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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <__globinit_sixtimes@plt>
  mov    $0x3,%esi
  mov    $0x14,%edi
  call   <__CrestCall@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <__CrestInt@plt>
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x15,%edi
  call   <__CrestLoad@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <mult@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  mov    %rax,%rsi
  mov    $0x17,%edi
  call   <__CrestHandleReturn@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x16,%edi
  call   <__CrestStore@plt>
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1a,%edi
  call   <__CrestLoad@plt>
  mov    $0x1f,%edx
  mov    $0x0,%esi
  mov    $0x19,%edi
  call   <__CrestLoad@plt>
  mov    -0x14(%rbp),%eax
  cmp    $0x1f,%eax
  sete   %al
  movzbl %al,%eax
  mov    %rax,%rdx
  mov    $0xc,%esi
  mov    $0x18,%edi
  call   <__CrestApply2@plt>
  mov    -0x14(%rbp),%eax
  cmp    $0x1f,%eax
  jne    L12
  mov    $0x1,%edx
  mov    $0x7,%esi
  mov    $0x1b,%edi
  call   <__CrestBranch@plt>
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    $0x1d,%edi
  call   <__CrestLoad@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1e,%edi
  call   <__CrestStore@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L18
L12:
  mov    $0x0,%edx
  mov    $0x8,%esi
  mov    $0x1c,%edi
  call   <__CrestBranch@plt>
  mov    $0x2,%edx
  mov    $0x0,%esi
  mov    $0x1f,%edi
  call   <__CrestLoad@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x20,%edi
  call   <__CrestStore@plt>
  movl   $0x2,-0x10(%rbp)
L18:
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x21,%edi
  call   <__CrestLoad@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x22,%edi
  call   <__CrestStore@plt>
  movl   $0x0,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x23,%edi
  call   <__CrestLoad@plt>
  mov    $0x24,%edi
  call   <__CrestReturn@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int main(void) {
  int a;
  int b;
  int __retres3;
  {
    __globinit_sixtimes();
    __CrestCall(20, 3);
    __CrestInt(&a);
    __CrestLoad(21, (unsigned long)(&a), (long long)a);
    a = mult(a);
    __CrestHandleReturn(23, (long long)a);
    __CrestStore(22, (unsigned long)(&a));
    __CrestLoad(26, (unsigned long)(&a), (long long)a);
    __CrestLoad(25, (unsigned long)0, (long long)31);
    __CrestApply2(24, 12, (long long)(a == 31));
    if (a == 31) {
      __CrestBranch(27, 7, 1);
      __CrestLoad(29, (unsigned long)0, (long long)1);
      __CrestStore(30, (unsigned long)(&b));
      b = 1;
    } else {
      __CrestBranch(28, 8, 0);
      __CrestLoad(31, (unsigned long)0, (long long)2);
      __CrestStore(32, (unsigned long)(&b));
      b = 2;
    }
    __CrestLoad(33, (unsigned long)0, (long long)0);
    __CrestStore(34, (unsigned long)(&__retres3));
    __retres3 = 0;
    __CrestLoad(35, (unsigned long)(&__retres3), (long long)__retres3);
    __CrestReturn(36);
    return (__retres3);
  }
}
```