Power of Cryptography - poj 2109
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 20351 | Accepted: 10284 |
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 nth. 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
3
1234
注:这题分类是贪心算法,但是看了discuss之后竟然发现用double一句可以AC,也是醉了
k^n=p
n=log(p)/log(k)
log(k)=log(p)/n
2^log(k)=2^(log(p)/n)
k=p^(1/n)
附:float,double,long double的范围
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<iostream>
#include<math.h>
using namespace std; int main()
{
double n,p;
while(cin>>n>>p)
cout<<pow(p,1.0/n)<<endl;
return ;
}
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