题目：
分析：
某OIer兔崽子的此题代码中的三个函数名：dfs、ddfs、dddfs（充满毒瘤的气息
显然，行与行之间、列与列之间是互相独立的。考虑背包，用(f[k][i][j])表示用前(k)种颜色占了(i)行(j)列的方案数，(g[i][j])表示用颜色(k)占据(i)行(j)列的方案数，(c[i])表示颜色为(i)的棋子数，就有如下方程：
[f[k][i][j]=sum _{a=0}^i sum_{b=0}^j f[k-1][i-a][j-b] imes g[a][b] imes C_{n-(i-a)}^a imes C_{m-(j-b)}^b(abgeq c[i])
]
(g[i][j])在算的时候注意要减去有空行或空列的情况（枚举有多少行、列不是空的）。注意要(a=i)且(b=j)的情况要跳过：
[g[i][j]=C_{ij}^{c[k]}-sum_{a=0}^i sum_{b=0}^j g[a][b] imes C_i^a imes C_j^b(ijgeq c[k])
]
代码：
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
using namespace std;
namespace zyt
{
template<typename T>
inline void read(T &x)
{
char c;
bool f = false;
x = 0;
do
c = getchar();
while (c != '-' && !isdigit(c));
if (c == '-')
f = true, c = getchar();
do
x = x * 10 + c - '0', c = getchar();
while (isdigit(c));
if (f)
x = -x;
}
template<typename T>
inline void write(T x)
{
static char buf[20];
char *pos = buf;
if (x < 0)
putchar('-'), x = -x;
do
*pos++ = x % 10 + '0';
while (x /= 10);
while (pos > buf)
putchar(*--pos);
}
typedef long long ll;
const int N = 40, T = 20, p = 1e9 + 9;
int n, m, c, arr[T], C[N * N][N * N], f[T][N][N], g[N][N];
void init()
{
for (int i = 0; i < N * N; i++)
{
C[i][0] = 1;
for (int j = 1; j <= i; j++)
C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % p;
}
}
int work()
{
init();
read(n), read(m), read(c);
for (int i = 1; i <= c; i++)
read(arr[i]);
f[0][0][0] = 1;
for (int k = 1; k <= c; k++)
{
memset(g, 0, sizeof(g));
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
{
if (i * j >= arr[k])
{
g[i][j] = C[i * j][arr[k]];
for (int a = 0; a <= i; a++)
for (int b = 0; b <= j; b++)
{
if (a != i || b != j)
g[i][j] = (g[i][j] - (ll)g[a][b] *
C[i][a] % p * C[j][b] % p + p) % p;
}
}
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
for (int a = 1; a <= i; a++)
for (int b = 1; b <= j; b++)
if (a * b >= arr[k])
f[k][i][j] = (f[k][i][j] +
(ll)f[k - 1][i - a][j - b] * g[a][b] % p *
C[n - (i - a)][a] % p * C[m - (j - b)][b]) % p;
}
int ans = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
ans = (ans + f[c][i][j]) % p;
write(ans);
return 0;
}
}
int main()
{
return zyt::work();
}