Description

The Great Dodgers company has recently developed a brand-new playing machine.

You put a coin into the machine and pull the handle. After that it chooses some integer number. If the chosen number is zero you win a jackpot. In the other case the machine tries to divide the chosen number by the lucky numbers \(p_1 , p_2 , \cdots , p_n\) . If at least one of the remainders is zero --- you win.

Great Dodgers want to calculate the probability of winning on their machine. They tried to do it, but failed. So Great Dodgers hired you to write a program that calculates the corresponding probability.

Unfortunately, probability theory does not allow you to assume that all integer numbers have equal probability. But one mathematician hinted you that the required probability can be approximated as the following limit:

\[\lim_{k \to \infty}\frac{S_k}{2k+1}
\]

Here \(S_k\) is the number of integers between \(-k\) and \(k\) that are divisible by at least one of the lucky numbers.

Input

Input file contains \(n\) --- the number of lucky numbers \((1 \le n \le 16)\), followed by \(n\) lucky numbers \((1 \le p_i \le 10^9)\).

Output

It is clear that the requested probability is rational. Output it as an irreducible fraction.

On the first line of the output file print the numerator of the winning probability. On the second line print its denominator. Both numerator and denominator must be printed without leading zeroes. Remember that the fraction must be irreducible.

Sample Input

2

4 6

Sample Output

1

3

第一次用java写程序,为了不打高精度(因为必须涉及高精除),代码几乎是蒯的。\(2^N\)枚举,相互求\(lcm\),贡献即为\(\frac{1}{lcm}\),然后用容斥原理合答案即可。

ACMjava1H速成戳这里

import java.math.*;
import java.util.*;
public class Main
{
static BigInteger d,ret,temp,yy;
static int n,dd;
static boolean mark = true;
static BigInteger[] a = new BigInteger[20];
public static void main(String[] args)
{
Scanner in = new Scanner (System.in);
n = in.nextInt();
temp = BigInteger.ONE;
ret = BigInteger.ZERO;
for (int i = 0;i < n;++i)
{
int k = in.nextInt();
a[i] = BigInteger.valueOf(k);
d = temp.gcd(a[i]);
temp = temp.multiply(a[i]).divide(d);
}
for (int i = 1;i < (1<<n);++i)
{
mark = false; yy = BigInteger.ONE;
for (int j = 0;j < n;++j) if (((1 << j) & i) > 0) { mark = !mark; d = a[j].gcd(yy); yy = yy.multiply(a[j]).divide(d); }
if (mark) ret = ret.add(temp.divide(yy));
else ret = ret.subtract(temp.divide(yy));
}
d = ret.gcd(temp);
System.out.println(ret.divide(d));
System.out.println(temp.divide(d));
}
}

POJ2103 Jackpot的更多相关文章

  1. UVa 10684 - The jackpot

    题目大意:给一个序列,求最大连续和. 用sum[i]表示前i个元素之和,那么以第i个元素结尾的最大连续和就是sum[i]-sum[j] (j<i)的最大值,也就是找前i-1个位置sum[]的最小 ...

  2. 【2005-2006 ACM-ICPC, NEERC, Moscow Subregional Contest】Problem J. Jack-pot

    简单dfs,差分一下A数组和建出字典树能写得更方便,若不这么做代码时就会像我一样难受. #include<cstdio> #include<cstring> #include& ...

  3. Web测试的常用测试用例与知识

    1. Web测试中关于登录的测试 2. 搜索功能测试用例设计 3. 翻页功能测试用例 4. 输入框的测试 5. Web测试的常用的检查点 6. 用户及权限管理功能常规测试方法 7. Web测试之兼容性 ...

  4. web测试常用的用例及知识

      1.      Web测试中关于登录的测试... 1 2.      搜索功能测试用例设计... 2 3.      翻页功能测试用例... 3 4.      输入框的测试... 5 5.    ...

  5. An Introduction to Interactive Programming in Python (Part 1) -- Week 2_1 练习

    # Practice Exercises for Functions # Solve each of the practice exercises below. # 1.Write a Python ...

  6. Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)C. Bear and Poker

                                                  C. Bear and Poker                                     ...

  7. iOS开发之 几本书

    <object_c 编程之道书> <iOS 7 UI Transition Guide> iOS开发指南:从零基础到App Store上架[国内第一本iOS架构设计图书,涵盖i ...

  8. C Primer Plus(第五版)7

    第 7 章 C 控制语句:分支和跳转 在本章中你将学习下列内容: · 关键字:if(如果),else(否则),switch(切换),continue(继续),break(中断), case(情况),d ...

  9. Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) A. Bear and Poker 分解

    A. Bear and Poker Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/573/pro ...

随机推荐

  1. php 5.3+ 连接mssql

    php5.3+里已经没有mssql的dll扩展了,需要使用SQL Server Driver for PHP 这里有两个版本有两个版本支持不同的php版本. 1.SQL Server Driver f ...

  2. .net中压缩和解压缩的处理

    最近在网上查了一下在.net中进行压缩和解压缩的方法,方法有很多,我找到了以下几种: 1.利用.net自带的压缩和解压缩方法GZip 参考代码如下: //======================= ...

  3. SQLServer2012分离出的数据库存放路径

    分离出的数据库没有保存位置提示,经常会导致分离出的数据库找不到  以下是分离出的数据库默认位置: C:\Program Files\Microsoft SQL Server\MSSQL11.MSSQL ...

  4. C#对于sql server数据库的简单操作

    1.在用windows模式登陆sql server 数据库 简历一个student的数据库,然后新建查询: create table student ( id int auto_increment p ...

  5. iOS svn版本回退 cornerstone

    http://blog.csdn.net/x32sky/article/details/46866899   IOS开发中,SVN如何恢复到某一个版本(以Cornerstone为例) Cornerst ...

  6. 浅谈break 、continue、return,goto四种语句的区别。

    浅谈break .continue.return三种语句的区别: break,continue,return这三个具有跳转功能的语句在c语言中经常被用到,近期身边有些小伙伴总是把它们的用法搞乱,在这里 ...

  7. C#2.0至4.0 的一些特性

    罗列清单备查 一.C#2.0 1. Partial class 分部类 file1.cs using System; public partial class MyClass { public voi ...

  8. 如何使用Json-lib

    数组与List.Collection等都用JSONArray解析 boolean[] boolArray = new boolean[]{true,false,true}; JSONArray jso ...

  9. getDrawingRect,getHitRect,getLocalVisibleRect,getGlobalVisibleRect

    本文主要大体讲下getHitRect().getDrawingRect().getLocalVisibleRect().getGlobalVisibleRect. getLocationOnScree ...

  10. 以前写过的ajax基础案例(王欢-huanhuan)

    //load方法的使用   $('#loadBtn').click(function(){    //responseTxt 包含来自请求的结果数据 //statusTxt 包含请求的状态(" ...