1 //Accepted    176 KB    0 ms 2 //预处理出来0--20的部分 3 //dp 4 //其中用到-1的值 5 //把所有的点左移一下 6 #include <cstdio> 7 #include <cstring> 8 #include <iostream> 9 using namespace std;10 const int imax_n = 22;11 int dp[imax_n][imax_n][imax_n];12 int n;13 void Dp()14 {15     memset(dp,0,sizeof(dp));16     for (int i=0;i<imax_n;i++)17     for (int j=0;j<imax_n;j++)18     for (int k=0;k<=1;k++)19     dp[i][j][k]=1;20     for (int i=0;i<=1;i++)21     for (int j=0;j<imax_n;j++)22     for (int k=0;k<imax_n;k++)23     dp[i][j][k]=1;24     for (int i=0;i<imax_n;i++)25     for (int j=0;j<=1;j++)26     for (int k=0;k<imax_n;k++)27     dp[i][j][k]=1;28     for (int i=2;i<imax_n;i++)29     {30         for (int j=2;j<imax_n;j++)31         {32             for (int k=2;k<imax_n;k++)33             {34                 if (i<j && j<k)35                 dp[i][j][k]=dp[i][j][k-1]+dp[i][j-1][k-1]-dp[i][j-1][k];36                 else37                 dp[i][j][k]=dp[i-1][j][k]+dp[i-1][j-1][k]+dp[i-1][j][k-1]-dp[i-1][j-1][k-1];38             }39         }40     }41 }42 int getAns(int a,int b,int c)43 {44     if (a<=0 || b<=0 || c<=0) return 1;45     if (a>20 || b>20 || c>20) return dp[21][21][21];46     return dp[a+1][b+1][c+1];47 }48 int main()49 {50     Dp();51     int a,b,c;52     while (scanf("%d%d%d",&a,&b,&c) && !(a==-1 && b==-1 && c==-1))53     {54         printf("w(%d, %d, %d) = %d\n",a,b,c,getAns(a,b,c));55     }56     return 0;57 }