HDU 5810 Balls and Boxes（盒子与球）

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++】

 

#include <cstdio>
__int64 GCD(__int64 a, __int64 b){
    __int64 c;
    while (c = a%b) a = b, b = c;
    return b;
}
int main(){
    __int64 n, m, g;
    while (scanf("%I64d%I64d", &n, &m), n + m){
        n *= m - 1; m *= m;
        g = GCD(n, m);
        printf("%I64d/%I64d
", n / g, m / g);
    }
    return 0;
}