| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 20346 | Accepted: 10281 |
Description
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
Output
Sample Input
2 16 3 27 7 4357186184021382204544
Sample Output
4 31234
k^n=p,则p^(1/n)=k。且函数可以直接用pow(x,y)去求x^y。
收获:巩固了一下基础。启发了一下思维。
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
#include<cstdio> #include<iostream> #include<cmath> using namespace std; int main() { double n,p; while(cin>>n>>p) { cout<<pow(p,1/n)<<endl; } return 0; }
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