HDU-5194-DZY Loves Balls(BestCoder Round # 35 )

DZY Loves Balls

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

Total Submission(s): 374 Accepted Submission(s): 205



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

HintCase 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

Source

BestCoder Round #35

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BestCoder解析：

Problem A - DZY Loves Balls
考虑期望的可加性。第i(1≤i<n+m)个位置上出现0，第i+1个位置上出现1的概率是mn+m×nn+m−1，那么答案自然就是∑i=1n+m−1mn+m×nn+m−1=nmn+m
如果你不能马上想到上述的简便的方法，也可以选择暴力枚举所有01串，也是可以AC的。最后一步你需要再计算一下gcd，十分简便。

n个黑球，m个白球,1表示取出的是黑球，0表示取出的是白球。

这题最好的方法就是找出题目的规律，即第i(1<=i<n+m)个位置上出现0，第i+1个位置上出现1的概率是 m/(n+m) * n/(n+m-1) .

因为要求‘01’串在S串中出现的位置，即满足这个条件，但是'0'不可以在末尾，'1'不可以在开头.对于前面n+m-1个位置都可以为0.

第i+1个位置上出现1的概率是：(出现黑球（出现1）的概率)*(出现在第i个位置上的概率)。

但是一共有n+m-1个位置上可以出现1，所以便是所有位置上的概率想加.

等价于= m/(n+m) * n/(n+m-1) * (n+m-1) = n*m/(n+m).

import java.io.*;
import java.util.*;

public class Main
{

public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
while (input.hasNext())
{
int n = input.nextInt();
int m = input.nextInt();
int temp = GCD(n * m, n + m);
System.out.println(n * m / temp + "/" + (n + m) / temp);
}
}

public static int GCD(int x, int y)
{
if (x < y)
return GCD(y, x);
while (x % y != 0)
{
int temp = x % y;
x = y;
y = temp;
}
return y;
}
}