^* E4 J" b
i=0;! `9 _( `6 L; Y( P3 z$ a* ?: O
while((s[i]=getchar())!='\n')i++;
* I" t6 V) T: c7 h$ ] p=s;
: X1 V/ g. }; k2 A while(*p!='\n')2 ^- _+ r. I+ q" b- w1 x( q4 t' y
{if(*p>='a'&&*p<='z')3 A9 ^/ e) ]: `* Y# v+ i4 H- C
++sle;* P; M$ t) m6 v# L M
else if(*p>='A'&&*p<='Z')
$ `# b% V$ e3 R* `1 d+ c ++cle;
7 _+ \1 f! A, u) d: `) i- X& U else if(*p==' ')& e9 o* W+ q. E2 Q
++wsp;
o" E" V& V4 U; ~( ]7 C) u else if(*p>='0'&&*p<='9')$ ~2 B, h! ?7 I; O3 {6 G8 e2 S2 i) i
++di;
8 @% ~2 E j- c) s/ M$ l3 P else
3 J" ]2 p: A- c3 q" K/ j% \, K ++ot;4 n1 ^8 c% O- H
p++;0 L6 ?6 K$ B7 @7 p! Z: x
}, R. v; k! S* j8 o( y3 @% ^
printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot);
5 J+ \, c; q( f3 I$ d4 D3 Z% u, l4 \}
2 L+ D7 r% R+ g+ \$ r9 J( ~10.94 _8 ^7 Q8 N5 s) [2 M/ S$ p
main()
& ]5 P0 A L% W) q* i) t$ X& ~{int a[3][3],*p,i;! K% \3 ]3 q/ o, r
for(i=0;i<3;i++)
5 J1 z' t; M6 L: g9 f! X scanf("%d,%d,%d",a[i][0],a[i][1],a[i][2]);
/ Q# i7 Z2 d- w2 t q! G! n3 k* i p=a;/ j% q7 y4 T+ W- w; `$ K( O5 E. }
move(p);
8 {9 B4 d3 V& q7 I2 j! R# m% P for(i=0;i<3;i++)5 j" K& L6 f* J q# \- x4 Y
printf("%d %d %d\n",a[i][0],a[i][1],a[i][2]);- f7 }: S; n4 _ @, b" F
}% G# B; X$ }/ b) p7 @! l
move(pointer)
" B$ A9 H. I) Uint *pointer;# [* R, ]8 \) L: S+ a5 Y9 t
{int i,j,t;1 V8 e. b1 a3 B# O: p7 p
for(i=0;i<2;i++)
8 ]* m! D8 b' a for(j=i+1;j<3;j++)0 t+ b; @' U: J
{t=*(pointer+3*i+j);
+ r) y, R2 x* K: {% x) T *(pointer+3*i+j)=*(pointer+3*j+i);
+ Z9 J! L+ K9 \, x *(pointer+3*j+i)=t;& U- I* I L8 E- y* K( R
}' z" w* T/ k9 F8 `) H' {
}
- k' d9 b$ K: e8 s' ^; R' C1 c10.109 c8 R0 _$ {9 k$ a* i- E- F. U
main()( a' a2 K5 n. D$ e1 C0 V
{int a[5][5],*p,i,j;$ U6 i) }! w) W
for(i=0;i<5;i++). N' m9 A, M- y! s; e, X6 ~/ ?
for(j=0;j<5;j++)# H4 p- f/ A- s& K% u0 [* V
scanf("%d",&a[i][j]);, R9 \( j2 D8 P: D& r' p- W
p=a;
5 d' S# L K. W; ?# ~0 C1 y change(p);( d4 G# {/ V- J7 b. d9 s" Q
for(i=0;i<5;i++)
* p1 f. ?3 {$ M {printf("\n");
( W8 h. A1 R2 N: Q9 ~& s for(j=0;j<5;j++)
& ?, m! x. M! T1 W% y3 X" M3 z printf("%8d",a[i][j]);
" @5 T' V: S/ D, T1 B" w }2 I* A' A. ^' @( m2 {, I
}
) T) ]" F. z- m: b; v: Dchange(p)* G- C( D. K7 m6 @! Z: p* f
int *p;# I& ~0 G5 T. I( P3 h
{int i,j,change;7 ^# L# x3 X/ j, L( ]! c
int *pmax,*pmin;
' b4 A- k% N% k% }1 d pmax=p;. [7 D& U/ ^% u# @+ J
pmin=p;
( n5 j" ]. b* C0 q) M: Z for(i=0;i<5;i++)/ O& [+ Y' B5 @- ?3 V# B: A3 [$ {
for(j=0;j<5;j++); N1 N- s! }. K1 R4 [
{if(*pmax<*(p+5*i+j))pmax=p+5*i+j;; p, H$ `$ g4 R& x1 E
if(*pmin>*(p+5*i+j))pmin=p+5*i+j;1 b. l! m0 C. C$ A2 \
}
3 O, _$ d3 W' U6 `. u [ ` change=*(p+12); `6 B$ [" b X4 I% F
*(p+12)=*pmax;4 J7 o1 p3 u4 t' d+ q
*pmax=change;
/ _7 e" [6 n4 b% K$ u$ C2 s change=*p;
5 d( \. q3 ]3 C* k" b+ G z *p=*pmin;
! m( e9 b0 {( B4 Y5 i *pmin=change;
+ E# c7 c# F1 Q: n4 m5 \- ] pmin=p+1;
2 Q# P# w6 {6 M3 o' l8 S K/ E" m for(i=0;i<5;i++) D) F, d' g0 m4 Z$ Y
for(j=0;j<5;j++)& g4 _' [' {! ?
