hdu-5810 Balls and Boxes(概率期望)

题目链接：

Balls and Boxes

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 65536/65536 K (Java/Others)

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

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：

推期望公式的题,我推的过程跟题解的不太一样;

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

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=3e5+10;
const int maxn=2e3+14;
const double eps=1e-12;

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

int main()
{
    LL n,m;
    while(1)
    {
        read(n);read(m);
        if(n==0&&m==0)break;
        if(m==1)printf("0/1\n");
        else
        {
            LL temp=gcd(n*(m-1),m*m);
            printf("%lld/%lld\n",n*(m-1)/temp,m*m/temp);
        }
    }

    return 0;
}