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poj 2109 Power of Cryptography
http://poj.org/problem?id=2109
题意：kn = p，已知n，p，求k。1<=n<= 200, 1<=p<10101， 1<=k<=109
分析：我也是被吓到了，这么大的数据，可是看了 Discuss说一句话就可以AC,我就想到了k=pow(p,1.0/n);不过k，p，n必须是double才可以，
至于为什么我查了一下：
类型 长度 (bit) 有效数字 绝对值范围
float 32 6~7 10^(-37) ~ 10^38
double 64 15~16 10^(-307) ~10^308
long double 128 18~19 10^(-4931) ~ 10 ^ 4932
另外别人说这题可以是二分加高精度，我也不会做。也是无语了。
Power of Cryptography
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 19419 | Accepted: 9804 |
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 k n =p.
Sample Input
2 163 277 4357186184021382204544
Sample Output
431234
poj 2109 Power of Cryptography
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