CF-27E - Number With The Given Amount Of Divisors（枚举+dfs）

最新推荐文章于 2019-08-02 13:50:37 发布

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该问题要求找到具有给定数量因子的最小正整数，保证答案不超过10^18。输入是一个整数n（1≤n≤1000）。解决方案涉及枚举和深度优先搜索（DFS），由于2^64会导致溢出，因此枚举范围限制在64内，一个数的因子数可以表示为各个质因数指数加1的乘积。

E - Number With The Given Amount Of Divisors

Crawling in process... Crawling failed Time Limit:2000MSMemory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 27E

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

Output

Input

Output

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";
    }
}