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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x18(%rbp)
  movl   $0x64,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x24(%rbp),%eax
  test   %eax,%eax
  jns    L50
  mov    -0x24(%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    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x24(%rbp),%eax
  neg    %eax
  mov    %eax,-0x24(%rbp)
  jmp    L59
L50:
  mov    -0x24(%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    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L59:
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  jmp    L8
L63:
  mov    -0x20(%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,-0x20(%rbp)
  addl   $0x1,-0x10(%rbp)
L8:
  cmpl   $0x9,-0x20(%rbp)
  jg     L63
  movl   $0x0,-0x1c(%rbp)
  jmp    L56
L30:
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %edx,%ecx
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %ecx,%eax
  sub    %edx,%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L56:
  mov    -0x1c(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L30
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L53
  call   <__stack_chk_fail@plt>
L53:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D1
type_infered: string
name: disassemble_data
arguments: label: D48
type_infered: string
name: disassemble_data
arguments: label: D12
type_infered: string
name: disassemble_data
arguments: label: D52
type_infered: string

tool: D1: Enter an integer > 

tool: D48: %d

tool: D12: %d


tool: D52: That's all, have a nice day!

assistant: ```c
int main(void) {
  int x, y, i;
  int step1 = 10, step2 = 100;
  int num;
  int max = 0;
  printf("Enter an integer > ");
  scanf("%d", &x);
  if (x < 0) {
    printf("%d\n", x % 10);
    x = x * -1;
  } else {
    printf("%d\n", x % 10);
  }
  y = x;
  while (y >= 10) {
    y /= 10;
    max++;
  }
  for (i = 0; i < max; i++) {
    num = ((x % step2 - x % step1) / step1);
    printf("%d\n", num);
    step2 *= 10;
    step1 *= 10;
  }
  printf("That's all, have a nice day!\n");
  return (0);
}
```