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 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputGiven 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\n",num); return 0; }