CF 27E Number With The Given Amount Of Divisors

6

12

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>

using namespace std;

#define ll long long
const int MAXN=10010;;

ll ans;
int n;

int prm[10]={2,3,5,7,11,13,17,19,23,29};

void dfs(int a,int b,ll tmp)
{
    if(tmp<ans && a == n)ans=tmp;
    if(a>=n )return;
    for(int i=1;i<=64;i++)
    {
        if(tmp>ans/prm[b])break;//改为tmp*prm[b]>ans---------wa.........因为longlong * int 可能超出longlong的界限
        tmp*=prm[b];
        dfs(a*(i+1),b+1,tmp);
    }
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        ans=LONG_LONG_MAX;
        dfs(1,0,1);
        printf("%lld\n",ans);
    }

    return 0;
}

//prm范围为啥10都够了？我不明白