Large [Luo Gu P1373] [explanations] and a small sum to flee uim

This question can easily see is a dp (dp hurdle because it is in looking)

A little think about yy can be a good state:

Number of f [i] [j] [k] [0/1] Representative reached the point (i, j), the amount of the liquid membrane phase difference k (small-uim a = k), small a / uim taken last case

A pit:

F [ 850 ] [ 850 ] [ 20 is ] [ 2 ]: Accepted 100  2 .00s 103 .31MB 
F [ 850 ] [ 850 ] [ 25 ] [ 2 ]: unaccepted 85  . 1 .11s 125 .00MB // MLE three points

Transfer also do well: a point may only coming from the left or top

int tmp=(c-mp[i][j]+k)%k;
f[i][j][c][0]+=(f[i-1][j][tmp][1]%mod+f[i][j-1][tmp][1]%mod)%mod;
f[i][j][c][0]%=mod;            
tmp=(c+mp[i][j])%k;
f[i][j][c][1]+=(f[i-1][j][tmp][0]%mod+f[i][j-1][tmp][0]%mod)%mod;
f[i][j][c][1]%=mod;

Wherein the liquid film on a point of difference between the representative tmp

Pit II: liquid volume may be greater than the film at the time of input k

Pit III: can start from any point, should each f [i] [j] [mp [i] [j]] [0] value of 1

Four pit: c-mp [i] [j] may be negative, if not written tmp = (c-mp [i] [j] + k)% k but tmp = (c-mp [i] [j ])% k, then,

Unaccepted 10 2.01s 103.01MB

Not much to say, look at the code

. 1 #include <bits / STDC ++ H.>
 2  the using  namespace STD;
 . 3  int n-, m, K, MP [ 850 ] [ 850 ];
 . 4  // pit a: bomb space 
. 5  int F [ 850 ] [ 850 ] [ 20 is ] [ 2 ], SUM;
 . 6  #define MOD 1000000007
 . 7  int main () {
 . 8      iOS :: sync_with_stdio ( 0 );
 . 9      CIN >> >> m >> n- K;
 10      K ++ ;
 . 11      // read 
12 is      for( Int I = . 1 ; I <= n-; I ++ ) {
 13 is          for ( int J = . 1 ; J <= m; J ++ ) {
 14              CIN >> MP [I] [J];
 15              // pit II: film liquid an amount of 
16              MP [I] [J] =% K;
 . 17              // pit three: the starting point 
18 is              F [I] [J] [MP [I] [J]] [ 0 ] = . 1 ;
 . 19          }
 20 is      }
 21 is      // the DP 
22 is      for ( int I = . 1 ; I <= n-; I ++ ) {
 23 is          for(int j=1;j<=m;j++){
24             for(int c=0;c<k;c++){
25                 //坑四：取膜 
26                 int tmp=(c-mp[i][j]+k)%k;
27                 f[i][j][c][0]+=(f[i-1][j][tmp][1]%mod+f[i][j-1][tmp][1]%mod)%mod;
28                 f[i][j][c][0]%=mod;
29                 
30                 tmp=(c+mp[i][j])%k;
31                 f[i][j][c][1]+=(f[i-1][j][tmp][0]%mod+f[i][j-1][tmp][0]%mod)%mod;
32                 f[i][j][c][1]%=mod;
33             }
34             sum=(sum+f[i][j][0][1])%mod;
35         }
36     }
37     cout<<sum<<endl;
38     return 0;
39 }

End Sahua

Large [Luo Gu P1373] [explanations] and a small sum to flee uim

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