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    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x20(%rbp)
  movl   $0x1,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
L116:
  lea    -0x48(%rbp),%rdx
  lea    -0x4c(%rbp),%rax
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x4c(%rbp),%edx
  mov    -0x48(%rbp),%eax
  cmp    %eax,%edx
  jne    L97
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L55
  jmp    L107
L97:
  movl   $0x0,-0x44(%rbp)
  jmp    L8
L12:
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x44(%rbp),%eax
  cltq
  mov    %edx,-0x40(%rbp,%rax,4)
  addl   $0x1,-0x44(%rbp)
L8:
  cmpl   $0x5,-0x44(%rbp)
  jle    L12
L24:
  mov    -0x38(%rbp),%edx
  mov    -0x4c(%rbp),%eax
  cmp    %eax,%edx
  jne    L18
  mov    -0x34(%rbp),%edx
  mov    -0x48(%rbp),%eax
  cmp    %eax,%edx
  jne    L18
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L68
L18:
  mov    -0x4c(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mov    -0x48(%rbp),%eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  mov    -0x38(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mov    -0x34(%rbp),%eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  divsd  %xmm2,%xmm0
  comisd %xmm1,%xmm0
  jbe    L105
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
L105:
  mov    -0x4c(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mov    -0x48(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  mov    -0x38(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mov    -0x34(%rbp),%eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  divsd  %xmm2,%xmm1
  comisd %xmm1,%xmm0
  jbe    L40
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%eax
  mov    %eax,-0x40(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x3c(%rbp)
L40:
  mov    -0x40(%rbp),%edx
  mov    -0x30(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x38(%rbp)
  mov    -0x3c(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x34(%rbp)
  jmp    L24
L68:
  jmp    L116
L107:
  call   <__stack_chk_fail@plt>
L55:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D37
datatype: string
name: retrieve_data
arguments: data_label: D31
datatype: string
name: retrieve_data
arguments: data_label: D18
datatype: string
name: retrieve_data
arguments: data_label: D0
datatype: string

tool: D37: %d %d

tool: 

tool: D18: L

tool: D0: R

assistant: ```c
int main() {
  int m, n, i, arr[6], init[6] = {0, 1, 1, 1, 1, 0};
  while (1) {
    scanf("%d %d", &m, &n);
    if (!(m ^ n))
      return 0;
    for (i = 0; i < 6; ++i)
      arr[i] = init[i];
    while (1) {
      if (m == arr[2] && n == arr[3]) {
        puts("");
        break;
      }
      if ((double)m / n < (double)arr[2] / arr[3]) {
        printf("L");
        arr[4] = arr[2];
        arr[5] = arr[3];
      }
      if ((double)m / n > (double)arr[2] / arr[3]) {
        printf("R");
        arr[0] = arr[2];
        arr[1] = arr[3];
      }
      arr[2] = arr[0] + arr[4];
      arr[3] = arr[1] + arr[5];
    }
  }
}
```