if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;
) m A( e5 k0 S" b$ }' h change=*(p+4);6 q/ n) l" y9 W5 |. Z
*(p+4)=*pmin;2 p( q) h9 O6 D" a) A. K/ i, x
*pmin=change;/ Y* j! s4 c% r- U P
pmin=p+1;
7 ^% B. ]' u1 x for(i=0;i<5;i++)# w5 e! K& a1 O
for(j=0;j<5;j++)( L4 f j: y; z. v; t) n& L
if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))& @7 Z# j$ G. v; a
pmin=p+5*i+j;/ e' Z$ A8 D9 a- h& A& [$ m" w
change=*(p+20);# r5 l( A q/ n' x' }5 h
*(p+20)=*pmin;
V9 E9 C% L. r0 Z6 j *pmin=change;1 X2 [ t' {( |! ], p3 }' }
pmin=p+1;$ f3 R' x' D# Y) I
for(i=0;i<5;i++)
/ _8 @0 X( x% t N5 j& l4 q for(j=0;j<5;j++)
* L2 v( j& e- N, f% F0 ` if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20))
5 H: a8 _& ^. p' d% Y" \2 k &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;" K) N1 g: R1 N8 r3 n+ |
change=*(p+24);
& B% ~4 {9 F. s1 ~ *(p+24)=*pmin;) x4 v, j2 j5 G" y& v2 @. U
*pmin=change;
4 I5 m, G" r' v8 l}! @" F5 Z' Z' r! K
10.11
* C& I. q# E0 H3 U4 C$ b9 Imain()
- G1 w: B* d8 g; R{int i;
, |' y5 t. j- M4 \4 U: S2 a char *p,str[10][10];7 a) s: W4 a8 Z3 ^5 V' N+ {4 [
for(i=0;i<10;i++)# r2 v3 `7 ^' T
scanf("%s",str[i]);
$ j" H. K6 X! a; ?7 R p=str;
* k: Y! W6 L( y$ x8 ~ sort(p);
4 a. P$ J& R2 M& L for(i=0;i<10;i++)$ V# Y# Y! Z2 P+ T$ d0 k3 X
printf("%s\n",str[i]);( p8 z3 Y; w0 d1 J* D3 X
}
$ M0 `, l- ~" X9 X$ Y( n- qsort(p)
4 a) \* {6 R5 F. Schar *p;, Z) J% C p6 e/ r; N C
{int i,j;/ _3 Q1 U/ J. Y
char s[10],*smax,*smin;
( z& L- i8 o) C& n6 | for(i=0;i<10;i++)
6 i3 V! p" w& i5 |5 G+ t {smax=p+10*i;8 r. \+ c4 L' g$ b6 |
for(j=i+1;j<10;j++). E# D1 {9 j: N5 y
{smin=p+10*j;' d5 p# w# ^; c! _! ^: Q
if(strcmp(smax,smin)>0)5 o8 d7 A# S- {2 H6 r- }$ p6 T
{strcpy(s,smin);
. W* K* t; g% @3 g# }8 u strcpy(smin,smax);
/ Y! E, e- `' { M- c7 s strcpy(smax,s);
- p. |; a4 M$ p" S }
6 w. W+ U) P" ^+ q$ [. ?9 ]1 k }
) d, X- ~" i1 \. \( L! ~ }
& Z9 D! R8 W* P# o2 o8 O4 X}7 I4 q# `% O; p: T% _
10.12( F+ z' y5 Z4 d2 p+ h" O. Q
#define MAX 20- v, P! ]2 c( F6 E, S+ d
main()+ C2 Y: g/ n2 ~; H' i) Q, j7 R7 u
{int i;
. Y" w0 A2 _9 F/ _: v# k char *pstr[10],str[10][MAX]; R0 {1 a4 j+ Z9 T/ p% V
for(i=0;i<10;i++)
9 J2 e3 v! ~) q& m3 v. K W1 r pstr[i]=str[i];
+ g2 [9 N8 Q! U7 S3 q7 | for(i=0;i<10;i++)( I7 D; C: Z& H# D" O8 z ]& w( t
scanf("%s",pstr[i]);
/ `0 L# @3 i5 l3 Q; k" U sort(pstr);, Q/ j+ ~4 I' q3 D. i. [- z
for(i=0;i<10;i++)
) B; Z) Y7 ^" r( u7 h- c9 G- f printf("%s\n",pstr[i]);
, |- v! Z( i2 K4 i+ _, E) l) L, B}- G1 l& n; V+ T! x; s, O
sort(pstr)- a, @/ R9 \# U4 f( W6 x
char *pstr[10];* s" G6 c6 Y" @; f- `
{int i,j;
1 b1 c* X! u, N& }+ ]6 b char *p;% ^# t7 S- ^0 i
for(i=0;i<10;i++)* E9 t% Y" W( s6 R) Y3 ?
