POJ 2109.Power of Cryptography

题目:http://poj.org/problem?id=2109

AC代码(C++):

#include <iostream>#include <algorithm>#include <stdio.h>#include <vector>#include <queue>#include <math.h>#include <string>#include <string.h>#include <bitset>using namespace std;  int main(void){    double n,p;    while(cin>>n>>p)cout<<pow(p,1.0/n)<<endl;}

总结:

基本功还是不够扎实, 首先double的范围是-1.7*10(-308)～1.7*10(308), 其次是n√p可以写成p^(1/n).