CF27E Number With The Given Amount Of Divisors

显然质数越小越好，越小的质数次方越大越好，写个dfs爆搜一下

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const int N=4e5+10;
const int inf=0x3f3f3f3f;
int n;
ll ans=1000000000000000001;
int prime[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
void dfs(ll now,int cur,int p,int cnt) {
    if(cnt>n) return;
    if(now>ans) return;
    if(cur>15) return;
    if(cnt==n){
        ans=now;return;
    }
    for(int i=1;i<=p;++i) {
        dfs(now*=prime[cur],cur+1,i,cnt*(i+1));
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin>>n;
    dfs(1ll,0,64,1);
    cout<<ans<<endl;
}