user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jne    L45
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L45
L87:
  mov    -0xc(%rbp),%eax
  neg    %eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    $0x9,%eax
  jg     L45
  mov    -0xc(%rbp),%eax
  neg    %eax
  mov    %eax,%esi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L45:
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  js     L87
  jmp    L74
L50:
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jne    L74
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L57
L74:
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jg     L50
L57:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D69
datatype: string
name: extract_symbolic_data
arguments: tag: D20
datatype: string
name: extract_symbolic_data
arguments: tag: D34
datatype: string
name: extract_symbolic_data
arguments: tag: D18
datatype: string
name: extract_symbolic_data
arguments: tag: D32
datatype: string
name: extract_symbolic_data
arguments: tag: D24
datatype: string

tool: D69: Enter an integer > 

tool: D20: %d

tool: D34: 
0
That's all, have a nice day!

tool: D18: 
%d

tool: D32: 
%dThat's all, have a nice day!


tool: D24: 
That's all, have a nice day!

assistant: ```c
int main() {
  int num;
  printf("Enter an integer > ");
  scanf("%d", &num);
  if (num == 0) {
    printf("\n0\nThat's all, have a nice day!\n");
  }
  while (num < 0) {
    num = -num;
    printf("\n%d", num % 10);
    num = num / 10;
    if (num < 10) {
      printf("\n%dThat's all, have a nice day!\n", -num);
    }
  }
  while ((num <= 9999999999) && (num > 0)) {
    printf("\n%d", num % 10);
    num = num / 10;
    if (num == 0) {
      printf("\nThat's all, have a nice day!\n");
      break;
    }
  }
  return (0);
}
```