user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_register_a_value_for_crazy>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  mov    %edi,%esi
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x1,(%rsp)
  movabs $0x100000000,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x200000000,%rax
  mov    %rax,0x10(%rsp)
  add    $0x2,%rax
  mov    %rax,0x18(%rsp)
  movl   $0x1,0x20(%rsp)
  call   <crazy@plt>
  xor    %r9d,%r9d
  cmp    %ebp,%eax
  je     L150
  mov    %ebx,%esi
  mov    $0x7354,%edi
  call   <crazy@plt>
  mov    $0x7354,%r9d
  cmp    %ebp,%eax
  je     L150
  mov    %ebx,%esi
  mov    $0xe6a8,%edi
  call   <crazy@plt>
  mov    $0xe6a8,%r9d
  cmp    %ebp,%eax
  je     L150
  mov    %ebx,%esi
  mov    $0xe6a7,%edi
  call   <crazy@plt>
  mov    $0xe6a7,%r9d
  cmp    %ebp,%eax
  je     L150
  mov    %ebx,%esi
  mov    $0xe6a6,%edi
  call   <crazy@plt>
  cmp    %ebp,%eax
  je     L152
  mov    $0xa,%r11d
  mov    $0x1,%r10d
  xor    %r9d,%r9d
  mov    $0xaaaaaaab,%ecx
  jmp    L113
  cs nopw 0x0(%rax,%rax,1)
L20:
  imul   %rcx,%rbp
  imul   %rcx,%r8
  imul   %rcx,%rdi
  imul   %r10d,%eax
  shr    $0x21,%rbp
  lea    (%r10,%r10,2),%r10d
  shr    $0x21,%r8
  mov    %rbp,%rbx
  shr    $0x21,%rdi
  mov    %r8,%rbp
  add    %eax,%r9d
  mov    %rdi,%r12
  sub    $0x1,%r11d
  je     L150
L113:
  mov    %ebp,%r8d
  mov    %r12d,%edi
  mov    %r8,%rdx
  imul   %rcx,%rdx
  shr    $0x21,%rdx
  lea    (%rdx,%rdx,2),%eax
  mov    %ebp,%edx
  mov    %ebx,%ebp
  sub    %eax,%edx
  mov    %rdi,%rax
  imul   %rcx,%rax
  shr    $0x21,%rax
  lea    (%rax,%rax,2),%esi
  mov    %r12d,%eax
  sub    %esi,%eax
  mov    %rbp,%rsi
  imul   %rcx,%rsi
  shr    $0x21,%rsi
  lea    (%rsi,%rsi,2),%esi
  sub    %esi,%ebx
  lea    (%rbx,%rbx,2),%esi
  lea    (%rsi,%rax,1),%ebx
  cmp    %edx,(%rsp,%rbx,4)
  je     L20
  mov    %esi,%eax
  cmp    %edx,(%rsp,%rax,4)
  je     L125
  lea    0x1(%rsi),%eax
  cmp    %edx,(%rsp,%rax,4)
  je     L80
  lea    0x2(%rsi),%eax
  cmp    %edx,(%rsp,%rax,4)
  je     L156
  mov    $0xffff,%r9d
L150:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x30,%rsp
  mov    %r9d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L125:
  xor    %eax,%eax
  jmp    L20
  nop
L152:
  mov    $0xe6a6,%r9d
  jmp    L150
  nopl   0x0(%rax,%rax,1)
L80:
  mov    $0x1,%eax
  jmp    L20
  nopw   0x0(%rax,%rax,1)
L156:
  mov    $0x2,%eax
  jmp    L20
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int get_register_a_value_for_crazy(unsigned int source, unsigned int destination, unsigned int current_register_a_value) {
  unsigned int crz[] = {1, 0, 0, 1, 0, 2, 2, 2, 1};
  unsigned int result = 0;
  unsigned int position = 0;
  unsigned int shift = 1;
  if (crazy(0, source) == destination)
    return 0;
  if (crazy(29524, source) == destination)
    return 29524;
  if (crazy(59048, source) == destination)
    return 59048;
  if (crazy(59048 - 1, source) == destination)
    return 59048 - 1;
  if (crazy(59048 - 2, source) == destination)
    return 59048 - 2;
  while (position < 10) {
    unsigned int k;
    unsigned int i = source % 3;
    unsigned int j = destination % 3;
    unsigned int cv = current_register_a_value % 3;
    unsigned int solution = 4;
    if (crz[cv + 3 * i] == j) {
      solution = cv;
    } else {
      for (k = 0; k < 3; k++) {
        if (crz[k + 3 * i] == j) {
          solution = k;
          break;
        }
      }
    }
    if (solution > 2)
      return 0xFFFF;
    result += solution * shift;
    shift *= 3;
    position++;
    source /= 3;
    destination /= 3;
    current_register_a_value /= 3;
  }
  return result;
}
```