2016SDAU课程练习一1012 Problem M

最新推荐文章于 2016-07-06 23:44:14 发布

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本文介绍了一个关于高效计算整数根的编程问题，包括输入输出要求、示例及AC代码。此外，还提供了float、double和long double等数据类型的位长、有效数字与绝对值范围。

Problem M

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 60000/30000K (Java/Other)

Total Submission(s) : 50 Accepted Submission(s) : 27

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

没啥好说的了，本来还想用模拟，想想不会干脆用常规方法做了，结果ac了。。。

还有无限循环。。。

到网上找了一下几个数据类型的范围分享一下：

类型长度 (bit) 有效数字绝对值范围

float 32 6~7 10^(-37) ~ 10^38

double 64 15~16 10^(-307) ~10^308

long double 128 18~19 10^(-4931) ~ 10 ^ 4932

ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#include<algorithm>
using namespace std;
int main(){
int n;
double p;
while(cin>>n){
cin>>p;
cout<<pow(p,1.0/n)<<endl;
}
return 0;
}