Power of Cryptography
题目链接:http://poj.org/problem?id=2109
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).
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<10
101 and there exists an integer k, 1<=k<=10
9 such that k
n = 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
题意:给定N,P,求K,(kn = p)。
题解:1、K<=10^9,二分答案即可。
2、公式两边取对数,nlog(k)=log(p),得k = exp(log(p)/n),所以可以用double直接求答案
AC代码:
#include <iostream>
#include <cmath>
using namespace std;
double n,p;
int main(int argc, char const *argv[])
{
while(cin>>n>>p)
{
cout<<pow(p,1/n)<<endl;
}
return 0;
}

本文介绍了一种通过二分查找及对数运算求解整数根的方法,针对给定的整数n和p,找到满足kn=p的k值。利用二分查找技巧,结合对数性质简化计算过程。
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