{for(j=i+1;j<10;j++)
! ~( Q% ]7 q5 e( K7 w7 b+ D {if(strcmp(*(pstr+i),*(pstr+j))>0)3 _! C5 k0 ?' j! g
{p=*(pstr+i);
! g2 }! d I! w3 A *(pstr+i)=*(pstr+j);
6 a- B5 i+ e# I. i- Y6 J u3 a6 R *(pstr+j)=p;) E" z) ?' ~# j$ J, P
}
; H- W+ I" R8 L }
' n8 @* m3 _) v q8 J }8 g9 J6 Y9 h, O% s# T: t! |
}
- U8 ]* ~' B) P* C9 `, X$ h10.13
6 L6 b( R" i) t#include"math.h"
# \" e4 D5 z. G4 {main()
0 Z4 ]8 x9 X( f* Y8 z) T{int n=20;
r, w' G. A( L, w: M6 ]% ^ float a,b,a1,b1,a2,b2,c,(*p)(),jiff();
: n3 I, z2 I/ w/ V' Q scanf("%f,%f",&a,&b);" G' U3 k- q! G4 T+ t: M( e/ R" s
scanf("%f,%f",&a1,&b1); k \8 X4 z, ]) w" S) ?
scanf("%f,%f",&a2,&b2);/ ~' ]0 ]6 S9 E# n2 z4 `
p=sin;+ \+ L4 `3 W/ o5 b5 v$ h( g
c=jiff(a,b,n,p);
|/ m% |4 ]2 x6 F2 { printf("sin=%f\n",c);9 B9 u1 [( ? n+ f5 o0 x. X* x
p=cos;
1 w& j. Q p* a+ T; R c=jiff(a1,b1,n,p);
, a' t, T9 ~5 s2 g1 e printf("cos=%f\n",c);1 g6 z* f: b: k; `+ P
p=exp;
a' _) m% }- J& X c=jiff(a2,b2,n,p);) J' _. n+ J; ~, l9 C5 a
printf("exp=%f\n",c);
& I& r2 U- p) M, P. _( }}
( D) ]+ l' p4 w: i/ lfloat jiff(a,b,n,p)% Q2 |$ O* J; O% G7 l
float a,b,(*p)();- [5 N& F1 r; B$ K& Q1 c: O, Q
int n;
- I4 m; X; l2 C{int i;
, `# g: _. k! H float x,f,h,area;6 r9 ^; W7 Z/ X Q2 ~. {
h=(b-a)/n;
6 v& D- k5 O: b! } x=a;
0 I" D, P: T$ p! x9 { area=0;
1 J/ G- H, F3 o9 p) {9 R: U1 N- K9 _ for(i=1;i<=n;i++)
/ I1 B E1 a7 T% d {x=x+h;
0 T6 c; i" x! b! o area=area+(*p)(x)*h;
; b N% k% N" s( |7 p( N- P }; b7 c& M; _) ]& G( @7 `
return(area);2 {/ W7 j+ ^: h! d" A* V% w
}
0 r# C9 i7 a6 e$ [10.14
9 U) R3 y$ H) fmain()1 |$ k$ [, _ u& [
{int i,n,num[20];# S1 J! N$ g' z2 f# T, ?0 o% O
char *p; c2 K5 g6 q4 p' ^
scanf("%d",&n);
4 i2 S& t. M/ v7 I- E5 O! F ~ for(i=0;i scanf("%d",&num[i]);
+ Z0 L h9 Q; A5 m: P1 O$ ^: Y/ I p=num;# e& w" u" S! c5 K" }# {
sort(p,n);
t' o; k' H! g, f9 Q! ?5 ~4 S3 z# Z for(i=0;i printf("%8d",num[i]);% \ | }% N3 ^- D; w( H
}
- L3 B. A* W& g9 \1 G& l gsort(p,m)+ c6 r! O. O; u' _# M- J i
char *p;8 [0 w H' A6 k Z! D: ?" |9 C
int m;7 Y9 q. o( g$ L& P3 s! g
{int i;
9 S- _; H/ u& E: Y char change,*p1,*p2;# b! Q+ e. J e- P- L2 q& U; G
for(i=0;i {p1=p+i;
( X6 S3 ?" T8 R. T8 Q, i p2=p+(m-1-i);
0 [3 B* h2 G* Z2 F$ \& C( @ change=*p1;
( G+ n& U7 G3 F *p1=*p2;0 y/ e( h$ B8 h" j; n
*p2=change;
2 ~' y, Y% |: j! \4 S( A7 D' W }
6 ?) k; m- d& z5 Z/ h' W+ u5 _}
$ j' ~1 A% g9 n: }0 P5 i- G" Q10.15! w: I" b* p. {
main()- x1 i0 E; W& L/ P- G
{int i,j,*pnum,num[4];
I& }* u, b- k& Q2 S7 F float score[4][5],aver[4],*psco,*pave;9 Z% {5 U$ S' r; P- @% j
char course[5][10],*pcou;/ d9 C/ ~3 q. e" u
pcou=course[0];
( g% ?) z: Z5 M1 }9 Y: l6 X for(i=0;i<5;i++)
\6 a; V/ N/ D& q/ W+ } scanf("%s",pcou+10*i);
