user: 这段汇编代码用C语言表示是什么样的:

```asm
<backSubstitution>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x28,%rsp
  mov    %rsi,0x8(%rsp)
  call   <zeros@plt>
  mov    %rax,%r10
  lea    -0x1(%rbx),%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jle    L117
  mov    0x8(%rsp),%rax
  mov    %r10,0x18(%rsp)
  movslq %ebx,%r12
  mov    %ebx,%r11d
  movl   $0x0,(%rsp)
  mov    0x4(%rsp),%r14d
  xor    %r8d,%r8d
  xor    %r13d,%r13d
  add    $0x8,%rax
  lea    0x0(,%r12,8),%r9
  mov    %rax,0x10(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L108:
  add    $0x1,%r13d
  cmp    %r13d,%ebx
  jle    L114
  movslq (%rsp),%rcx
  mov    0x8(%rsp),%rdi
  lea    0x1(%r8),%r15
  mov    0x10(%rsp),%rsi
  lea    (%rcx,%r8,1),%rax
  lea    (%rdi,%rax,8),%r10
  movslq %r11d,%rax
  lea    (%rax,%r8,1),%rdx
  sub    %rax,%rcx
  lea    (%rdi,%rdx,8),%rdi
  mov    %r14d,%edx
  add    %rax,%rdx
  add    %r8,%rdx
  lea    (%rsi,%rdx,8),%rdx
  mov    %r15,%rsi
  nopl   (%rax)
L113:
  movsd  (%rdi),%xmm0
  movsd  (%r10),%xmm1
  movapd %xmm0,%xmm2
  divsd  %xmm1,%xmm2
  cmp    %r8d,%ebx
  jle    L54
  mov    %rdi,%rax
  jmp    L65
  nopl   0x0(%rax,%rax,1)
L119:
  movsd  (%rax),%xmm0
  movsd  (%rax,%rcx,8),%xmm1
L65:
  mulsd  %xmm2,%xmm1
  add    $0x8,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rax)
  cmp    %rdx,%rax
  jne    L119
L54:
  mulsd  0x0(%rbp,%r8,8),%xmm2
  add    %r9,%rdi
  add    %r9,%rdx
  sub    %r12,%rcx
  movsd  0x0(%rbp,%rsi,8),%xmm0
  subsd  %xmm2,%xmm0
  movsd  %xmm0,0x0(%rbp,%rsi,8)
  add    $0x1,%rsi
  cmp    %esi,%ebx
  jg     L113
L25:
  add    %ebx,(%rsp)
  mov    %r15,%r8
  sub    $0x1,%r14d
  add    %ebx,%r11d
  cmp    0x4(%rsp),%r13d
  jne    L108
  mov    0x18(%rsp),%r10
L107:
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rsi
  lea    0x0(,%r12,8),%rdi
  movsd  -0x8(%rbp,%rdi,1),%xmm0
  imul   %ebx,%eax
  movslq %eax,%rdx
  add    %r12,%rdx
  divsd  -0x8(%rsi,%rdx,8),%xmm0
  mov    %ebx,%esi
  movsd  %xmm0,-0x8(%r10,%rdi,1)
  sub    $0x2,%esi
  js     L91
  mov    %ebx,%edx
  sub    %ebx,%eax
  mov    0x8(%rsp),%r11
  movslq %esi,%rsi
  neg    %edx
  cltq
  mov    0x4(%rsp),%ecx
  movslq %edx,%rdx
  lea    -0x8(,%rdx,8),%r9
  lea    (%rax,%rsi,1),%rdx
  shl    $0x3,%rsi
  lea    (%r11,%rdx,8),%r8
  lea    (%r11,%rax,8),%rdx
  nop
L93:
  movsd  0x0(%rbp,%rsi,1),%xmm1
  cmp    %ecx,%ebx
  jle    L11
  movslq %ecx,%rax
  nopl   (%rax)
L94:
  movsd  (%rdx,%rax,8),%xmm0
  mulsd  (%r10,%rax,8),%xmm0
  add    $0x1,%rax
  subsd  %xmm0,%xmm1
  cmp    %eax,%ebx
  jg     L94
L11:
  divsd  (%r8),%xmm1
  sub    %rdi,%rdx
  add    %r9,%r8
  movsd  %xmm1,(%r10,%rsi,1)
  sub    $0x8,%rsi
  sub    $0x1,%ecx
  jne    L93
L91:
  add    $0x28,%rsp
  mov    %r10,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L114:
  lea    0x1(%r8),%r15
  jmp    L25
L117:
  movslq %ebx,%r12
  jmp    L107
```

assistant: ```c
double *backSubstitution(int n, double *matrix, double b[]) {
  int i, j, k;
  double aux, temp;
  double *x = zeros(n);
  for (i = 0; i < (n - 1); i++) {
    for (j = (i + 1); j < n; j++) {
      aux = *(matrix + n * j + i) / *(matrix + n * i + i);
      for (k = i; k < n; k++) {
        *(matrix + n * j + k) -= (aux * (*(matrix + n * i + k)));
      }
      b[j] -= (aux * b[i]);
    }
  }
  x[n - 1] = b[n - 1] / *(matrix + n * (n - 1) + n - 1);
  for (i = (n - 2); i >= 0; i--) {
    temp = b[i];
    for (j = (i + 1); j < n; j++) {
      temp -= (*(matrix + n * i + j) * x[j]);
    }
    x[i] = temp / *(matrix + n * i + i);
  }
  return x;
}
```