HDU 4832 Chess

思路：把行列的情況分別dp求出來，然後列舉行用幾行。豎用幾行，然後相乘累加起來就是答案
程式碼：

#include <stdio.h>
#include <string.h>
#include <iostream>
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;
}