user: 请从给定的汇编指令推导出C源代码：

```asm
<nChancellors>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <findnextSlot@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L112
L14:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  js     L81
L16:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <getPosition@plt>
  cmp    %eax,-0x1c(%rbp)
  jle    L81
  mov    -0x4(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <checkConflict@plt>
  test   %rax,%rax
  jne    L16
L81:
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <getPosition@plt>
  cmp    %eax,-0x1c(%rbp)
  jle    L42
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  js     L42
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jge    L64
  mov    -0x4(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <findnextSlot@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L112
L64:
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <printBoard@plt>
  jmp    L112
L42:
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jns    L90
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jne    L90
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <printBoard@plt>
L90:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jle    L110
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
L110:
  subl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jns    L112
  subl   $0x1,-0x4(%rbp)
L112:
  cmpl   $0x0,-0x4(%rbp)
  jns    L14
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int nChancellors(int *board, int n) {
  int solutions = 0;
  int x = findnextSlot(board, n, 0);
  while (x >= 0) {
    if (board[x] >= 0)
      do
        board[x]++;
      while (getPosition(board, x) < n && checkConflict(board, n, x));
    if (getPosition(board, x) < n && board[x] >= 0) {
      if (x < n - 1)
        x = findnextSlot(board, n, x);
      else
        printBoard(board, n, ++solutions);
    } else {
      if (board[n - 1] < 0 && x == n - 1)
        printBoard(board, n, ++solutions);
      if (board[x] > 0)
        board[x] = 0;
      if (board[--x] < 0)
        x--;
    }
  }
  return solutions;
}
```