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2023-2024:teams:cute_red_meow:codeforces2

目录

Meow

A

toby:

一个构造题…… 按照要求依次构造 3:0 3:1 3:2 2:3 1:3 0:3 的情况即可。

虽然简单,但是构造起来情况有点多,略微繁琐。

E

toby:

构造题,构造一个 城堡 的移动方案。

点开查看我的移动方案

点开查看我的移动方案

const char* pos[] = {

  "a1","a2","a3","a4","a5","a6","a7","a8",
  "b8","b7","b6","b5","b4","b3","b2","b1",
  "c1","c2","c3","c4","c5","c6","c7","c8",
  "d8","d7","d6","d5","d4","d3","d2","d1",
  "e1","e2","e3","e4","e5","e6","e7","e8",
  "f8","f7","f6","f5","f4","f3","f2","f1",
  "g1","g2","g3","g4","g5","g6","g8","g7",
  "h7","h6","h5","h4","h3","h2","h1","h8"

};

F

Red:

数论水题。

G

Red:

签到,暴力。

K

yuki & red:

想了一堆假的做法

toby:

维护区间的 free 和 busy 就可以了。然后 merge 也很容易。可惜没时间了,赛后写的。

维护内容是: 最后一段 busy 前的 free 时间,和最后一段 busy 持续到哪个时间。修改只会影响 log 个区间。

merge 方法是: 把前一段的 busy - r 用来填后一段的 free,如果没填完则大区间 busy 是后区间的 busy,free 是累加。否则就把 busy 往后挪。

L

yuki:

签到喵

M

toby:

依然是构造题。构造一个 3D 的图,使得和给定有向图连通性一致。

只需要构造一个每层可以放 9 个数的,而且隔开的,而且和下一层的每个数都可以连在一起的,就可以了。

但是写起来尤其繁琐啊喂!

Dirty: 有时候可以跨一步从一个地方到另一个地方,所以每层数与数之间的隔板多加一层就可以了。

2023-2024/teams/cute_red_meow/codeforces2.txt · 最后更改: 2023/08/15 01:06 由 yuki