2016多校训练Contest7: 1002 Hearthstone hdu5810

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.

E[V]=E[​m​​∑​i=1​m​​(X​i​​−​X​¯​​)​2​​​​]=E[(X​i​​−​X​¯​​)​2​​]=E[X​i​2​​−2X​i​​​X​¯​​+​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[X​i​2​​]−2​X​¯​​E[X​i​​]+E[​X​¯​​​2​​]=E[X​i​2​​]−2​X​¯​​​2​​+​X​¯​​​2​​=E[X​i​2​​]−​m​2​​​​n​2​​​​

所以关键是要求出E[X_i^2]E[X​i​2​​]. 我们用随机变量Y_jY​j​​来表示第j个球是否在第i个盒子中，如果在则Y_j = 1Y​j​​=1，否则Y_j = 0Y​j​​=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[X​i​2​​]=E[(∑​j=1​n​​Y​j​​)​2​​]=E[∑​j=1​n​​Y​j​2​​]+2E[∑​j=1​n​​∑​k=1,k≠j​n​​Y​j​​Y​k​​]=nE[Y​j​2​​]+n(n−1)E[Y​j​​Y​k​​]

=\frac{n}{m}+\frac{n(n-1)}{m^2}=​m​​n​​+​m​2​​​​n(n−1)​​

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]=​m​​n​​+​m​2​​​​n(n−1)​​−​m​2​​​​n​2​​​​=​m​2​​​​n(m−1)​​

官方题解如上。。因为是队友推的我具体不太清楚。。

#include <stdio.h>#include <algorithm>using namespace std;typedef __int64 ll;ll gcd(ll n,ll m){    ll r;    while(m){        r = n%m;        n = m;        m = r;    }    return n;}int main(void){    ll a,b;    ll n,m;    while(scanf("%I64d%I64d",&n,&m)!=EOF){        if(n==0&&m==0){            break;        }        a = n*(m-1);        b = m*m;        ll g = gcd(a,b);        printf("%I64d/%I64d\n",a/g,b/g);    }    return 0;}