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HDU 4228

很明显可以转化为反素数的题目。由于有n种不同的方式,所以,数的约数可以为2*n或者2*n-1

#include <iostream>#include <cstdio>#include <cstring>#define LL __int64using namespace std;LL p[16]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};  LL n,k;void dfs(LL dep,LL tmp,LL num, LL &ans){	if(num>k) return;	if(num==k){		if(tmp<ans)		ans=tmp;		return ;	}	for(LL i=1;i<=60;i++){		if(tmp*p[dep]>ans) break;		tmp*=p[dep];		dfs(dep+1,tmp,num*(i+1),ans);	}}int main(){	while(scanf("%I64d",&n),n){		k=2*n;		LL ansi=(1LL<<60);	//	cout<<ansi<<endl;		dfs(0,1,1,ansi);	//	cout<<ansi<<endl;		k=2*n-1;		LL ansj=(1LL<<60);		dfs(0,1,1,ansj);	//	cout<<ansj<<endl;		printf("%I64d\n",ansi<ansj?ansi:ansj);	}	return 0;}

  

HDU 4228