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Project Euler 21–Amicable numbers

By Wu Yudong on August 12, 2018

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.

题目大意:

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>

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
Completed on Wed, 24 Jul 2013, 06:07

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