user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
L38:
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x28(%rbp),%eax
  test   %eax,%eax
  jle    L38
  mov    -0x28(%rbp),%eax
  cmp    $0x13,%eax
  jg     L38
  mov    -0x28(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x2,(%rax)
  jmp    L6
L105:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L48
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  je     L48
  subl   $0x1,-0x24(%rbp)
  jmp    L138
L48:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x24(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
L138:
  addl   $0x1,-0x24(%rbp)
L6:
  mov    -0x28(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L105
  movl   $0x0,-0x24(%rbp)
L21:
  mov    -0x28(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <next_generation@plt>
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L13
L93:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L63
  mov    $0x2e,%eax
  jmp    L98
L63:
  mov    $0x2a,%eax
L98:
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x20(%rbp)
L13:
  mov    -0x28(%rbp),%eax
  cmp    %eax,-0x20(%rbp)
  jl     L93
  mov    $0xa,%edi
  call   <putchar@plt>
  cmpl   $0x0,-0x1c(%rbp)
  je     L62
  addl   $0x1,-0x24(%rbp)
  cmpl   $0x3e7,-0x24(%rbp)
  jle    L21
  jmp    L143
L62:
  nop
L143:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x1a4,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L69
  call   <__stack_chk_fail@plt>
L69:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D33
type: string
name: unpack_data
arguments: data_label: D25
type: string
name: unpack_data
arguments: data_label: D32
type: string
name: unpack_data
arguments: data_label: D18
type: string

tool: D33: Enter N: 

tool: D25: %d%*c

tool: D32: %d

tool: D18: %c 

assistant: ```c
int main() {
  int *curr, *next, n, i = 0;
  do
    scanf("%d%*c", ({
            printf("Enter N: ");
            &n;
          }));
  while (n <= 0 || n >= 20);
  curr = (int *)malloc(sizeof(int) * n);
  next = (int *)malloc(sizeof(int) * n);
  curr[n] = 2;
  for (; i < n; i++) {
    scanf("%d", (curr + i));
    if (!(curr[i] == 0 || curr[i] == 1)) {
      i--;
      continue;
    }
    next[i] = curr[i];
  }
  i = 0;
  do {
    next_generation(curr, next, n);
    {
      int i = 0, flag = 0;
      for (; i < n; i++) {
        printf("%c ", (curr[i] == 0) ? '.' : '*');
        flag += curr[i];
        curr[i] = next[i];
      }
      printf("\n");
      if (!flag)
        break;
    }
  } while (++i < 1000);
  free(curr);
  free(next);
  return 420;
}
```