2 x; s2 a% J" d& |1 d9 _ printf("number");
* V4 c( }3 `8 G/ U+ h for(i=0;i<5;i++)
, ^" J1 x" Z* s" u8 D2 e printf(",%s",pcou+10*i);
3 x5 ~* G. F4 \5 L4 Q printf("\n");
3 ^0 r) M1 m7 U# A' K psco=score;) D) G! \! I6 P+ z8 K/ e% {) a
pnum=num;) j) s" U& S9 Y6 I) R5 k9 c% q
for(i=0;i<4;i++)
. L1 Y: Z+ U/ p! f, q {scanf("%d",pnum+i);
6 ?: F9 {( v1 p for(j=0;j<5;j++)
C& s! ?! v4 D, k& D9 R$ H scanf(",%f",psco+5*i+j);
- J- F" X& o- _ }7 T/ r* X2 \: p8 C, Z
pave=aver;- }% c4 S1 V6 G
printf("\n");
: v) {, e2 B4 Y6 g avsco(psco,pave);4 v1 T3 j8 H) B7 r; z8 ^2 U
avcour1(pcou,psco);$ o+ ?) ]! `$ O6 q8 `
printf("\n");
3 Q N2 v" a2 m fali2(pcou,pnum,psco,pave);
% o, E, e+ T; ] printf("\n");
# ?" A: o6 o% n. X6 J8 Q good(pcou,pnum,psco,pave);- q {; Q' z3 H" }; l& z4 T
}
/ F" y6 t4 k4 G5 G& Javsco(psco,pave)
+ p3 F4 P$ o* C: x$ U& A& u( F' W3 ~6 @& ifloat *psco,*pave;! N9 a8 }- m) Q& l( t( ]
{int i,j;
3 k7 Y4 }" j1 I* o float sum,average;, R, o/ Q, ?4 @8 j
for(i=0;i<4;i++)
8 ?2 J* r$ m' x! o. f8 f/ I/ c {sum=0;
* o' J: `$ u a( F for(j=0;j<5;j+). Z1 w& A' m5 q2 Y$ B! @8 A9 _) a( x
sum+=(*(psco+5*i+j));! b" o% |6 B7 }6 x
average=sum/5;! d" A. Q* W8 J% I
*(pave+i)=average;
# G6 d# d& A# q9 O }
$ c6 v; |, M! w9 g- o$ ]}
, M" U* M( F: a8 e( E; favcour1(pcou,psco)3 |: D. e' Q. s2 K+ B: d
char *pcou;! e6 F2 [4 J4 D- A, c+ x$ ~
float *psco;
4 S3 Z+ B: Z5 Y0 s% o/ J- |{int i;
( b; G; t; z% Z float sum,average1;% s9 x# V3 T! L% R/ `
sum=0;
% b+ v3 q' g/ k. Z' @ for(i=0;i<4;i++)
: m3 }' R' g' f sum+=(*(psco+5*i))
/ I8 ~) V8 B. G8 j& ]; K | average1=sum/4;
. \2 q: l% c# L" q5 Z' W printf("%s %5.2f\n",pcou,average1);* f) b& l; m% D- O0 G
}
0 R' |8 F2 k% y% v6 a6 pfali2(pcou,pnum,psco,pave)
- o1 e6 h7 u1 y6 ^: Pchar *pcou;0 \ ?) ?7 l" A, Q6 `9 r! P
int *pnum;
2 B" Y- d4 Y! E: }5 Kfloat *psco,*pave;; M' l9 a# d3 m1 ]$ |
{int i,j,k,label;4 [6 n2 o9 {7 v. \& x
printf("\nnumber\n");
1 k+ m- t& W3 L6 y6 e3 w1 S for(i=0;i<5;i++)# ^4 ]- A# V! p8 C* U" p# v
printf("%-8s",pcou+10*i);' A2 ~# r9 ^4 e, s
printf("\naverage\n");
8 A3 n6 _" D: C( s) ^: N for(i=0;i<4;i++)
E5 e: Y2 ?) m) F" \6 y7 {2 V {label=0;
3 E7 p0 h5 A; G' L Y for(j=0;j<5;j++)* S! i4 _6 S2 X5 Y5 w0 K7 }
if(*(psco+5*i+j)<60.0)label++;
2 [9 b5 C. W% a { if(label>=2)
5 [2 s4 T2 T- f: D1 W3 U( i {printf("%-8d",*(pnum+i));( Z9 Q, O3 a! }+ G4 u4 I
for(k=0;k<5;k++)/ d- i7 b6 q9 n; ^5 J, y- }5 d
printf("%-8.2f",*(psco+5*i+k));4 a2 Q, ]& f; k3 y& q
printf("%-8.2f",*(pave+i));
b3 |5 c0 l" K7 N }
4 F, F; x: O4 _1 B' E: i W } D; m1 G, h+ |' `
}
( y' m+ j) M. q5 jgood(pcou,pnum,psco,pave)
4 ^# W {7 [, ?8 X% \0 _char *pcou;8 ^) _$ I; k" B7 p
int *pnum;
$ q0 F( t L5 t- [float *psco,*pave;( Z" k5 D9 Q: f" C6 U. ?
