HDU 5194 DZY Loves Balls

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 n black balls and m white balls in the big box.

Now, DZY starts to randomly pick out the balls one by one. It forms a sequence S. If at the i-th operation, DZY takes out the black ball, Si=1, otherwise Si=0.

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

 

Input

The input consists several test cases. (TestCase≤150)

The first line contains two integers, n, m(1≤n,m≤12)

 

Output

For each case, output the corresponding result, the format is p/q(p and q are coprime)

 

Sample Input

1 1
2 3

 

Sample Output

1/2
6/5

Hint
Case 1: S='01' or S='10', so the expected times = 1/2 = 1/2
Case 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：

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string>
#include<string.h>
#include<math.h>
#include<map>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
#define MAX 0x3f3f3f3f
#define MIN -0x3f3f3f3f
#define PI 3.14159265358979323
#define N 1005
int 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;
}