user: Translate the below assembly code into C programming language:

```asm
<queens>:
  endbr64
  push   %rbp
  lea    (%rdi,%rdi,1),%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x988,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x978(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L26
  movslq %edi,%rdx
  xor    %esi,%esi
  lea    0x194(%rsp),%rdi
  shl    $0x2,%rdx
  call   <memset@plt>
L79:
  lea    -0x2(%rbp),%eax
  cmp    $0x1,%ebp
  lea    0x658(%rsp),%rdi
  lea    0x4(,%rax,4),%rdx
  mov    $0x4,%eax
  cmovle %rax,%rdx
  xor    %esi,%esi
  call   <memset@plt>
  mov    $0x1,%eax
  sub    %ebx,%eax
  cmp    %eax,%ebx
  jle    L49
  lea    -0x1(%rbx),%edx
  xor    %esi,%esi
  sub    %eax,%edx
  cltq
  lea    0x4(,%rdx,4),%rdx
  lea    0x4bc(%rsp,%rax,4),%rdi
  call   <memset@plt>
L49:
  mov    %rsp,%rsi
  xor    %edx,%edx
  lea    0x4bc(%rsp),%rdi
  xor    %r8d,%r8d
L8:
  lea    0x1(%rdx),%ecx
  cmp    %ecx,%ebx
  jge    L81
  add    $0x1,%r8d
L64:
  test   %edx,%edx
  je     L119
  movslq -0x4(%rsi),%rcx
  sub    $0x4,%rsi
  movl   $0x0,0x190(%rsp,%rcx,4)
  mov    %rcx,%rax
  lea    (%rcx,%rdx,1),%ecx
  movslq %ecx,%rcx
  movl   $0x0,0x650(%rsp,%rcx,4)
  mov    %eax,%ecx
  sub    %edx,%ecx
  movslq %ecx,%rcx
  movl   $0x0,0x4bc(%rsp,%rcx,4)
  mov    %edx,%ecx
L85:
  add    $0x1,%eax
  cmp    %eax,%ebx
  jge    L126
  lea    -0x1(%rcx),%edx
  jmp    L64
  nopl   0x0(%rax)
L81:
  mov    $0x1,%eax
L126:
  movslq %eax,%r9
  mov    0x190(%rsp,%r9,4),%edx
  test   %edx,%edx
  jne    L85
  lea    (%rcx,%rax,1),%edx
  movslq %edx,%rdx
  mov    0x650(%rsp,%rdx,4),%ebp
  test   %ebp,%ebp
  jne    L85
  mov    %eax,%r10d
  sub    %ecx,%r10d
  movslq %r10d,%r10
  lea    (%rdi,%r10,4),%r10
  mov    (%r10),%r11d
  test   %r11d,%r11d
  jne    L85
  movl   $0x1,0x650(%rsp,%rdx,4)
  add    $0x4,%rsi
  mov    %ecx,%edx
  mov    %eax,-0x4(%rsi)
  movl   $0x1,0x190(%rsp,%r9,4)
  movl   $0x1,(%r10)
  jmp    L8
L119:
  mov    0x978(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L96
  add    $0x988,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L26:
  cmp    $0x1,%ebp
  jle    L49
  jmp    L79
L96:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int queens(int n) {
  int dtab[100];
  int freetab1[100 + 1];
  int freetab2[2 * 100 + 1];
  int freetab3a[2 * 100 - 1];
  {
    int i;
    for (i = 1; i <= n; i++) {
      freetab1[i] = 0;
    }
  }
  {
    int i;
    for (i = 2; i <= 2 * n; i++) {
      freetab2[i] = 0;
    }
  }
  {
    int i;
    for (i = -(n - 1); i < n; i++) {
      (&freetab3a[100 - 1])[i] = 0;
    }
  }
  {
    int counter = 0;
    int i = 0;
    int *Dptr = &dtab[0];
  entry:
    i++;
    if (i > n) {
      counter++;
    } else {
      int j;
      for (j = 1; j <= n; j++) {
        if (freetab1[j] == 0 && freetab2[j + i] == 0 && (&freetab3a[100 - 1])[j - i] == 0) {
          freetab1[j] = 1;
          freetab2[j + i] = 1;
          (&freetab3a[100 - 1])[j - i] = 1;
          *Dptr++ = j;
          goto entry;
        comeback:
          j = *--Dptr;
          freetab1[j] = 0;
          freetab2[j + i] = 0;
          (&freetab3a[100 - 1])[j - i] = 0;
        }
      }
    }
    i--;
    if (i > 0)
      goto comeback;
    return counter;
  }
}
```