Codeforces Beta Round #27 (Codeforces format, Div. 2) E. Number With The Given Amount Of Divisors 反素数

E. Number With The Given Amount Of Divisors
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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
思路:反素数;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define esp 0.00000000001
#define pi 4*atan(1)
const int N=1e5+10,M=2e7+10,inf=1e9+10,mod=1e9+9;
const ll INF=1e18;
int p[N]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,53,59};
ll x;
ll num;
void dfs(int pos,ll ans,ll sum,int pre)
{
    if(sum>x)return;
    if(sum==x)
    num=min(ans,num);
    for(int i=1;i<=pre;i++)
    {
        if(INF/ans<p[pos])break;
        dfs(pos+1,ans*=p[pos],sum*(i+1),i);
    }
}
int main()
{
    scanf("%lld",&x);
    num=INF;
    dfs(0,1,1,63);
    printf("%lld
",num);
    return 0;
}
原文地址:https://www.cnblogs.com/jhz033/p/5793068.html