# 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:
* 总数目包含题目给出的那个示例。
* 乘数和被乘数交换后作为同一方案来看待。
以下错误的是?
## aop
### before
```c
#include
using namespace std;
```
### after
```c
```
## 答案
```c
int bei[10];
map mp;
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;
}
```
## 选项
### A
```c
const int N = 1e3 + 5;
int a[15] = {9, 2, 1, 3, 8, 5, 6, 7, 4}, res = 0, vis[12], temp1;
bool fun(int x)
{
int t1 = 0, t2 = 0;
for (int i = x, j = 1; i >= 0; i--, j *= 10)
{
t1 = a[i] * j + t1;
}
for (int i = 8, j = 1; i > x; i--, j *= 10)
{
t2 = a[i] * j + t2;
}
temp1 = t1 * t2;
int temp = temp1;
memset(vis, 0, sizeof(vis));
while (temp)
{
vis[temp % 10] = 1;
temp /= 10;
}
for (int i = 1; i <= 9; i++)
{
if (!vis[i])
return false;
}
return true;
}
int main()
{
for (int i = 0; i < 9; i++)
a[i] = i + 1;
set st;
do
{
for (int i = 0; i < 8; i++)
{
if (fun(i))
{
st.insert(temp1);
res++;
}
}
} while (next_permutation(a, a + 9));
cout << res / 2 << '\n';
return 0;
}
```
### B
```c
int st[10], res[10], book[10], cnt;
int cal(int l, int r)
{
int t = 0;
for (int i = l; i <= r; i++)
{
t = t * 10 + res[i];
}
return t;
}
bool check(int k)
{
for (int i = 0; i < 10; i++)
book[i] = 0;
while (k)
{
book[k % 10]++;
k /= 10;
}
bool flag = true;
for (int i = 1; i <= 9; i++)
{
if (book[i] != 1)
flag = false;
}
return flag;
}
void dfs(int k)
{
if (k == 9)
{
for (int i = 0; i < 8; i++)
{
int a = cal(0, i);
int b = cal(i + 1, 8);
if (check(a * b))
{
cnt++;
}
}
return;
}
for (int i = 1; i <= 9; i++)
{
if (!st[i])
{
st[i] = 1;
res[k] = i;
dfs(k + 1);
st[i] = 0;
}
}
}
int main()
{
dfs(0);
cout << cnt / 2;
return 0;
}
```
### C
```c
typedef long long LL;
const int N = 15;
bool vis[N];
int ans;
bool check_2(LL sumb)
{
bool st[N];
for (int i = 1; i <= 9; i++)
{
st[i] = 0;
}
while (sumb)
{
st[sumb % 10] = true;
sumb = sumb / 10;
}
bool flag = false;
for (int i = 1; i <= 9; i++)
{
if (!st[i])
{
return false;
}
}
return true;
}
bool check_1(int suma, int sumc)
{
LL sumb = (LL)suma * sumc;
if (check_2(sumb))
{
return true;
}
return false;
}
void dfs_c(int u, int suma, int sumc)
{
if (u > 9)
{
return;
}
if (check_1(suma, sumc))
{
ans++;
return;
}
for (int i = 1; i <= 9; i++)
{
if (!vis[i])
{
vis[i] = true;
dfs_c(u + 1, suma, sumc * 10 + i);
vis[i] = false;
}
}
}
void dfs_a(int u, int suma)
{
dfs_c(u, suma, 0);
for (int i = 1; i <= 9; i++)
{
if (!vis[i])
{
vis[i] = true;
dfs_a(u + 1, suma * 10 + i);
vis[i] = false;
}
}
}
int main()
{
dfs_a(0, 0);
cout << ans / 2 << endl;
return 0;
}
```