HDU5810 Balls and Boxes

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 12 20 0

 

Sample Output

0/11/2
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(∑(x_i-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 longconst 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;}