結果

問題 No.500 階乗電卓
コンテスト
ユーザー qwewe
提出日時 2025-05-14 13:03:17
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,355 bytes
コンパイル時間 232 ms
コンパイル使用メモリ 83,008 KB
実行使用メモリ 54,136 KB
最終ジャッジ日時 2025-05-14 13:04:52
合計ジャッジ時間 1,926 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 AC * 3
other AC * 7 WA * 13
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ソースコード

diff # 

import sys

# Set the modulus for the last 12 digits
MOD = 1000000000000

def solve():
    """
    Reads the input N, calculates N! mod 10^12, and prints the result.
    """
    # Read the input integer N
    n_str = sys.stdin.readline()
    # Strip potential whitespace and convert to integer
    try:
        n = int(n_str.strip())
    except ValueError:
        # Handle potential errors if input is not a valid integer
        # Although constraints say N is a positive integer, adding a check is robust.
        print("Invalid input")
        return

    # Constraint check (though the problem guarantees 1 <= N)
    if n <= 0:
        # Factorial is not defined for non-positive integers in this context
        # Or handle as per specific requirements if needed.
        # Based on the problem, N is always a positive integer.
        # This branch might not be strictly necessary but adds safety.
        print("Input must be a positive integer.")
        return

    # Optimization:
    # If N is large enough, N! will have many trailing zeros.
    # The number of trailing zeros in N! is determined by the number of factors of 5.
    # Number of factors of 5 = floor(N/5) + floor(N/25) + floor(N/125) + ...
    # We need the result modulo 10^12 = (2*5)^12 = 2^12 * 5^12.
    # If N! has at least 12 factors of 5 and 12 factors of 2 (which it will if it has 12 factors of 5),
    # then N! is divisible by 10^12.
    # Let's find the smallest N for which the number of factors of 5 is >= 12.
    # N=5: 1
    # N=10: 2
    # N=15: 3
    # N=20: 4
    # N=25: 5 + 1 = 6
    # N=30: 7
    # N=35: 8
    # N=40: 9
    # N=45: 10
    # N=50: 10 + 2 = 12
    # So, for N >= 50, N! mod 10^12 is 0.
    if n >= 50:
        print(0)
    else:
        # For N < 50, calculate N! mod 10^12 iteratively.
        result = 1
        for i in range(1, n + 1):
            result = (result * i) % MOD
            # Optimization: if result becomes 0, it will stay 0
            # This happens if i >= 25 and the number of factors of 5 reaches 12
            # and enough factors of 2 are accumulated.
            # However, with N < 50, this check might not save much time
            # and the main N>=50 check already handles the guaranteed zero case.
            # if result == 0:
            #    break
        print(result)

# Run the solve function
solve()
0