DZY Loves Balls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 807 Accepted Submission(s): 439

Problem Description

There are

black balls and

white balls in the big box.

Now, DZY starts to randomly pick out the balls one by one. It forms a sequence

. If at the

-th operation, DZY takes out the black ball,

Si=1

, otherwise

Si=0

DZY wants to know the expected times that '01' occurs in

Input

The input consists several test cases. (

TestCase≤150

)

The first line contains two integers,

m(1≤n,m≤12)

Output

For each case, output the corresponding result, the format is

p/q

(

and

are coprime)

Sample Input

1 12 3

Sample Output

1/26/5            Hint    Case 1: S='01' or S='10', so the expected times = 1/2 = 1/2Case 2: S='00011' or S='00101' or S='00110' or S='01001' or S='01010' or S='01100' or S='10001' or S='10010' or S='10100' or S='11000',so the expected times = (1+2+1+2+2+1+1+1+1+0)/10 = 12/10 = 6/5

题目大意：

给你n个黑球，m个白球，黑球标记为1，白球标记为0，问在所有的组合当中一共出现了多少个“01”串。

解题思路：

用概率统计的角度讲，这就是一个n重的伯努利试验。首先，确定一个随机变量。

设置为Xi，则在Xi位置上出现白球，并在X(i+1)位置上出现黑球的概率是p=(m/(n+m))*(n/(n+m-1))。这就是出现01串的概率，否则其他的情况概记为q=1-p。

即Xi只有两种状态，出现01串为1，否则为0。如下图所示。

这就是一个最为简单的二项分布了，以为Xi的取值是从1-(m+n-1)的。

所以有

得E(X)=(m/(m+n))*(n/(n+m-1))*(n+m-1)=(m*n)/(m+n)。

程序里面需要求的就是m*n和m+n的最大公约数化简了。

源代码：

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              using namespace std;#define MAX 0x3f3f3f3f#define MIN -0x3f3f3f3f#define PI 3.14159265358979323#define N 1005int gcd(int n, int m){ int temp; if (m > n) swap(m, n); if (m == 1 || n == 1) return 1; temp = n%m; while (temp != 0) { n = m; m = temp; temp = n%m; } return m;}int main(){ int n, m; int num; int nn, mm; while (scanf("%d%d", &n, &m) != EOF) { nn = m*n; mm = m + n; num = gcd(nn, mm); printf("%d/%d\n", nn / num, mm / num); } return 0;}