user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<check_cyclic>:
  endbr64
  push   %rbp
  pxor   %xmm3,%xmm3
  push   %rbx
  movdqa %xmm3,%xmm6
  movdqa %xmm3,%xmm4
  sub    $0x48,%rsp
  movdqu (%rdi),%xmm2
  movdqa D14(%rip),%xmm5
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movslq 0x10(%rdi),%rbp
  movslq 0x14(%rdi),%rbx
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm5,%xmm6
  punpckldq %xmm2,%xmm1
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  pcmpgtd %xmm1,%xmm4
  imul   $0x51eb851f,%rbp,%rbp
  movdqa %xmm1,%xmm0
  mov    %esi,%eax
  movdqa %xmm6,%xmm7
  pmuludq %xmm5,%xmm0
  sar    $0x1f,%eax
  pmuludq %xmm1,%xmm7
  imul   $0x51eb851f,%rbx,%rbx
  pmuludq %xmm5,%xmm4
  sar    $0x25,%rbp
  sub    %eax,%ebp
  imul   $0x64,%ebp,%eax
  sar    $0x25,%rbx
  mov    %ebp,0x10(%rsp)
  paddq  %xmm7,%xmm4
  sub    %eax,%esi
  mov    %edi,%eax
  psllq  $0x20,%xmm4
  sar    $0x1f,%eax
  paddq  %xmm4,%xmm0
  movdqa %xmm2,%xmm4
  sub    %eax,%ebx
  punpckhdq %xmm2,%xmm4
  imul   $0x64,%ebx,%eax
  mov    %ebx,0x14(%rsp)
  pcmpgtd %xmm4,%xmm3
  pmuludq %xmm4,%xmm6
  pmuludq %xmm5,%xmm4
  sub    %eax,%edi
  movdqa %xmm3,%xmm1
  pmuludq %xmm5,%xmm1
  paddq  %xmm6,%xmm1
  psllq  $0x20,%xmm1
  paddq  %xmm1,%xmm4
  movdqa %xmm2,%xmm1
  psrad  $0x1f,%xmm1
  shufps $0xdd,%xmm4,%xmm0
  psrad  $0x5,%xmm0
  psubd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  movd   %xmm0,%r11d
  movaps %xmm0,(%rsp)
  pslld  $0x1,%xmm1
  paddd  %xmm0,%xmm1
  pslld  $0x3,%xmm1
  paddd  %xmm0,%xmm1
  pslld  $0x2,%xmm1
  psubd  %xmm1,%xmm2
  pshufd $0x55,%xmm2,%xmm1
  movd   %xmm2,%r10d
  movd   %xmm1,%r9d
  movdqa %xmm2,%xmm1
  punpckhdq %xmm2,%xmm1
  pshufd $0xff,%xmm2,%xmm2
  movd   %xmm1,%r8d
  pshufd $0x55,%xmm0,%xmm1
  movd   %xmm2,%edx
  movd   %xmm1,%eax
  movdqa %xmm0,%xmm1
  punpckhdq %xmm0,%xmm1
  add    %r11d,%eax
  pshufd $0xff,%xmm0,%xmm0
  movd   %xmm1,%ecx
  add    %eax,%ecx
  movd   %xmm0,%eax
  add    %ecx,%eax
  lea    (%r9,%r10,1),%ecx
  add    %r8d,%ecx
  add    %ebp,%eax
  add    %edx,%ecx
  add    %ebx,%eax
  add    %esi,%ecx
  add    %edi,%ecx
  cmp    %ecx,%eax
  jne    L37
  movdqa D2(%rip),%xmm0
  mov    $0x5,%ecx
  mov    $0x1,%eax
  movl   $0x5,0x34(%rsp)
  movups %xmm0,0x24(%rsp)
L129:
  cmp    $0xffffffff,%eax
  je     L102
  cmp    %r10d,(%rsp,%rax,4)
  je     L84
L102:
  movslq 0x28(%rsp),%rax
  cmp    $0xffffffff,%eax
  je     L149
  cmp    %r10d,(%rsp,%rax,4)
  je     L150
L149:
  movslq 0x2c(%rsp),%rax
  cmp    $0xffffffff,%eax
  je     L38
  cmp    %r10d,(%rsp,%rax,4)
  je     L42
L38:
  movslq 0x30(%rsp),%rax
  cmp    $0xffffffff,%eax
  je     L58
  cmp    %r10d,(%rsp,%rax,4)
  je     L107
L58:
  movslq 0x34(%rsp),%rax
  cmp    $0xffffffff,%eax
  je     L37
  cmp    %r10d,(%rsp,%rax,4)
  je     L48
L37:
  xor    %eax,%eax
L103:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L75
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L48:
  mov    %edi,%r10d
  mov    $0x5,%eax
  nopl   0x0(%rax)
L158:
  movl   $0xffffffff,0x20(%rsp,%rax,4)
  sub    $0x1,%ecx
  jne    L31
  xor    %eax,%eax
  cmp    %r10d,%r11d
  sete   %al
  jmp    L103
L84:
  mov    %r9d,%r10d
  mov    $0x1,%eax
  jmp    L158
L150:
  mov    %r8d,%r10d
  mov    $0x2,%eax
  jmp    L158
L107:
  mov    %esi,%r10d
  mov    $0x4,%eax
  jmp    L158
L42:
  mov    %edx,%r10d
  mov    $0x3,%eax
  jmp    L158
L75:
  call   <__stack_chk_fail@plt>
L31:
  movslq 0x24(%rsp),%rax
  jmp    L129
```

tool_calls: name: retrieve_data
arguments: label: D14
type_guessed: byte[16]
name: retrieve_data
arguments: label: D2
type_guessed: byte[16]

tool: {"D14": [0x1f,0x85,0xeb,0x51,0x1f,0x85,0xeb,0x51,0x1f,0x85,0xeb,0x51,0x1f,0x85,0xeb,0x51]}

tool: {"D2": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int check_cyclic(int *nums) {
  int l1[] = {0, 0, 0, 0, 0, 0};
  int l2[] = {0, 0, 0, 0, 0, 0};
  for (int i = 0; i < 6; i++) {
    l1[i] = nums[i] / 100;
    l2[i] = nums[i] - 100 * (nums[i] / 100);
    if (0)
      printf("%d\t%d\t%d\n", nums[i], l1[i], l2[i]);
  }
  if (l1[0] + l1[1] + l1[2] + l1[3] + l1[4] + l1[5] != l2[0] + l2[1] + l2[2] + l2[3] + l2[4] + l2[5])
    return 0;
  int idx[] = {0, 1, 2, 3, 4, 5};
  int flag;
  int n = 0;
  for (int i = 0; i < 6 - 1; i++) {
    flag = 0;
    for (int j = 1; j < 6; j++) {
      if (idx[j] != -1) {
        if (l2[n] == l1[idx[j]]) {
          flag = 1;
          idx[j] = -1;
          n = j;
          break;
        }
      }
    }
    if (flag == 0)
      return 0;
  }
  if (l2[n] == l1[0])
    return 1;
  return 0;
}
```