UVA POJ 2109 Power of Cryptography的解题方法有哪些？

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《密码学力量概述》当前密码学研究涉及（包括但不限于）大素数和在这些素数中计算数的幂。该领域的工作已导致将数论的结果应用于实践。

Power of Cryptography

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

Source

México and Central America 2004

#include<stdio.h> #include<math.h> #define N 1000000 int main() { double n,m; double t; int i; while(scanf("%lf%lf",&m,&n)!=EOF) { for(i=1;i<N;i++) { t = pow(i,m); if(t==n) { break; } } printf("%d\n",i); } return 0; }

2

#include<stdio.h> #include<math.h> int main() { double p,n; while(scanf("%lf%lf",&n,&p)!=EOF) { printf("%.lf\n",pow(p,1/n)); } }

Time Limit:1000MS

Memory Limit:30000K

Total Submissions:18102

Accepted:9135