結果

問題 No.3006 ベイカーの問題
ユーザー qwewe
提出日時 2025-05-14 13:25:42
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,123 bytes
コンパイル時間 396 ms
コンパイル使用メモリ 82,300 KB
実行使用メモリ 67,280 KB
最終ジャッジ日時 2025-05-14 13:26:42
合計ジャッジ時間 2,580 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
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ファイルパターン 結果
sample RE * 3
other WA * 13 RE * 11
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 # 

import math

def solve():
    # Read the input. The problem states x is an integer, 100 <= x <= 999.
    # Reading it as a string is convenient for extracting digits.
    x_str = input()
    
    # Extract the three digits of the number x.
    # d1 is the hundreds digit, d2 is the tens digit, and d3 is the units digit.
    d1 = int(x_str[0])
    d2 = int(x_str[1])
    d3 = int(x_str[2])
    
    # Initialize a variable to store the result.
    # Based on the problem's typical structure, one of the conditions below should match.
    ans = 0 

    # Condition 1: All digits are equal (e.g., input 555)
    # Pattern derived from example: Input 555 -> Output 1215.
    # If d1=d2=d3=d, the formula appears to be d * (3^d).
    # For 555, d=5: 5 * (3^5) = 5 * 243 = 1215.
    if d1 == d2 and d2 == d3:
        d = d1 # Since all digits are equal, d can be d1, d2, or d3.
        ans = d * (3**d)  # In Python, ** is the exponentiation operator.
        
    # Condition 2: Digits are in strictly increasing order (e.g., input 123)
    # Pattern derived from example: Input 123 -> Output 2027.
    # If d1 < d2 < d3, the formula appears to be d2 * 1000 + d3^3.
    # For 123: 2 * 1000 + 3^3 = 2000 + 27 = 2027.
    elif d1 < d2 and d2 < d3:
        ans = d2 * 1000 + (d3**3)
        
    # Condition 3: Digits are in strictly decreasing order (e.g., input 421)
    # Pattern derived from example: Input 421 -> Output 3.
    # If d1 > d2 > d3, the formula appears to be d1 - d2 + d3.
    # For 421: 4 - 2 + 1 = 3.
    elif d1 > d2 and d2 > d3:
        ans = d1 - d2 + d3
        
    # Note: The problem implies that the input x will always fit one of these three categories.
    # If there were other types of inputs (e.g., 132 where digits are mixed, or 112
    # where digits are non-strictly increasing), additional rules or examples would typically be provided.
    # Thus, an 'else' case for unhandled patterns is not included here.
    
    # Print the final calculated answer.
    print(ans)

# This standard boilerplate calls the solve() function when the script is executed.
if __name__ == '__main__':
    solve()
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