# 9数算式

观察如下的算式：
```
9213 x 85674 = 789314562
```
左边的乘数和被乘数正好用到了1~9的所有数字，每个1次。
而乘积恰好也是用到了1~9的所有数字，并且每个1次。

请你借助计算机的强大计算能力，找出满足如上要求的9数算式一共有多少个？

注意：

1. 总数目包含题目给出的那个示例。
2. 乘数和被乘数交换后作为同一方案来看待。

## aop

### before

```cpp
#include <bits/stdc++.h>
using namespace std;
int bei[10];
map<long long, int> mp;
```
### after

```cpp

```

## 答案

```cpp
int main()
{
    int a[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int res = 0;
    do
    {
        for (int i = 1; i < 9; i++)
        {
            memset(bei, 0, sizeof(bei)); 
            long long int ans, left = 0, right = 0, t = 0, x = 0, y = 0;
            for (int j = 0; j <= i; j++)
            {
                left = left * 10 + a[j];
            }
            x = left;
            x = x * 10;
            for (int k = i + 1; k < 9; k++)
            {
                right = right * 10 + a[k];
                x = x * 10 + a[k];
            }
            y = right;
            y = y * 10;
            for (int j = 0; j <= i; j++)
                y = y * 10 + a[j];

            ans = left * right;
            long long int ff = ans;
            while (ans > 0)
            {
                int x = ans % 10;
                ans = ans / 10;
                if (bei[x] == 0 && x != 0)
                {
                    bei[x] = 1;
                    t++;
                }
            }
            if (t == 9 && mp.count(x) == 0 && mp.count(y) == 0)
            {
                res++;
                mp[x] = 1;
                mp[y] = 1;
            }
        }
    } while (next_permutation(a, a + 9)); 
    cout << res << endl;
    return 0;
}
```
## 选项


### A

```cpp
int main()
{
    int a[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int res = 0;
    while (next_permutation(a, a + 9))
    {
        for (int i = 1; i < 9; i++)
        {
            memset(bei, 0, sizeof(bei));
            long long int ans, left = 0, right = 0, t = 0, x = 0, y = 0;
            for (int j = 0; j <= i; j++)
            {
                left = left * 10 + a[j];
            }
            x = left;
            x = x * 10;
            for (int k = i + 1; k < 9; k++)
            {
                right = right * 10 + a[k];
            }
            y = right;
            y = y * 10;
            for (int j = 0; j <= i; j++)
                y = y * 10 + a[j];

            ans = left * right;
            long long int ff = ans;
            while (ans >= 0)
            {
                int x = ans % 10;
                ans = ans / 10;
                if (bei[x] == 0 && x != 0)
                {
                    bei[x] = 1;
                    t++;
                }
            }
            if (mp.count(x) == 0 && mp.count(y) == 0)
            {
                res++;
                mp[x] = 1;
                mp[y] = 1;
            }
        }
    };
    cout << res << endl;
    return 0;
}
```

### B

```cpp
int main()
{
    int a[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int res = 0;
    do
    {
        for (int i = 1; i < 9; i++)
        {
            memset(bei, 0, sizeof(bei)); 
            long long int ans, left = 0, right = 0, t = 0, x = 0, y = 0;
            for (int j = 0; j <= i; j++)
            {
                left = left * 10 + a[j];
            }
            x = left;
            x = x * 10;
            for (int k = i + 1; k < 9; k++)
            {
                right = right * 10 + a[k];
                x = x * 10 + a[k];
            }
            y = right;
            y = y * 10;
            for (int j = 0; j <= i; j++)
                y = y * 10 + a[j];

            long long int ff = ans;
            while (ans > 0)
            {
                int x = ans % 10;
                ans = ans / 10;
                if (bei[x] == 0 && x != 0)
                {
                    bei[x] = 1;
                    t++;
                }
            }
            if (t == 9 && mp.count(x) == 0 && mp.count(y) == 0)
            {
                res++;
                mp[x] = 1;
                mp[y] = 1;
            }
        }
    } while (next_permutation(a, a + 9)); 
    cout << res << endl;
    return 0;
}
```

### C

```cpp
int main()
{
    int a[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int res = 0;
    while (next_permutation(a, a + 9))
    {
        for (int i = 1; i < 9; i++)
        {
            memset(bei, 0, sizeof(bei));
            long long int ans, left = 0, right = 0, t = 0, x = 0, y = 0;
            for (int j = 0; j <= i; j++)
            {
                left = left * 10 + a[j];
            }
            x = left;
            x = x * 10;
            for (int k = i + 1; k < 9; k++)
            {
                right = right * 10 + a[k];
            }
            y = right;
            y = y * 10;
            for (int j = 0; j <= i; j++)
                y = y * 10 + a[j];

            ans = left * right;
            long long int ff = ans;
            while (ans > 0)
            {
                int x = ans % 10;
                ans = ans / 10;
                if (bei[x] == 0 && x != 0)
                {
                    bei[x] = 1;
                    t++;
                }
            }
            if (t == 9 && mp.count(x) == 0 && mp.count(y) == 0)
            {
                res++;
                mp[x] = 1;
                mp[y] = 1;
            }
        }
    };
    cout << res << endl;
    return 0;
}
```