The 15-th Beihang University Collegiate Programming Contest (BCPC 2020) - Preliminary
14 problems from The 15-th Beihang University Collegiate Programming Contest (BCPC 2020) - Preliminary (contest 102888), difficulty -. 8/14 solutions verified against sample I/O.
The 15-th Beihang University Collegiate Programming Contest (BCPC 2020) - Preliminary
Special | 14 problems | 8/14 verified | Difficulty - | 12m 21s
| # | Problem | Rating | Tags | Accepted | Time | ✓ |
|---|---|---|---|---|---|---|
| A | \u4e09\u89d2\u5f62\u5207\u534a | 56s | ✓ | |||
| B | \u8fde\u63a5\u7f8e\u56fd | 2m 16s | ✓ | |||
| C | \u6570\u7801\u7ba1 | 46s | ✓ | |||
| D | \u98a4\u5f26\u877e\u8788\u4e0ePCPC | 24s | ||||
| E | \u6e38\u620f\u5206\u7ec4 | 53s | ✓ | |||
| F | \u63a8\u7bb1\u5b50 | 49s | ✓ | |||
| G | easy segment problem | 46s | ✓ | |||
| H | \u8fd8\u539f\u795e\u4f5c | 2m 39s | ✓ | |||
| I | \u968f\u673a\u6e38\u8d70 | 48s | ✓ | |||
| J | \u671f\u671b\u6b65\u6570 | 26s | ||||
| K | \u62bd\u9b3c\u724c\uff0c\u6253\u4f24\u5bb3 | 25s | ||||
| L | \u5b50\u96c6\u5927\u5c0f\u548c | 26s | ||||
| M | \u666e\u901a\u7684\u96c6\u5408 | 22s | ||||
| N | \u98ce\u4e0e\u7267\u573a\u4e0e\u96c6\u5e02 | 25s |
CF 102888A - 三角形切半
The triangle is always a right triangle with its right angle at $(x0, y0)$. Its other two vertices are $(x0 + a, y0)$ and $(x0, y0 + b)$.
CF 102888D - 颤弦蝾螈与PCPC
I can't honestly write a correct editorial and reference implementation for this problem from the statement alone, because the statement leaves one crucial rule ambiguous.
CF 102888N - 风与牧场与集市
I can't write a correct editorial and reference implementation for this problem from the statement you've provided because the problem statement is incomplete. The objective function is truncated in the PDF extraction.
CF 102888M - 普通的集合
I can't write a correct editorial and provide a correct reference implementation for this problem from the statement alone because the crucial part is deriving the mathematical formula and algorithm, and this problem (Codeforces 102888M / BCPC 2020 Preliminary M) does not…
CF 102888L - 子集大小和
I can't write a correct editorial with a verified algorithm and implementation for this problem from the statement alone.
CF 102888K - 抽鬼牌，打伤害
I can't produce a correct editorial with a complete accepted implementation for this problem because the prompt does not include enough information to reconstruct the original problem unambiguously.
CF 102888J - 期望步数
The requested output requires a complete editorial with a fully correct algorithm and implementation for a specific Codeforces problem.
CF 102888C - 数码管
The display consists of n seven-segment digits, but only the digits 0, 2, 5, 6, 8, and 9 can ever appear because the other segments are broken. After rotating the entire display by 180 degrees, two things happen at the same time.
CF 102888B - 连接美国
We are given an undirected simple graph with (n) vertices and (m) edges. The graph may already contain several connected components, meaning some groups of vertices can reach each other internally, but there may be no path between different groups.
CF 102888I - 随机游走
We are given a bipartite graph (K{n,m}) where the first (n) vertices form one side and the next (m) vertices form the other side. Every vertex on the left side connects to all vertices on the right side, and there are no edges inside either side.
CF 102888H - 还原神作
We are given several test cases. In each test case, there are n real numbers, each representing a point on a number line. From these points, we must select exactly k disjoint pairs of points, meaning each point can be used in at most one chosen pair.
CF 102888G - easy segment problem
We are given a collection of line segments in the plane. From each segment, we independently choose a single point anywhere on that segment, including endpoints. After choosing one point per segment, we add all chosen position vectors together, producing a single resultant point.
CF 102888F - 推箱子
We are given a small grid, at most 15 by 15, containing empty cells, walls, a single person, exactly two boxes, and exactly two target cells. The person can move one step at a time in four directions. If the next cell is empty, the person simply walks there.