[uva 591] Box of Bricks – hwchang0417



[uva 591] Box of Bricks – hwchang0417

輸入n和n個高度的堆疊,試問最少需要多少移動次數,可以將所有堆疊堆成一樣高?

Programming Language: C

Execution time: 0 ms

Solution: 

統計所有堆疊的高度,求得平均後,計算所有堆疊的高度減去平均高度的和,表示每個堆疊要堆到平均高度所需的次數,由於題目是把高的堆疊堆到低的堆疊,所以移動次數等於和的1/2。


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