#include<stdio.h>
int main()
{
    int N,i,j,s;
    scanf("%d",&N);
    for(i=1;i<=N;i++)
    {
        s=0;
        for(j=1;j<i;j++)
        {
            if(i%j==0)
            {
                s=s+j;
            }
        }
        if(s==i)
        {
            printf("%d its fastors are %d\n",i,j);
        }
    }
    return 0;
}
怎么改进代码才能使每个因数都输出，如下
6 its fastors are 1 2 3
28 its fastors are 1 2 4 7 14
496 its fastors are 1 2 4 8 16 31 62 124 248
这是个完数题

#include<stdio.h>
int main()
{
    int N,i,j,s;
    int fac[10],k=0;
    scanf("%d",&N);
    for(i=1;i<=N;i++)
    {
        s=0;
        for(j=1;j<i;j++)
        {
            if(i%j==0)
            {
                s=s+j;
                fac[k]=j;
                k++;
            }
        }
        if(s==i)
        {
            printf("%d its fastors are:",i);
            for(k--;k>=0;k--)
                printf("%d ",fac[k]);
            printf("\n");
        }
        else
        {
            k=0;
        }
    }
    return 0;
}

30
6 its fastors are:3 2 1
28 its fastors are:14 7 4 2 1

领教了

2楼似乎是正解

#include<stdio.h>
int main()
{
    int n=0,i=0,sum;
    for(n=2;n<1000;n++){
        sum=0;
        for(i=1;i<n;i++){
            if(n%i==0){
                sum += i;
            }
        }
        if(sum == n){
            printf("%d its facotors are ",n);
            for(i=1;i<n;i++){
                if(n%i==0){
                    printf("%d ",i);
                }
            }
            printf("\n");
        }
    }
    return 0;
}

#include<stdio.h>
int main()
{
    int n=0,i=0,sum;
    for(n=2;n<1000;n++){
        sum=0;
        for(i=1;i<n;i++){
            if(n%i==0){
                sum += i;
            }
        }
        if(sum == n){
            printf("%d its facotors are ",n);
            for(i=1;i<n;i++){
                if(n%i==0){
                    printf("%d ",i);
                }
            }
            printf("\n");
        }
    }
    return 0;
}