user: 这段汇编代码用C语言表示是什么样的？

```asm
<go>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x38(%rbp)
  mov    -0x34(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x24(%rbp)
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L3
L13:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x20(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x18(%rbp)
L3:
  mov    -0x18(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L13
  mov    -0x10(%rbp),%eax
  add    %eax,-0x14(%rbp)
  jmp    L8
L23:
  mov    -0x20(%rbp),%edx
  mov    -0x34(%rbp),%eax
  add    %eax,%edx
  mov    -0x38(%rbp),%eax
  cmp    %edx,%eax
  cmovg  %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  sub    %edx,%eax
  imul   -0x24(%rbp),%eax
  add    %eax,-0x14(%rbp)
  negl   -0x24(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x20(%rbp)
L8:
  mov    -0x20(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L23
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint8_t go(int n, int *partials, int end) {
  int step = n * 2;
  int mode = 1;
  int lo = n - 1;
  int hi = lo + n;
  int rhi = hi + n;
  int res = 0;
  int res2 = 0;
  while (rhi < end) {
    int val = partials[hi] - partials[lo];
    int val2 = partials[rhi] - partials[hi];
    res += val;
    res2 -= val2;
    lo += step;
    hi += step;
    rhi += step;
  }
  res += res2;
  while (lo < end) {
    hi = ((lo + n) < (end) ? (lo + n) : (end));
    res += mode * (partials[hi] - partials[lo]);
    mode = -mode;
    lo += step;
  }
  return abs(res) % 10;
}
```