結果

問題 No.1624 三角形の反射
コンテスト
ユーザー qwewe
提出日時 2025-05-14 13:20:44
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 37 ms / 2,000 ms
コード長 2,075 bytes
コンパイル時間 393 ms
コンパイル使用メモリ 82,396 KB
実行使用メモリ 54,084 KB
最終ジャッジ日時 2025-05-14 13:22:33
合計ジャッジ時間 2,321 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 22
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ソースコード

diff # 

import math

a_str = input() # Input format e.g., "X.YYY"

# Parse the input string a_str to get an exact fraction for a.
# a = (integer part of a * 1000 + fractional part of a) / 1000
parts = a_str.split('.')
# parts[0] is the integer part string, parts[1] is the fractional part string (3 digits)
# Example: "0.124" -> parts[0]="0", parts[1]="124". numerator_val = 0*1000 + 124 = 124.
# Example: "12.345" -> parts[0]="12", parts[1]="345". numerator_val = 12*1000 + 345 = 12345.
# A simpler way to form the integer is to concatenate parts[0] and parts[1].
# "0.124" -> "0"+"124" = "0124" -> 124.
# "12.345" -> "12"+"345" = "12345" -> 12345.
numerator_val = int(parts[0] + parts[1])
denominator_val = 1000 # Since a has 3 decimal places

# Simplify the fraction a = numerator_val / denominator_val
common_divisor = math.gcd(numerator_val, denominator_val)

# (X, Y) is the coordinate of the target vertex in the unfolded plane.
# X corresponds to the denominator of the simplified fraction (Q_unfolded).
# Y corresponds to the numerator of the simplified fraction (P_unfolded).
Y = numerator_val // common_divisor  # P_unfolded
X = denominator_val // common_divisor # Q_unfolded

# Determine which vertex is hit (A, B, or C)
vertex_char = ''
if (X + Y) % 2 == 0:  # If X and Y are both even, or both odd
    vertex_char = 'A'
elif X % 2 != 0 and Y % 2 == 0:  # If X is odd and Y is even
    vertex_char = 'B'
elif X % 2 == 0 and Y % 2 != 0:  # If X is even and Y is odd
    vertex_char = 'C'

# Calculate the number of reflections using the derived formula:
# N_reflections = (X-1) + (Y-1) + (X+Y)//2 + abs(X-Y)//2
# (X-1): reflections from vertical lines (images of side AC)
# (Y-1): reflections from horizontal lines (images of side AB)
# (X+Y)//2: reflections from diagonals x+y=k (k odd)
# abs(X-Y)//2: reflections from diagonals x-y=k (k odd)
# Since X and Y are always >= 1 (a >= 0.001), X-1 and Y-1 are non-negative.
num_reflections = (X - 1) + (Y - 1) + \
                  (X + Y) // 2 + \
                  abs(X - Y) // 2

print(f"{vertex_char} {num_reflections}")
0