20241215 ◎

McDonald’s seasonal menu McRib was not very tasty.

• participated in Codeforces Round 993 (Div. 4)
  • The math problems are too difficult for me
• household chores

2040C. Ordered Permutations - 1200

Codeforces Round #992 (Div.2) Editorial

:thinking_face:

I can’t understand or replicate the solution, but the way they construct a permutation might be worth remembering.

Instead of generating all permutations n!, the solution picks two positions (the leftmost/rightmost blank spaces) for each number so that only 2^n permutations need to be checked.

2044A. Easy Problem - 800

The problem can be rephrased as counting how many different ways we can insert a partition between n identical blocks.

For example, consider the number 5 expressed as:

5 = 1 + 1 + 1 + 1 + 1

Some of possible ways to place a single partition are:

• 1 + 1 | 1 + 1 + 1 (a = 2, b = 3)
• 1 + 1 + 1 + 1 | 1 (a = 4, b = 1)

for _ in range(int(input())):
    print(int(input()) - 1)

2044B. Normal Problem - 800

for _ in range(int(input())):
    s: list[str] = list(input())
    srev: list[str] = s[::-1]
    for i in range(len(s)):
        if srev[i] == "p":
            srev[i] = "q"
        elif srev[i] == "q":
            srev[i] = "p"
    print("".join(srev))

_ lb

TODO:

• Register for Codeforces Global Round 28
• Review Codeforces Round 993 (Div. 4)

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