E. Number With The Given Amount Of Divisors

E. Number With The Given Amount Of Divisors
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Given 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；

（1）一个反素数的所有质因子必然是从2开始的连续若干个质数，因为反素数是保证约数个数为的这个数尽量小

（2）同样的道理，如果，那么必有

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<queue>
 6 #include<math.h>
 7 #include<set>
 8 #include<vector>
 9 #include<string.h>
10 using namespace std;
11 typedef long long LL;
12 bool prime[100];
13 int ans[100];
14 LL minn = 1e18+10;
15 LL maxx = 1e18;
16 void  dfs(LL n,int cn,int pre,int need,int cnt);
17 int main(void)
18 {
19     LL m,n;
20     int i,j;
21     int cn = 0;
22     memset(prime,0,sizeof(prime));
23     for(i = 2 ; i < 100 ; i++)
24     {
25         if(!prime[i])
26         {
27             ans[cn++] = i;
28             for(j = i; (i*j) <= 100 ; j++)
29             {
30                 prime[i*j]=true;
31             }
32         }
33     }
34     int ak = 16;
35     while(scanf("%lld",&n)!=EOF)
36     {
37         minn = 1e18+10;
38         dfs(1,0,63,n,1);
39         printf("%lld\n",minn);
40     }
41     return 0;
42 }
43 void  dfs(LL n,int cn,int pre,int need,int cnt)
44 {
45     if(cnt>need)return ;
46     if(cnt == need)
47     {
48          minn = min(minn,n);
49         return ;
50     }
51     LL ap = 1;
52     int i,j;
53     for(i = 1; i <= pre; i++)
54     {   if(maxx/ap>=ans[cn])
55         ap*=ans[cn];
56         else return ;
57         if(maxx/n >= ap)dfs(n*ap,cn+1,pre,need,cnt*(1+i));
58         else return ;
59 
60     }
61 }

 

posted @ 2016-08-30 13:22  sCjTyC  阅读(286)  评论(0)  编辑  收藏  举报