用set判断是否输出过 + 筛质数

n = int(input())

mp = {}

def isprime(x):
    if x < 2:
        return False
    for i in range(2,int(x ** 0.5) + 1):
        if x % i == 0:
            return False
    return True

for rank in range(1,n + 1):
    mp[input()] = rank

k = int(input())

flag = set()

while k > 0:
    k -= 1
    id = input()
    if id in flag:
        print(f"{id}: Checked")
        continue

    if id not in mp:
        print(f"{id}: Are you kidding?")
    else:
        rank = mp[id]
        if rank == 1:
            res = "Mystery Award"
        elif isprime(rank):
            res = "Minion"
        else:
            res = "Chocolate"
        print(f"{id}: {res}")
        flag.add(id)

判断是否是质数那一步可以优化一下，可以提前把数据范围内的质数先筛出来，然后判断是否是质数直接查表即可：

N = (int)(1e4 + 10)
n = int(input())

mp = {}

st = [False] * N
primes = []

def get_primes(n): # 线性筛质数
    for i in range(2,n + 1):
        if not st[i]:
            primes.append(i)

        j = 0
        while primes[j] * i <= n:
            st[primes[j] * i] = True
            if i % primes[j] == 0:
                break
            j += 1

get_primes(n)

for rank in range(1,n + 1):
    mp[input()] = rank

k = int(input())

flag = set()

while k > 0:
    k -= 1
    id = input()
    if id in flag:
        print(f"{id}: Checked")
        continue

    if id not in mp:
        print(f"{id}: Are you kidding?")
    else:
        rank = mp[id]
        if rank == 1:
            res = "Mystery Award"
        elif not st[rank]:
            res = "Minion"
        else:
            res = "Chocolate"
        print(f"{id}: {res}")
        flag.add(id)