user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<m1tr>:
  endbr64
  push   %r15
  mov    %r8d,%eax
  mov    %rdi,%r10
  mov    %r9d,%r8d
  push   %r14
  lea    -0x1(%rsi),%edi
  add    %eax,%eax
  add    %r8d,%r8d
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    0x40(%rsp),%edx
  mov    %edi,-0x14(%rsp)
  test   %edi,%edi
  jle    L73
  mov    0x38(%rsp),%edi
  lea    (%r12,%r12,1),%r9d
  lea    (%rcx,%rcx,1),%ebx
  mov    %esi,%r11d
  lea    (%rdx,%rdx,1),%r14d
  lea    (%r9,%rbx,1),%ebp
  mov    %eax,%r13d
  xor    %r12d,%r12d
  imul   %r14d,%edi
  cmp    $0x1,%edx
  jne    L98
  nopl   0x0(%rax,%rax,1)
L85:
  mov    %r13d,%eax
  add    $0x1,%r12d
  cltd
  idiv   %r8d
  cmp    %r12d,%r11d
  jle    L97
  mov    %edx,%r14d
  mov    %r12d,%r15d
  cs nopw 0x0(%rax,%rax,1)
L69:
  add    %ebx,%edx
  mov    %edx,%eax
  sub    %r8d,%eax
  cmp    %edx,%r8d
  cmovle %eax,%edx
  add    %r9d,%r14d
  mov    %r14d,%eax
  sub    %r8d,%eax
  cmp    %r14d,%r8d
  cmovle %eax,%r14d
  test   %edi,%edi
  jle    L14
  movslq %edx,%rax
  xor    %esi,%esi
  lea    (%r10,%rax,8),%rcx
  movslq %r14d,%rax
  lea    (%r10,%rax,8),%rax
  nopl   0x0(%rax)
L50:
  movsd  (%rax),%xmm1
  movupd (%rcx),%xmm0
  add    $0x2,%esi
  add    $0x10,%rcx
  add    $0x10,%rax
  movsd  %xmm1,-0x10(%rcx)
  movsd  -0x8(%rax),%xmm1
  movsd  %xmm1,-0x8(%rcx)
  movups %xmm0,-0x10(%rax)
  cmp    %esi,%edi
  jg     L50
L14:
  add    $0x1,%r15d
  cmp    %r15d,%r11d
  jne    L69
L97:
  add    %ebp,%r13d
  cmp    %r12d,-0x14(%rsp)
  jne    L85
L73:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L98:
  mov    %eax,-0x10(%rsp)
  movslq %r14d,%r12
  xor    %eax,%eax
  mov    %ebp,-0xc(%rsp)
  shl    $0x3,%r12
  mov    %eax,%ebp
L24:
  mov    -0x10(%rsp),%eax
  add    $0x1,%ebp
  mov    %ebp,%r15d
  cltd
  idiv   %r8d
  mov    %edx,%r13d
  cmp    %ebp,%r11d
  jle    L30
  nopl   (%rax)
L101:
  add    %ebx,%edx
  mov    %edx,%eax
  sub    %r8d,%eax
  cmp    %edx,%r8d
  cmovle %eax,%edx
  add    %r9d,%r13d
  mov    %r13d,%eax
  sub    %r8d,%eax
  cmp    %r13d,%r8d
  cmovle %eax,%r13d
  test   %edi,%edi
  jle    L66
  movslq %edx,%rax
  xor    %esi,%esi
  lea    (%r10,%rax,8),%rcx
  movslq %r13d,%rax
  lea    (%r10,%rax,8),%rax
  nopl   0x0(%rax)
L105:
  movsd  (%rax),%xmm1
  movupd (%rcx),%xmm0
  add    %r14d,%esi
  movsd  %xmm1,(%rcx)
  movsd  0x8(%rax),%xmm1
  movsd  %xmm1,0x8(%rcx)
  add    %r12,%rcx
  movups %xmm0,(%rax)
  add    %r12,%rax
  cmp    %edi,%esi
  jl     L105
L66:
  add    $0x1,%r15d
  cmp    %r15d,%r11d
  jne    L101
L30:
  mov    -0xc(%rsp),%edx
  add    %edx,-0x10(%rsp)
  cmp    -0x14(%rsp),%ebp
  jne    L24
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  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;
      }
    }
  }
}
```