#include <stdio.h>
void main()
{
    int a[10]={1,2,3,4,5,6,7,8,9,10};
    int i,j;
    int count;
    for (i=0;i<=9;i++)
    {
        count=0;
        for (j=1;j<=i;j++)
        {
            if (a[i]%a[j]==0)
                count++;
        }
        if (count==2)
                printf("%d ",a[i]);

    }
}                答案怎么是 4 ，9 ？求解啊 。，

#include <stdio.h>
#include<stdlib.h>
int main()
{
    int a[10]={1,2,3,4,5,6,7,8,9,10};
    int i,j;
    int count;
    for (i=0;i<=9;i++)
    {
        count=0;
        for (j=0;j<=i;j++)
        {
            if (a[i]%a[j]==0)
                count++;
        }
        if (count==2)
                printf("%d ",a[i]);

    }
   system("PAUSE");
    return 0;
}       

重a[0]开始%，一直到a[i]本身

你那样没对1进行取余，所以都少了1个因数，你得到的是有3个因数的

#include <stdio.h>
void main()
{
    int a[10]={1,2,3,4,5,6,7,8,9,10};
    int i,j;
    int count;
    for (i=0;i<=9;i++)
    {
        count=0;           
        for (j=1;j<=i;j++)
        {
            if (a[i]%a[j]==0)
                count++;
        }
        if (count==1)        //若count=2,则表示a[i]能被三个数整除，所以改为count=1，表示a[i]只能被两个数整除
                printf("%d ",a[i]);

    }
}

#include <stdio.h>
void main()
{
    int a[10]={1,2,3,4,5,6,7,8,9,10};
    int i,j;
    int count;
    for (i=0;i<=9;i++)
    {
        count=0;
        for (j=1;j<=a[i];j++)
        {
            if (a[i]%j==0)
                count++;
        }
        if (count==2)
                printf("%d ",a[i]);

    }
}