poj——2109——Power of Cryptography

最新推荐文章于 2019-02-25 10:55:37 发布

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最新推荐文章于 2019-02-25 10:55:37 发布

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分类专栏：二分法

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本文探讨了在计算机科学中解决整数幂次方根计算问题的算法，包括直接使用数学库函数pow进行快速计算，以及采用二分查找法实现精确求解。通过实例代码演示了如何在给定整数n和p的情况下找到满足kn=p的整数k，深入浅出地阐述了解决此类数学问题的高效方法。

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escription

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 nth. 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

题意：求n的几次方是m，用c++函数pow

#include<cstdio>
#include<cmath>
#include <algorithm>

using namespace std;
int main()
{
    double n, m;
    while(scanf("%lf%lf", &n, &m) != EOF)
        printf("%.0lf\n" ,pow(m, 1 / n));
    return 0;
}

二分法：（学习）

#include<iostream>
#include<cstdio>
#include<cmath>
#include <algorithm>
using namespace std;
int main()
{
    double n, m;
    while(scanf("%lf%lf", &n, &m) != EOF)
    {
        long long L=0;
        long long R=1000000005;
        long long mid;
        while(R-0.00000001>L)
        {
           mid=(L+R)/2;
            if(pow(mid,n)-m>0)
              R=mid;
            else if(pow(mid,n)-m<0)
              L=mid;
            else
            {
                cout<<mid<<endl;
                break;
            }
        }
    }
    return 0;
}