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Problem solution · Python

CCC 2025 S3 - Pretty Pens

CCC 2025 S3 - Pretty Pens: 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
98 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 2025 S3 - Pretty Pens, 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

  • 98 lines of Python from the credited upstream file ccc25s3.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 2025 S3 - Pretty Pens · PythonPython
Use this to learn the idea, then write your own version.
# OsunPerson# py3 solution for 25' s3 - Pretty Pens # CCC 2025 S3 - Pretty Pensmostpretty = 10**11leastpretty = 10**11total = 0 n, m, q =map(int, input().split()) cpu = [[0,0,0]]*nstorage = [[0, 0, 0]]*(q+1) prettypercolor = [0]*(m+1) #Map the most pretty pen for each color u = 1for i in range(n):    c, p = map(int, input().split())     if p > prettypercolor[c]:        prettypercolor[c] = p        cpu[i] = [c, p, 1]        if p < mostpretty:            mostpretty = p        for x in range(i):            if cpu[x][0] == c and cpu[x][2] == 1:                cpu[x][2] = 0                break    else:        cpu[i] = [c, p, 0] def main():    global mostpretty    leastpretty = 10**11    total = 0    #Map the leastpretty pretty pen to u = 2    for i in range(n):        if cpu[i][2] == 1 and cpu[i][1] < leastpretty:            leastpretty = cpu[i][1]            cpu[i][2]=2            for x in range(i):                if cpu[x][2] == 2:                    cpu[x][2] = 1                    break     #Map the Mostpretty pen to replace the least pretty pretty pen    for i in range(n):        if cpu[i][2] == 0 and cpu[i][1] > mostpretty:            mostpretty = cpu[i][1]            cpu[i][2]=3            for x in range(n):                if cpu[x][2] == 2:                    cpu[x][2] = 0            for x in range(i):                if cpu[x][2] == 3:                    cpu[x][2] = 0                    break     #Total the total pretty pens    for i in range(n):        if cpu[i][2] == 1 or cpu[i][2] == 2 or cpu[i][2] == 3:            total += cpu[i][1]    print(total) for i in range(q):    a, b, num = map(int, input().split())    storage[i] = [a, b, num] main()for i in range(q):    mostpretty = 10     if storage[i][0] == 1:        cpu[storage[i][1]-1][0] = storage[i][2]    else:        cpu[storage[i][1]-1][1] = storage[i][2]     for x in range(n):        cpu[x][2] = 0     for x in range(m):        prettypercolor[x+1] = 0     #Reset the most pretty pen for each color    for x in range(n):        if cpu[x][1] >= prettypercolor[cpu[x][0]]:            prettypercolor[cpu[x][0]] = cpu[x][1]            cpu[x][2] = 1            mostpretty = cpu[x][1]            for y in range(x):                if cpu[y][0] == cpu[x][0] and cpu[y][2] == 1:                    cpu[y][2] = 0                    break        else:            cpu[x][2] = 0     main()

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