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- Решение с тернарен оператор:
- #include <iostream>
- using namespace std;
- int main() {
- int n;
- cin >> n;
- for (int i = 0; i < n; i++) {
- for (int j = 0; j < n * 2; j++) {
- cout << ((i != 0 && i != n - 1 && j != 0 && j != n * 2 - 1) ? '/' : '*');
- }
- for (int k = 0; k < n; k++) {
- cout << (((i == n / 2 && n % 2 == 1) || (i + 1 == n / 2 && n % 2 == 0)) ? '|' : ' ');
- }
- for (int j = 0; j < n * 2; j++) {
- cout << ((i != 0 && i != n - 1 && j != 0 && j != n * 2 - 1) ? '/' : '*');
- }
- cout << endl;
- }
- return 0;
- }
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