数论 --- 简单计算

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 18018		Accepted: 9090

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

【题目来源】

México and Central America 2004

http://poj.org/problem?id=2109

【题目分析】

这道题目里面 p, n 没有超过 double 范围，所以可以直接算。

一开始看了分类说是贪心，还想了半天

各种数据类型范围：

unsigned   int   0～4294967295   
int   2147483648～2147483647 
unsigned long 0～4294967295
long   2147483648～2147483647
long long的最大值：9223372036854775807
long long的最小值：-9223372036854775808
unsigned long long的最大值：1844674407370955161

__int64的最大值：9223372036854775807
__int64的最小值：-9223372036854775808
unsigned __int64的最大值：18446744073709551615

#include<stdio.h>
#include<math.h>
int main()
{
    double m,n;
    while(scanf("%lf%lf",&m,&n)!=EOF)
    {
        printf("%.0lf
",pow(n,1.0/m));
    }
    return 0;
}