poj 2109 Power of Cryptography
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 16388 | Accepted: 8285 |
Description
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
Output
Sample Input
2 16
3 27
7 4357186184021382204544
Sample Output
4
3
1234
给你两个数n和p,那么让你求p的根号n的结果,即k^n = p,让我们求出K的值,我是看了网上的做法才知道原来还可以这样
#include<stdio.h>
#include<math.h>
int main()
{
double a, b;
while(scanf("%lf %lf", &a, &b) != EOF)
{
printf("%g\n", pow(b, 1.0/a));
}
return 0;
}
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