最新湖南工业大学C语言考试题目.doc

1、湖南工业大学C语言题目1、#include <conio.h> 改错 #include <stdio.h> int arrmax( int arr33) int i,j,max; max=arr00; /*found*/ for ( i=0; i<3; i+) 此处将1改为0，并且去掉等号 for ( j=0; j<3; j+) if (max < arrij) max = arrij; return (max); void main() int a33=4,4,34,7,3,12,5,6,5,i,j; /*found*/ printf("T

2、he max is: %dn",arrmax(a); 将arrmax(a00)改为arrmax(a) 2、#define M 10 #include<stdio.h> void main( ) int i,a100,t; /clrscr(); for(i=0;i<M;i+) /*found*/ ai=i+1; /*found*/ for(i=0;i<M/2;i+) aM-i-1=ai;for(i=0;i<M;i+) printf("%5d",ai); printf("n"); 3、#include <coni

3、o.h> 编辑 #include <stdio.h> int fun(int m) /*begin*/int a1,a2,a3,s;a1=m/100;a2=m%100/10;a3=m%10;s=a1+a2+a3;return(s); /*end*/ void main() int m,s; /clrscr(); printf("Enter m :n"); scanf("%d",&m); s=fun(m); printf("Result is: %dn",s);4、#include <conio.h>

4、; 改错 #include <stdio.h>main() int a10=10,4,2,7,3,12,5,34,5,9,i,s; clrscr(); s = 0; /*found*/ for ( i=0; i<10; i+) 去掉等号 if (ai % 2 = 0) /*found*/ s = s + ai; 将i改为ai printf("The result is: %dn", s); 5、#include <stdio.h> 填空 #include <conio.h> void main( ) int num,k; /clrs

5、cr() ; /*found*/ k=0; 给k赋值 printf("Please enter a number:") ; scanf("%d",&num) ; do /*found*/ k=k+num%10 ; num%10可求出num的各位 num/=10; while(num) ; printf("n%dn",k) ; 6、#include <conio.h> 改错 #include <stdlib.h> #include <math.h> void main() int i,j,x,

6、p; /clrscr(); x=1; /*found*/ j=12; while(j>1) 去掉等号 p=(x+1)*2; /*found*/ x=p; j-; printf("total is %dn",p); 7、#include "stdio.h" 改错 void main() long int f20=1,1; int i;/clrscr(); /*found*/ f0=1;f1=1; 赋初值 for(i=0;i<20;i+) fi+2=fi+fi+1; 保证前面两个数要输出，所以要这样改/*found*/ if(i%5=0) pri

7、ntf("n"); i%5=0保证一行5个数输出 printf("%8ld",fi); printf("n"); 8、#include <stdio.h> 编辑 #include <math.h> #include <stdlib.h> double fun(double x) /*begin*/ int q; double n,t,s; n=1; s=0; t=1; q=1; while(fabs(t)>=x) s=s+t; n=n+2; q=-q; t=q/n; return s; /*e

8、nd*/ void main() double x,sum; printf("Enter x:n"); scanf("%lf",&x); sum=fun(x);printf("sum=%.4fn",sum) NONO( ); 9、#include <conio.h> 改错 #include <stdio.h> int max; fun ( int arr,int n ) int pos,i; max = arr0; pos = 0; for ( i=1; i<n; i+) if (max <

9、arri) max = arri; /*found*/ pos= i; return (pos); void main() int a10=1,4,2,7,3,12,5,34,5,9,n; /clrscr(); /*found*/ n = fun(a,10); printf("The max is: %d ,pos is: %dn", max , n); 10、#include "stdio.h" 编辑 #include "math.h" #include "conio.h" #include "stdl

10、ib.h" void main() int m=6,n=45,t,i,temp;/*begin*/ temp=n*m; while(m!=0) i=n%m; n=m; m=i; t=temp/n; 注： 此时最大公约数是n,要通过此程序会求最大公约数 最小公倍数等于两个数的乘积除以它们的最大公约数 /*end*/ printf("The Lowest Common Multiple of %d and %d is %dn",m,n,t); NONO(m,n,t); 11、#include <math.h> 填空 #include <conio.h

11、> #include <stdio.h> main() int m,n,i,t; long int s=0; /clrscr(); scanf("%d,%d",&m,&n); if( m>n ) t=m; m=n; n=t; /*found*/ for (i=m;i<=n;i+) if ( i%3=0 && i%7=0 ) s += i; /*found*/ printf("Sum is : %ldn",s ); 12 #include <stdio.h> 填空 void main

12、() long s, t, sl=1; int d; /clrscr(); printf("nPlease enter s:"); scanf("%ld", &s); t = 0; while ( s > 0) d = s%10; /*found*/ if (d%2=0) 判断余数 t=d * sl + t; sl *= 10; /*found*/ s =s/10; 遇到奇数退出寻环后的处理 printf("The result is: %ldn", t); 13、#include "stdio.h"

