結果

問題 No.1975 Zigzag Sequence
コンテスト
ユーザー lam6er
提出日時 2025-03-20 20:38:11
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 290 ms / 2,000 ms
コード長 3,152 bytes
コンパイル時間 208 ms
コンパイル使用メモリ 81,932 KB
実行使用メモリ 132,892 KB
最終ジャッジ日時 2025-03-20 20:38:42
合計ジャッジ時間 7,499 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 * 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 # 

MOD = 10**9 + 7

def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    N = int(data[0])
    A = list(map(int, data[1:N+1]))
    
    if N < 3:
        print(0)
        return
    
    # Coordinate compression
    sorted_A = sorted(set(A))
    rank = {v:i for i, v in enumerate(sorted_A)}
    max_rank = len(sorted_A) - 1
    
    # Precompute pow2
    pow2 = [1] * (N + 1)
    for i in range(1, N + 1):
        pow2[i] = (pow2[i-1] * 2) % MOD
    
    class FenwickTree:
        def __init__(self, size):
            self.size = size
            self.tree = [0] * (self.size + 2)
        
        def update(self, idx, val):
            idx += 1  # 1-based index for Fenwick Tree
            while idx <= self.size + 1:
                self.tree[idx] = (self.tree[idx] + val) % MOD
                idx += idx & -idx
        
        def query(self, idx):
            # sum [0..idx] (inclusive of original rank, but Fenwick Tree uses 1-based)
            res = 0
            idx += 1  # 1-based index
            while idx > 0:
                res = (res + self.tree[idx]) % MOD
                idx -= idx & -idx
            return res
    
    # Preprocess left sum_x and sum_x_large
    left_ft = FenwickTree(max_rank)
    sum_left_small = [0] * N
    sum_left_large = [0] * N
    for i in range(N):
        r = rank[A[i]]
        # Sum of elements < A[i]
        sum_small = left_ft.query(r - 1)
        # Sum of elements > A[i] = total_sum - sum <= A[i]
        total_sum = left_ft.query(max_rank)
        sum_large = (total_sum - left_ft.query(r)) % MOD
        sum_left_small[i] = sum_small
        sum_left_large[i] = sum_large
        # Update Fenwick tree with current element's contribution (2^{i-1})
        val = pow2[i]  # 2^(i) since 0-based here. 2^{i} = 2^{(i+1) -1} ? Wait, for x, the exponent is x-1
        # Since in code, i is 0-based. For x, which is 1-based, i runs from 0 to N-1 in the loop (x from 1 to N)
        # So x = i+1. 2^{x-1} = 2^{i}
        left_ft.update(r, pow2[i])
    
    # Preprocess right sum_z and sum_z_large
    right_ft = FenwickTree(max_rank)
    sum_right_small = [0] * N
    sum_right_large = [0] * N
    for i in reversed(range(N)):
        r = rank[A[i]]
        # Sum of elements < A[i]
        sum_small = right_ft.query(r - 1)
        # Sum of elements > A[i] = total_sum - sum <= A[i]
        total_sum = right_ft.query(max_rank)
        sum_large = (total_sum - right_ft.query(r)) % MOD
        sum_right_small[i] = sum_small
        sum_right_large[i] = sum_large
        # Update Fenwick tree with current element's contribution (2^{n - (i+1)})
        # because i is 0-based. original z is i+1 (1-based)
        exp = N - (i + 1)
        val = pow2[exp]
        right_ft.update(r, val)
    
    total = 0
    for i in range(N):
        s1 = sum_left_small[i]
        s2 = sum_right_small[i]
        contrib_peak = (s1 * s2) % MOD
        s3 = sum_left_large[i]
        s4 = sum_right_large[i]
        contrib_valley = (s3 * s4) % MOD
        total = (total + contrib_peak + contrib_valley) % MOD
    
    print(total % MOD)

if __name__ == '__main__':
    main()
0