結果
| 問題 |
No.3083 One Two
|
| コンテスト | |
| ユーザー |
 qwewe
|
| 提出日時 | 2025-05-14 13:18:19 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
RE
|
| 実行時間 | - |
| コード長 | 6,392 bytes |
| コンパイル時間 | 248 ms |
| コンパイル使用メモリ | 82,304 KB |
| 実行使用メモリ | 66,944 KB |
| 最終ジャッジ日時 | 2025-05-14 13:19:12 |
| 合計ジャッジ時間 | 1,538 ms |
|
ジャッジサーバーID （参考情報） |
judge1 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | RE * 3 |
| other | RE * 7 |
ソースコード
import datetime
import sys
# Function to check if a year is a leap year according to the Gregorian calendar rules.
# A year is a leap year if it is divisible by 4, unless it is divisible by 100 but not by 400.
def is_leap(year):
"""Checks if a year is a leap year."""
return (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
def solve():
# Read integer inputs Y, N, D from standard input.
# Y: The year the students entered junior high school.
# N: The number of students.
# D: The number of days after April 1st, Y, 8 PM.
Y, N, D = map(int, sys.stdin.readline().split())
# Define the reference start date: April 1st of year Y.
start_date = datetime.date(Y, 4, 1)
# Calculate the target date d_D by adding D days to the start date.
# The datetime library automatically handles month ends, year ends, and leap years.
target_date = start_date + datetime.timedelta(days=D)
# Extract the year, month, and day of the target date.
Y_D = target_date.year
M_D = target_date.month
Day_D = target_date.day
# Determine if year Y is a leap year. L_Y will be 1 if leap, 0 otherwise.
L_Y = 1 if is_leap(Y) else 0
# Calculate the size of the set S1.
# S1 includes dates from January 1st to April 1st inclusive, in year Y.
# The number of days is Jan(31) + Feb(28 or 29) + Mar(31) + Apr(1) = 31 + (28+L_Y) + 31 + 1 = 91 + L_Y.
S1_size = 91 + L_Y
# Calculate the size of the set S2.
# S2 includes dates from April 2nd to December 31st inclusive, in year Y.
# Total days in year Y is K_Y = 365 + L_Y.
# S2_size = K_Y - S1_size = (365 + L_Y) - (91 + L_Y) = 274.
# This size is constant regardless of whether Y is a leap year.
S2_size = 274
min_val = 0 # Initialize minimum possible count of students still 12
max_val = 0 # Initialize maximum possible count of students still 12
# The logic depends on whether the target date d_D falls in the same year Y or the next year Y+1.
# This transition occurs after D=274 days (Dec 31 is 274 days after Apr 1).
if Y_D == Y:
# Case 1: Target date d_D is within year Y. This happens when D <= 274.
# Students still 12 are those from S1 plus those from S2 whose birthday is after d_D.
# k = size of the subset of S2 with dates strictly after d_D.
# The number of days from d_D (exclusive) to Dec 31.
# This equals the total days in S2 (274) minus the number of days from Apr 2 to d_D (inclusive), which is D.
# So, k = 274 - D.
k = 274 - D
# k1 = size of S1.
k1 = S1_size
# k2b = size of the subset of S2 with dates on or before d_D.
# These are dates from Apr 2 up to d_D. The number of such dates is exactly D.
k2b = D
# Calculate minimum value:
# The number of students still 12 is N_1 + N_{2, after d_D}. This can be rewritten as N - N_{2, before d_D}.
# To minimize this value, we maximize N_{2, before d_D}.
# The maximum number of students whose birthdays can fall into S_{2, before d_D} is min(N, k2b).
# So, min_val = N - min(N, k2b) = max(N - k2b, 0).
min_val = max(0, N - k2b)
# Calculate maximum value:
# The number of students still 12 is N_1 + N_{2, after d_D}.
# To maximize this value, we prioritize selecting students whose birthdays are in S1 or S_{2, after d_D}.
# The maximum number of such students is min(N, k1 + k).
max_val = min(N, k1 + k)
else: # Y_D == Y + 1
# Case 2: Target date d_D is in year Y+1. This happens when D > 274.
# Students still 12 are only those from S1 whose birthday (M, D) is calendrically after (M_D, Day_D).
# Calculate k1a: Number of dates in S1 that are calendrically after (M_D, Day_D).
# We iterate through all dates in S1 and count how many satisfy the condition.
k1a = 0 # Initialize count
# Start iteration from Jan 1, Y.
date_iter = datetime.date(Y, 1, 1)
# End iteration at Apr 1, Y (inclusive).
end_date_S1 = datetime.date(Y, 4, 1)
while True:
# Get month and day of the current date in the iteration.
current_M = date_iter.month
current_Day = date_iter.day
# Check if the current date (M, D) is calendrically after the target date's month and day (M_D, Day_D).
# A date (M1, D1) is after (M2, D2) if M1 > M2 or (M1 == M2 and D1 > D2).
is_after = False
if current_M > M_D:
is_after = True
elif current_M == M_D:
if current_Day > Day_D:
is_after = True
# If the current date is after (M_D, Day_D), increment the counter k1a.
if is_after:
k1a += 1
# Break the loop if we have processed the last date in S1 (April 1st).
if date_iter == end_date_S1:
break
# Move to the next day. datetime handles increments correctly across month/year boundaries and leap years.
date_iter += datetime.timedelta(days=1)
# k1 = Total size of S1.
k1 = S1_size
# k1b = Size of S1 subset on or before (M_D, Day_D). Calculated as total size minus the 'after' part.
k1b = k1 - k1a
# k2 = Total size of S2.
k2 = S2_size
# Calculate minimum value:
# The number of students still 12 is N_{1, after d_D} = N_{1a}.
# To minimize N_{1a}, we prioritize selecting students whose birthdays are from S_{1, before d_D} and S2.
# The total capacity outside S_{1, after d_D} is k1b + k2.
# The minimum number of students we must select from S_{1, after d_D} is max(0, N - (k1b + k2)).
min_val = max(0, N - (k1b + k2))
# Calculate maximum value:
# To maximize N_{1a}, we prioritize selecting students whose birthdays are from S_{1, after d_D}.
# The maximum number of students we can select from S_{1, after d_D} is min(N, k1a).
max_val = min(N, k1a)
# Print the calculated minimum and maximum values separated by a space.
print(f"{min_val} {max_val}")
# Execute the solve function when the script is run.
solve()