user: 这段汇编代码用C语言表示是什么样的？

```asm
<solve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movsd  (%rax),%xmm1
  divsd  %xmm1,%xmm0
  mov    -0x20(%rbp),%rax
  movsd  %xmm0,(%rax)
  movl   $0x1,-0x10(%rbp)
  jmp    L61
L14:
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L9
L11:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L9:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L11
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  subsd  -0x8(%rbp),%xmm0
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x10(%rbp)
L61:
  mov    -0x10(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L14
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  sub    $0x8,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L18
L32:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L34
L7:
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  %xmm2,%xmm1
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L34:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L7
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  sub    $0x8,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  subl   $0x1,-0x10(%rbp)
L18:
  cmpl   $0x0,-0x10(%rbp)
  jg     L32
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int solve(double **A, double *z, int n) {
  int i, k;
  double sum;
  z[0] /= A[0][0];
  for (i = 1; i < n; i++) {
    sum = 0.0;
    for (k = 0; k < i; k++)
      sum += A[k][i] * z[k];
    z[i] = (z[i] - sum) / A[i][i];
  }
  z[n - 1] /= A[n - 1][n - 1];
  for (i = n - 1; i >= 1; i--) {
    for (k = i; k < n; k++)
      z[i - 1] -= A[i - 1][k] * z[k];
    z[i - 1] /= A[i - 1][i - 1];
  }
  return 0;
}
```