user: 这段汇编代码用C语言表示是什么样的：

```asm
<reverse_burrows_wheeler_transform>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x36f000(%rsp),%r11
L66:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L66
  sub    $0x6b8,%rsp
  mov    %edi,%r8d
  mov    %rsi,%r9
  mov    %rdx,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x36f6a8(%rsp)
  xor    %eax,%eax
  lea    -0x4(%rdi),%eax
  add    %rsi,%rax
  mov    (%rax),%r11d
  mov    -0x4(%rax),%edi
  mov    %rsp,%rax
  lea    0x404(%rsp),%rcx
  mov    %rax,%rdx
L32:
  movl   $0x0,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L32
  mov    %r8d,%ebx
  sub    $0x8,%ebx
  je     L112
  mov    0x400(%rsp),%ebp
  lea    -0x8(%r8),%esi
  mov    $0x0,%edx
  mov    $0x0,%r12d
  mov    $0x1,%r13d
  jmp    L87
L126:
  movzbl (%r9,%rdx,1),%ecx
  addl   $0x1,(%rsp,%rcx,4)
L28:
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  je     L137
L87:
  cmp    %edx,%r11d
  jne    L126
  add    $0x1,%ebp
  mov    %r13d,%r12d
  jmp    L28
L137:
  test   %r12b,%r12b
  je     L112
  mov    %ebp,0x400(%rsp)
L112:
  lea    0x410(%rsp),%rdx
  lea    0x814(%rsp),%rsi
  mov    $0x0,%ecx
L136:
  mov    %ecx,(%rdx)
  add    (%rax),%ecx
  movl   $0x0,(%rax)
  add    $0x4,%rdx
  add    $0x4,%rax
  cmp    %rsi,%rdx
  jne    L136
  test   %ebx,%ebx
  je     L88
  mov    %ebx,%ebx
  mov    $0x0,%eax
  jmp    L92
L118:
  add    $0x1,%rax
  cmp    %rbx,%rax
  je     L88
L92:
  cmp    %eax,%r11d
  je     L118
  movzbl (%r9,%rax,1),%ecx
  mov    (%rsp,%rcx,4),%edx
  mov    %edx,%esi
  add    0x410(%rsp,%rcx,4),%esi
  mov    %eax,0x820(%rsp,%rsi,4)
  add    $0x1,%edx
  mov    %edx,(%rsp,%rcx,4)
  jmp    L118
L88:
  cmp    $0x9,%r8d
  je     L21
  mov    %r10,%rax
  lea    -0xa(%r8),%edx
  lea    0x1(%r10,%rdx,1),%rcx
L12:
  mov    %edi,%edi
  movzbl (%r9,%rdi,1),%edx
  mov    %dl,(%rax)
  mov    0x820(%rsp,%rdi,4),%edi
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L12
L21:
  mov    0x36f6a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x36f6b8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverse_burrows_wheeler_transform(unsigned int inputBlockLength, unsigned char *inputBlockData, unsigned char *outputBlockData) {
  unsigned int inputDataFrequency[257];
  unsigned int runningTotal[257];
  unsigned int transformationVector[900001];
  unsigned int *temp = (unsigned int *)&inputBlockData[inputBlockLength - 4];
  unsigned int last = *temp;
  unsigned int first = *(temp - 1);
  for (unsigned int i = 0; i < 257; i++) {
    inputDataFrequency[i] = 0;
  }
  for (unsigned int i = 0; i < inputBlockLength - 8; i++) {
    if (i == last) {
      inputDataFrequency[256]++;
    } else {
      inputDataFrequency[inputBlockData[i]]++;
    }
  }
  unsigned int sum = 0;
  for (unsigned int i = 0; i < 257; i++) {
    runningTotal[i] = sum;
    sum += inputDataFrequency[i];
    inputDataFrequency[i] = 0;
  }
  for (unsigned int i = 0; i < inputBlockLength - 8; i++) {
    if (i == last) {
    } else {
      transformationVector[inputDataFrequency[inputBlockData[i]] + runningTotal[inputBlockData[i]]] = i;
      inputDataFrequency[inputBlockData[i]]++;
    }
  }
  unsigned int count = 0;
  unsigned int i = first;
  for (unsigned int j = 0; j < inputBlockLength - 9; j++) {
    outputBlockData[count++] = inputBlockData[i];
    i = transformationVector[i];
  }
}
```