请教高手“完数”怎么求啊！

[tyranter]

完数的算法思想啊，急急急！！！
1000以内得完数用程序输出！

[rookie001]

bool iscomdata(int nValue)
{
int i;
int nCount = 1;
int nHalf = (nValue+2)/2;

for(i=2;i<nHalf;i++)
{
if(nValue%i==0)
nCount+=i;
}
if(nCount==nValue)
return true;
return false;
}

[rookie001]

没时间想...帮你找了个...再用一个循环...逐个判断并输出就好了~~

[magipan]

弄个函数求一个数的因子之和

int fun1(int a)
{
	int sum=0,i;
  for(i=1;i<=a/2;i++)
  {
		if(a%i==0)
		{
			sum+=i;
		}
	}
	return sum;
}

[t1397018]

#include<stdio.h>
#define N 1000
void main()
{
   int i,yz,count=0;
   for(i=2;i<=N;i++)
   {
	   for(yz=1;yz<=(i/2+1);yz++)
	   {
	       if(i%yz==0)
			   count+=yz;
	   }
	   if(count==i)
		   printf("%d\n",i);
	   count=0;
   }
}

[hb0800092]

#include<stdio.h>
int main()
{
int i,sum;
for (i = 2; i < 10000; i++)
{
sum = 0;
for (int j = 1; j <= i/2; j++)
{
if (i%j == 0)
sum += j;
}
if (sum == i)
printf("%d ",i);
}
return 0;
}

[猫已经找不回了]

http://download.csdn.net/source/469495
懒得很，去下载吧

[feng4206yu]

#include "stdio.h"
#include "conio.h"

int main()
{
    int n, m, sum;

    for (n = 1; n <= 1000; n++)
    {
        sum = 0;

        for (m = 1; m <= n / 2; m++)
            if (n % m == 0) 
                sum += m;

        if (n == sum)
        {
            printf("%d its factors are", n);
            for (m = 1; m <= n / 2; m++)
                if (n % m == 0) 
                    printf("%d,", m);
            printf("\b \n");
        }
    }

    getch();

    return 0;
}

[tyranter]

谢谢各位大虾！受教！！！...