{int i,j,k,label;) V1 I3 h' i; R; s' v
printf("number");
U7 F& r3 f1 m1 F for(i=0;i<5;i++)5 E( s3 V6 x9 J+ f" y: R
printf("%-8s",pcou+10*i);
; N8 S( T( K7 g3 u printf("average");2 l( F9 n4 j3 c3 W D w, Z, k
for(i=0;i<4;i++)
* a' [$ L! n6 ?: V {label=0; S) ~; L5 [- `2 q. u
for(j=0;j<5;j++)
' E2 K( t8 J/ E8 r* g if(*(psco+5*i+j)>=85.0)label++;% P8 L0 C: B/ I$ K, ^
if((label>=5)||(*(pave+i)>=90))+ {8 n3 v# S/ ]2 H i& D" C, N; V7 K" I
{printf("%-8d",*(pnum+i));8 Q+ v) p+ @" G8 Z5 q' ]
for(k=0;k<5;k++)0 ~5 m! z0 d) s9 V+ i5 ]
printf("%-8.2f",*(psco+5*i+k));; {9 v f) n) R
printf("%-8.2f",*(pave+i));: A7 Y. C k! [$ a! r0 x
}
# L5 p: t: U* E2 u8 Z }
4 \. C b' b" r: Q}
6 \6 B1 _7 I) H2 r, V2 z' x" Q/ j10.164 |6 p% ] w# Q1 U( ^" {, W! [! ~
#include"stdio.h"+ j, o9 W9 b' H/ o: M; a8 R
main()( W; v- [. z7 i/ a6 Q) {
{char str[50],*pstr;
7 t& V. ~, \, M/ b5 j% U/ ? int i,j,k,m,e10,digit,ndigit,a[10],*pa;
2 ]( b. j0 S: I0 t* P6 i gets(str);7 D6 }' T5 T8 P) r0 q/ G
pstr=str;
% l3 `( d$ \- D2 _ D pa=a;
# F" J$ F% D% S8 s ndigit=0;
+ C& Z/ ?0 n( k. v i=j=0;
( k% i7 d' o3 S5 V( d% O% F while(*(pstr+i)!='\0')
' ^, j, k! s- ` {if((*(pstr+i)>='0')&&(*(pstr+i)<='9'))# j0 i5 I( T0 @
j++;
[- q. ?9 l S" a1 s N9 n" ]% A- \ else W8 i' A w; D$ }* F7 |
{if(j>0)* ^: z; ]* K6 i3 r$ V# h
{digit=*(pstr+i-1)-48;! k2 A0 r( P+ [9 ]0 m3 W
k=1;- V7 J. J$ s! @- n, l
while(k {e10=1;
$ k. B+ U/ N+ T+ P7 R) A( s for(m=1;m<=k;m++)
& W1 T3 _; {1 E- V" O" a0 w e10=e10*10;
' d( k- }7 C) q! J0 y; s digit+=(*(pstr+i-1-k)-48)*e10;
2 P) ]+ X+ d: c$ ^) D2 C, \+ b9 k k++;
8 `% F% N7 K; x* S }
' }( N6 P, {( \/ a3 F *pa=digit;
3 p0 f) m" h- |3 ]7 {, w ndigit++;: h# u' U$ `: f f9 _, L/ R9 z
pa++;
6 g5 V8 x# [3 _7 I, ~8 U5 x+ A j=0;
8 j' U0 A7 k9 ~. O }
2 {6 `, Z$ g( h3 f! w( u+ W. s0 r }$ z6 C8 B4 K" D# T- T
i++;9 [$ I( @) x4 y" U; U) @
}7 Q6 ]3 ~9 j k1 d( l3 G/ Q+ R
if(j>0)
; b* V: h. E: {! ?: m. z; m8 D {digit=*(pstr+i-1)-48;4 H+ I; g& U$ g" m9 I: N
k=1; n+ ?' p9 \; `2 `$ r$ @
while(k {e10=1;0 e" k* o, u& O/ Y
for(m=1;m<=k;m++)
% v; a# Q5 j- V0 F! s e10=e10*10;7 y$ z/ Z: ~6 ^
digit+=(*(pstr+i-1-k)-48)*e10;
' u2 \9 Q8 S% x# [3 z k++;
& c: _& r' q4 K/ Y" }$ ` }: y8 x5 l8 t4 j8 B8 f5 h
*pa=digit;* F" \! L9 N% v6 h
ndigit++;- ]& p; ~2 i- D/ s" R7 `3 `
j=0;/ W7 g* U5 a; ^5 Z: o( m* a
} ! R/ l) F$ D1 m+ N" h+ |
printf("ndigit=%d\n",ndigit);% k: o' _, j! w8 B, M6 I
j=0;
( C v& x2 m" U! N) d) q/ {+ S pa=a;" k/ T$ t& N! z% n4 P) C7 l
for(j=0;j printf("%d",*(pa+j));
- M3 v- y8 H( P; {% j4 I& h}
/ E# Y" h$ B/ I, G/ x$ a. R10.17
: `* y" e- ?# qmain()
- @, S# [. A: S! ?{int m;
" Y+ w( y4 S f g- v char str1[20],str2[20],*p1,*p2;7 X0 t0 V2 N, Y6 N0 K
scanf("%s",str1);
! Z( A/ L5 J U& H) M3 J. X- I$ h scanf("%s",str2);
) C9 U+ w9 P, V% Z p1=str1;+ M( r- X3 y% v! I2 o
p2=str2;2 f5 k. a, ~1 L3 r3 c& J
m=strcmp(p1,p2);
% j/ s9 T9 R9 n! I1 j printf("%d\n",m);
