結果

問題 No.696 square1001 and Permutation 5
ユーザー lam6er
提出日時 2025-03-31 17:34:45
言語 PyPy3
(7.3.15)
結果
MLE  
実行時間 -
コード長 1,550 bytes
コンパイル時間 268 ms
コンパイル使用メモリ 82,560 KB
実行使用メモリ 850,800 KB
最終ジャッジ日時 2025-03-31 17:35:38
合計ジャッジ時間 17,804 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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 -- * 2
other MLE * 1 -- * 11
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ソースコード

diff #
プレゼンテーションモードにする

import sys
class FenwickTree:
    def __init__(self, size):
        self.n = size
        self.tree = [0] * (self.n + 1)  # 1-based indexing
    def update(self, idx, delta):
        # Update the idx-th element by delta (1-based index)
        while idx <= self.n:
            self.tree[idx] += delta
            idx += idx & -idx
    def query(self, idx):
        # Query the prefix sum from 1 to idx (1-based index)
        res = 0
        while idx > 0:
            res += self.tree[idx]
            idx -= idx & -idx
        return res
def main():
    input = sys.stdin.read().split()
    n = int(input[0])
    p = list(map(int, input[1:n+1]))
    
    if n == 0:
        print(1)
        return
    
    # Precompute factorial array using optimized method
    max_fact = 1
    for i in range(1, n):
        max_fact *= i
    
    fact = [0] * n
    if n >= 1:
        fact[n-1] = max_fact  # (n-1)! for the largest term
    
    for i in range(n-2, -1, -1):
        if i+1 == 0:
            fact[i] = 0
        else:
            fact[i] = fact[i+1] // (i+1)
    
    # Initialize Fenwick Tree with all elements present
    ft = FenwickTree(n)
    for i in range(1, n+1):
        ft.update(i, 1)
    
    total = 0
    for i in range(n):
        num = p[i]
        # Count numbers less than 'num' that are still present
        cnt = ft.query(num - 1)
        k = n - 1 - i
        total += cnt * fact[k]
        # Remove 'num' from available numbers
        ft.update(num, -1)
    
    print(total + 1)
if __name__ == "__main__":
    main()
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