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Problem solution · Python

CCC 2023 S3 - Palindromic Poster

CCC 2023 S3 - Palindromic Poster: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
103 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For CCC 2023 S3 - Palindromic Poster, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 103 lines of Python from the credited upstream file ccc23s3.py.
  • The implementation visibly relies on ordered lookup.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2023 S3 - Palindromic Poster · PythonPython
Use this to learn the idea, then write your own version.
# By Oscar Zhou, Abbey Park High School# This solution still requires commented explanations. Please feel free to add them! # https://dmoj.ca/problem/ccc23s3 import sysinput = sys.stdin.readline n, m, r, c = map(int, input().split()) def output(graph):    for item in graph:        print("".join(item)) graph = [["a"] * m for i in range(n)]if r == n and c == m:    output(graph)    sys.exit(0)if r == 0 and c == 0:    for i in range(1, m):        graph[0][i] = "b"    for i in range(1, n):        graph[i][0] = "b"    output(graph)    sys.exit(0) # Test case 2if n == 2 and m == 2:    if (r == 1 and c == 2) or (r == 2 and c == 1):        print("IMPOSSIBLE")        sys.exit(0) for i in range(r):    for j in range(m):        graph[i][j] = "b"for i in range(n):    for j in range(c):        graph[i][j] = "b"if r == 0:    for i in range(c, m):        graph[0][i] = "c"if c == 0:    for i in range(r, n):        graph[i][0] = "c"if r == n and c == 0:    for i in range(m):        graph[0][i] = "a"if c == m and r == 0:    for i in range(n):        graph[i][0] = "a"if r == n and c != m:    for i in range(m):        graph[0][i] = "a"    if m % 2 == 0:        if c % 2 == 1:            print("IMPOSSIBLE")            sys.exit(0)        else:            for j in range(c//2):                for row in range(1, n):                    graph[row][j] = "a"                    graph[row][m-j-1] = "a"    else:        if c % 2 == 0:            for j in range(c//2):                for row in range(1, n):                    graph[row][j] = "a"                    graph[row][m-j-1] = "a"        else:            for j in range(m//2, m//2-c//2-1, -1):                for row in range(1, n):                    graph[row][j] = "a"                    graph[row][m-j-1] = "a"    output(graph)    sys.exit(0) if r != n and c == m:    for i in range(n):        graph[i][0] = "a"    if n % 2 == 0:        if r % 2 == 1:            print("IMPOSSIBLE")            sys.exit(0)        else:            for j in range(r//2):                for row in range(1, m):                    graph[j][row] = "a"                    graph[n-j-1][row] = "a"    else:        if r % 2 == 0:            for j in range(r//2):                for row in range(1, m):                    graph[j][row] = "a"                    graph[n-j-1][row] = "a"        else:            for j in range(n//2, n//2-r//2-1, -1):                for row in range(1, m):                    graph[j][row] = "a"                    graph[n-j-1][row] = "a"    output(graph)    sys.exit(0) output(graph)

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