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back to boardShow all messages Hide all messagesConsidering that d[i][j] indicate the number of i bricks divided into j columns steps.So if we cut away the downmost brick of every steps, and we can get d[i][j] = d[i-j][j-1](if the 1st step only have one brick) + d[i-j][j](if the 1st step have more than one brick). Thus q = d[n][2->max_j] really smart! how can you get that dude? Hello, Thank you for your hint, it was very useful for me
and yes, your solution is awesome! really nice idea, i like it. But how does this transition ensures that the next column always has more bricks than the current column? |
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