（Problem 21）Amicable numbers

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a  b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

 

原题大意：

d(n)定义为n 的所有真因子（小于 n 且能整除 n 的整数）之和。
如果 d(a) = b 并且 d(b) = a, 且 a  b, 那么 a 和 b 就是一对相亲数（amicable pair），并且 a 和 b 都叫做亲和数（amicable number）。

例如220的真因子是 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 和 110; 因此 d(220) = 284. 284的真因子是1, 2, 4, 71 和142; 所以d(284) = 220.

计算10000以下所有亲和数之和。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<ctype.h>
#include<stdlib.h>
#include<stdbool.h>

int FactorSum(int n)  //计算n的所有小于n的因素和
{
	int i;
	int sum=1;
	for(i=2; i<=n/2; i++)
	{
		if(n%i==0)
			sum+=i;
	}
	return sum;
}

int main()
{
	int t,i=2;
	int sum=0;
	while(i<10000)
	{
		t=FactorSum(i);
		if(t!=i && FactorSum(t)==i) 
			sum+=i;
		i++;
	}
	printf("%d\n",sum);
	return 0;
}

 

Answer:

31626