user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  lea    D61(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x3,%eax
  jle    L67
  lea    D114(%rip),%rdi
  test   $0x1,%al
  jne    L2
  call   <puts@plt>
  lea    D124(%rip),%rdi
  call   <puts@plt>
  lea    D96(%rip),%rdi
  call   <puts@plt>
  lea    D7(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x4,0x4(%rsp)
  jle    L40
  mov    $0x5,%ebx
  lea    D38(%rip),%r12
  lea    D20(%rip),%rbp
  nopl   0x0(%rax)
L3:
  lea    0x1(%rbx),%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  add    $0x2,%ebx
  call   <puts@plt>
  cmp    %ebx,0x4(%rsp)
  jge    L3
L40:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <puts@plt>
  lea    D124(%rip),%rdi
  call   <puts@plt>
  lea    D106(%rip),%rdi
  call   <puts@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x5,0x4(%rsp)
  jle    L40
  mov    $0x6,%ebx
  lea    D38(%rip),%r12
  lea    D20(%rip),%rbp
  xchg   %ax,%ax
L41:
  lea    0x1(%rbx),%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  add    $0x2,%ebx
  call   <puts@plt>
  cmp    %ebx,0x4(%rsp)
  jge    L41
  jmp    L40
L67:
  lea    D101(%rip),%rdi
  call   <puts@plt>
  jmp    L40
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D61
type_infered: string
name: retrieve_data
arguments: data_tag: D114
type_infered: string
name: retrieve_data
arguments: data_tag: D124
type_infered: string
name: retrieve_data
arguments: data_tag: D96
type_infered: string
name: retrieve_data
arguments: data_tag: D7
type_infered: string
name: retrieve_data
arguments: data_tag: D38
type_infered: string
name: retrieve_data
arguments: data_tag: D20
type_infered: string
name: retrieve_data
arguments: data_tag: D106
type_infered: string
name: retrieve_data
arguments: data_tag: D11
type_infered: string
name: retrieve_data
arguments: data_tag: D1
type_infered: string
name: retrieve_data
arguments: data_tag: D101
type_infered: string

tool: D61: %d

tool: D114: YES

tool: D124: 1 * 2 = 2

tool: D96: 2 * 3 = 6

tool: D7: 6 * 4 = 24

tool: D38: %d - %d = 1


tool: D20: 24 * 1 = 24

tool: D106: 3 + 4 = 7

tool: D11: 7 + 5 = 12

tool: D1: 12 * 2 = 24

tool: D101: NO

assistant: ```c
int main() {
  int n, i;
  scanf("%d", &n);
  if (n < 4) {
    printf("NO\n");
  } else {
    if (n % 2 == 0) {
      printf("YES\n");
      printf("1 * 2 = 2\n");
      printf("2 * 3 = 6\n");
      printf("6 * 4 = 24\n");
      for (i = 5; i <= n; i += 2) {
        printf("%d - %d = 1\n", i + 1, i);
        printf("24 * 1 = 24\n");
      }
    } else {
      printf("YES\n");
      printf("1 * 2 = 2\n");
      printf("3 + 4 = 7\n");
      printf("7 + 5 = 12\n");
      printf("12 * 2 = 24\n");
      for (i = 6; i <= n; i += 2) {
        printf("%d - %d = 1\n", i + 1, i);
        printf("24 * 1 = 24\n");
      }
    }
  }
  return 0;
}
```