user: 这段汇编代码用C语言表示是什么样的:

```asm
<split_and_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdx,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  cmp    $0x1,%esi
  je     L93
  mov    %rdi,%r13
  mov    %esi,%edi
  mov    %esi,%ebx
  shr    $0x1f,%edi
  add    %esi,%edi
  sar    %edi
  call   <floor@plt>
  mov    %ebx,%edx
  xor    %r10d,%r10d
  mov    %rsp,%rdi
  shr    $0x1f,%edx
  mov    %eax,%r14d
  lea    (%rbx,%rdx,1),%ecx
  and    $0x1,%ecx
  sub    %edx,%ecx
  cmp    $0x1,%ecx
  sete   %r10b
  add    %eax,%r10d
  movslq %r10d,%r9
  lea    0x0(,%r9,4),%r8
  lea    0xf(%r8),%rdx
  mov    %rdx,%rsi
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rdi
  and    $0xfffffffffffffff0,%rsi
  cmp    %rdi,%rsp
  je     L17
L102:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdi,%rsp
  jne    L102
L17:
  and    $0xfff,%esi
  sub    %rsi,%rsp
  test   %rsi,%rsi
  je     L13
  orq    $0x0,-0x8(%rsp,%rsi,1)
L13:
  cltq
  mov    %rsp,%rsi
  mov    %rsp,%r12
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L92
L70:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L70
L92:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L85
L107:
  mov    %r8,-0x78(%rbp)
  mov    %rsp,%r15
  test   %r10d,%r10d
  jle    L36
  lea    0x0(,%r9,4),%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %ecx,-0x70(%rbp)
  mov    %r10d,-0x6c(%rbp)
  mov    %r9,-0x68(%rbp)
  mov    %rdx,-0x60(%rbp)
  call   <memcpy@plt>
  mov    -0x78(%rbp),%r8
  mov    -0x60(%rbp),%rdx
  mov    %r15,%rdi
  lea    0x0(%r13,%r8,1),%rsi
  call   <memcpy@plt>
  mov    -0x70(%rbp),%ecx
  mov    -0x6c(%rbp),%r10d
  mov    -0x68(%rbp),%r9
L36:
  cmp    $0x1,%ecx
  jne    L65
  mov    0x0(%r13,%r9,4),%eax
  mov    %eax,(%r12,%r9,4)
L65:
  mov    %r10d,%esi
  lea    -0x50(%rbp),%rdx
  mov    %r12,%rdi
  mov    %r10d,-0x60(%rbp)
  movq   $0x0,-0x40(%rbp)
  call   <split_and_count>
  lea    -0x48(%rbp),%rdx
  mov    %r14d,%esi
  mov    %r15,%rdi
  call   <split_and_count>
  sub    $0x8,%rsp
  mov    -0x60(%rbp),%esi
  mov    %r14d,%ecx
  push   %r13
  mov    %r15,%rdx
  lea    -0x40(%rbp),%r9
  mov    %ebx,%r8d
  mov    %r12,%rdi
  call   <merge@plt>
  mov    -0x48(%rbp),%rax
  pop    %rdx
  add    -0x50(%rbp),%rax
  pop    %rcx
  add    -0x40(%rbp),%rax
L93:
  mov    -0x58(%rbp),%rbx
  mov    %rax,(%rbx)
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L85:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L107
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int split_and_count(int *items, int n, long *inv) {
  if (n == 1) {
    *inv = 0;
    return 0;
  }
  int i, g, h;
  g = (int)floor(n / 2);
  h = g;
  if (n % 2 == 1) {
    h = h + 1;
  }
  int left_hand[h];
  int right_hand[g];
  for (i = 0; i < h; i++) {
    left_hand[i] = items[i];
    right_hand[i] = items[h + i];
  }
  if (n % 2 == 1) {
    left_hand[h] = items[h];
  }
  long x, y, z;
  z = 0;
  split_and_count(left_hand, h, &x);
  split_and_count(right_hand, g, &y);
  merge(left_hand, h, right_hand, g, n, &z, items);
  *inv = x + y + z;
  return 0;
}
```