Balls and Boxes

                                                                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                 Total Submission(s): 797    Accepted Submission(s): 526

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.
 

 

Author
SYSU
 

 

Source
2016 Multi-University Training Contest 
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题意：把n个球放到m个盒子里面,求上面这个式子的期望;
思路：打表找规律，发现答案=n*(m-1)/m*m 求gcd即可
证明：
E(V)=1/m*E(∑(xi-x)2)=E((x-n/m)2)=E(x2)-2*n/m*E(x)+n2/m2
E(x)=n/m;E(x2)=D(x)+[E(x)]2;变成二项分布了,D(x)=n*(m-1)/m2
所以带到上面的式子中就变成了E(v)=n*(m-1)/m2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

LL n,m;

LL gcd(LL x,LL y)
{
    return x?gcd(y%x,x):y;
}

int main()
{
    while(~scanf("%lld%lld",&n,&m)&&(n+m))
    {
        if(m==1) {printf("0/1\n");continue;}
        n=(m-1)*n;
        m=m*m;
        printf("%lld/%lld\n",n/gcd(n,m),m/gcd(n,m));
    }
    return 0;
}