solution.md

# 地宫取宝

X 国王有一个地宫宝库，是 n×m 个格子的矩阵，每个格子放一件宝贝，每个宝贝贴着价值标签。
地宫的入口在左上角，出口在右下角。
小明被带到地宫的入口，国王要求他只能向右或向下行走。
走过某个格子时，如果那个格子中的宝贝价值比小明手中任意宝贝价值都大，小明就可以拿起它（当然，也可以不拿）。
当小明走到出口时，如果他手中的宝贝恰好是 k 件，则这些宝贝就可以送给小明。
请你帮小明算一算，在给定的局面下，他有多少种不同的行动方案能获得这 k 件宝贝。

#### 输入格式

第一行 3 个整数，n,m,k，含义见题目描述。
接下来 n 行，每行有 m 个整数 Ci 用来描述宝库矩阵每个格子的宝贝价值。

#### 输出格式

输出一个整数，表示正好取 k 个宝贝的行动方案数。
该数字可能很大，输出它对 1000000007 取模的结果。

#### 数据范围

```
1≤n,m≤50,
1≤k≤12,
0≤Ci≤12
```
#### 输入样例1：

```
2 2 2
1 2
2 1
```
#### 输出样例1：

```
2
```
#### 输入样例2：

```
2 3 2
1 2 3
2 1 5
```
#### 输出样例2：

```
14
```

## aop

### before

```cpp
#include <iostream>
using namespace std;
const int N = 55;
const int MOD = 1e9 + 7;
int dp[N][N][13][14];
int g[N][N];
int n, m, k;
```
### after

```cpp

```

## 答案

```cpp
int main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> g[i][j];
            g[i][j]++;
        }
    }
    dp[1][1][1][g[1][1]] = 1;
    dp[1][1][0][0] = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            for (int u = 0; u <= k; u++)
            {
                for (int v = 0; v <= 13; v++)
                {
                    dp[i][j][u][v] = (dp[i][j][u][v] + dp[i - 1][j][u][v]) % MOD;
                    dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j - 1][u][v]) % MOD;
                    if (u > 0 && v == g[i][j])
                    {
                        for (int c = 0; c < v; c++)
                        {
                            dp[i][j][u][v] = (dp[i][j][u][v] + dp[i - 1][j][u - 1][c]) % MOD;
                            dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j - 1][u - 1][c]) % MOD;
                        }
                    }
                }
            }
        }
    }
    int res = 0;
    for (int i = 0; i <= 13; i++)
        res = (res + dp[n][m][k][i]) % MOD;
    cout << res << endl;
    return 0;
}

```
## 选项


### A

```cpp
int main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> g[i][j];
            g[i][j]++;
        }
    }
    dp[1][1][1][g[1][1]] = 1;
    dp[1][1][0][0] = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            for (int u = 0; u <= k; u++)
            {
                for (int v = 0; v <= 13; v++)
                {
                    dp[i][j][u][v] = (dp[i][j][u][v] + dp[i - 1][j][u][v]) % MOD;
                    dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j - 1][u][v]) % MOD;
                    if (u > 0 && v == g[i][j])
                    {
                        for (int c = 0; c < v; c++)
                        {
                            dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j][u - 1][c]) % MOD;
                            dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j - 1][u - 1][c]) % MOD;
                        }
                    }
                }
            }
        }
    }
    int res = 0;
    for (int i = 0; i <= 13; i++)
        res = (res + dp[n][m][k][i]) % MOD;
    cout << res << endl;
    return 0;
}
```

### B

```cpp
int main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> g[i][j];
            g[i][j]++;
        }
    }
    dp[1][1][1][g[1][1]] = 1;
    dp[1][1][0][0] = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            for (int u = 0; u <= k; u++)
            {
                for (int v = 0; v <= 13; v++)
                {
                    dp[i][j][u][v] = (dp[i][j][u][v] + dp[i - 1][j][u][v]) % MOD;
                    dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j - 1][u][v]) % MOD;
                    if (u > 0 && v == g[i][j])
                    {
                        for (int c = 0; c < v; c++)
                        {
                            dp[i][j][u][v] = (dp[i][j][u][v] + dp[i - 1][j][u - 1][c]) % MOD;
                            dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j][u - 1][c]) % MOD;
                        }
                    }
                }
            }
        }
    }
    int res = 0;
    for (int i = 0; i <= 13; i++)
        res = (res + dp[n][m][k][i]) % MOD;
    cout << res << endl;
    return 0;
}
```

### C

```cpp
int main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> g[i][j];
            g[i][j]++;
        }
    }
    dp[1][1][1][g[1][1]] = 1;
    dp[1][1][0][0] = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            for (int u = 0; u <= k; u++)
            {
                for (int v = 0; v <= 13; v++)
                {
                    dp[i][j][u][v] = (dp[i][j][u][v] + dp[i - 1][j][u][v]) % MOD;
                    dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j - 1][u][v]) % MOD;
                    if (u > 0 && v == g[i][j])
                    {
                        for (int c = 0; c < v; c++)
                        {
                            dp[i][j][u][v] = (dp[i][j][u][v] + dp[i - 1][j][u][c]) % MOD;
                            dp[i][j][u][v] = (dp[i][j][u][v] + dp[i][j - 1][u - 1][c]) % MOD;
                        }
                    }
                }
            }
        }
    }
    int res = 0;
    for (int i = 0; i <= 13; i++)
        res = (res + dp[n][m][k][i]) % MOD;
    cout << res << endl;
    return 0;
}
```