1 条题解
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1
#include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int Mod=9901; int dp[210][110],n,k; int main(){ scanf("%d%d",&n,&k); for (int i=1;i<=k;i++)dp[1][i]=1; for (int tk=1;tk<=k;tk++) for (int i=3;i<=n;i+=2) for (int j=1;j<i;j+=2) (dp[i][tk]+=dp[j][tk-1]*dp[i-j-1][tk-1])%=Mod; printf("%d",(dp[n][k]-dp[n][k-1]+Mod)%Mod); return 0; } #include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int Mod=9901; int dp[210][110],n,k; int main(){ scanf("%d%d",&n,&k); for (int i=1;i<=k;i++)dp[1][i]=1; for (int tk=1;tk<=k;tk++) for (int i=3;i<=n;i+=2) for (int j=1;j<i;j+=2) (dp[i][tk]+=dp[j][tk-1]*dp[i-j-1][tk-1])%=Mod; printf("%d",(dp[n][k]-dp[n][k-1]+Mod)%Mod); return 0; } #include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int Mod=9901; int dp[210][110],n,k; int main(){ scanf("%d%d",&n,&k); for (int i=1;i<=k;i++)dp[1][i]=1; for (int tk=1;tk<=k;tk++) for (int i=3;i<=n;i+=2) for (int j=1;j<i;j+=2) (dp[i][tk]+=dp[j][tk-1]*dp[i-j-1][tk-1])%=Mod; printf("%d",(dp[n][k]-dp[n][k-1]+Mod)%Mod); return 0; }
- 1
信息
- ID
- 575
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 7
- 标签
- 递交数
- 13
- 已通过
- 10
- 上传者