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
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define rep(i, a, b) for (int i = a; i <= b; ++i) 4 5 const int N = 801; 6 const long long Mo = 1000000007; 7 8 int n, m, k, a[N][N], dp[N][N][17][2]; 9 10 int main() { 11 scanf("%d%d%d", &n, &m, &k); ++k; 12 rep(i, 1, n) rep(j, 1, m) { 13 scanf("%d", &a[i][j]); 14 dp[i][j][a[i][j] % k][0] = 1; 15 } 16 17 rep(i, 1, n) rep(j, 1, m) rep(p, 0, k) { 18 dp[i][j][p][0] = (dp[i][j][p][0] + dp[i - 1][j][(p - a[i][j] + k) % k][1]) % Mo; 19 dp[i][j][p][0] = (dp[i][j][p][0] + dp[i][j - 1][(p - a[i][j] + k) % k][1]) % Mo; 20 dp[i][j][p][1] = (dp[i][j][p][1] + dp[i - 1][j][(p + a[i][j]) % k][0]) % Mo; 21 dp[i][j][p][1] = (dp[i][j][p][1] + dp[i][j - 1][(p + a[i][j]) % k][0]) % Mo; 22 } 23 24 long long ans = 0; 25 rep(i, 1, n) rep(j, 1, m) { 26 ans = (ans + dp[i][j][0][1]) % Mo; 27 } 28 29 printf("%lld\n", ans); 30 31 return 0; 32 }