SDAU课程练习1012

最新推荐文章于 2017-08-20 08:21:40 发布

窦小雨

最新推荐文章于 2017-08-20 08:21:40 发布

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分类专栏：贪心

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Problem 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

题目大意：

给定n m，现在n^p=m; 求p

思路：

如果是扫描查找的话，需要自己写出大数相乘的规则，显然是比较麻烦的。所以用数学知识转化一下的话，就比较简单了， p=m ^ 1/n 这个题是借鉴的程队的思路。嘿嘿。不过需要注意的是，2^3=8 8^1/2 ≈ 3 约等于这里就要注意四舍五入。

感想：

一开始感觉挺难得，可能需要定义大数乘法规则，再用二分快速查找。但是转化一下完全用不到啊！！因为double类型完全可以储存比较精确的数值。

AC代码：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int main()
{
    //freopen("r.txt", "r", stdin);
    double n,p;
    int t;
    while(cin>>n)
    {
        cin>>p;
        t=pow(p,1/n)+0.5;
        cout<<t<<endl;
    }
}