Power of Cryptography
生活随笔
收集整理的這篇文章主要介紹了
Power of Cryptography
小編覺得挺不錯的,現在分享給大家,幫大家做個參考.
//只用一行核心代碼就可以過的天坑題目............= =
題目:
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).
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 4357186184021382204544Sample Output
4 3 1234 哎~不多說了,代碼如下: #include<iostream> #include<cmath> using namespace std; int main() {double n,p;while(cin>>n>>p){cout<<pow(p,1/n)<<endl;}return 0; }代碼來源:http://blog.csdn.net/zcube/article/details/8545523
轉載于:https://www.cnblogs.com/teilawll/p/3204786.html
總結
以上是生活随笔為你收集整理的Power of Cryptography的全部內容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: 12 求1+2+...+n
- 下一篇: centos 安装 erlang