HDU 5810 Balls and Boxes（盒子与球）

最新推荐文章于 2019-08-21 10:53:13 发布

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最新推荐文章于 2019-08-21 10:53:13 发布

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原文链接：http://www.cnblogs.com/Simon-X/p/5757953.html

本文介绍了HDU 5810 Balls and Boxes的题目背景及解决方法，通过理解实验中球等概率分配到盒子的过程，采用二项分布原理求解方差的期望值。

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HDU 5810 Balls and Boxes（盒子与球）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Description

题目描述

Mr. Chopsticks is interested in random phenomena, and he conducts an experiment to study randomness. In the experiment, he throws n balls into m boxes in such a manner that each ball has equal probability of going to each boxes. After the experiment, he calculated the statistical variance V asWhere is the number of balls in the ith box, and is the average number of balls in a box.

Your task is to find out the expected value of V.

Chopsticks先生突然对随机现象来了兴趣，还做了个实验来研究随机性。实验中，他将n个球等概率丢进m个盒子里。然后，他用下面的式子计算方差V

表示第i个盒子中球的数量，表示每个盒子中球的平均数。

你的任务就是找出V的期望。

Input

输入

The input contains multiple test cases. Each case contains two integers n and m (1 <= n, m <= 1000 000 000) in a line.

The input is terminated by n = m = 0.

多组测试用例。每个测试用例有一行两个整数n和m(1 <= n, m <= 1000 000 000)。

n = m = 0 时，输入结束。

Output

输出

For each case, output the result as A/B in a line, where A/B should be an irreducible fraction. Let B=1 if the result is an integer.

对于每个用例，输出一行结果A/B，A/B为不可约分数。结果为整数时，令B=1。

Sample Input - 输入样例	Sample Output - 输出样例
2 1 2 2 0 0	0/1 1/2

Hint

提示

In the second sample, there are four possible outcomes, two outcomes with V = 0 and two outcomes with V = 1.

在第二个样例中，有4种可能结果，两种V = 0和一种V = 1。

【题解】

类似二项分布的实验，得到所有可能的方差，再对方差取期望……等等，这不就是求二项分布的方差吗？（脑子一抽：所有可能 + 取期望 = 等概率）

二项分布D(X) = np(1-p)

p = 1/m 带入D(X) 得

【代码 C++】

 1 #include <cstdio>
 2 __int64 GCD(__int64 a, __int64 b){
 3     __int64 c;
 4     while (c = a%b) a = b, b = c;
 5     return b;
 6 }
 7 int main(){
 8     __int64 n, m, g;
 9     while (scanf("%I64d%I64d", &n, &m), n + m){
10         n *= m - 1; m *= m;
11         g = GCD(n, m);
12         printf("%I64d/%I64d\n", n / g, m / g);
13     }
14     return 0;
15 }