結果

問題 No.211 素数サイコロと合成数サイコロ (1)
コンテスト
ユーザー qwewe
提出日時 2025-05-14 12:59:48
言語 PyPy3
(7.3.17)
コンパイル:
pypy3 -mpy_compile _filename_
実行:
pypy3 _filename_
結果
MLE  
(最新)
AC  
(最初)
実行時間 -
コード長 2,610 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 230 ms
コンパイル使用メモリ 95,336 KB
実行使用メモリ 78,464 KB
最終ジャッジ日時 2026-07-10 15:33:59
合計ジャッジ時間 3,847 ms
ジャッジサーバーID
（参考情報） 		judge2_0 / judge1_0
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
other MLE * 33
権限があれば一括ダウンロードができます

ソースコード

diff #
raw source code 

# -*- coding: utf-8 -*-
import sys

# Function to read a single integer from standard input
def get_int():
    """Reads a line from stdin, removes leading/trailing whitespace, and converts it to an integer."""
    return int(sys.stdin.readline().strip())

# Main function containing the problem-solving logic
def solve():
    """Solves the Prime and Composite Dice problem."""
    
    # Read the target product K from input
    k = get_int()

    # Define the faces of the prime die (素数サイコロ)
    # These are the first 6 prime numbers.
    prime_faces = [2, 3, 5, 7, 11, 13]
    
    # Define the faces of the composite die (合成数サイコロ)
    # These are the first 6 composite numbers greater than 1.
    composite_faces = [4, 6, 8, 9, 10, 12]

    # Initialize a counter for outcomes where the product of the faces equals K
    favorable_outcomes_count = 0
    
    # Calculate the total number of possible outcomes when rolling both dice.
    # Each die has 6 faces, and the rolls are independent.
    # The total number of possible pairs (prime_face, composite_face) is 6 * 6 = 36.
    total_outcomes = len(prime_faces) * len(composite_faces) # This will be 36

    # Iterate through each possible face value of the prime die
    for p_face in prime_faces:
        # For each prime face value, iterate through each possible face value of the composite die
        for c_face in composite_faces:
            # Check if the product of the current pair of face values equals the target K
            if p_face * c_face == k:
                # If the product matches K, increment the count of favorable outcomes
                favorable_outcomes_count += 1

    # Calculate the probability.
    # Probability = (Number of favorable outcomes) / (Total number of outcomes)
    # Use float() conversion for total_outcomes to ensure floating-point division, 
    # which is necessary for calculating the probability accurately.
    # Since total_outcomes is guaranteed to be 36 (non-zero), we don't need to check for division by zero.
    probability = favorable_outcomes_count / float(total_outcomes)

    # Print the calculated probability.
    # The problem requires the output to be accurate up to an absolute error of 10^-12.
    # Formatting the output to 17 decimal places using an f-string is a standard way 
    # to ensure sufficient precision for competitive programming tasks.
    print(f"{probability:.17f}")

# Standard Python entry point check. 
# This ensures that the solve() function is called only when the script is executed directly.
if __name__ == '__main__':
    solve()
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