FZU 1607：Greedy division

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本博客探讨了一个基于数论的复杂问题——黄金金币的贪婪分配，并提供了高效的素数分解算法解决方案。通过深入分析，揭示了如何利用数学原理解决实际问题。

Problem 1607 Greedy division

Accept: 339 Submit: 1301
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

Oaiei has inherited a large sum of wealth recently; this treasure has n pieces of golden coins. Unfortunately, oaiei can not own this wealth alone, and he must divide this wealth into m parts (m>1). He can only get one of the m parts and the m parts have equal coins. Oaiei is not happy for only getting one part of the wealth. He would like to know there are how many ways he can divide the wealth into m parts and he wants as many golden coins as possible. This is your question, can you help him?

Input

There are multiply tests, and that will be 500000. For each test, the first line is a positive integer N(2 <= N <= 1000000), N indicates the total number of golden coins in the wealth.

Output

For each test, you should output two integer X and Y, X is the number of ways he can divide the wealth into m parts where m is large than one, Y is the number of golden coins that he can get from the wealth.

Sample Input

568

Sample Output

1 13 33 4

Hint

There are huge tests, so you should refine your algorithm.

Source

FOJ月赛-2008年5月

//跟着AC大牛博客中一篇数论文章的索引做题，本题其实是一道很水的素数分解问题！

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 1005

bool x[maxn];
int prime[maxn],m=0;
void Init()
{
	int i,tmp;
	memset(x,false,sizeof(x));
	x[0]=x[1]=true;
	for(i=2;i<maxn;i++)
	{
		if(!x[i])
		{
			prime[++m]=i;
			tmp=i*2;
			while(tmp<maxn)
			{
				x[tmp]=true;
				tmp+=i;
			}
		}
	}
}

int main()
{
	Init();
	int n,i,k,ans,num,tmp;
	while(~scanf("%d",&n))
	{
		ans=1;
		num=tmp=n;
		for(i=1;i<=m&&prime[i]*prime[i]<=n;i++)
		{
			if(n&&n%prime[i]==0)
			{
				if(prime[i]<num)
					num=prime[i];
				k=0;
				while(n&&n%prime[i]==0)
				{
					k++;
					n/=prime[i];
				}		
				ans*=(k+1);
			}
		}
		if(n>1)
			ans*=2;
		printf("%d %d\n",ans-1,tmp/num);
	}
	return 0;	
}