bzoj1801: [ahoi2009]chess Chinese chess

Dp.

If the pressure DP, you can only get 50 points. and positive solution is both better than the shape of the pressure to write, but also the positive solution.

F[I][J][K] Indicates that there is a J column in line I has a chess piece, there are two pieces of k column, and then the DP transfer is good (the equation is too difficult to write, do not write ...) ）

100*100*mod may explode int, so a long long is used.

#include <cstdio>#include<algorithm>#include<cstring>using namespaceStd;typedefLong LongLL;Const intMAXN = -+Ten;ConstLL mod =9999973; LL F[MAXN][MAXN][MAXN]; LL N,m,res;intMain () {scanf ("%lld%lld",&n,&m); Memset (F,0,sizeof(f)); f[0][0][0]=1;  for(LL i=1; i<=n;i++) {         for(LL j=0; j<=m;j++)         for(LL k=0; k<=m;k++)if(j+k<=m) {LL&h =F[i][j][k]; H= (h+f[i-1][J][K])%MoD; if(j>=1) h= (h+f[i-1][j-1][k]* (m-j-k+1))%MoD; if(k>=1) h= (h+f[i-1][j+1][k-1]* (j+1))%MoD; if(j>=2) h= (h+f[i-1][j-2][k]* ((m-j-k+1) * (m-j-k+2)/2))%MoD; if(k>=1) h= (h+f[i-1][j][k-1]* (m-j-k+1) *j)%MoD; if(k>=2) h= (h+f[i-1][j+2][k-2]* ((j+2) * (j+1)/2))%MoD; }} Res=0;  for(intI=0; i<=m;i++)      for(intj=0; j<=m;j++)if(i+j<=m) res= (Res+f[n][i][j])%MoD; printf ("%lld\n", RES); return 0;}

bzoj1801: [ahoi2009]chess Chinese chess