Power of Cryptography

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).
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

题目大意:输入k,p,输出n,使kn==p

解题思路:

    看Discuss说用double能水过 于是。。。。

    PS:据说要用到二分法+高精度 等以后有空再研究一下吧

Code:

 #include<cstdio>
#include<cmath>
int main()
{
double n,m;
while(scanf("%lf%lf",&n,&m)!=EOF)
printf("%.0lf\n",pow(m,/n));
return ;
}

POJ2109——Power of Cryptography的更多相关文章

  1. [POJ2109]Power of Cryptography

    [POJ2109]Power of Cryptography 试题描述 Current work in cryptography involves (among other things) large ...

  2. POJ-2109 Power of Cryptography(数学或二分+高精度)

    题目链接: https://vjudge.net/problem/POJ-2109 题目大意: 有指数函数 k^n = p , 其中k.n.p均为整数且 1<=k<=10^9 , 1< ...

  3. Power of Cryptography(用double的泰勒公式可行分析)

    Power of Cryptography Time limit: 3.000 seconds http://uva.onlinejudge.org/index.php?option=com_onli ...

  4. 贪心 POJ 2109 Power of Cryptography

    题目地址:http://poj.org/problem?id=2109 /* 题意:k ^ n = p,求k 1. double + pow:因为double装得下p,k = pow (p, 1 / ...

  5. poj 2109 Power of Cryptography

    点击打开链接 Power of Cryptography Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16388   Ac ...

  6. UVA 113 Power of Cryptography (数学)

    Power of Cryptography  Background Current work in cryptography involves (among other things) large p ...

  7. Poj 2109 / OpenJudge 2109 Power of Cryptography

    1.Link: http://poj.org/problem?id=2109 http://bailian.openjudge.cn/practice/2109/ 2.Content: Power o ...

  8. POJ 2109 :Power of Cryptography

    Power of Cryptography Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 18258   Accepted: ...

  9. POJ 2109 -- Power of Cryptography

    Power of Cryptography Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 26622   Accepted: ...

随机推荐

  1. ZigBee协议学习之网络层

    ZigBee的体系结构中,底层采用IEEE 802.15.4的物理层和媒介层,再次基础上,ZigBee联盟建立了自己的网络层(NWL)和应用层框架. ZigBee网络层的主要功能包括设备的连接和断开. ...

  2. oracle 中proc和oci操作对缓存不同处理

    oracle 中proc和oci操作对缓存不同处理

  3. Java权限讲解

    Java访问权限就如同类和对象一样,在Java程序中随处可见. Java的访问权限,根据权限范围从大到小为:public > protected > package > privat ...

  4. Windows Phone 8.1 页面导航

    1. Windows Phone 8.1 的应用框架 一个应用拥有 1 个 Window,一个 Window 包含 1 个 Frame,一个 Frame 包含 多个 Page. 获取 Frame 的方 ...

  5. MySQL学习笔记之数据存储类型

    说明:本文是作者对MySQL数据库数据存储类型的小小总结. Numeric Type (数字类型) 1.TINYINT.SMALLINT.MEDIUMINT.INT.BIGINT主要根据存储字节长度不 ...

  6. 用css3写出来的进度条

    夜深了,废话不多说,先上代码: <style> * { box-sizing: border-box } .wrapper { width: 350px; margin: 200px au ...

  7. JS中的!=、== 、!==、===的用法和区别。

    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 var num = 1;   var str = '1';   var test = 1;   t ...

  8. Python 中的引用和类属性的初步理解

    最近对Python 的对象引用机制稍微研究了一下,留下笔记,以供查阅. 首先有一点是明确的:「Python 中一切皆对象」. 那么,这到底意味着什么呢? 如下代码: #!/usr/bin/env py ...

  9. jsp分页代码之pageUtil类

    pageUtil类负责得到每页的开始数和结束数 package control; public class PageUtil { private int pageSize;//每页显示的条数 priv ...

  10. Inside of Jemalloc

    INSIDE OF JEMALLOCThe Algorithm and Implementation of Jemalloc author: vector03mail:   mmzsmm@163.co ...