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(XiX¯)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[mi=1m(XiX¯)2]=E[(XiX¯)2]=E[Xi22XiX¯+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[Xi2]2X¯E[Xi]+E[X¯2]=E[Xi2]2X¯2+X¯2=E[Xi2]m2n2

所以关键是要求出E[X_i^2]E[Xi2]. 我们用随机变量Y_jYj来表示第j个球是否在第i个盒子中,如果在则Y_j = 1Yj=1,否则Y_j = 0Yj=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[Xi2]=E[(j=1nYj)2]=E[j=1nYj2]+2E[j=1nk=1,kjnYjYk]=nE[Yj2]+n(n1)E[YjYk]

=\frac{n}{m}+\frac{n(n-1)}{m^2}=mn+m2n(n1)

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]=mn+m2n(n1)m2n2=m2n(m1)

#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;
}











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