user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  lea    D18(%rip),%rdi
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r12
  call   <puts@plt>
  lea    D95(%rip),%rdi
  call   <puts@plt>
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D98(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L22
  lea    0x10(%rsp),%rbp
  xor    %ebx,%ebx
  lea    D4(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L57:
  mov    %ebx,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L57
L22:
  lea    D75(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L25
  xor    %ebx,%ebx
  lea    0x10(%rsp),%r13
  lea    D105(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L12:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L12
L25:
  lea    D82(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L28
  movslq %edx,%rax
  lea    D105(%rip),%rbp
  shl    $0x2,%rax
  add    %rax,%r12
  lea    0x10(%rsp,%rax,1),%rbx
  lea    -0x1(%rdx),%eax
  shl    $0x2,%rax
  sub    %rax,%r12
  nopl   (%rax)
L8:
  mov    -0x4(%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  sub    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L8
L28:
  lea    D20(%rip),%rdi
  call   <puts@plt>
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L66
  add    $0x1b0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L66:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D18
type_infered: string
name: unpack_data
arguments: label: D0
type_infered: string
name: unpack_data
arguments: label: D95
type_infered: string
name: unpack_data
arguments: label: D38
type_infered: string
name: unpack_data
arguments: label: D98
type_infered: string
name: unpack_data
arguments: label: D4
type_infered: string
name: unpack_data
arguments: label: D75
type_infered: string
name: unpack_data
arguments: label: D105
type_infered: string
name: unpack_data
arguments: label: D82
type_infered: string
name: unpack_data
arguments: label: D20
type_infered: string

tool: D18: 

Leer N valores en un arreglo e imprimirlo en orden inverso:

tool: D0: %d

tool: D95: ------------------------------------------------------------------------

tool: D38: Numero de elementos (N) :

tool: D98: Ingrese %d elementos en el arreglo :


tool: D4: element - %d : 

tool: D75: 
Arreglo original : 

tool: D105: % 5d

tool: D82: 

Los valores en orden inverso son :

tool: D20: 


assistant: ```c
void main() {
  int i, n, a[100];
  printf("\n\nLeer N valores en un arreglo e imprimirlo en orden inverso:\n");
  printf("------------------------------------------------------------------------\n");
  printf("Numero de elementos (N) :");
  scanf("%d", &n);
  printf("Ingrese %d elementos en el arreglo :\n", n);
  for (i = 0; i < n; i++) {
    printf("element - %d : ", i);
    scanf("%d", &a[i]);
  }
  printf("\nArreglo original : \n");
  for (i = 0; i < n; i++) {
    printf("% 5d", a[i]);
  }
  printf("\n\nLos valores en orden inverso son :\n");
  for (i = n - 1; i >= 0; i--) {
    printf("% 5d", a[i]);
  }
  printf("\n\n");
}
```