HDU 5810 Balls and Boxes two distributions

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 DescriptionMr. Chopsticks is interested in the random phenomena, and he conducts an experiment to study randomness. In the experiment, he throws n balls into m boxes in such a manner so each ball have equal probability of going to each B Oxes. 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.
 

InputThe input contains multiple test cases. Each case contains the integers n and m (1 <= N, M <=) in a line.
The input is terminated by n = m = 0. 

Outputfor each case, output the result as A/b in A line, where A/b should is an irreducible fraction. Let b=1 if the result was an integer. 

Sample Input2 0 

Sample Output0/11/2Hintin the second sample, there is four possible outcomes, and the other outcomes with V = 0 and both outcomes with V = 1. 

AuthorSysu 

Sourcemulti-university Training Contest 7 

recommendwange2014 | We have carefully selected several similar problems for you: 5820 5819 5817 5816 5815 Test Instructions: give you n ball, m box, the probability that each ball falls into each box is so possible, the expectation of variance is obtained.

 #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 scanfconst int mod=1000000007;const int n=1e3+10;ll gcd (ll a,ll b) {if (b==0) retur    n 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;} 

　Analysis: The game is a feeling of what distribution, but learn a lot and forget, finally Baidu a bit, only to find that can be two distribution to do;

For each box, the probability of each ball falling into it is p=1/m;

So a total of n balls P (x=k) =c (n,k) *p^k* (1-p) ^ (n-k), apparently two distributions;

Two-item distribution mathematical expectation E (x) =NP (n is the number of experiments, p is the probability of each test ball falling into the box);

Variance d (x) =NP (1-p)

In this subject D (x) =n/m* (1-1/m) =n* (m-1)/(m^2);

Then because each box is equal, the variance is the degree of confusion that describes the data, so the variance of multiple equal boxes is the same as a single box variance

HDU 5810 Balls and Boxes two distributions