HDU 4832 (DP + Count problem)

HDU 4832 Chess

Train of Thought: Find the rows and columns in dp respectively, and then use several lines in one session, and use several lines in the vertical layout. Then, multiply them to accumulate the answer.
Code:

#include 
  
   #include 
   
    #include 
    
     using namespace std;typedef long long ll;const ll MOD = 9999991;const int N = 1005;int t, n, m, k, x, y;ll dp1[N][N], dp2[N][N], C[N][N];int main() {    for (int i = 0; i <= 1000; i++) {    C[i][0] = C[i][i] = 1;    for (int j = 1; j < i; j++) {        C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;    }    }    int cas = 0;    scanf("%d", &t);    while (t--) {    scanf("%d%d%d%d%d", &n, &m, &k, &x, &y);    memset(dp1, 0, sizeof(dp1));    memset(dp2, 0, sizeof(dp2));    dp1[0][x] = dp2[0][y] = 1;    for (int i = 1; i <= k; i++) {        for (int j = 1; j <= n; j++) {        if (j >= 2)             dp1[i][j] = (dp1[i][j] + dp1[i - 1][j - 2]) % MOD;        if (j >= 1)            dp1[i][j] = (dp1[i][j] + dp1[i - 1][j - 1]) % MOD;        dp1[i][j] = (dp1[i][j] + dp1[i - 1][j + 1]) % MOD;        dp1[i][j] = (dp1[i][j] + dp1[i - 1][j + 2]) % MOD;        }    }    for (int i = 1; i <= k; i++) {        for (int j = 1; j <= m; j++) {        if (j >= 2)             dp2[i][j] = (dp2[i][j] + dp2[i - 1][j - 2]) % MOD;        if (j >= 1)            dp2[i][j] = (dp2[i][j] + dp2[i - 1][j - 1]) % MOD;        dp2[i][j] = (dp2[i][j] + dp2[i - 1][j + 1]) % MOD;        dp2[i][j] = (dp2[i][j] + dp2[i - 1][j + 2]) % MOD;        }    }    ll heng[N], shu[N];    memset(heng, 0, sizeof(heng));    memset(shu, 0, sizeof(shu));    for (int i = 1; i <= n; i++)        for (int kk = 0; kk <= k; kk++)        heng[kk] = (heng[kk] + dp1[kk][i]) % MOD;    for (int i = 1; i <= m; i++)        for (int kk = 0; kk <= k; kk++)        shu[kk] = (shu[kk] + dp2[kk][i]) % MOD;    ll ans = 0;    for (int i = 0; i <= k; i++) {        ans = (ans + (heng[i] * shu[k - i] % MOD) * C[k][i] % MOD) % MOD;    }    printf("Case #%d:\n", ++cas);    cout << ans << endl;    }    return 0;}