USACO 2.3 Cow Pedigrees (nocows)

//Main idea: //Dynamic Programming//dp[i][j] denote the number of binary trees whose node number is i and height is less or equal to j;//state transition equation://dp[i][j] = dp[m][j-1]*dp[i-1-m][j-1];//dp[m][j-1] denote the left subtree with m nodes and no more than j-1 height ;//and dp[i-1-m][j-1] denote the right subtree i-1-m nodes(the root take 1 node and left subtree take m),//no more than j-1 height;//So the number of birnary trees with i nodes and j height is dp[i][j]-dp[i][j-1];/*ID: haolink1PROG: nocowsLANG: C++*///#include <iostream>#include <fstream>using namespace std;int dp[200][100];int main(){    ifstream fin("nocows.in");    int node_num, hight;    fin >> node_num >> hight;    //boundary conditions    for(int j = 1; j <= hight; j++){        dp[1][j] = 1;    }    for(int i = 2; i <= node_num; i++){        for(int j = 2; j <= hight; j++){            for(int m = 1; m <= i-2; m++){                //mod dp[i][j] by 9901 in case it exceed the int max positive value  and become negative;                dp[i][j] = (dp[i][j] + dp[m][j-1]*dp[i-1-m][j-1])%9901;            }        }    }    ofstream fout("nocows.out");    //Note: after mod operation, dp[node_num][hight] may be less than dp[node_num][hight-1];    //So we add 9901 at the end;    fout << (dp[node_num][hight] - dp[node_num][hight-1] + 9901)%9901<<endl;    return 0;}