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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0xe8(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0xe4(%rbp)
  jmp    L5
L89:
  mov    -0xe4(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xd0(%rbp),%rdx
  mov    -0xe4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0xe4(%rbp)
L5:
  mov    -0xe8(%rbp),%eax
  cmp    %eax,-0xe4(%rbp)
  jl     L89
  movl   $0x0,-0xe0(%rbp)
  jmp    L38
L118:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0xe0(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xdc(%rbp)
  jmp    L75
L30:
  mov    -0xe0(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%edx
  mov    -0xdc(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jne    L77
  mov    -0xdc(%rbp),%eax
  mov    %eax,-0xd8(%rbp)
  jmp    L96
L73:
  mov    -0xd8(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%edx
  mov    -0xd8(%rbp),%eax
  cltq
  mov    %edx,-0xd0(%rbp,%rax,4)
  addl   $0x1,-0xd8(%rbp)
L96:
  mov    -0xe8(%rbp),%eax
  cmp    %eax,-0xd8(%rbp)
  jl     L73
  mov    -0xe8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xe8(%rbp)
L77:
  addl   $0x1,-0xdc(%rbp)
L75:
  mov    -0xe8(%rbp),%eax
  cmp    %eax,-0xdc(%rbp)
  jl     L30
  addl   $0x1,-0xe0(%rbp)
L38:
  mov    -0xe8(%rbp),%eax
  cmp    %eax,-0xe0(%rbp)
  jl     L118
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0xd4(%rbp)
  jmp    L20
L103:
  mov    -0xd4(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xd4(%rbp)
L20:
  mov    -0xe8(%rbp),%eax
  cmp    %eax,-0xd4(%rbp)
  jl     L103
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type: string
name: parse_data
arguments: tag: D18
type: string
name: parse_data
arguments: tag: D23
type: string
name: parse_data
arguments: tag: D56
type: string
name: parse_data
arguments: tag: D47
type: string
name: parse_data
arguments: tag: D15
type: string

tool: D0: Dati un numar n natural:

tool: D18: %d

tool: D23: a[%d]= 

tool: D56: Vectorul unicat:

tool: D47: %d 

tool: D15: 
Exit...

assistant: ```c
int main() {
  int a[50];
  int n;
  printf("Dati un numar n natural:\n");
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    printf("a[%d]= ", i);
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; ++i) {
    printf("\n");
    for (int j = i + 1; j < n; ++j) {
      if (a[i] == a[j]) {
        for (int h = j; h < n; ++h) {
          a[h] = a[h + 1];
        }
        --n;
      }
    }
  }
  printf("Vectorul unicat:\n");
  for (int i = 0; i < n; ++i) {
    printf("%d ", a[i]);
  }
  printf("\nExit...\n");
  return 0;
}
```