poj2109 Power of Cryptography 

 1 ///2014.3.2
 2 ///poj2109
 3 
 4 /**
 5  *解法来自csdn上“ζёСяêτ - 小優YoU”的博客
 6  *原博客链接：
 7  * http://blog.csdn.net/lyy289065406/article/details/6642602
 8  */
 9 
10 #include <iostream>
11 #include <cstdio>
12 #include <cmath>
13 using namespace std;
14 
15 int main( )
16 {
17 //    freopen("in","r",stdin);
18 //    freopen("out","w",stdout);
19 
20     double n,p;
21     while( scanf("%lf%lf",&n,&p)!=EOF ){
22         cout<<pow(p,1.0/n)<<endl;
23     }
24     return 0;
25 }

 

posted @ 2014-03-02 21:01  basement_boy  阅读(131)  评论(0)  编辑  收藏  举报