码迷,mamicode.com
首页 > 其他好文 > 详细

Project Euler:Problem 21 Amicable numbers

时间:2015-06-01 16:43:34      阅读:126      评论:0      收藏:0      [点我收藏+]

标签:c++   project euler   

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.



求满足这个要求的数的和



#include <iostream>
#include<map>
using namespace std;

int count(int a)
{
	int res = 0;
	for (int i = 1; i < a; i++)
	{
		if (a%i == 0)
			res += i;
	}
	return res;
}


int main()
{
	map<int, int>mp;
	for (int i = 1; i < 10000; i++)
	{
		mp[i] = count(i);
	}
	int res = 0;
	for (int i = 1; i < 10000; i++)
	{
		if (mp[i] != i&&mp[i] < 10000 && mp[mp[i]] == i)
		{
			//cout << i << " " << mp[i] << endl;
			res+=i;
		}
	}
	cout << res << endl;
	system("pause");
	return 0;
}


Project Euler:Problem 21 Amicable numbers

标签:c++   project euler   

原文地址:http://blog.csdn.net/youb11/article/details/46311467

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!