HDU5810（2016多校第七场）——Balls and Boxes（组合数学，概率）

Balls and Boxes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 29    Accepted Submission(s): 21

Problem Description

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

V=∑mi=1(Xi−X¯)2m

where  Xi  is the number of balls in the ith box, and  X¯  is the average number of balls in a box.
Your task is to find out the expected value of V.

 

Input

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.

 

Output

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.

 

Sample Input

    
    
    
    

     
     
     
     2 1
2 2
0 0
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     0/1
1/2


     
     
     
     

      
      
      
      

       
       
       
       Hint
      
      
      
      
In the second sample, there are four possible outcomes, two outcomes with V = 0 and two outcomes with V = 1.

     
     
     
     
     
     
     
      
    
    
    
    

要注意球和盒子都是不一样的，位置不同就是不同的情况。

队友推的公式。

其实我并不太懂，附上官方题解吧。

E[V]=E[​m​​∑​i=1​m​​(X​i​​−​X​¯​​)​2​​​​]=E[(X​i​​−​X​¯​​)​2​​]=E[X​i​2​​−2X​i​​​X​¯​​+​X​¯​​​2​​]

= E[X_i^2] - 2\bar XE[X_i] + E[\bar X^2] = E[X_i^2] - 2\bar X^2 + \bar X^2 = E[X_i^2] - \frac{n^2}{m^2}=E[X​i​2​​]−2​X​¯​​E[X​i​​]+E[​X​¯​​​2​​]=E[X​i​2​​]−2​X​¯​​​2​​+​X​¯​​​2​​=E[X​i​2​​]−​m​2​​​​n​2​​​​

所以关键是要求出E[X_i^2]E[X​i​2​​]. 我们用随机变量Y_jY​j​​来表示第j个球是否在第i个盒子中，如果在则Y_j = 1Y​j​​=1，否则Y_j = 0Y​j​​=0. 于是

E[X_i^2] = E[(\sum_{j=1}^{n} Y_j)^2] = E[\sum_{j=1}^{n} Y_j^2] + 2E[\sum_{j=1}^{n} \sum_{k=1,k\neq j}^{n} Y_jY_k] = nE[Y_j^2] + n(n-1)E[Y_jY_k]E[X​i​2​​]=E[(∑​j=1​n​​Y​j​​)​2​​]=E[∑​j=1​n​​Y​j​2​​]+2E[∑​j=1​n​​∑​k=1,k≠j​n​​Y​j​​Y​k​​]=nE[Y​j​2​​]+n(n−1)E[Y​j​​Y​k​​]

=\frac{n}{m}+\frac{n(n-1)}{m^2}=​m​​n​​+​m​2​​​​n(n−1)​​

因此，

E[V] = \frac{n}{m} + \frac{n(n-1)}{m^2} - \frac{n^2}{m^2} = \frac{n(m-1)}{m^2}E[V]=​m​​n​​+​m​2​​​​n(n−1)​​−​m​2​​​​n​2​​​​=​m​2​​​​n(m−1)​​

#include
#include
#include
using namespace std;
const int maxn=1000;
typedef unsigned long long ULL;
ULL A,B,n,m;
ULL GCD(ULL a,ULL b)
{
    if(b==0){
        return a;
    }
    else{
        return GCD(b,a%b);
    }
}
void solve(void)
{

}
int main()
{
    while(scanf("%I64u %I64u",&n,&m)!=EOF&&n&&m){
        ULL tem=GCD(m*n-n,m*m);
        A=(m*n-n)/tem;
        B=(m*m)/tem;
        printf("%I64u/%I64u\n",A,B);
    }
    return 0;
}