9 }" A; t0 N: a) h+ X}
! v a: a; ` j7 o" t3 L D. [& `strcmp(p1,p2)
2 c P* o; E8 o- jchar *p1,*p2;9 l, ]5 C7 \' _9 }6 T4 w
{int i=0;; m# V, h# E/ @, p
while(*(p1+i)==*(p2+i))
% x p! @+ P( e if(*(p+i++)=='\0')return(0);. P4 J1 B0 u' l1 M4 ]5 W+ N
return(*(p1+i)-*(p2+i));
: Q1 g! ]. q. S3 `7 P$ p; e}& R6 `% w$ b" K
10.18
4 o4 r& S! p3 c! umain()
" H2 Q. p0 t N" O3 J2 [3 g$ S+ g y{static char *mname[13]={"illeagl","January","February","March",
5 t: i8 S) {! V "April","May","June","July","August","September","October",0 D& f1 c1 l5 Y" i1 c$ N% ]1 x9 a
"November","December"};! G" @9 l: l. i; j3 W
int n;
7 [1 c a& _! l scanf("%d",&n);. }8 G" h, P, D' t$ {+ S
if((n>=1)&&(n<=12))% A+ V3 H, V4 X' h7 s3 T5 F/ b
printf("%s\n",*(mname+n));/ q9 Q O) ~ L( _# q& `+ q: q: }
else/ a7 G2 q3 @ k/ _ F
printf("error");- q" y+ _. a* Q6 I" F' ]- t: } C9 |
}% ]( Y. ^. K4 c+ j! J' D3 V
10.20
! H1 @! J6 f- z: {main()5 x8 I8 y8 W& R/ r
{int i;" k' i6 \( e7 {; I
char **p,*pstr[5],str[5][10];6 w4 W G$ \! [, E
for(i=0;i<5;i++)
0 {; @0 F3 n% h4 h0 L pstr[i]=str[i];# u: y; W1 M9 }& y# P. s. @ O7 P# N
for(i=0;i<5;i++)
# g, m8 N2 b, x3 ~: u3 ~# e scanf("%s",pstr[i]);
+ c' z! T6 f: [" l, Q, p: i" P$ v p=pstr;. C7 R" s' }: Y& C! [
sort(p);5 |6 r R, V$ [- I
for(i=0;i<5;i++) P+ X3 o j- g% ~. {
printf("%s\n",pstr[i]);
3 `9 A/ y0 j2 X0 E. d3 u; ^}
+ v8 b- I% F+ `9 a" S7 C6 d1 C' asort(p)* n( q% O1 |. a5 D
char **P;
# J# k& f: u# s# w{int i,j;; I" T0 l8 L- Q7 o; y
char *pchange;
- G8 H0 n5 `4 t4 w- e4 n for(i=0;i<5;i++)
4 g3 o( \8 w8 Y) {( \ f {for(j=i+1;j<5;j++)
t8 k- }1 r# I% } {if(strcmp(*(p+i),*(p+j))>0)
/ T/ N) r; i0 @, o/ @ {pchange=*(p+i);/ O8 i: o0 \1 M1 m& d
*(p+i)=*(p+j);
) ?! |* G+ P: w: p) K6 P7 y *(p+j)=pchange;
! u. d4 k- W- U% X$ O }
; w: Y, ?& m" o }8 T( G' d" b ?$ [
}
0 O/ o, J, z S/ H6 {}" S, P1 a/ q( S( k2 N
10.215 D+ u0 }; O$ h# c9 l2 Z
main()
+ q3 A/ j! j: w% M{int i,n,digit[20],**p,*pstr[20];
4 H% ?% c/ I( A% o8 n0 f scanf("%d",&n);. G& H6 I" V# u% _0 E
for(i=0;i pstr[i]=&digit[i];" K2 T- ^! ]! p- Z# @8 g |
for(i=0;i scanf("%d",pstr[i]);
7 d2 l5 u) Z) g4 Q" W p=pstr;! K( A8 m6 @( \6 `, j7 G
sort(p,n);
N9 m# u$ d4 J( q$ _) D for(i=0;i printf("%d ",*pstr[i]);
* h; [. \0 `/ J* n+ _}
( x( i* n: G- C K( ~0 d& Asort(p,n), a* _$ E0 Q8 T; k
int **p,n;
' e6 |* Y( y1 E+ j( A1 O6 o{int i,j,*pchange;
6 E" _4 O* c5 D( | for(i=0;i {for(j=i+1;j {if(**(p+i)>**(p+j))
4 T; T. ?: _) @( ] {pchange=*(p+i);
% ^8 ^6 H4 k, i0 n% R* x. e *(p+i)=*(p+j);7 t. R, p4 e5 D+ ?5 W7 d
*(p+j)=pchange;5 t( u4 \8 z7 @0 G S
}
9 |7 m9 C. G; i' | }
) M9 j& i% L4 \- `3 u }
% p, y: h0 r& h- E}6 t. g, b" Y: {% z" T
第十一章 结构体与共用体7 x& I- W& O9 ^
11.1
. C0 n$ `/ C2 ~! i$ Jstruct
! n+ l1 X* B' X# R+ n8 A3 [ {int year;
5 D" v% y; r/ h int month;3 m0 L8 `6 A7 S0 q6 L
int day;- T3 O4 l1 y+ U' r# z9 L" G- J
}date;+ v! P3 W( Z( q+ V) h. |7 O# \( R
