2016SDAU课程练习一 1012

Problem M

Problem DescriptionCurrent 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：给定m，你；n的p次幂等于m，求p

 

思路：这个题借鉴了一下别人的思路，用一下pow函数问题就迎刃而解了。要注意四舍五入的问题，浮点数向整数转化时，采用的是直接丢弃小数部分的方式，所以结果会偏小，如果要四舍五入的话，运算结果要加上0.5。

 

感想：感觉一开始把问题复杂了，其实就是个求幂的问题

AC代码：

#include <cstdio>
#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<math.h>
#include<string.h>
#include<map>
#include<set>
#include<vector>
using namespace std;
int main()
{
    double n,p;
    int i;
    while(cin>>n)
    {
        cin>>p;
        i=pow(p,1/n)+0.5;
        cout<<i<<endl;
    }
}