user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<LPC_inverse_pred_gain_QA>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  and    $0x1,%r15d
  lea    -0x1(%rdx),%r14d
  push   %r13
  shl    $0x6,%r15
  push   %r12
  add    %rsi,%r15
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  movl   $0x40000000,(%rdi)
  mov    %rsi,0x18(%rsp)
  test   %r14d,%r14d
  jle    L57
  movslq %r14d,%r8
  lea    0x0(,%r8,4),%rbp
  cs nopw 0x0(%rax,%rax,1)
L48:
  lea    (%r15,%rbp,1),%rdx
  mov    (%rdx),%r13d
  lea    0xfff0(%r13),%eax
  cmp    $0x1ffe0,%eax
  ja     L43
  shl    $0xf,%r13d
  mov    $0x3fffffff,%r12d
  mov    $0x2e,%esi
  mov    %rdx,0x10(%rsp)
  neg    %r13d
  movslq %r13d,%r13
  mov    %r13,%rax
  imul   %r13,%rax
  sar    $0x20,%rax
  sub    %eax,%r12d
  mov    %r12d,%edi
  call   <SKP_INVERSE32_varQ@plt>
  movslq %r12d,%rcx
  mov    %r14d,%esi
  mov    %eax,%r9d
  movslq (%rbx),%rax
  and    $0x1,%esi
  mov    %r9d,%edi
  shl    $0x6,%rsi
  mov    %r9d,0xc(%rsp)
  imul   %rcx,%rax
  sar    $0x20,%rax
  shl    $0x2,%eax
  mov    %eax,(%rbx)
  mov    0x18(%rsp),%rax
  lea    (%rax,%rsi,1),%r12
  call   <SKP_Silk_CLZ32@plt>
  mov    0xc(%rsp),%r9d
  mov    0x10(%rsp),%rdi
  xor    %edx,%edx
  sub    $0x1,%eax
  mov    %eax,%ecx
  shl    %cl,%r9d
  mov    $0x10,%ecx
  movslq %r9d,%r9
  sub    %eax,%ecx
  nopl   (%rax)
L12:
  movslq -0x4(%rdi),%r10
  mov    (%r15,%rdx,4),%eax
  sub    $0x4,%rdi
  imul   %r13,%r10
  sar    $0x20,%r10
  add    %r10d,%r10d
  sub    %r10d,%eax
  cltq
  imul   %r9,%rax
  sar    $0x20,%rax
  shl    %cl,%eax
  mov    %eax,(%r12,%rdx,4)
  add    $0x1,%rdx
  cmp    %edx,%r14d
  jg     L12
  sub    $0x4,%rbp
  sub    $0x1,%r14d
  je     L18
  mov    %r12,%r15
  jmp    L48
L57:
  mov    %r15,%r12
L18:
  mov    (%r12),%eax
  lea    0xfff0(%rax),%edx
  cmp    $0x1ffe0,%edx
  jbe    L20
L43:
  add    $0x28,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  shl    $0xf,%eax
  mov    $0x3fffffff,%edx
  neg    %eax
  cltq
  imul   %rax,%rax
  sar    $0x20,%rax
  sub    %eax,%edx
  movslq %edx,%rax
  movslq (%rbx),%rdx
  imul   %rdx,%rax
  sar    $0x20,%rax
  shl    $0x2,%eax
  mov    %eax,(%rbx)
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int LPC_inverse_pred_gain_QA(int *invGain_Q30, int A_QA[2][16], const int order) {
  int k, n, headrm;
  int rc_Q31, rc_mult1_Q30, rc_mult2_Q16, tmp_QA;
  int *Aold_QA, *Anew_QA;
  Anew_QA = A_QA[order & 1];
  *invGain_Q30 = (1 << 30);
  for (k = order - 1; k > 0; k--) {
    if ((Anew_QA[k] > ((int)((0.99975) * ((long long)1 << (16)) + 0.5))) || (Anew_QA[k] < -((int)((0.99975) * ((long long)1 << (16)) + 0.5)))) {
      return 1;
    }
    rc_Q31 = -((Anew_QA[k]) << (31 - 16));
    rc_mult1_Q30 = (0x7FFFFFFF >> 1) - (int)((((long long)((rc_Q31)) * ((rc_Q31)))) >> (32));
    ;
    ;
    rc_mult2_Q16 = SKP_INVERSE32_varQ(rc_mult1_Q30, 46);
    *invGain_Q30 = (((int)((((long long)((*invGain_Q30)) * ((rc_mult1_Q30)))) >> (32))) << (2));
    ;
    ;
    Aold_QA = Anew_QA;
    Anew_QA = A_QA[k & 1];
    headrm = SKP_Silk_CLZ32(rc_mult2_Q16) - 1;
    rc_mult2_Q16 = ((rc_mult2_Q16) << (headrm));
    for (n = 0; n < k; n++) {
      tmp_QA = Aold_QA[n] - (((int)((((long long)((Aold_QA[k - n - 1])) * ((rc_Q31)))) >> (32))) << (1));
      Anew_QA[n] = (((int)((((long long)((tmp_QA)) * ((rc_mult2_Q16)))) >> (32))) << (16 - headrm));
    }
  }
  if ((Anew_QA[0] > ((int)((0.99975) * ((long long)1 << (16)) + 0.5))) || (Anew_QA[0] < -((int)((0.99975) * ((long long)1 << (16)) + 0.5)))) {
    return 1;
  }
  rc_Q31 = -((Anew_QA[0]) << (31 - 16));
  rc_mult1_Q30 = (0x7FFFFFFF >> 1) - (int)((((long long)((rc_Q31)) * ((rc_Q31)))) >> (32));
  *invGain_Q30 = (((int)((((long long)((*invGain_Q30)) * ((rc_mult1_Q30)))) >> (32))) << (2));
  ;
  ;
  return 0;
}
```