13、 填空#include "string.h" void main() char s10; int i; long int n; /clrscr(); /*found*/ n=0; 赋初值 scanf("%s",s); for(i=0;i<strlen(s);i+) /*found*/ n=n*10+si-'0' 将字符串转换成整数的方法 printf("n=%ldn",n); 14、#include "stdio.h" 编辑 #include "math.h" #inclu

14、de "stdlib.h" double fun(int n) /*begin*/double y=1.0;int i;for(i=2;i<=n;i+)y+=1.0/(i*i);return (y) /*end*/ void main() double s; int n; /clrscr(); printf("Enter s:n"); scanf("%d",&n); s=fun(n); printf("s= %.3fn",s); NONO( ); 15、一样的.16、#include "st

15、dio.h" 编辑 #include "math.h" #include "stdlib.h" double fun(double x) /*begin*/跟第8题一样 这个我也不知道了 /*end*/ void main() double x,sum; printf("Enter x:n"); scanf("%lf",&x); sum=fun(x); printf("sum=%.4fn",sum);17、#include <conio.h> 编辑 #include

16、 <stdio.h> int fun(int a33) /*begin*/ int max=a00; int i,j; for(i=0;i<3;i+) for(j=0;j<3;j+) if(max<aij) max=aij; /*end*/ void main() int a33=1,2,3,4,9,5,7,8,6; int i,j,max; / clrscr(); printf("array is:n"); for(i=0;i<3;i+) for(j=0;j<3;j+) printf("%5d",aij); p

17、rintf("n"); max=fun(a); printf("Result is: %dn",max); 18、#include <conio.h> 改错 #include <stdio.h> void main() int a33=4,4,34,37,3,12,5,6,5,i,j,max,min; /clrscr(); max = min = a00; for ( i=0; i<3; i+) /*found*/ for ( j=0; j<3; j+) 将1改为0 if ( max < aij ) max =

18、aij; /*found*/ if (min >aij) 小于号改为大于号 min = aij; printf("The max is: %dn", max); printf("The min is: %dn", min); 19、#include <conio.h> 填空 #include <stdio.h> double average( int arr , int n ) int k=0,i; double s; s = 0; for ( i=0 ;i<n; i+) /*found*/ if (arri % 2

19、=1) 是恒等号 s = s + arri; k+; return (s/k) ; void main() int a12=10,4,2,7,3,12,5,34,5,9,21,18; double s;/ clrscr(); /*found*/ s = average(a,12); 和第一题结果类似 printf("The result is: %.2fn", s); 20、#include "stdio.h" 编辑 #include "math.h" #include "conio.h" #include &qu

20、ot;stdlib.h" void main() int m=76,n=40,t,i; /*begin*/i=n*m; int r; while(n!=0) r=m%n; m=n; n=r; t=m;/*end*/ printf("The Highest Common Divisor of %d and %d is %dn",m,n,t); NONO(m,n,t); NONO( int m,int n,int t) FILE *f; f=fopen("D:exam01700311PROGOUT.DAT","w"); fpri

21、ntf(f,"Maximal Common Divisor Of %d and %d is %dn",m,n,t); fclose(f); 21、#include <math.h> 填空 #include <conio.h> #include <stdio.h> long int fun(int x) long int s=1; int i; for (i=1;i<=x;i+) s*=i; /*found*/ return (s); void main() long int m=12,n=2,s1,s2,s3; /clrscr();

22、 s1=fun(m); s2=fun(n); s3=fun(m-n); /*found*/ printf("Cmn is : %ld",s1/（s2*s3)）; 22、#include "stdio.h" 编辑 #include "math.h" #include "conio.h" #include "stdlib.h" void main() int m=1234,a,b,c,d; /*begin*/a=m%10;b=m%100/10; 这里补充的4行是求四位数各个数位的表示方法c=m%10

23、00/100;d=m/1000; /*end*/ printf("m=%d,%d %d %d %dn",m,a,b,c,d); NONO(m,a,b,c,d); NONO( x,a,b,c,d ) int x,a,b,c,d; FILE *f; f=fopen("D:exam01701219PROGOUT.DAT","w"); fprintf(f,"x=%d: %3d%3d%3d%3d#n",x,a,b,c,d); fclose(f); 23、 #include <math.h> #include <conio.h> #include <stdio.h> void main() int a10=2,3,5,7,8,9,10,11,12,13,i,j,k,count; / clrscr(); count = 0; for ( i=0; i<10; i+) k=sqrt(ai); for ( j=2; j<=k; j+)if (ai % j = 0) break; /*found*/ if(j>k) count+; /*found*/printf("prime number(s) is(are): %d"