user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1d0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L84
L26:
  movl   $0x1,-0x1b8(%rbp)
  jmp    L155
L50:
  mov    -0x1b8(%rbp),%eax
  cltq
  mov    -0x1b8(%rbp),%edx
  mov    %edx,-0xe0(%rbp,%rax,4)
  addl   $0x1,-0x1b8(%rbp)
L155:
  cmpl   $0x32,-0x1b8(%rbp)
  jle    L50
  movl   $0x0,-0x1b8(%rbp)
  movl   $0x0,-0x1c0(%rbp)
  movl   $0x1,-0xe0(%rbp)
  movl   $0x0,-0x1b4(%rbp)
  mov    -0x1c4(%rbp),%eax
  mov    %eax,-0x1bc(%rbp)
L4:
  mov    -0x1b8(%rbp),%eax
  cltq
  mov    -0xe0(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L86
  cmpl   $0x0,-0x1c0(%rbp)
  je     L27
  mov    -0x1b4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1b4(%rbp)
  mov    -0x1b8(%rbp),%edx
  movslq %edx,%rdx
  mov    -0xe0(%rbp,%rdx,4),%edx
  cltq
  mov    %edx,-0x1b0(%rbp,%rax,4)
  mov    -0x1b8(%rbp),%eax
  cltq
  movl   $0x0,-0xe0(%rbp,%rax,4)
  movl   $0x0,-0x1c0(%rbp)
  subl   $0x1,-0x1bc(%rbp)
  jmp    L86
L27:
  cmpl   $0x1,-0x1bc(%rbp)
  je     L65
  movl   $0x1,-0x1c0(%rbp)
L86:
  mov    -0x1c4(%rbp),%eax
  cmp    %eax,-0x1b8(%rbp)
  jne    L80
  movl   $0x1,-0x1b8(%rbp)
L80:
  addl   $0x1,-0x1b8(%rbp)
  jmp    L4
L65:
  nop
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x1bc(%rbp)
  jmp    L108
L127:
  cmpl   $0x0,-0x1bc(%rbp)
  je     L123
  mov    -0x1bc(%rbp),%eax
  cmp    -0x1b4(%rbp),%eax
  jge    L123
  mov    $0x2c,%edi
  call   <putchar@plt>
L123:
  mov    -0x1bc(%rbp),%eax
  cltq
  mov    -0x1b0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x1bc(%rbp)
L108:
  mov    -0x1bc(%rbp),%eax
  cmp    -0x1b4(%rbp),%eax
  jl     L127
  mov    -0x1b8(%rbp),%eax
  cltq
  mov    -0xe0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L84:
  lea    -0x1c4(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  test   %eax,%eax
  je     L15
  mov    -0x1c4(%rbp),%eax
  test   %eax,%eax
  jne    L26
L15:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L82
  call   <__stack_chk_fail@plt>
L82:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D22
type_infered: string
name: unpack_data
arguments: tag: D27
type_infered: string
name: unpack_data
arguments: tag: D23
type_infered: string
name: unpack_data
arguments: tag: D15
type_infered: string

tool: D22: Discarded cards:

tool: D27:  %d

tool: D23: 
Remaining card: %d


tool: D15: %d

assistant: ```c
int main() {
  int n, thrn[51], card[52], cut, rn, i, indt;
  while (scanf("%d", &n) && n) {
    for (i = 1; i < 51; i++)
      card[i] = i;
    i = 0;
    cut = 0;
    card[0] = 1;
    indt = 0;
    rn = n;
    while (1) {
      if (card[i]) {
        if (cut) {
          thrn[indt++] = card[i];
          card[i] = 0;
          cut = 0;
          rn--;
        } else {
          if (rn == 1)
            break;
          cut = 1;
        }
      }
      if (i == n)
        i = 1;
      i++;
    }
    for (printf("Discarded cards:"), rn = 0; rn < indt; rn++) {
      if (rn && rn < indt)
        putchar(',');
      printf(" %d", thrn[rn]);
    }
    printf("\nRemaining card: %d\n", card[i]);
  }
  return 0;
}
```