Canadian Computing Competition: 2021 Stage 1, Junior #5, Senior #2
A new and upcoming artist has a unique way to create checkered patterns. The idea is to
use an -by-
canvas which is initially entirely black. Then the artist repeatedly chooses
a row or column and runs their magic brush along the row or column. The brush changes
the colour of each cell in the row or column from black to gold or gold to black.
Given the artist's choices, your job is to determine how much gold appears in the pattern
determined by these choices.
Input Specification
The first line of input will be a positive integer . The second line of input will be a positive
integer
. The third line of input will be a positive integer
. The remaining input will be
lines giving the choices made by the artist. Each of these lines will either be
R
followed
by a single space and then an integer which is a row number, or C
followed by a single space
and then an integer which is a column number. Rows are numbered top down from to
.
Columns are numbered left to right from
to
.
The following table shows how the available marks are distributed.
Subtask | Description | |||
---|---|---|---|---|
only one cell, and up to | ||||
only one row, and up to | ||||
up to up to | ||||
up to up to |
Output Specification
Output one non-negative integer which is equal to the number of cells that are gold in the pattern determined by the artist's choices.
Sample Input 1
3
3
2
R 1
C 1
Output for Sample Input 1
4
Explanation of Output for Sample Input 1
After running the brush along the first row, the canvas looks like this:
GGG
BBB
BBB
Then after running the brush along the first column, four cells are gold in the final pattern determined by the artist's choices:
BGG
GBB
GBB
Sample Input 2
4
5
7
R 3
C 1
C 2
R 2
R 2
C 1
R 4
Output for Sample Input 2
10
Explanation of Output for Sample Input 2
Ten cells are gold in the final pattern determined by the artist's choices:
BGBBB
BGBBB
GBGGG
GBGGG
Comments
Small hint that i think people may need, for the last cases you would need to think outside the box to make it in the time limit, instead of changing the graph think about the amount of changes you need to make this.
R 1
R 1
=
R 0
think about that as a hint so for time don't think 2d array think 2 arrays.
really useful and thank u. I've been thinking about this problem for a long time. lol
Is this possible in python ?
Yes https://dmoj.ca/problem/ccc21s2/rank/?language=PY2&language=PY3&language=PYPY&language=PYPY3