srm 554 dv2 1000pt

好暴力的dp啊。。有木有更简单的方法呢？

# include <cstdio>
# include <algorithm>
# include <string>
# include <cstring>
# include <iostream>
# define MOD 1234567891
using namespace std;
typedef long long ll;
ll dp[50][10][4][4][4][4];
class TheBrickTowerHardDivTwo
{
public:
    int find(int C, int K, int H)
    {
        ll n,k,a,b,c,d,x,y,z,i,ans=0ll;
        for (a=0;a<C;++a)
            for (b=0;b<C;++b)
                for (c=0;c<C;++c)
                    for (d=0;d<C;++d)
                    {
                        ll p = (a==b)+(b==c)+(c==d)+(a==d);
                        dp[1][p][a][b][c][d] = 1;
                    }    
        for (n=2;n<=H;++n)
            for (k=0;k<=K;++k)
                for (a=0;a<C;++a)
                    for (b=0;b<C;++b)
                        for (c=0;c<C;++c)
                            for (d=0;d<C;++d)
                                for (x=0;x<C;++x)
                                    for (y=0;y<C;++y)
                                        for (z=0;z<C;++z)
                                            for (i=0;i<C;++i)
                                            {
                                                ll p = (a==b)+(b==c)+(c==d)+(a==d);
                                                p += ((a==x)+(b==y)+(c==z)+(d==i));
                                                if (p<=k)
                                                {
                                                    dp[n][k][a][b][c][d] += dp[n-1][k-p][x][y][z][i];    
                                                    dp[n][k][a][b][c][d] %= MOD;
                                                }
                                            }    
    for (k=0;k<=K;++k)
        for (n=1;n<=H;++n)                                
            for (a=0;a<C;++a)
                for (b=0;b<C;++b)
                    for (c=0;c<C;++c)
                        for (d=0;d<C;++d)
                        {
                            ans += dp[n][k][a][b][c][d];
                            ans %= MOD;
                        }
        return ans;
    }
};