user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x2,%eax
  jle    L99
  lea    -0x3(%rax,%rax,1),%edi
  mov    $0x1,%ecx
  mov    $0x0,%esi
  jmp    L121
L183:
  add    $0x1,%eax
  cmp    %eax,%edx
  jne    L183
  lea    (%rcx,%rsi,1),%esi
L28:
  add    $0x2,%ecx
  cmp    %edi,%ecx
  je     L149
L121:
  mov    %ecx,%edx
  mov    $0x0,%eax
  test   %ecx,%ecx
  jg     L183
  jmp    L28
L99:
  mov    $0x0,%esi
L149:
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L73
  mov    $0x0,%ebx
L142:
  mov    $0x20,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jg     L142
L73:
  mov    $0x2a,%edi
  call   <putchar@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L200
  mov    $0x0,%ebx
L84:
  mov    $0x20,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jg     L84
L200:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%ebx
  cmp    $0x2,%ebx
  jle    L12
  mov    $0x1,%r12d
  mov    $0x1,%ebp
L80:
  sub    %ebp,%ebx
  sub    $0x1,%ebx
  je     L185
L111:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L111
L185:
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    %r12d,%r13d
  test   %r12d,%r12d
  jle    L0
  mov    $0x0,%ebx
L208:
  mov    $0x23,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %r13d,%ebx
  jne    L208
L0:
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    %ebp,%ebx
  not    %ebx
  add    0x4(%rsp),%ebx
  je     L31
L25:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L25
L31:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebp
  mov    0x4(%rsp),%ebx
  add    $0x2,%r12d
  lea    -0x1(%rbx),%eax
  cmp    %ebp,%eax
  jg     L80
L12:
  mov    0x4(%rsp),%eax
  add    %eax,%eax
  cmp    $0x1,%eax
  jle    L83
  mov    $0x0,%ebx
L182:
  mov    $0x2a,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax,%rax,1),%eax
  cmp    %ebx,%eax
  jg     L182
L83:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L152
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L152:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D10
datatype: string
name: extract_symbolic_data
arguments: data_tag: D9
datatype: string

tool: D10: %d

tool: D9: %d


assistant: ```c
int main() {
  int i, n, j, h = 0;
  scanf("%d", &n);
  for (j = 1; j < n - 1; j++)
    for (i = 0; i < j * 2 - 1; i++)
      h++;
  printf("%d\n", h);
  for (i = 0; i < n - 1; i++)
    printf(" ");
  printf("*");
  for (i = 0; i < n - 1; i++)
    printf(" ");
  printf("\n");
  for (j = 1; j < n - 1; j++) {
    for (i = n - j - 1; i != 0; i--)
      printf(" ");
    printf("*");
    for (i = 0; i < j * 2 - 1; i++)
      printf("#");
    printf("*");
    for (i = n - j - 1; i != 0; i--)
      printf(" ");
    printf("\n");
  }
  for (i = 0; i < n * 2 - 1; i++)
    printf("*");
  return 0;
}
```