10943-how Do you add? (Partition method)

This problem can be simplified to the number of the N ball in the K box, because each box can be empty, so we might as well add a K ball, so that we can at least ensure that a box has a ball, and then use the partition method to divide n + K ball, that is, n + K-1 to choose K-1 position C (n + K-1,K-1)

The solution method can be recursive.

Other than that:

  number of permutations and combinations  of C (m,n) =a (m,n)/n!

C (m,0) = 1;

0! = 1;

Code:

#include <cstdio> #include <cmath>using namespace std;typedef long long ll;const int MAXN = 222;const int Mod
    = 1000000; LL c[maxn][maxn];void init () {    c[1][0] = 1; C[1][1] = 1;    for (int i = 2; I <=, i++) {for        (int j = 0; J <= I; j + +) {            if (j)                c[i][j] = c[i-1][j-1] + c[i-1][j ];            else                c[i][j] = 1;            C[I][J]%= mod;}}}    void Debug () {}int main () {    init ();    int n,m;    while (scanf ("%d%d", &n,&m)! = EOF) {        if (!n &&!m) break;        LL ans = c[n + m-1][m-1];        printf ("%lld\n", ans);    }    return 0;}

10943-how Do you add? (Partition method)