BestCoder Round #35（第一题解题报告（组合数学（期望的可加性）或暴力枚举）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5194

DZY Loves Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 500    Accepted Submission(s): 276

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 
    
    
    
    
     
   
   
   
   

 

Source

BestCoder Round #35

解题思想：

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。

AC code：

#include<iostream>
#include<stdio.h>
#include<map>
#include<vector>
#include<set>
#include<cstdlib>
#include<string.h>
#include<algorithm>
#include<cmath>
#define MAXN 1000010
using namespace std;
int gcd(int a,int b)
{
	if(a<b)
	{
		int t=a;
		a=b;
		b=t;
	}
	int r=a%b;
	while(r)
	{
		a=b;
		b=r;
		r=a%b;
	}
	return b;
}
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int a=n*m;
		int b=n+m;
		int t=gcd(a,b);
		a/=t;
		b/=t;
		printf("%d/%d\n",a,b);	
	}
	return 0;
}