POJ 2109 Power of Cryptography（坑到没朋友）

Power of Cryptography

Time Limit: 1000MS Memory Limit: 30000KB 64bit IO Format: %lld & %llu

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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<10 ¹⁰¹ and there exists an integer k, 1<=k<=10 ⁹ such that k ⁿ = 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

   看到这道题立马想到了大数高精度，一看数据，必须二分，然后我就开始写了。。。。。此处省略一万字。。。。。

#include<stdio.h>#include<math.h>int main(){    double x,y;    while(~scanf("%lf%lf",&x,&y))    {        printf("%.0lf\n",pow(y,1/x));    }    return 0;}

看到这个coder后 我相信你们能懂我的心情，至于为什么，double不可以进行开方运算，但是我们可以倒过来运算啊，求p的n/1次方啊。一次运算得到的也是int型控制一下输出就行了。