user: 请从给定的汇编指令推导出C源代码:

```asm
<reverse_burrows_wheeler_transform>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x36f000(%rsp),%r11
L34:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L34
  sub    $0x6b0,%rsp
  mov    $0x80,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x36f6a8(%rsp)
  xor    %eax,%eax
  lea    -0x4(%rdi),%eax
  mov    %rdx,%r10
  mov    %rsp,%rdx
  add    %rsi,%rax
  mov    %edi,%r8d
  mov    %rsi,%r9
  mov    %rdx,%rdi
  mov    (%rax),%r11d
  mov    -0x4(%rax),%esi
  xor    %eax,%eax
  mov    %r8d,%ebp
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  sub    $0x8,%ebp
  je     L61
  mov    %ebp,%edi
  xor    %r12d,%r12d
  xor    %ebx,%ebx
  jmp    L113
  nopw   0x0(%rax,%rax,1)
L121:
  movzbl (%r9,%rax,1),%ecx
  add    $0x1,%rax
  addl   $0x1,(%rsp,%rcx,4)
  cmp    %rdi,%rax
  je     L91
L113:
  cmp    %eax,%r11d
  jne    L121
  add    $0x1,%rax
  add    $0x1,%ebx
  mov    $0x1,%r12d
  cmp    %rdi,%rax
  jne    L113
L91:
  test   %r12b,%r12b
  je     L61
  mov    %ebx,0x400(%rsp)
L61:
  mov    %rdx,%rdi
  lea    0x410(%rsp),%rax
  xor    %edx,%edx
  lea    0x814(%rsp),%rcx
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %edx,(%rax)
  add    $0x4,%rax
  add    (%rdi),%edx
  add    $0x4,%rdi
  movl   $0x0,-0x4(%rdi)
  cmp    %rcx,%rax
  jne    L20
  mov    %ebp,%ebx
  xor    %eax,%eax
  test   %ebp,%ebp
  je     L1
L99:
  cmp    %eax,%r11d
  je     L73
  movzbl (%r9,%rax,1),%ecx
  mov    (%rsp,%rcx,4),%edx
  mov    %edx,%edi
  add    $0x1,%edx
  add    0x410(%rsp,%rcx,4),%edi
  mov    %eax,0x820(%rsp,%rdi,4)
  mov    %edx,(%rsp,%rcx,4)
L73:
  add    $0x1,%rax
  cmp    %rbx,%rax
  jne    L99
L1:
  cmp    $0x9,%r8d
  je     L95
  lea    -0xa(%r8),%edx
  mov    %r10,%rax
  lea    0x1(%r10,%rdx,1),%rcx
  xchg   %ax,%ax
L9:
  movzbl (%r9,%rsi,1),%edx
  add    $0x1,%rax
  mov    0x820(%rsp,%rsi,4),%esi
  mov    %dl,-0x1(%rax)
  cmp    %rax,%rcx
  jne    L9
L95:
  mov    0x36f6a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L118
  add    $0x36f6b0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L118:
  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];
  }
}
```