user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x738,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x2728(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  lea    D10(%rip),%r15
  lea    D29(%rip),%r14
  jmp    L102
L26:
  movslq %esi,%rsi
  sub    $0x1,%eax
  mov    %al,0x10(%rsp,%rsi,1)
L69:
  lea    0xc(%rsp),%rdi
  mov    %r14,%rsi
  call   <sprintf@plt>
  movzbl 0xc(%rsp),%eax
  mov    %al,-0x4(%r13)
  movzbl 0xd(%rsp),%eax
  mov    %al,-0x3(%r13)
  movzbl 0xe(%rsp),%eax
  mov    %al,-0x2(%r13)
  sub    $0x1,%rbx
  sub    $0x1,%ebp
  cmp    $0xffffffff,%ebp
  je     L135
L4:
  lea    -0x1(%rbx),%rdi
  mov    $0x0,%esi
  call   <strtod@plt>
  mov    %eax,%r12d
  mov    %rbx,%r13
  movb   $0x0,-0x1(%rbx)
  lea    -0x4(%rbx),%rdi
  mov    $0x0,%esi
  call   <strtod@plt>
  lea    (%r12,%r12,4),%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  jns    L69
  add    $0x3e8,%edx
  mov    %ebp,%esi
  movslq %ebp,%rax
  movzbl 0x10(%rsp,%rax,1),%eax
  cmp    $0x30,%al
  jne    L26
  mov    %rbx,%rcx
L22:
  movb   $0x39,-0x5(%rcx)
  sub    $0x1,%esi
  sub    $0x1,%rcx
  movzbl -0x5(%rcx),%eax
  cmp    $0x30,%al
  je     L22
  jmp    L26
L135:
  mov    $0x4,%ebp
L113:
  lea    0x10(%rsp),%rdi
  mov    $0x0,%esi
  call   <strtod@plt>
  movslq %eax,%rdx
  imul   $0x78787879,%rdx,%rdx
  sar    $0x23,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    $0x4,%ecx
  add    %ecx,%edx
  cmp    %edx,%eax
  jne    L117
  lea    D32(%rip),%rdi
  call   <printf@plt>
L102:
  lea    0x10(%rsp),%rsi
  mov    %r15,%rdi
  call   <scanf@plt>
  cmpb   $0x30,0x10(%rsp)
  je     L103
  movslq %ebp,%rcx
  lea    0x1(%rbp),%eax
  cltq
  lea    0xf(%rsp),%rdx
  cmpb   $0x0,0x10(%rsp,%rcx,1)
  je     L17
L1:
  mov    %eax,%ebp
  add    $0x1,%rax
  cmpb   $0x0,(%rdx,%rax,1)
  jne    L1
L17:
  cmp    $0x4,%ebp
  jle    L113
  movslq %ebp,%rax
  lea    0x10(%rsp,%rax,1),%rbx
  sub    $0x5,%ebp
  jmp    L4
L117:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L102
L103:
  mov    0x2728(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  mov    $0x0,%eax
  add    $0x2738,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L82:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D10
type_infered: string
name: parse_data
arguments: label: D29
type_infered: string
name: parse_data
arguments: label: D32
type_infered: string
name: parse_data
arguments: label: D24
type_infered: string

tool: D10: %s


tool: D29: %03d

tool: D32: 1


tool: D24: 0


assistant: ```c
int main(int argc, const char *argv[]) {
  int l = 0, e, n, i;
  char num[10000], temp[3], new[4];
  while (scanf("%s\n", num), num[0] != '0') {
    while (num[l] != '\0') {
      l++;
    }
    while (l > 4) {
      e = strtod(&(num[l - 1]), NULL);
      num[--l] = '\0';
      n = strtod(&(num[l - 3]), NULL);
      n = n - (e * 5);
      if (n < 0) {
        n = 1000 + n;
        i = l - 4;
        while (num[i] == '0') {
          num[i] = '9';
          i--;
        }
        num[i] = num[i] - 1;
      }
      sprintf(new, "%03d", n);
      num[l - 3] = new[0];
      num[l - 2] = new[1];
      num[l - 1] = new[2];
    }
    int result = (int)(strtod(num, NULL));
    if (!(result % 17))
      printf("1\n");
    else
      printf("0\n");
  }
  return 0;
}
```