APCS Jan 2022 - h083 Q3 [Python]

Problem statement:

A fortune-stick container holds M sticks, each labeled with a string of lowercase English letters. Draw two sticks with strings S and T. If their concatenation can be split into two identical halves, the result is an auspicious answer indicating divine approval; otherwise, it indicates disapproval or no answer. For example, piep and ie form piepie, which splits into pie and pie, while foo and bar do not satisfy the condition. Interestingly, if S followed by T works, T followed by S also works: both piepie and iepiep split into identical halves. Count how many pairs among the M sticks produce an auspicious answer, counting each combination only once.

m = int(input())
c = set()

for _ in range(m):
    c.add(input())


total = 0
for i in c:
    for idx in range((len(i) - 1 )//2):
        idx += 1
        if i[:idx] == i[-idx:] and i[idx:-idx] in c:
            total += 1

print(total)

Solution report

Loop over every string i in the set and use index idx to check whether the two ends match.
If they match and the middle substring remaining after removing both ends is also in set c, this is a valid pair.
Increment total for each pair found. Print total to output the number of qualifying pairs.

Important points

String slicing: For abyyyab, check matching prefixes and suffixes by looping through half the string: if i[:idx] == i[-idx:].
Searching with set(): Store all strings in Python’s built-in set(). Check if i[idx:-idx] in c to ensure the middle substring is also a string in the container.

Key concepts:

Enumeration
HASH
SET

[✔️] You may use the code directly if needed.
[✔️] Please credit the source when sharing.

This article was translated from Traditional Chinese by AI. Meaning or wording may differ; please refer to the Traditional Chinese original as the authoritative version.

APCS Jan 2022 - h083 Q3 [Python]

Problem statement:

A fortune-stick container holds M sticks, each labeled with a string of lowercase English letters. Draw two sticks with strings S and T. If their concatenation can be split into two identical halves, the result is an auspicious answer indicating divine approval; otherwise, it indicates disapproval or no answer. For example, piep and ie form piepie, which splits into pie and pie, while foo and bar do not satisfy the condition. Interestingly, if S followed by T works, T followed by S also works: both piepie and iepiep split into identical halves. Count how many pairs among the M sticks produce an auspicious answer, counting each combination only once.

m = int(input()) c = set() for _ in range(m): c.add(input()) total = 0 for i in c: for idx in range((len(i) - 1 )//2): idx += 1 if i[:idx] == i[-idx:] and i[idx:-idx] in c: total += 1 print(total)

Solution report

Loop over every string i in the set and use index idx to check whether the two ends match.

If they match and the middle substring remaining after removing both ends is also in set c, this is a valid pair.

Increment total for each pair found. Print total to output the number of qualifying pairs.

Important points

String slicing: For abyyyab, check matching prefixes and suffixes by looping through half the string: if i[:idx] == i[-idx:].

Searching with set(): Store all strings in Python’s built-in set(). Check if i[idx:-idx] in c to ensure the middle substring is also a string in the container.

Key concepts:

Enumeration

HASH

SET

[✔️] You may use the code directly if needed.

[✔️] Please credit the source when sharing.

This article was translated from Traditional Chinese by AI. Meaning or wording may differ; please refer to the Traditional Chinese original as the authoritative version.