user: 这段汇编代码用C语言表示是什么样的？

```asm
<m1tr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %ecx,-0x44(%rbp)
  mov    %r8d,-0x48(%rbp)
  mov    %r9d,-0x4c(%rbp)
  shll   -0x40(%rbp)
  shll   -0x44(%rbp)
  shll   -0x48(%rbp)
  shll   -0x4c(%rbp)
  shll   0x18(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L24
L1:
  mov    -0x40(%rbp),%edx
  mov    -0x44(%rbp),%eax
  add    %edx,%eax
  imul   -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0x48(%rbp),%eax
  add    %edx,%eax
  cltd
  idivl  -0x4c(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L9
L43:
  mov    -0x44(%rbp),%eax
  add    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x4c(%rbp),%eax
  jl     L13
  mov    -0x4c(%rbp),%eax
  sub    %eax,-0x18(%rbp)
L13:
  mov    -0x40(%rbp),%eax
  add    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x4c(%rbp),%eax
  jl     L31
  mov    -0x4c(%rbp),%eax
  sub    %eax,-0x14(%rbp)
L31:
  movl   $0x0,-0x1c(%rbp)
  jmp    L65
L51:
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%ecx
  mov    -0x1c(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x14(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%ecx
  mov    -0x1c(%rbp),%eax
  add    %ecx,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x14(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x14(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    0x18(%rbp),%eax
  add    %eax,-0x1c(%rbp)
L65:
  mov    0x10(%rbp),%eax
  imul   0x18(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L51
  addl   $0x1,-0x20(%rbp)
L9:
  mov    -0x3c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jg     L43
  addl   $0x1,-0x24(%rbp)
L24:
  mov    -0x3c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x24(%rbp)
  jl     L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void m1tr(double *a, int n, int nn1, int nn2, int off, int max, int rep, int rstr) {
  int i, j, k, ind1, ind2;
  double tmp1, tmp2;
  nn1 *= 2;
  nn2 *= 2;
  off *= 2;
  max *= 2;
  rstr *= 2;
  for (i = 0; i <= n - 2; i++) {
    ind1 = (i * (nn1 + nn2) + off) % max;
    ind2 = ind1;
    for (j = i + 1; j <= n - 1; j++) {
      ind1 = ind1 + nn2;
      if (ind1 >= max)
        ind1 -= max;
      ind2 = ind2 + nn1;
      if (ind2 >= max)
        ind2 -= max;
      for (k = 0; k < rep * rstr; k += rstr) {
        tmp1 = a[ind1 + k];
        tmp2 = a[ind1 + k + 1];
        a[ind1 + k] = a[ind2 + k];
        a[ind1 + k + 1] = a[ind2 + k + 1];
        a[ind2 + k] = tmp1;
        a[ind2 + k + 1] = tmp2;
      }
    }
  }
}
```