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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x1c(%rbp)
  lea    -0x2c(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x30(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L20
L98:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    -0x28(%rbp),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x28(%rbp)
L20:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jl     L98
  mov    -0x2c(%rbp),%eax
  mov    -0x30(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x18(%rbp),%xmm0
  mov    -0x30(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  divss  %xmm1,%xmm0
  movd   %xmm0,%eax
  movd   %eax,%xmm0
  call   <round@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x18(%rbp),%xmm0
  mov    -0x30(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  divss  %xmm1,%xmm0
  movd   %xmm0,%eax
  movd   %eax,%xmm0
  call   <abs@plt>
  mov    %eax,-0x20(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jle    L1
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x28(%rbp)
  jmp    L46
L23:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L24
L84:
  mov    -0x24(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%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)
  addl   $0x1,-0x24(%rbp)
L24:
  mov    -0x30(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x24(%rbp)
  jl     L84
  mov    -0x30(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x28(%rbp)
L46:
  mov    -0x30(%rbp),%eax
  sub    -0x18(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jl     L23
L1:
  movl   $0x0,-0x28(%rbp)
  jmp    L32
L97:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x28(%rbp)
L32:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jl     L97
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L94
  call   <__stack_chk_fail@plt>
L94:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_guessed: string
name: parse_data
arguments: tag: D28
type_guessed: string
name: parse_data
arguments: tag: D7
type_guessed: string

tool: D0: %d%d

tool: D28: Reverse Loop

tool: D7: %d 

assistant: ```c
int main() {
  int n, k, kactual, q, i, j, g, temp, loc;
  int *arr = NULL;
  int absolute = 0;
  float rounder = 0.0f;
  scanf("%d%d", &n, &k);
  arr = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    arr[i] = i + 1;
  }
  kactual = k % n;
  rounder = round((float)kactual / (float)n);
  absolute = abs((float)kactual / (float)n);
  if (kactual > 0) {
    printf("Reverse Loop");
    for (i = 0; i < (n - kactual); i++) {
      temp = arr[0];
      for (j = 0; j < n - 1; j++) {
        arr[j] = arr[j + 1];
      }
      arr[n - 1] = temp;
    }
  }
  for (i = 0; i < n; i++) {
    printf("%d ", arr[i]);
  }
  return 0;
}
```