Difference between revisions of "2018 Multi-University, Nowcoder Day 3"
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Solved by ultmaster. 00:49 (+) | 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 == | == 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 (+)