贪心数学公式

最新推荐文章于 2025-06-30 23:06:35 发布

山农数学姜赫

最新推荐文章于 2025-06-30 23:06:35 发布

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分类专栏：贪心算法

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本文介绍了一种算法，用于高效计算整数幂次方的正根，该算法适用于1到200范围内的整数n和1到10^101范围内的整数p，其中p为某个整数k的n次幂。通过输入一对整数n和p，程序能够找到满足条件的整数k。

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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
前一个数的几次根号下。
借鉴某雨
#include<iostream>
#include<string.h>
#include<set>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<math.h>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
using namespace std;
int main()
{
    double n;
   double p;
    int i;
    int j;
    int sum;
    while(cin>>n)
    {
        cin>>p;
        sum=pow(p,1/n)+0.5;
      cout<<sum<<endl;
    }
    return 0;
}
 //数学公式 借鉴 窦和雨的 经验