結果

問題 No.444 旨味の相乗効果
コンテスト
ユーザー lam6er
提出日時 2025-03-31 17:25:25
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 861 ms / 2,500 ms
コード長 3,177 bytes
コンパイル時間 149 ms
コンパイル使用メモリ 82,160 KB
実行使用メモリ 79,952 KB
最終ジャッジ日時 2025-03-31 17:26:31
合計ジャッジ時間 4,700 ms
ジャッジサーバーID
（参考情報） 		judge4 / 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 * 5
other AC * 23
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
    n, c = map(int, sys.stdin.readline().split())
    a = list(map(int, sys.stdin.readline().split()))
    
    # Step 1: Compute elementary symmetric sums e_1 ... e_n mod MOD
    e = [0] * (n + 1)
    e[0] = 1
    for num in a:
        num_mod = num % MOD
        # Iterate backwards to prevent overwriting
        for k in range(n, 0, -1):
            e[k] = (e[k] + num_mod * e[k-1]) % MOD
    
    # e now contains e_0 to e_n; e[1] to e[n] are needed
    
    # Step 2: Compute S_0 to S_{n-1}
    S = [0] * (n)
    S[0] = 1
    for current_c in range(1, n):
        s = 0
        for k in range(1, current_c + 1):
            # Compute the sign (-1)^(k+1) mod MOD
            sign = 1 if (k + 1) % 2 == 0 else -1
            term = (e[k] * sign) % MOD
            term = (term * S[current_c - k]) % MOD
            s = (s + term) % MOD
        S[current_c] = s
    
    # Now handle S_c
    if c < n:
        sc = S[c]
    else:
        # Need to build the transition matrix and use matrix exponentiation
        # Build the transition matrix of size n x n
        mat = [[0] * n for _ in range(n)]
        # First row: coefficients from e_1 to e_n with signs
        for k in range(1, n+1):
            sign = 1 if (k + 1) % 2 == 0 else -1
            coeff = (e[k] * sign) % MOD
            mat[0][k-1] = coeff
        # Other rows: identity matrix shift
        for i in range(1, n):
            mat[i][i-1] = 1
        
        # Compute exponent and power the matrix
        exponent = c - (n - 1)
        # Function to multiply matrices with mod
        def multiply(A, B):
            res = [[0] * n for _ in range(n)]
            for i in range(n):
                for k in range(n):
                    if A[i][k]:
                        for j in range(n):
                            res[i][j] = (res[i][j] + A[i][k] * B[k][j]) % MOD
            return res
        
        # Matrix exponentiation by squaring
        def matrix_pow(mat, power):
            result = [[0] * n for _ in range(n)]
            for i in range(n):
                result[i][i] = 1
            while power > 0:
                if power % 2 == 1:
                    result = multiply(result, mat)
                mat = multiply(mat, mat)
                power //= 2
            return result
        
        powered_mat = matrix_pow(mat, exponent)
        
        # Initial vector is [S_{n-1}, S_{n-2}, ..., S_0]
        initial_vec = S[:n][::-1]  # reversed
        # Multiply matrix with vector
        new_vec = [0] * n
        for i in range(n):
            for k in range(n):
                new_vec[i] = (new_vec[i] + powered_mat[i][k] * initial_vec[k]) % MOD
        sc = new_vec[0]
    
    # Compute sum a_i^c mod MOD
    sum_aic = 0
    for num in a:
        num_mod = num % MOD
        if num_mod == 0 and c > 0:
            # a_i is 0, a_i^c is 0 if c >=1
            term = 0
        else:
            term = pow(num_mod, c, MOD)
        sum_aic = (sum_aic + term) % MOD
    
    answer = (sc - sum_aic) % MOD
    # Ensure positive
    answer = (answer + MOD) % MOD
    print(answer)
    
if __name__ == '__main__':
    main()
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