CodeForces 27 E.Number With The Given Amount Of Divisors（dfs）

Description

给出一整数n，求有n个因子的最小正整数，保证答案不会超过1018

Input

一个正整数n(1≤n≤1000)

Output

输出有n个因子的最小正整数

Sample Input

4

Sample Output

6

Solution

答案不超过1018​，直接对前十几个素因子暴搜即可

Code

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<queue>#include<map>#include<set>#include<ctime>using namespace std;typedef long long ll;typedef pair<int,int>P;const int INF=0x3f3f3f3f,maxn=101;int n,p[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47};ll ans;void dfs(int pos,int tn,ll tans){    if(n==tn)    {        ans=min(ans,tans);        return ;    }    for(int i=1;i<=60;i++)    {        if(tn*(i+1)>n||tans*p[pos]>ans)break;        tans*=p[pos];         dfs(pos+1,tn*(i+1),tans);    }}int main(){    while(~scanf("%d",&n))    {        ans=1e18;        dfs(0,1,1);        printf("%I64d\n",ans);    }    return 0;}