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KubEF
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Oct 12, 2025
| 1) queen1(перебор) | ||
| - Изначально поставить ферзя n вариантов, далее (n-1) вариантов и т.д. O(n!) | ||
| - Для всех перестановок проверяется диагональ: два вложенных цикла это примерно O(n**2) | ||
| - Общая сложность O(n!*n**2) |
Collaborator
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Обычно в markdown для формул используют latex: $O(n! \cdot n^2)$
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| return 0 | ||
| return 1 |
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| """ | ||
| def queen(n): | ||
| cols = [None]*n | ||
| k = 0 | ||
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| def diagonal(row, col): | ||
| for r in range(row): | ||
| if cols[r] == col or abs(cols[r] - col) == abs( | ||
| r - row | ||
| ): # вертикаль или диагональ совпадают | ||
| return 0 | ||
| return 1 | ||
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| def solve(row=0): | ||
| nonlocal k | ||
| if row == n: | ||
| k += 1 | ||
| return | ||
| for col in range(n): | ||
| if diagonal(row, col): | ||
| cols[row] = col # ставим ферзя | ||
| solve(row + 1) # идём в следующую строку | ||
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| solve() | ||
| return k |
Collaborator
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Предполагается, что вы придумаете сильно более быстрый алгоритм. Я дал подсказки некоторым из ваших одногруппников, проявите софт скиллы)
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add a task about queens