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这样做。枚举法 精度,将flout转换成int,全部数×100 枚举范围,max(ck) >= t >= 1 从大到小枚举,当出现第一个 c1/t+c2/t+...cn/t>=request的时候输出t(还原为浮点数)就是答案 可以用二分法加速。 Followed by: Post your reply here: |
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