user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<euler_problem_5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x9d8,-0x4(%rbp)
L62:
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x2e8ba2e9,%rax,%rax
  shr    $0x20,%rax
  sar    %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
  add    %edx,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  jne    L85
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x2aaaaaab,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  jne    L85
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x4ec4ec4f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  jne    L85
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0xe,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  test   %edx,%edx
  jne    L85
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0xffffffff88888889,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x3,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x4,%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L85
  mov    -0x4(%rbp),%eax
  and    $0xf,%eax
  test   %eax,%eax
  jne    L85
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x78787879,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x4,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L85
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x38e38e39,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %ecx,%edx
  sub    %eax,%edx
  test   %edx,%edx
  jne    L85
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x6bca1af3,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  add    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  mov    %ecx,%edx
  sub    %eax,%edx
  test   %edx,%edx
  jne    L85
  mov    -0x4(%rbp),%eax
  jmp    L104
L85:
  addl   $0x14,-0x4(%rbp)
  jmp    L62
L104:
  pop    %rbp
  ret
```

assistant: ```c
int euler_problem_5() {
  int i;
  for (i = 2520;; i += 20) {
    if (i % 11 == 0 && i % 12 == 0 && i % 13 == 0 && i % 14 == 0 && i % 15 == 0 && i % 16 == 0 && i % 17 == 0 && i % 18 == 0 && i % 19 == 0) {
      return i;
    }
  }
}
```