hdu 5810 Balls and Boxes 二项分布

Balls and Boxes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 260    Accepted Submission(s): 187

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 7

 

 

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题意：给你n个球,m个盒子，每个球落入每个盒子的概率是等可能的，求方差的期望值。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef  long long  ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=1e3+10;

ll gcd(ll a,ll b)
{
    if(b==0) return a;
    else return gcd(b,a%b);
}

int main()
{
    ll n,m;
    while(~scanf("%lld%lld",&n,&m)&&(n||m))
    {
      ll fenzi=n*(m-1),fenmu=m*m;
      while(1)
      {
          ll k=gcd(fenzi,fenmu);
          if(k==1) break;
          fenzi/=k;fenmu/=k;
      }
      printf("%lld/%lld\n",fenzi,fenmu);
    }
    return 0;
}

　分析：比赛时就感觉是个什么分布，，但是学的很多又忘了，最后百度了一下，才发现可以二项分布做；

对于每个盒子，每个球落入其中的概率是p=1/m；

那么总共n个球p(x=k)=C（n,k）*p^k*(1-p)^(n-k),显然的二项分布；

二项分布数学期望E(x)=np(n是实验次数，p是每次试验球落入盒子的概率)；

方差D（x）=np(1-p)

本题中D（x）=n/m*(1-1/m)=n*(m-1)/（m^2）;

 然后因为每个盒子是平等的，方差又是描述数据的混乱程度，所以多个均等的盒子的方差与单个盒子方差

是一样的