E - Number With The Given Amount Of Divisors
Crawling in process...Crawling failedTime Limit:2000MSMemory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

Given the number n, find the smallest positive integer which has exactlyn divisors. It is guaranteed that for the givenn 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.

Sample Input

Input
4
Output
6
Input
6
Output
12

思路：枚举，2^64会爆，所以最多枚举到64，一个数的因子数为（p1+1）*（P2+1）...

#include<iostream>
#include<cstring>
using namespace std;
const long long oo=1000000000000000009;
const int p[]={2,3,5,7,11,13,17,19,23,29};
int n;
long long ans;
void dfs(int dep,long long tmp,int x)
{ if(dep>n)return;
  if(dep==n&&ans>tmp){ans=tmp;return;}
  for(int i=1;i<=64;i++)///最多，2^64会爆
  { if(ans/p[x]<tmp)break;
    dfs(dep*(i+1),tmp*=p[x],x+1);
  }
}
int main()
{
    while(cin>>n)
    {
     ans=oo;
      dfs(1,1,0);
      cout<<ans<<"\n";
    }
}



The article write by nealgavin