main()
# E9 i! Q+ a1 _{int days;6 F8 f$ K' V' T
scanf("%d,%d,%d",&date.year,&date.month,&date.day);
" ~% y$ A) c1 {- g" U switch(date.month)
* u/ m; G; F+ h: _) U- v {case 1:days=date.day;break;( j1 o5 s$ e6 H2 R
case 2:days=date.day+31;break;
! Z8 J. Q' e) \! l case 3:days=date.day+59;break;0 V4 Z# W0 u M8 d( J
case 4:days=date.day+90;break;! x3 ?, V. a3 s" g' m# k3 U5 z
case 5:days=date.day+120;break;
* \( ^( A, o! t. H4 J. z2 z case 6:days=date.day+151;break;% q& c/ e; w/ l
case 7:days=date.day+181;break;
^2 u7 N- I! @ case 8:days=date.day+212;break;- E6 K9 q# [9 ~! \
case 9:days=date.day+243;break;
* W; a6 T4 x3 [( ~+ W' q* M case 10:days=date.day+273;break;
0 Q) ?4 z1 f; F# g% @/ A case 11:days=date.day+304;break;/ ^' @2 h" L- a" I Y+ v4 |
case 12:days=date.day+334;break;
6 e4 h1 B2 K: I( |7 V0 D }0 h- C, K2 j( O, k
if((date.year%4==0&&date.year%100!=0||date.year%400==0)
/ _4 B9 a! {3 i- h7 k &&date.month>=3)
8 L1 d! P" ]/ k days+=1;7 K2 F8 t7 W3 c7 U" W
printf("days=%d\n",days); P# R& p! {" L) [5 h
}% I& l, X8 R3 y* t }& F) K. i9 i9 v1 |
11.2; @) u; n5 @, X
struct dt4 l, }1 V" D* b
{int year;
; w/ \8 m) c. z8 ?5 l9 v int month;
+ c6 h5 c8 l1 U) q8 C7 Z int day;
" C7 y" }& j- }, h }date; L j% F0 L5 J( `2 ?, x: a7 j
main()
, J# i% V n1 i% R3 ?$ K8 V+ G9 e{" v8 H- s- T N/ I3 f* h
scanf("%d,%d,%d",&date.year,&date.month,&date.day);
/ ?5 W5 w2 ?& k0 \1 C4 c printf("\n%d\n",days(date.year,date.month,date.day));* I1 j1 q/ P/ I1 t3 Q
}: i' f4 e) g. a0 `8 z z* o
days(year,month,day); B/ o! e- L5 y, X- o x
int year,month,day; w' f4 U2 n. ?
{int daysum=0,i;
$ h% s( b" s5 a" H0 J static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}2 t) o( ^# I4 x2 v$ [1 _3 B$ I
for(i=1;i daysum+=daytab[i];
6 Z6 S) {/ W/ D daysum+=day;
I" W7 }) E- W2 z3 _. T if((year%4==0&&year%100!=0||year%400==0)&&month>=3)( h8 F# B+ w9 P
daysum+=1;
7 J& n. B# O( T- K return(daysum);8 }: A& [1 H' O3 p6 Z
}# `' A% D1 X) s( t- O
11.3* j1 C" w) Z6 P0 e
11.4
" z4 d/ E$ k/ B#define N 5% X' j& {7 C$ t" z
struct student
l4 X$ T$ @& W0 C {char num[6];- `: J* q( J6 t% {6 d- Q
char name[8];8 Z1 u! T, j# m4 A; h1 _' D/ ?
int score[4];3 v0 X1 Z" ?" M
}stu[N];# M3 Y6 ]8 }* y) ]1 C; M
main()
$ K6 K7 X5 R, T, R{% o- Y% ]4 b' m: |
input(stu);, Z3 ?( \, p t4 m, ]+ ~
print(stu);
2 F5 O" c- V: |}6 _9 W# _6 F' D( Q
input(stu)
% T: _2 F# y' ^9 G* H- w! Q( }2 x0 Astruct student stu[];) t# _: q8 x K
{int i,j;
$ @- i% b) g4 x$ P: R for(i=0;i {printf("number");: C* U% `4 f' S; G+ C5 {
scanf("%s",stu[i].num);! v; I. T9 s. b, _
printf("name");
& f, F! [' X: p/ p/ t9 O; j) ^, @ scanf("%s",stu[i].name);! y1 M' b3 o5 Z. F! |4 F& k
for(j=0;j<3;j++)
d9 G3 [2 q$ z, @+ P {printf("\nscore\n");% H5 H; d% d- _! v
scanf("%d",&stu[i].score[j]);% q! L; V2 w9 @0 X w; \9 X0 K
}
+ X, `7 L. `+ n, l printf("\n");9 s, g) T& O7 Z6 [1 a( B9 ?
