POJ 2109 - Power of Cryptography

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 22902		Accepted: 11555

Description

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered
to be only of theoretical interest. 

This problem involves the efficient computation of integer roots of numbers. 

Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the nth. power, for an integer k (this integer is
what your program must find).
Input

The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10101 and there exists an integer k, 1<=k<=109 such that kn = p.
Output

For each integer pair n and p the value k should be printed, i.e., the number k such that
kn = p.
Sample Input

2 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234

Source

México and Central America 2004

Analysis
本题大意是给定一个n和p，求k，使得k的n次方等于p，这个题卡了我好几个小时，直到经过了别人的提醒才恍然大悟，原来用我们万能的power函数就可以轻松解决，只要求p的1/n次方即可，不过要注意，数据要用double类型，输出的时候用%.0lf控制输出整数部分。
[b]Source
Code[/b]

#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
double a,b;
while(~scanf("%lf %lf",&a,&b))
printf("%.0lf\n",pow(b,1/a));
return 0;
}