第1个回答 2017-11-07
2 个关键:
2位数字的随机数: a[i]= 10 + rand() % 90;
10位或个位 含5 的 并高于平均值的 数:
if (a[i]>ave && ( a[i]%5==0 || (a[i]/10) %5 == 0)) {b[n]=a[i];n++;}
程序:
#include <stdio.h>
#include <time.h>
int main()
{
int a[50],b[50];
int i,j,n,t;
float ave=0;
srand((unsigned)time(NULL));
for (i=0;i<50;i++){
a[i]= 10 + rand() % 90;
ave = ave + a[i];
}
ave = ave / 50.0;
for (i=0;i<50;i++) printf("%d ",a[i]);
printf("\nave=%g\n",ave);
n=0;
for (i=0;i<50;i++){
if (a[i]>ave && ( a[i]%5==0 || (a[i]/10) %5 == 0)) {b[n]=a[i];n++;}
}
for (i=0;i<n-1;i++) for (j=i+1;j<n;j++)
if (b[j]>b[i]) {t=b[i];b[i]=b[j];b[j]=t;};
for (i=0;i<n;i++) printf("%d ",b[i]);
return 0;
}本回答被提问者采纳