D - Hachi https://atcoder.jp/contests/abc181/tasks/abc181_d

abc181_d.py

def solv(S):
    if len(S) <= 2:
        return int(S) % 8 == 0 or int(S[::-1]) % 8 == 0
    else:  # 3桁以上の場合：3桁の8の倍数が作れるかを判定
        # 各数字の出現回数
        digit_count = [0] * 10
        for ch in S:
            digit_count[int(ch)] += 1

        # 3桁の8の倍数をチェック
        for multiple in range(104, 1000, 8):
            need_count = [0] * 10
            for ch in str(multiple):
                need_count[int(ch)] += 1

            # 作れるか判定
            possible = True
            for d in range(10):
                if need_count[d] > digit_count[d]:
                    possible = False
                    break

            if possible:
                return True

        return False


S = input()
print("Yes") if solv(S) else print("No")