user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x4,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L137
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  imul   %edx,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L15
L137:
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  imul   %eax,%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x20(%rbp)
L15:
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x1c(%rbp)
  jmp    L97
L27:
  movl   $0x0,-0x18(%rbp)
  jmp    L132
L95:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L10
  mov    $0x2a,%edi
  call   <putchar@plt>
  jmp    L13
L10:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x18(%rbp)
  je     L51
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  sub    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jne    L45
L51:
  mov    $0x2a,%edi
  call   <putchar@plt>
  jmp    L13
L45:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x18(%rbp)
  jge    L42
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  sub    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jle    L42
  mov    $0x23,%edi
  call   <putchar@plt>
  jmp    L13
L42:
  mov    $0x20,%edi
  call   <putchar@plt>
L13:
  addl   $0x1,-0x18(%rbp)
L132:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L95
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x1c(%rbp)
L97:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L27
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D17
type_infered: string
name: retrieve_data
arguments: data_label: D10
type_infered: string

tool: D17: %d

tool: D10: %d


assistant: ```c
int main() {
  int h, base, count;
  scanf("%d", &h);
  base = 1 + (h - 1) * 2;
  int base_hash = base - 4;
  int h_hash = h - 2;
  if (h_hash % 2 == 0)
    count = (base_hash + 1) * (h_hash / 2);
  else
    count = (base_hash - 1) * ((h_hash - 1) / 2) + base_hash;
  printf("%d\n", count);
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < base; j++) {
      if (i == h - 1)
        printf("*");
      else if (j == (base / 2) + i || j == (base / 2) - i)
        printf("*");
      else if (j < (base / 2) + i && j > (base / 2) - i)
        printf("#");
      else
        printf(" ");
    }
    printf("\n");
  }
  return 0;
}
```