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;
}
poj 2109 Power of Cryptography的更多相关文章
- 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 / ...
- POJ 2109 -- Power of Cryptography
Power of Cryptography Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 26622 Accepted: ...
- POJ 2109 Power of Cryptography 数学题 double和float精度和范围
Power of Cryptography Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 21354 Accepted: 107 ...
- poj 2109 Power of Cryptography (double 精度)
题目:http://poj.org/problem?id=2109 题意:求一个整数k,使得k满足kn=p. 思路:exp()用来计算以e为底的x次方值,即ex值,然后将结果返回.log是自然对数,就 ...
- POJ - 2109 Power of Cryptography(高精度log+二分)
Current work in cryptography involves (among other things) large prime numbers and computing powers ...
- POJ 2109 Power of Cryptography【高精度+二分 Or double水过~~】
题目链接: http://poj.org/problem?id=2109 参考: http://blog.csdn.net/code_pang/article/details/8263971 题意: ...
- POJ 2109 Power of Cryptography 大数,二分,泰勒定理 难度:2
import java.math.BigInteger; import java.util.Scanner; public class Main { static BigInteger p,l,r,d ...
- 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 ...
- POJ 2109 :Power of Cryptography
Power of Cryptography Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 18258 Accepted: ...
随机推荐
- Storm0.9.4安装 - OPEN 开发经验库
Storm0.9.4安装 - OPEN 开发经验库 bin/zkServer.sh start /home/guym/down/kafka_2.8.0-0.8.0/config/zookeeper.p ...
- 使用bind(this)的情况
1.setInterval().setTimeout()的回调函数,一定要加.bind(this)方法. 原因是:在setInterval()中定义的回调函数,是在同步代码执行完后,随着事件触发来异步 ...
- mysql 远程访问授权
给mysql改了密码了,程序就不让登录了: mysql 网外链接 Access denied for user 这不能忍啊!咋办?授权呗! 命令行: GRANT ALL PRIVILEGES ON * ...
- Angular学习(4)- 数组双向梆定
示例: <!DOCTYPE html> <html ng-app="MyApp"> <head> <title>Study 4< ...
- JS中exec函数与match函数的区别与联系
总结: 正则规则的声明,两种方法: exec是RegExp类的匹配方法 match是字符串类的匹配方法 var reg = /aaa/g; var reg = new RegExp("aaa ...
- Ansible之playbook
简介 playbook是一个非常简单的配置管理和多主机部署系统.可作为一个适合部署复杂应用程序的基础.playbook可以定制配置,可以按指定的操作步骤有序执行,支持同步和异步方式.playbook是 ...
- Python 处理多层嵌套列表
>>> movies =[ "the holy grail", 1975,"terry jones",91, ["graham ch ...
- apidoc,一个相当不错的文档生成器
http://apidocjs.com/ 例子:myapp目录下的MyCode.java /** * * @api {get} /company/list 获取公司信息 * @apiName 获取公司 ...
- Linux下查看和添加环境变量
转自:http://blog.sina.com.cn/s/blog_688077cf01013qrk.html $PATH:决定了shell将到哪些目录中寻找命令或程序,PATH的值是一系列目录,当您 ...
- 一探前端开发中的JS调试技巧
前言 调试技巧,在任何一项技术研发中都可谓是必不可少的技能.掌握各种调试技巧,必定能在工作中起到事半功倍的效果.譬如,快速定位问题.降低故障概率.帮助分析逻辑错误等等.而在互联网前端开发越来越重要的今 ...