Difference between revisions of "2018 Multi-University, Nowcoder Day 3"

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Solved by ultmaster. 00:49 (+)
 
Solved by ultmaster. 00:49 (+)
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== Problem F ==
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Upsolved by zerol.
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题意:询问一列数的某一段的所有子序列形成的数字的 SOD 分布,同时支持修改一个位置上的数。
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题解:SOD(sum of digits) 这种运算不改变 模 base-1 的值。线段树维护区间的答案和区间中 0 的个数,合并代价 16^2。( SOD(0) = 0, SOD(v) = (v - 1) % (base - 1) + 1)
  
 
== Problem H ==
 
== Problem H ==

Revision as of 11:26, 26 July 2018

Problem A

Solved by kblack. 01:16 (+1)

题意:容量是四元组,求限制内最大价值和。

题解:将物品分为两半,每半用 $2^{18}$ 枚举,然后耗费 $36^4$ 更新一遍,处理没有用完限制的部分。

Problem B

Unsolved. (-2)

Problem C

Solved by zerol. 00:45 (+)

题意:每次把数列中一个区间的数剪切到数列的开头,最后打印这个数列。

题解:模板题。treap / splay / rope(pb_ds)

Problem D

Unsolved. (-6)

Problem E

Solved by ultmaster. 00:49 (+)

Problem F

Upsolved by zerol.

题意:询问一列数的某一段的所有子序列形成的数字的 SOD 分布,同时支持修改一个位置上的数。

题解:SOD(sum of digits) 这种运算不改变 模 base-1 的值。线段树维护区间的答案和区间中 0 的个数,合并代价 16^2。( SOD(0) = 0, SOD(v) = (v - 1) % (base - 1) + 1)

Problem H

Solved by kblack. 00:19 (+)

温暖的枚举签到。

Problem I

Solved by ultmaster. 04:24 (+)

Problem J

Solved by ultmaster. 02:51 (+)