SRM 554 div2

The competition was still smooth. The only pity was that 1000pt was not made out. I did not think clearly before writing it. I wrote anything when I thought of it, which led to the loss of AK opportunities ....

250pt:

Question, not explained

500pt:

Same as above

1000pt:

Question: You can have up to four colors of bricks, each of which has an infinite number (1*1*1 Unit Cube ), construct a 2*2 * H (H <= h) prism, and the number of solutions with the same color logarithm of two adjacent cubes smaller than or equal to K

You can design such a status to describe the problem.

DP [H] [I] [J] [k] [l] [x]

Represent the height h in sequence, the color (I, J, K, L) of the top four cubes, the total color logarithm X, you can also compress the four colors into a thought binary. I just opened six dimensions.

The transfer process is easy. The key is to design the status ~

#include<cstdio>#include<cstring>#include<string>#include<vector>#include<set>#include<algorithm>using namespace std;typedef long long lld;const int mod = 1234567891;lld dp[50][4][4][4][4][8];class TheBrickTowerHardDivTwo{    public :    int find(int n, int K, int H)    {        memset(dp,0,sizeof(dp));        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                for(int k=0;k<n;k++)                  for(int l=0;l<n;l++)                    {                      int tmp=(i==j)+(i==k)+(k==l)+(j==l);                      if(tmp<=K) dp[1][i][j][k][l][tmp]=1;                    }         for(int h=2;h<=H;h++)           for(int x=0;x<=K;x++)            for(int i=0;i<n;i++)              for(int j=0;j<n;j++)                for(int k=0;k<n;k++)                  for(int l=0;l<n;l++)                    for(int a0=0;a0<n;a0++)                     for(int a1=0;a1<n;a1++)                      for(int a2=0;a2<n;a2++)                         for(int a3=0;a3<n;a3++)                          {                            int tmp=(i==j)+(i==k)+(k==l)+(j==l)+(a0==i) +(a1==j)+(a2==k)+(a3==l);                            if(x+tmp<=K && dp[h-1][a0][a1][a2][a3][x])                             dp[h][i][j][k][l][x+tmp]=(dp[h-1][a0][a1][a2][a3][x]+dp[h][i][j][k][l][x+tmp])%mod;                          }                          lld ans=0;                          for(int h=1;h<=H;h++)                           for(int x=0;x<=K;x++)                            for(int i=0;i<n;i++)                             for(int j=0;j<n;j++)                              for(int k=0;k<n;k++)                                for(int l=0;l<n;l++)                                {                                    ans+=dp[h][i][j][k][l][x];                                    ans%=mod;                                }                                return (int)ans;    } };