网易第一题ac 以及第三题90% 代码

第一题 给定一个数组，是否存在一个x，使得数组中某些元素加上x，某些元素减去x之后，数组中的所有元素都相等 ac

# -*- coding=utf-8 -*-
def check_equal(seq):
    length = len(seq)
    if length <= 2:
        return 'YES'

    if length != 3:
        return 'NO'

    seq.sort()
    a, b, c = seq
    if b - a == c - b:
        return 'YES'
    else:
        return 'NO'


if __name__ == "__main__":
    seq_nums = int(input())
    for i in range(seq_nums):
        length = int(input())
        seq = list(map(int, input().split()))
        seq = list(set(seq))
        print(check_equal(seq))

第三题 求给定所有数对中，能够整除每个数对中最少一个元素的最大质数x 通过了90%

# -*- coding=utf-8 -*-
from math import sqrt


def get_max_gcd(seq):
    # 得到质因数上线
    min_value = min(seq, key=lambda x: max(x[0], x[1]))
    min_value = min(min_value)
    zhi_list = get_zhi(min_value)
    for zhi in zhi_list:
        flag = True
        for item in seq:
            if not check_equal(item, zhi):
                flag = False
                break
        if flag:
            return zhi

    return -1


def check_equal(item, zhi):
    for _ in item:
        if _ % zhi == 0:
            return True
    return False


def get_zhi(value):
    sqrt_value = int(sqrt(value)) + 1
    res = [2]
    i = 3
    while i <= value:
        flag = True
        j = 0
        while j < len(res) and res[j] <= sqrt_value:
            if i % res[j] == 0:
                flag = False
                break
            else:
                j += 1
        if flag:
            res.append(i)
        i += 1
    return res[::-1]


if __name__ == "__main__":
    n = int(input())
    seq = []
    for i in range(n):
        a, b = map(int, input().split())
        seq.append((a, b))
    print(get_max_gcd(seq))