一个合法的身份证号码由17位地区、日期编号和顺序编号加1位校验码组成。校验码的计算规则如下：

首先对前17位数字加权求和，权重分配为：{7，9，10，5，8，4，2，1，6，3，7，9，10，5，8，4，2}；然后将计算的和对11取模得到值Z；最后按照以下关系对应Z值与校验码M的值：

Z：0 1 2 3 4 5 6 7 8 9 10
M：1 0 X 9 8 7 6 5 4 3 2

现在给定一些身份证号码，请你验证校验码的有效性，并输出有问题的号码。

输入格式：

输入第一行给出正整数N（≤100）是输入的身份证号码的个数。随后N行，每行给出1个18位身份证号码。

输出格式：

按照输入的顺序每行输出1个有问题的身份证号码。这里并不检验前17位是否合理，只检查前17位是否全为数字且最后1位校验码计算准确。如果所有号码都正常，则输出All passed。

输入样例1：

4
320124198808240056
12010X198901011234
110108196711301866
37070419881216001X

输出样例1：

12010X198901011234
110108196711301866
37070419881216001X

输入样例2：

2
320124198808240056
110108196711301862

输出样例2：

All passed

编程语言：Python

解题思想：

1.使用列表和字典来存放权重、校验码的对应关系

2.首先判断前17位是否都是数字，使用isnumeric方法，如果是，则求出权重和并判断与校验码是否匹配

3.设置一个标记，在整个循环过程中，只要有一个身份证号不正确，就改变标记，为最后的输出提供判断

时间复杂度：O(N)，虽然是双层循环，但内循环为固定常数

代码如下：

NOI = int(input())
Weights = [7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2]  # 每一位号码对应的权重
CheckCode = {  # 利用字典来存放Z值与校验码的对应关系
    0: '1',
    1: '0',
    2: 'X',
    3: '9',
    4: '8',
    5: '7',
    6: '6',
    7: '5',
    8: '4',
    9: '3',
    10: '2'
}
 
Judge = True  # 所有身份证号是否都正确的标记，默认为是
for i in range(NOI):
    Identity = input()  # 输入身份证号
    Z = 0
    if Identity[:17].isnumeric() == True:  # 该身份证号的前17位全为数字
        for j in range(17):  # 求权重和
            Z += int(Identity[j]) * Weights[j]
        Z %= 11
        if CheckCode[Z] != Identity[17]:  # Z值对应校验码与身份证号最后一位不匹配
            print(Identity)
            Judge = False
    else:
        Judge = False
        print(Identity)
 
if Judge == True:  # 所有身份证号都正确
    print("All passed")

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