Power of Cryptography - poj 2109

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 n^th. 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 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234注:这题分类是贪心算法，但是看了discuss之后竟然发现用double一句可以AC，也是醉了k^n=pn=log(p)/log(k)log(k)=log(p)/n2^log(k)=2^(log(p)/n)k=p^(1/n)附：float，double，long double的范围类型          长度 (bit)           有效数字          绝对值范围float            32                6~7            10^(-37) ~ 10^38double           64                15~16          10^(-307) ~10^308long double      128               18~19          10^(-4931) ~ 10 ^ 4932

 1 #include<iostream>
 2 #include<math.h>
 3 using namespace std;
 4
 5
 6 int main()
 7 {
 8     double n,p;
 9     while(cin>>n>>p)
10         cout<<pow(p,1.0/n)<<endl;
11     return 0;
12 }