poj 2109 Power of Cryptography
2024-10-15 15:02:04
Power of Cryptography
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 16388 | Accepted: 8285 |
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 nth. power, for an integer k (this integer is
what your program must find).
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
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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