HDU 4832(DP+计数问题)
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HDU 4832 Chess
思路:把行列的情况分别dp求出来。然后枚举行用几行,竖用几行,然后相乘累加起来就是答案
代码:
#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:
", ++cas);
cout << ans << endl;
}
return 0;
}
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