# 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; } ```