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Godrik0
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Dec 10, 2025
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| available = ((1 << n) - 1) & ~(cols | diag1 | diag2) | ||
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| while available: | ||
| pos = available & -available | ||
| available -= pos | ||
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| solve(row + 1, cols | pos, (diag1 | pos) << 1, (diag2 | pos) >> 1) |
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Хотелось бы видеть комментарии, поясняющие эту сложную логику.
Битовые операции достаточно тяжело воспринимать вот так, остается только догадываться или вручную вычислять что же делает каждая из них.
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