C语言程序设计_配套习题_答案

《C语言程序设计》配套习题集

答案

李忠月 2010-5-10

目录

概述....................................................................................................................................2 类型、运算符与表达式...................................................................................................19 分支结构..........................................................................................................................21 循环结构..........................................................................................................................26 函数..................................................................................................................................36 数组..................................................................................................................................46 C预处理器.......................................................................................................................59 指针..................................................................................................................................60 结构..................................................................................................................................65 位运算..............................................................................................................................76 文件..................................................................................................................................77

概述

一、

1

单项选择题。

2

3

4

5

6

7

8

9

10

D

11

B

12

A

13

A

14

C

15

D

16

B

17

A

18

B

19

B

20

A

21

D

C

A

B

C

C

C

C

C

A 二、

程序设计题。

1.

#include<stdio.h> int main() { printf("%d\n", EOF); return 0; } 2.

#include<stdio.h> int main() { int a, b;/*变量的声明*/ scanf("%d%d", &a, &b);/*输入两个整数*/ /*求两个整数的和, 乘积, 差和商, 并且输出*/ printf("%d %d %d %d\n", a+b, a*b, a-b, a/b); return 0; } 3.

#include <stdio.h> #include <math.h> int main ()

{ double x1,y1,x2,y2; double a,b,c; scanf("%lf %lf %lf %lf", &x1,&y1,&x2,&y2); a = (x1-x2)*(x1-x2); b = (y1-y2)*(y1-y2); c = sqrt(a+b); printf("%.2lf\n",c); return 0; } 4.

/* count blanks, tabs, and newlines */ #include<stdio.h> int main() { int c, nb=0, nt=0, nl=0; while((c=getchar())!=EOF){ if(c==' ') ++nb; if(c=='\t') ++nt; if(c=='\n') ++nl; } printf("%d %d %d\n", nb, nt, nl); return 0; } 5.

#include<stdio.h> #include<math.h> int main() { int i=1; while( !(i%3==1 && i%5==3 && i%7==5 && i%9==7)) i++; printf("%d\n", i); return 0; }

6.

#include <stdio.h> #include <math.h> int main() { int i; double sum=0; for(i=2; i<=10; i++) sum=sum+sqrt(i); printf("%.10f\n", sum); return 0; } 7.

#include<stdio.h> int main() { float s=100, h=100; int i; h=s/2; for(i=2; i<=10; i++){ s=s+2*h; h=h/2; } printf("%.2f\n",s); return 0; } 8.

#include<stdio.h> #include<math.h> int main() { float s=1, t=1, i=3; while(fabs(1/t)>=1e-6) { t = -t * (i-1) * i; s = s + 1/t; i += 2; } printf("%f\n", s); return 0;

} 9.

#include<stdio.h> #include<math.h> int main() { double t=1, sum=1, pi=0; int i=1, k=1; while(fabs(t)>=1e-6){ i = i+2; t = 1.0/i; k=-k; sum = sum + k*t; } pi=4*sum; printf("%f\n", pi); return 0; } 10.

#include<stdio.h> #include<math.h> int main() { double x=27; int n=2; while( !( pow(27, 1.0/(2*n))<1.00001 && pow(27, 1.0/(2*(n-1)))>1.00001) ) n++; printf("%d\n", n); return 0; } 11.

#include <stdio.h> #include <math.h> int main() { float y=1.05; int n=1; while(!(pow(1.05, n)<1e6 && pow(1.05, n+1)>1e6)) n++; printf("%d,%.3f\n", n, pow(1.05, n));

} 12.

#include <stdio.h> #include <math.h> int main() { double s=0, a=81; int i; for(i=1; i<=30; i++){ s=s+a; a=sqrt(a); } printf("%.3f\n",s); return 0; } 13.

#include <stdio.h> int main() { int a, b, i; int asum, bsum;

for(a=6; a<=5000; a++) { asum=0; for(i=1; i<a; i++)/*求a的因子之和asum*/ if(a%i==0) asum=asum+i; b=asum;/*a的因子和等于b*/ bsum=0; for(i=1; i<b; i++)/*求b的因子之和bsum*/ if(b%i==0) bsum=bsum+i; /*判断b的因子和是否等于a, 并且a不等于b*/ if(bsum==a && a!=b) printf("%d,%d\n", a, b); }

} 14.

#include<stdio.h> #include<math.h> int main() { int x, y, x1, y1; double z, z1; z1=10*cos(0-4)+5*sin(0-2); x1=0; y1=0; for(x=0; x<=10; x++) for(y=0; y<=10; y++) { z=10*cos(x-4)+5*sin(y-2); if(z<z1){ z1=z; x1=x; y1=y; } } printf("%d,%d\n", x1, y1); return 0; } 15.

#include <stdio.h> int main() { int a,b,c; int count=0; for(a=1; a<=100; a++) for(b=1; b<=100; b++) for(c=1; c<=100; c++) if((a*a+b*b)==c)/*注意不要写成c/(a*a+b*b)==1*/ count++; printf("%d\n", count); return 0; } 16.

#include<stdio.h> #include<math.h> int main() { int x, y, z, rslnum=0; int k=(int)sqrt(2000); for(x=-k; x<=k; x++) for(y=-k; y<=k; y++) for(z=-k; z<=k; z++) if(x*x+y*y+z*z==2000) rslnum++; printf("%d\n",rslnum); return 0; } 17.

#include<stdio.h> #include<math.h> float f(float); int main() { float max, …… 此处隐藏：4241字，全部文档内容请下载后查看。喜欢就下载吧 ……