Problem 23 Non-abundant sums （盈数）

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最新推荐文章于 2019-02-11 13:03:46 发布

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分类专栏： Project Euler ACM_其他数论 ACM_STL题文章标签： Problem 23 Non-abundant sums 盈数

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本文探讨了完美数、不足数及丰数的概念，并通过数学分析表明所有大于28123的整数都可以表示为两个丰数之和。提供了一个算法实现，用于找出所有不能表示为两个丰数之和的正整数之和。

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Non-abundant sums

Problem 23

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <fstream>
#include <cctype>

using namespace std;
bool isabundant (int n)
{
	long long sum=1;
	int k=n;
	for(int i=2;i<=k;i++)
	{
		int p=1;
		while(k%i==0)
		{
			p*=i;
			k/=i;
		}
		sum*=(p*i-1)/(i-1);
	}
	return sum-n>n;
}
vector<int>abundant;
int main()
{
	int flag[28123]={0};
    int ans=0;
	for(int i=1;i<=28123;i++)
	{
		if(isabundant(i))//abundant数 
		{
			abundant.push_back(i);
		}
	}
	for(int i=0;i<abundant.size();i++)
	for(int j=i;j<abundant.size();j++)
	{
		if(abundant[i]+abundant[j]<28123)
		{
			flag[abundant[i]+abundant[j]]=1;
		}
		else break;
	}
	for(int i=0;i<28123;i++)
	{
		if(flag[i]==0)
		ans+=i;
	}
	cout<<ans<<endl;
	return 0;
}