poj 2109 Power of Cryptography

                                                                                                                                           Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 20362		Accepted: 10289

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 16
3 27
7 4357186184021382204544

Sample Output

4
3

/*
**二分查找不过这个题内有个BUG
*/


#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;

int main()
{
    //freopen("in.txt","r",stdin);
    std::ios::sync_with_stdio(false);
    double n,p;
    while(cin>>n>>p)
    {
        LL left,right,mid;
        right   = INF;
        left = 0;
        while(left <= right)
        {
            mid = (left + right)/2;
            double tt = pow(mid,n);
            if(tt == p)
            {
                cout< p)
                right = mid;
            if(tt < p)
                left = mid+1;
        }
    }
    return 0;
}

/*
**这就是bug
*/
#include 
#include 

using namespace std;

int main()
{
    double n,p;
    while(cin>>n>>p)
    {
        cout<