HDU 5810 Balls and Boxes

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(XiX¯)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.

公式题,打表找了规律

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 10;
int T, n, m;

LL gcd(LL x, LL y)
{
    return x%y ? gcd(y, x%y) : y;
}

void solve()
{
    int f[N + 1][N + 1], c[N + 1][N + 1];
    c[0][0] = 1;
    rep(i, 1, N)
    {
        c[i][0] = c[i][i] = 1;
        rep(j, 1, i - 1) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
    }
    rep(i, 1, N)
    {
        printf("%d: ", i);
        int p = 1;
        rep(j, 0, N)
        {
            f[i][j] = 0;
            int g = 1;
            rep(k, 0, j)
            {
                f[i][j] += c[j][k] * (f[i - 1][k] + (j - k)*(j - k)*g);
                g *= i - 1;
            }
            //printf("%d ", f[i][j]);
            int q = i * f[i][j] - j * j * p;
            p *= i;
            printf("%d/%d ", q / gcd(q, p*i), p*i / gcd(q, p*i));
        }
        printf("\n");
    }
}

int main()
{
    //scanf("%d", &T);
    while (scanf("%d%d", &n, &m), n&&m)//T--)
    {
        LL x = 1LL * n*(m - 1), y = 1LL * m*m;
        printf("%lld/%lld\n", x / gcd(x, y), y / gcd(x, y));
    }
    return 0;
}


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