結果

問題 No.1025 Modular Equation
コンテスト
ユーザー lam6er
提出日時 2025-04-16 15:51:22
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 2,811 bytes
コンパイル時間 414 ms
コンパイル使用メモリ 81,376 KB
実行使用メモリ 76,152 KB
最終ジャッジ日時 2025-04-16 15:52:26
合計ジャッジ時間 9,750 ms
ジャッジサーバーID
（参考情報） 		judge3 / 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 AC * 3
other AC * 6 TLE * 1 -- * 25
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 sys
MOD = 10**9 + 7

def main():
    p, n, k, b = map(int, sys.stdin.readline().split())
    a_list = list(map(int, sys.stdin.readline().split()))
    
    # Separate zero and non-zero a_i
    zero_a = []
    non_zero_a = []
    for a in a_list:
        if a == 0:
            zero_a.append(a)
        else:
            non_zero_a.append(a)
    m = len(zero_a)
    n_prime = len(non_zero_a)
    
    # Precompute d = gcd(k, p-1)
    d = 0
    if p > 1:
        d = pow(k, 1, p-1)
        d = gcd(d, p-1)
    else:
        d = 1  # p=2, p-1=1
    
    # Precompute primitive root (not necessary here, but we can compute k-th power residues)
    # For each non-zero a_i, compute the set S_i of residues v where a_i x^k ≡ v mod p has solutions
    # For a_i !=0, f_i[v] = 1 if v ==0, else d if v*inv_ai is a k-th power residue, else 0
    # So for each non-zero a_i, precompute inv_ai and the set of v where v*inv_ai is a k-th power residue
    # But instead of precomputing S_i, we can compute f_i[v] on the fly
    
    # Precompute for each non-zero a_i, the frequency array
    # But to optimize, precompute for each a_i, the inv_ai and then for each v, compute c = v * inv_ai mod p
    # and check if c is a k-th power residue
    
    # Function to compute the frequency array for a given a_i
    def compute_f(a):
        if a == 0:
            return [0]*p  # handled separately
        inv_a = pow(a, p-2, p)
        f = [0]*p
        # Compute d = gcd(k, p-1)
        if p == 1:
            d_p = 1
        else:
            k_mod = k % (p-1)
            d_p = gcd(k_mod, p-1)
        # For each v in 0..p-1
        for v in range(p):
            c = (v * inv_a) % p
            if c == 0:
                f[v] = 1
            else:
                # Check if c is a k-th power residue
                # Compute c^((p-1)/d_p) mod p. If it's 1, then yes.
                if pow(c, (p-1)//d_p, p) == 1:
                    f[v] = d_p
                else:
                    f[v] = 0
        return f
    
    # Compute the frequency arrays for non-zero a_i
    freq_list = []
    for a in non_zero_a:
        freq = compute_f(a)
        freq_list.append(freq)
    
    # Now, compute the DP
    dp = [0] * p
    dp[0] = 1
    for freq in freq_list:
        new_dp = [0] * p
        for s in range(p):
            if dp[s] == 0:
                continue
            for v in range(p):
                if freq[v] == 0:
                    continue
                new_s = (s + v) % p
                new_dp[new_s] = (new_dp[new_s] + dp[s] * freq[v]) % MOD
        dp = new_dp
    
    # Handle the zero terms
    pow_p_m = pow(p, m, MOD)
    answer = (dp[b] * pow_p_m) % MOD
    print(answer)

def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

if __name__ == '__main__':
    main()
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