usaco Cow Pedigrees

dp，有点懂也有点不懂，晕.....

code：

/*ID: yueqiqLANG: C++TASK: nocows*/#include <set>#include <map>#include <ctime>#include <queue>#include <cmath>#include <stack>#include <limits>#include <vector>#include <bitset>#include <string>#include <cstdio>#include <cstring>#include <fstream>#include <string.h>#include <iostream>#include <algorithm>#define ls rt<<1#define rs rt<<1|1#define Si set<int>#define LL long long#define pb push_back#define PS printf(" ")#define Vi vector<int>#define LN printf("\n")#define SD(a) scanf("%d",&a)#define PD(a) printf("%d",a)#define SET(a,b) memset(a,b,sizeof(a))#define FF(i,a) for(int i(0);i<(a);i++)#define FD(i,a) for(int i(a);i>=(1);i--)#define FOR(i,a,b) for(int i(a);i<=(b);i++)#define FOD(i,a,b) for(int i(a);i>=(b);i--)#define readf freopen("nocows.in","r",stdin)#define writef freopen("nocows.out","w",stdout)const double pi = acos(-1.0);const int maxn = 50;const int BigP = 99999999;const int INF = 99999999;const int dx[]={0,1,0,-1};const int dy[]={1,0,-1,0};using namespace std;int N,K,dp[500][500];int main (){    readf;    writef;    SD(N);SD(K);    FOR(i,1,K) dp[1][i]=1;    FOR(i,3,N) FOR(j,2,K) FOR(k,1,i-2){        dp[i][j]=(dp[i][j]+dp[k][j-1]*dp[i-k-1][j-1])%9901;    }    PD((dp[N][K]-dp[N][K-1]+9901)%9901);LN;return 0;}