user: 这段汇编代码用C语言表示是什么样的？

```asm
<rfftf1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,0x18(%rsp)
  mov    0x4(%r8),%eax
  mov    %eax,0x1c(%rsp)
  test   %eax,%eax
  jle    L110
  mov    %rcx,%r13
  lea    -0x1(%rdi),%ebx
  cltq
  lea    (%r8,%rax,4),%r12
  mov    $0x1,%r8d
  mov    $0x1,%ebp
  mov    %rsi,0x8(%rsp)
  mov    %rdx,0x10(%rsp)
  mov    %edi,%r14d
  jmp    L173
L36:
  mov    0x8(%rsp),%rcx
  mov    0x10(%rsp),%rdx
  jmp    L161
L171:
  cmp    $0x5,%esi
  jne    L116
  lea    (%rdi,%rbx,1),%eax
  lea    (%rdi,%rax,1),%esi
  cltq
  movslq %ebx,%r8
  lea    (%rdi,%rsi,1),%r9d
  movslq %r9d,%r9
  lea    0x0(%r13,%r9,8),%r9
  push   %r9
  movslq %esi,%rsi
  lea    0x0(%r13,%rsi,8),%rsi
  push   %rsi
  lea    0x0(%r13,%rax,8),%r9
  lea    0x0(%r13,%r8,8),%r8
  mov    %r10d,%esi
  call   <radf5@plt>
  add    $0x10,%rsp
  mov    %r15d,%r8d
  jmp    L15
L99:
  lea    (%rdi,%rbx,1),%eax
  movslq %eax,%r8
  movslq %ebx,%rsi
  sub    $0x8,%rsp
  add    %edi,%eax
  cltq
  lea    0x0(%r13,%rax,8),%rax
  push   %rax
  lea    0x0(%r13,%r8,8),%r9
  lea    0x0(%r13,%rsi,8),%r8
  mov    %r10d,%esi
  call   <radf4@plt>
  add    $0x10,%rsp
  mov    %r15d,%r8d
L15:
  add    $0x1,%ebp
  sub    $0x4,%r12
  cmp    %ebp,0x1c(%rsp)
  jl     L160
L173:
  mov    0x4(%r12),%esi
  mov    %r14d,%ecx
  mov    %r14d,%eax
  cltd
  idiv   %esi
  mov    %eax,%r10d
  mov    %eax,%r14d
  mov    0x18(%rsp),%eax
  cltd
  idiv   %ecx
  mov    %eax,%edi
  lea    -0x1(%rsi),%eax
  imul   %edi,%eax
  sub    %eax,%ebx
  test   %r8d,%r8d
  sete   %r15b
  movzbl %r15b,%r15d
  je     L36
  mov    0x10(%rsp),%rcx
  mov    0x8(%rsp),%rdx
L161:
  cmp    $0x4,%esi
  je     L99
  jg     L171
  cmp    $0x2,%esi
  je     L131
  cmp    $0x3,%esi
  jne    L116
  lea    (%rdi,%rbx,1),%esi
  movslq %esi,%rsi
  movslq %ebx,%rax
  lea    0x0(%r13,%rsi,8),%r9
  lea    0x0(%r13,%rax,8),%r8
  mov    %r10d,%esi
  call   <radf3@plt>
  mov    %r15d,%r8d
  jmp    L15
L131:
  movslq %ebx,%rax
  lea    0x0(%r13,%rax,8),%r8
  mov    %r10d,%esi
  call   <radf2@plt>
  mov    %r15d,%r8d
  jmp    L15
L116:
  mov    %r10d,%ecx
  imul   %edi,%ecx
  cmp    $0x1,%edi
  je     L197
  test   %r8d,%r8d
  jne    L102
L41:
  sub    $0x8,%rsp
  movslq %ebx,%rax
  lea    0x0(%r13,%rax,8),%rax
  push   %rax
  mov    0x18(%rsp),%r9
  mov    0x20(%rsp),%r8
  mov    %r10d,%edx
  call   <radfg@plt>
  add    $0x10,%rsp
  mov    $0x0,%r8d
  jmp    L15
L197:
  test   %r8d,%r8d
  jne    L41
L102:
  sub    $0x8,%rsp
  movslq %ebx,%rax
  lea    0x0(%r13,%rax,8),%rax
  push   %rax
  mov    0x20(%rsp),%r9
  mov    0x18(%rsp),%r8
  mov    %r10d,%edx
  call   <radfg@plt>
  add    $0x10,%rsp
  mov    $0x1,%r8d
  jmp    L15
L160:
  mov    0x8(%rsp),%r15
  mov    0x10(%rsp),%r14
  cmp    $0x1,%r8d
  je     L110
  mov    0x18(%rsp),%eax
  test   %eax,%eax
  jle    L110
  mov    %eax,%edx
  mov    $0x0,%eax
L133:
  movsd  (%r14,%rax,8),%xmm0
  movsd  %xmm0,(%r15,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L133
L110:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void rfftf1(int n, double c[], double ch[], const double wa[], const int ifac[13 + 2]) {
  int i;
  int k1, l1, l2, na, kh, nf, ip, iw, ix2, ix3, ix4, ido, idl1;
  double *cinput, *coutput;
  nf = ifac[1];
  na = 1;
  l2 = n;
  iw = n - 1;
  for (k1 = 1; k1 <= nf; ++k1) {
    kh = nf - k1;
    ip = ifac[kh + 2];
    l1 = l2 / ip;
    ido = n / l2;
    idl1 = ido * l1;
    iw -= (ip - 1) * ido;
    na = !na;
    if (na) {
      cinput = ch;
      coutput = c;
    } else {
      cinput = c;
      coutput = ch;
    }
    switch (ip) {
    case 4:
      ix2 = iw + ido;
      ix3 = ix2 + ido;
      radf4(ido, l1, cinput, coutput, &wa[iw], &wa[ix2], &wa[ix3]);
      break;
    case 2:
      radf2(ido, l1, cinput, coutput, &wa[iw]);
      break;
    case 3:
      ix2 = iw + ido;
      radf3(ido, l1, cinput, coutput, &wa[iw], &wa[ix2]);
      break;
    case 5:
      ix2 = iw + ido;
      ix3 = ix2 + ido;
      ix4 = ix3 + ido;
      radf5(ido, l1, cinput, coutput, &wa[iw], &wa[ix2], &wa[ix3], &wa[ix4]);
      break;
    default:
      if (ido == 1)
        na = !na;
      if (na == 0) {
        radfg(ido, ip, l1, idl1, c, ch, &wa[iw]);
        na = 1;
      } else {
        radfg(ido, ip, l1, idl1, ch, c, &wa[iw]);
        na = 0;
      }
    }
    l2 = l1;
  }
  if (na == 1)
    return;
  for (i = 0; i < n; i++)
    c[i] = ch[i];
}
```