user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r15
  lea    D43(%rip),%rdi
  lea    D52(%rip),%r15
  push   %r14
  push   %r13
  lea    D19(%rip),%r13
  push   %r12
  push   %rbp
  lea    D31(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%r12
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jg     L16
  jmp    L51
  nopl   0x0(%rax)
L40:
  lea    D62(%rip),%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jle    L51
L16:
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmpq   $0x0,0x10(%rsp)
  je     L40
  mov    $0x9,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    0x10(%rsp),%r14
  cmp    %r14,%rax
  jle    L85
L95:
  mov    $0x9,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    $0x6,%esi
  mov    $0xa,%edi
  mov    %rax,%r8
  mov    %r14,%rax
  cqto
  idiv   %r8
  mov    %rdx,0x10(%rsp)
  call   <pow@plt>
  mov    0x10(%rsp),%r14
  cmp    %r14,%rax
  jle    L117
L98:
  mov    $0x6,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    $0x3,%esi
  mov    $0xa,%edi
  mov    %rax,%r8
  mov    %r14,%rax
  cqto
  idiv   %r8
  mov    %rdx,0x10(%rsp)
  call   <pow@plt>
  mov    0x10(%rsp),%r14
  cmp    %r14,%rax
  jle    L78
L74:
  mov    $0x3,%esi
  mov    $0xa,%edi
  add    $0x1,%ebx
  call   <pow@plt>
  mov    %rax,%r8
  mov    %r14,%rax
  cqto
  idiv   %r8
  mov    %rdx,%rdi
  mov    %rdx,0x10(%rsp)
  call   <num@plt>
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L16
L51:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L90
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L85:
  mov    $0x9,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %rax,%r8
  mov    %r14,%rax
  cqto
  idiv   %r8
  mov    %rax,%rdi
  call   <num@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    0x10(%rsp),%r14
  jmp    L95
L78:
  mov    $0x3,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %rax,%r8
  mov    %r14,%rax
  cqto
  idiv   %r8
  mov    %rax,%rdi
  call   <num@plt>
  mov    %r15,%rdi
  call   <printf@plt>
  mov    0x10(%rsp),%r14
  jmp    L74
  nopl   0x0(%rax)
L117:
  mov    $0x6,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %rax,%r8
  mov    %r14,%rax
  cqto
  idiv   %r8
  mov    %rax,%rdi
  call   <num@plt>
  lea    D44(%rip),%rdi
  call   <printf@plt>
  mov    0x10(%rsp),%r14
  jmp    L98
L90:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D43
data_type: string
name: disassemble_data
arguments: label: D52
data_type: string
name: disassemble_data
arguments: label: D19
data_type: string
name: disassemble_data
arguments: label: D31
data_type: string
name: disassemble_data
arguments: label: D62
data_type: string
name: disassemble_data
arguments: label: D25
data_type: string
name: disassemble_data
arguments: label: D44
data_type: string

tool: D43: %d

tool: D52: Thousand 

tool: D19: 


tool: D31: %Ld

tool: D62: Zero


tool: D25: Billion 

tool: D44: Million 

assistant: ```c
int main() {
  int t, i, j;
  long long n;
  scanf("%d", &t);
  for (i = 0; i < t; i++) {
    scanf("%Ld", &n);
    if (n == 0)
      printf("Zero\n");
    else {
      if (n >= pow(10, 9)) {
        num(n / (long long)pow(10, 9));
        printf("Billion ");
      }
      n = n % (long long)pow(10, 9);
      if (n >= pow(10, 6)) {
        num(n / (long long)pow(10, 6));
        printf("Million ");
      }
      n = n % (long long)pow(10, 6);
      if (n >= pow(10, 3)) {
        num(n / (long long)pow(10, 3));
        printf("Thousand ");
      }
      n = n % (long long)pow(10, 3);
      num(n);
      printf("\n");
    }
  }
  return 0;
}
```