luogu1373_小a和uim之大逃离 多维dp

传送门

巧妙之处在于dp的设计只用设计差值即可，因此不会mle，枚举的顺序问题也解决了

 #include <bits/stdc++.h>
 using namespace std;
 #define rep(i, a, b) for (int i = a; i <= b; ++i)

 const int N = ;
 const long long Mo = ;

 int n, m, k, a[N][N], dp[N][N][][];

 int main() {
     scanf("%d%d%d", &n, &m, &k); ++k;
     rep(i, , n) rep(j, , m) {
         scanf("%d", &a[i][j]);
         dp[i][j][a[i][j] % k][] = ;
     }

     rep(i, , n) rep(j, , m) rep(p, , k) {
         dp[i][j][p][] = (dp[i][j][p][] + dp[i - ][j][(p - a[i][j] + k) % k][]) % Mo;
         dp[i][j][p][] = (dp[i][j][p][] + dp[i][j - ][(p - a[i][j] + k) % k][]) % Mo;
         dp[i][j][p][] = (dp[i][j][p][] + dp[i - ][j][(p + a[i][j]) % k][]) % Mo;
         dp[i][j][p][] = (dp[i][j][p][] + dp[i][j - ][(p + a[i][j]) % k][]) % Mo;
     }

     long long ans = ;
     rep(i, , n) rep(j, , m) {
         ans = (ans + dp[i][j][][]) % Mo;
     }

     printf("%lld\n", ans);

     return ;
 }