user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    D61(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  call   <scanf@plt>
  mov    0x18(%rsp),%eax
  cmp    $0x14,%eax
  jg     L180
  xor    %ebp,%ebp
  lea    0x1c(%rsp),%r14
  test   %eax,%eax
  jg     L127
  jmp    L141
  nopl   0x0(%rax)
L85:
  mov    0x18(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebp,%eax
  je     L187
L75:
  lea    0x20(%rsp),%rbx
  mov    %r12d,(%rbx,%rbp,4)
  add    $0x1,%rbp
  cmp    %ebp,0x18(%rsp)
  jle    L135
L127:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x1c(%rsp),%r12d
  cmp    $0x1,%r12d
  ja     L156
  test   %rbp,%rbp
  jne    L85
L187:
  cmp    $0x1,%r12d
  jne    L75
  lea    D16(%rip),%rdi
  call   <printf@plt>
  jmp    L151
L156:
  lea    D25(%rip),%rdi
  mov    $0x1,%r12d
  call   <printf@plt>
L151:
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  add    $0xd8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L135:
  movl   $0x3e8,0xc(%rsp)
  lea    D33(%rip),%r15
  lea    D63(%rip),%r13
  lea    D62(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L166:
  mov    0x18(%rsp),%eax
  xor    %r14d,%r14d
  xor    %ebp,%ebp
  test   %eax,%eax
  jle    L65
  nopl   (%rax)
L60:
  mov    (%rbx,%r14,4),%edx
  test   %edx,%edx
  jne    L43
  mov    %r12,%rdi
  add    $0x1,%ebp
  add    $0x1,%r14
  call   <printf@plt>
  cmp    %r14d,0x18(%rsp)
  jg     L60
L65:
  mov    %r15,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%edx
  cmp    %ebp,%edx
  je     L82
  lea    0x70(%rsp),%rsi
  mov    %rbx,%rdi
  call   <next_generation@plt>
  subl   $0x1,0xc(%rsp)
  jne    L166
L82:
  xor    %r12d,%r12d
  jmp    L151
  nopw   0x0(%rax,%rax,1)
L43:
  mov    %r13,%rdi
  add    $0x1,%r14
  call   <printf@plt>
  cmp    %r14d,0x18(%rsp)
  jg     L60
  jmp    L65
L180:
  lea    D40(%rip),%rdi
  mov    $0x1,%r12d
  call   <printf@plt>
  jmp    L151
L141:
  lea    D72(%rip),%rdi
  mov    $0x1,%r12d
  call   <printf@plt>
  jmp    L151
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D61
type: string
name: extract_symbolic_data
arguments: data_label: D16
type: string
name: extract_symbolic_data
arguments: data_label: D25
type: string
name: extract_symbolic_data
arguments: data_label: D33
type: string
name: extract_symbolic_data
arguments: data_label: D63
type: string
name: extract_symbolic_data
arguments: data_label: D62
type: string
name: extract_symbolic_data
arguments: data_label: D40
type: string
name: extract_symbolic_data
arguments: data_label: D72
type: string

tool: D61: %d

tool: D16: 
First and last cell should always be 0!

tool: D25: 
Expected 0 or 1!

tool: D33: 


tool: D63: *

tool: D62: .

tool: D40: Length should be max 20!

tool: D72: Length should be 1 at least!

assistant: ```c
int main() {
  int length, n, current[20], next[20], countforZeros, i, j;
  scanf("%d", &length);
  if (length > 20) {
    printf("Length should be max 20!");
    return 1;
  }
  if (length <= 0) {
    printf("Length should be 1 at least!");
    return 1;
  }
  for (i = 0; i < length; i++) {
    scanf("%d", &n);
    if (n == 0 || n == 1) {
      if ((i == 0 || i == (length - 1)) && n == 1) {
        printf("\nFirst and last cell should always be 0!");
        return 1;
      }
      current[i] = n;
    } else {
      printf("\nExpected 0 or 1!");
      return 1;
    }
  }
  for (i = 0; i < 1000; i++) {
    countforZeros = 0;
    for (j = 0; j < length; j++) {
      if (current[j] == 0) {
        countforZeros++;
        printf(".");
      } else
        printf("*");
    }
    printf("\n");
    if (countforZeros == length)
      break;
    next_generation(current, next, length);
  }
  return 0;
}
```