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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)
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Description |
题目描述 |
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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 as 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
你的任务就是找出V的期望。 |
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Input |
输入 |
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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。
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Output |
输出 |
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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.
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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。
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Sample Input - 输入样例 |
Sample Output - 输出样例 |
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2 1 |
0/1 |
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Hint |
提示 |
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In the second sample, there are four possible outcomes, two outcomes with V = 0 and two outcomes with V = 1.
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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 }
HDU 5810 Balls and Boxes（盒子与球）
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原文地址：http://www.cnblogs.com/Simon-X/p/5757953.html
Where 
is the number of balls in the ith box, and 
is the average number of balls in a box.