查验身份证

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

4
320124198808240056
12010X198901011234
110108196711301866
37070419881216001X

12010X198901011234
110108196711301866
37070419881216001X

2
320124198808240056
110108196711301862

#include<stdio.h>
int main()
{
  int n, i, sum = 0;
    int z;
    char dui[17] = { '1','0','X','9','8','7','6','5','4','3','2' };
//因为x并非是一个数字，所以就不能用整形数组，就应该用一个字符数组
    int qz[17] = { 7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2 };
  scanf("%d", &n);
  int count = 1;
    char id[100][19];
    for (i = 0; i < n; i++)
    {
      scanf("%s", id[i]);
    }
    for (i = 0; i < n; i++)
    {
            int count = 1;
            sum=0;
            z=0;
      for (int k = 0; k < 18; k++)
      {
        sum += (id[i][k] - '0') * qz[k];
//(id[i][k] - '0')计算每个字符的大小为int型
      }
     z = sum % 11;
    if (dui[z] != id[i][17])
    {
      printf("%s", id[i]);
      printf("\n");
      count = 0;
    }
  }
  if (count)
  {
    printf("All passed");
  }
  return 0;
}