#include<stdio.h>
int main()
{
    int a,b,i,j,sum1=0,sum2=0;
    for(a=6;a<=5000;a++)
    {
        for(i=1;i<a/2;i++)
        {
            if (a%i==0)
            {
                sum1+=i;
             }
        }
        for(b=6;b<=5000;b++)
        {   
            for(j=1;j<b/2;j++)
            {
                if (b%j==0)
                {
                    sum2+=j;
                }
            }
   
           if (b=sum1&&a==sum2&&a!=b)
           {
             printf("%d与%d是1对亲密数\n",a,b);
           }
        }
    }

    return 0;
}

循环条件应该是i<=a/2，第二个应该是从1到sum1/2循环。

[ 本帖最后由 xzlxzlxzl 于 2015-3-24 14:43 编辑 ]

来个笨办法

#include <stdlib.h>
#include <stdio.h>

#define N 5000

int main(void) {
    int i, number = 0, friend = 0, temp = 0;

    while(++number <= N) {
        for(i = 1; i < number / 2 + 1; i++) 
            if(number % i == 0)
                friend += i;

        if(number >= friend)
            continue;

        for(i = 1; i < friend / 2 + 1; i++)
            if(friend % i == 0) 
                temp += i;

        if(number == temp)
            printf("(%d\t%d)\t", number, friend);

        friend = temp = 0;
    }

    return 0;
}