dp,有点懂也有点不懂,晕.....
code:
/* ID: yueqiq LANG: 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; }