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poj2109Power of Cryptography 大数开根

Power of Cryptography
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 20679   Accepted: 10468

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
1234

Source

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>

using namespace std;

int main()
{
	double m;
	int n;
	while(cin>>n>>m)
	{
		printf("%g\n",pow(m,1.0/n));
       //n^a=m->pow(n,a)=m->pow(m,1.0/n)=a;
	}//cout<<pow(p,1.0/n);  //指数的倒数就是开n次方
}
/*二分+高精
#include<iostream>
#include<stdio.h>
#include<math.h>
#define eps 0.0000000001

using namespace std;

double n,m,k;
void work()
{
    long long left,right,mid;
    left = 0;
    right = 1000000002;
    while(left + eps < right)
    {
        mid = (left + right) / 2;
        if(pow(mid,n)-m > 0)
            right = mid;
        else if(pow(mid,n) - m < 0)
            left = mid;
        else
        {
            printf("%.0ld\n",mid) ;
            break;
        }
    }
}
int main()
{
    while(cin>>n>>m)
        work();
    return 0;
}
*/

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