POJ 2109 Power of Cryptography

不知道那里贪心了。。。学长说贪心。

注意：数据类型范围，还是不太了解；

原题链接：http://poj.org/problem?id=2109

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13983		Accepted: 7136

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

Source

México and Central America 2004

各种数据类型范围：

来自kb的博客：http://www.cnblogs.com/kuangbin/archive/2011/07/23/2115001.html

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

//Accepted	212K	0MS	C++	212B	2012-09-18 20:47:04
#include<cstdio>
#include<cmath>
int main()
{
    double n,p;
    double ans;
    while(scanf("%lf%lf",&n,&p)!=EOF)
    {
        ans=pow(p,1/n);
        printf("%.0lf\n",ans);
    }
    return 0;
}