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C语言上机试题：9
1．题目：编一个Ｃ程序，它能读入一串浮点数（以－９９９９．０为结束标记，－９９９９．０不算在内，相邻的两个数用空格隔开），计算并输出这些数的平均值以及这些数中大于、小于平均值的数的个数。(注：可执行文件名为e9_1.exe)
程序设计：
#include "math.h"
main()
{float a[100],ave=0.0;
int ma=0,mi=0,qe=0,i=-1,n;
printf("Please enter float number:\n");
do {i++;
scanf("%f",%26amp;a[i]);
}
while (fabs(a[i]+9999.0)%26gt;1e-4);
n=i;
for (i=0;i%26lt;n;i++)
ave+=a[i];
ave/=n;
for (i=0;i%26lt;n;i++)
if (a[i]%26lt;ave) mi++;
else if (a[i]%26gt;ave) ma++;
else qe++;
printf("%26lt; %d,= %d,%26gt; %d\n",mi,qe,ma);
}
2．题目：编一个Ｃ程序，它能读入文件ｆ１．ｃ和ｆ２．ｃ中的所有整数，并把这些数按从小到大的次序写到文件ｆ３．ｃ中，文件中的相邻两个整数都用空格或换行符隔开。(注：可执行文件名为e9_2.exe)
程序设计：
#include "stdio.h"
main()
{FILE *fp1,*fp2,*fp3;
int a[100],n,i,j,k,x;
if ((fp1=fopen("f1.c","wb"))==NULL)
{printf("f1.c can not open!\n");
exit(0);
}
x=9999;
while (x!=-9999)
{scanf("%d",%26amp;x);
if (x!=-9999) fwrite(%26amp;x,sizeof(int),1,fp1);
}
fclose(fp1);
if ((fp2=fopen("f2.c","wb"))==NULL)
{printf("f2.c can not open!\n");
exit(0);
}
x=9999;
while (x!=-9999)
{scanf("%d",%26amp;x);
if (x!=-9999) fwrite(%26amp;x,sizeof(int),1,fp2);
}
fclose(fp2);
if ((fp1=fopen("f1.c","rb"))==NULL)
{printf("f1.c can not open!\n");
exit(0);
}
if ((fp2=fopen("f2.c","rb"))==NULL)
{printf("f2.c can not open!\n");
exit(0);
}
if ((fp3=fopen("f3.c","wb"))==NULL)
{printf("f3.c can not open!\n");
exit(0);
}
i=-1;
while (!feof(fp1))
fread(%26amp;a[++i],sizeof(int),1,fp1);
fclose(fp1);
i--;
rewind(fp2);
while (!feof(fp2))
fread(%26amp;a[++i],sizeof(int),1,fp2);
fclose(fp2);
n=i;
for (k=1;k%26lt;n;k++)
for (j=n-1,i=0;i%26lt;n-k;i++,j--)
if (a[j]%26lt;a[j-1])
{x=a[j];a[j]=a[j-1];a[j-1]=x;}
fwrite(a,sizeof(int),n,fp3);
fclose(fp3);
if ((fp3=fopen("f3.c","rb"))==NULL)
{printf("f3.c can not open!\n");
exit(0);
}
while (!feof(fp3))
{fread(%26amp;x,sizeof(int),1,fp3);
printf("%d ",x);
}
printf("\n");
}
3．题目：编一C程序，它能读入一正整数n（n%26lt;10），再输出如下图形：
共 n 行.。(注：可执行文件名为e9_3.exe)
程序设计：
main()
{int n,i,k,j;
printf("Please enter n(%26lt;10)=");
scanf("%d",%26amp;n);
printf("\n");
for (i=0;i%26lt;n;i++)
{printf("\n");
for (k=0;k%26lt;i;k++)
printf("%c",' ');
for (j=1;j%26lt;n+1;j++)
printf("%d",j);
}
printf("\n");
}
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