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(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.
 
公式题，打表找了规律
#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#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, rusing 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;}