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另一种思路，不知道对不对找到最小的flag*50 flag*50剪掉所有的 排序 然后找最小的和两个最大的， 每次两个最大的剪掉最小的那个数 gray-=min 最后如果gray够了结果就是flag不够就是flag-gray/(n/3); 可以用有两个方向的堆实现，但是我不会 Followed by: Post your reply here: |
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