1. 程式人生 > >【Number With The Given Amount Of Divisors 】【CodeForces - 27E】(反素數)

【Number With The Given Amount Of Divisors 】【CodeForces - 27E】(反素數)

題目:

Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.

Input

The first line of the input contains integer n (1 ≤ n ≤ 1000).

Output

Output the smallest positive integer with exactly n

 divisors.

Examples

Input

4

Output

6

Input

6

Output

12

解題報告:給你一個數字n,求最小的數字(他的因子數目是n的數)。反素數的模板題,之前沒有接觸過,第一次覺得沒有思路,後來慢慢就理解了,關於反素數的求解是參考了大佬的部落格這裡這裡!!!,使用dfs進行遍歷,增加滿足最小成立條件的優化。

ac程式碼:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;

const int maxn =1e5+100;

ll n;
ll ans=1e18;
ll num[]={0,2,3,5,7,11,13,17,19,23,29};

void fin(ll v,ll nn,ll p)
{
	if(nn>n)
	{
		return ;	
	}
	if(nn==n)
	{
		if(v<ans)
			ans=v;
		return ;
	}
	for(int i=1;i<=64;i++)
	{
		v*=num[p];
		if(v>=ans) break;
		if(nn<n)
		{
			fin(v,nn*(i+1),p+1);
		}
	}
}
int main()
{
	scanf("%lld",&n);
	if(n==1)
	{
		printf("1\n");
		return 0;
	}
	fin(1ll,1ll,1ll);
	printf("%lld\n",ans);
	return 0;
}