user: 这段汇编代码用C语言表示是什么样的?

```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    $0x748,%rsp
  xor    %ebx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x2738(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rax
  mov    %rax,0x8(%rsp)
  lea    0x1f(%rsp),%rax
  mov    %rax,(%rsp)
  nopl   (%rax)
L14:
  mov    0x8(%rsp),%rsi
  lea    D30(%rip),%rdi
  call   <scanf@plt>
  cmpb   $0x30,0x20(%rsp)
  je     L83
  movslq %ebx,%rdx
  lea    0x1(%rbx),%eax
  cmpb   $0x0,0x20(%rsp,%rdx,1)
  cltq
  je     L87
  nopl   0x0(%rax,%rax,1)
L51:
  mov    (%rsp),%rdx
  mov    %eax,%ebx
  add    $0x1,%rax
  cmpb   $0x0,(%rdx,%rax,1)
  jne    L51
L87:
  cmp    $0x4,%ebx
  jle    L10
  mov    0x8(%rsp),%rax
  movslq %ebx,%r8
  lea    0x1c(%rsp),%rbp
  sub    $0x5,%ebx
  lea    (%rax,%r8,1),%r15
  jmp    L93
  nopw   0x0(%rax,%rax,1)
L131:
  mov    %r10d,%edx
  lea    D0(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  movzbl 0x1c(%rsp),%eax
  mov    %al,-0x4(%r14)
  movzbl 0x1d(%rsp),%eax
  mov    %al,-0x3(%r14)
  movzbl 0x1e(%rsp),%eax
  mov    %al,-0x2(%r14)
  sub    $0x1,%ebx
  jb     L33
L93:
  mov    %r15,%r13
  sub    $0x1,%r15
  xor    %esi,%esi
  mov    %r15,%rdi
  mov    %r13,%r14
  call   <strtod@plt>
  movb   $0x0,-0x1(%r13)
  lea    -0x4(%r13),%rdi
  xor    %esi,%esi
  mov    %eax,%r12d
  call   <strtod@plt>
  mov    %eax,%r10d
  lea    (%r12,%r12,4),%eax
  sub    %eax,%r10d
  jns    L131
  movslq %ebx,%rax
  add    $0x3e8,%r10d
  mov    %ebx,%ecx
  movzbl 0x20(%rsp,%rax,1),%eax
  cmp    $0x30,%al
  jne    L86
  nopw   0x0(%rax,%rax,1)
L127:
  movb   $0x39,-0x5(%r13)
  movzbl -0x6(%r13),%eax
  sub    $0x1,%r13
  sub    $0x1,%ecx
  cmp    $0x30,%al
  je     L127
L86:
  movslq %ecx,%rcx
  sub    $0x1,%eax
  mov    %al,0x20(%rsp,%rcx,1)
  jmp    L131
  nopl   0x0(%rax)
L33:
  mov    $0x4,%ebx
L10:
  mov    0x8(%rsp),%rdi
  xor    %esi,%esi
  call   <strtod@plt>
  imul   $0xf0f0f0f1,%eax,%eax
  add    $0x7878787,%eax
  cmp    $0xf0f0f0e,%eax
  ja     L29
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L14
  nopl   0x0(%rax)
L29:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  jmp    L14
L83:
  mov    0x2738(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x2748,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D30
type: string
name: retrieve_data
arguments: data_label: D0
type: string
name: retrieve_data
arguments: data_label: D14
type: string
name: retrieve_data
arguments: data_label: D9
type: string

tool: D30: %s


tool: D0: %03d

tool: D14: 1


tool: D9: 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;
}
```