}7 P/ M, V3 M8 w1 o9 `
}9 q5 B8 }- v c# [+ |
print(stu)
( ]' x: ?/ e# z* h+ m* Pstruct student stu[];# Q K4 Z; L1 X
{int i,j;" l8 `* R2 q- [" a2 b
printf("\nnumber name score1 score2 score3 \n");; \) Z6 i: D* u; Z; g: t7 P
for(i=0;i {printf("%8s%10s",stu[i].num,stu[i].name);# a! n& S& O) j- ?. T, w
for(j=0;j<3;j++)
, H& C/ Z) p+ z! t& D1 d( e$ e+ W printf("%7d",stu[i].score[j]);
, E$ y+ r6 l5 E9 b# X" s* l printf("\n");
+ V& c& x0 K+ W7 { }
" n% m$ k p1 }}' V; J0 k6 X) e- b/ p
11.56 n9 E1 b% b4 w" ^! @( u
struct student
" X2 ]8 }) L& M% ? {char num[6];
: e! p9 r. j! q3 H7 ^1 G char name[8];
1 I" S- n, K; a0 {3 u- H int score[4];
) W) {! ?$ m0 d3 r* T float avr;
+ b% i* R8 m' l; [ }stu[5];/ {/ Z4 e" f/ [8 s6 R/ }& t9 S- t/ O
main()
/ X' X3 _& r, C3 T5 ?+ d9 U{int i,j,max,maxi,sum;) q' C1 @ x" [$ X8 H
float average;4 K! B6 Y: p5 T( N( n
for(i=0;i<5;i++)
5 x2 O9 d( _5 ?, ]; o {printf("number");( |5 \# Z$ @8 f) `
scanf("%s",stu[i].num);( f) b. Z% _/ \9 x) |2 u) y
printf("name");3 b0 u; G, f; y" m# m: q
scanf("%s",stu[i].name);) r4 S R9 C" \4 B, _$ L
for(j=0;j<3;j++)
( e# l+ ?5 k4 ~9 d$ C/ f {printf("\nscore\n");
5 d, y) s y Z% { scanf("%d",&stu[i].score[j]);- O) v) k0 t. {3 m1 J, I
}
: b% P5 N1 S/ ^1 K+ O$ {' F }
1 s8 \" u6 Q6 W9 U! Z7 W average=0;5 J% E( g& R3 O& l5 @2 q7 D6 H/ e+ D
max=0;/ o1 R% G7 ^# s0 G; j b8 U
maxi=0;3 N m7 {7 {$ O9 c
for(i=0;i<5;i++)
- V: G; {: i7 G+ U3 a* y9 C {sum=0;
7 H& n0 C; d/ {9 M3 A' q for(j=0;j<3;j++)0 |# |0 d" `% L" Z9 v5 y. U$ M: \
sum+=stu[i].score[j];
, [1 ?. [& N! u, X8 ] stu[i].avr=sum/3.0;
2 h5 {; v5 M$ I+ H" K' i average+=stu[i].avr;
: \' l/ }# X) n, p7 r if(sum>max)4 \/ P: [. r o& ]/ k1 C4 S' M
{max=sum;- |0 h$ D4 x2 `( I- H3 q8 q
maxi=i;6 h+ b7 z: C. I. A6 _
}$ D m ^" d, A) Y- D
}- x! N0 l8 G& ]( q# l0 E8 C2 r7 O
average/=5;
7 Q0 F/ J) n" \6 h2 A printf("number name score1 score2 score3 average\n");9 t4 m' K+ u5 S6 B$ z
for(i=0;i<5;i++)
: `! W6 F" M7 Y+ w3 a6 b L4 X {printf("%8s%10s",stu[i].num,stu[i].name);
, W$ h- A; a, M6 Q for(j=0;j<3;j++): p; R, } \1 l3 P
printf("%7d",stu[i].score[j]);
$ \+ D6 R, w! j& j! y, u6 F3 ?) S printf("%6.2f\n",stu[i].avr);
# ?! e' A( B0 [, m, x1 { }
# }! C+ a3 E4 V& ]+ m5 J" b( E% z printf("average=%5.2f\n",average);3 X; r/ d) T" e7 x
printf("The best student is %s,sum=%d\n",stu[maxi].name,max);3 D; q5 Y* \) |3 S/ c& S
}
* F5 b" ~; I' n/ D! F, F, I
: j0 D* D) f# H6 B. X1 O: C |
zan
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