user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D5(%rip),%rdi
  push   %r13
  lea    D34(%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    D97(%rip),%rdi
  call   <puts@plt>
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L36
  lea    0x10(%rsp),%rbp
  xor    %ebx,%ebx
  lea    D52(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L67:
  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     L67
L36:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L16
  xor    %ebx,%ebx
  lea    0x10(%rsp),%r13
  lea    D54(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L8:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L8
L16:
  lea    D61(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L58
  movslq %edx,%rax
  lea    D54(%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)
L32:
  mov    -0x4(%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  sub    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L32
L58:
  lea    D102(%rip),%rdi
  call   <puts@plt>
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0x1b0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
data_type: string
name: unpack_data
arguments: label: D34
data_type: string
name: unpack_data
arguments: label: D97
data_type: string
name: unpack_data
arguments: label: D20
data_type: string
name: unpack_data
arguments: label: D22
data_type: string
name: unpack_data
arguments: label: D52
data_type: string
name: unpack_data
arguments: label: D6
data_type: string
name: unpack_data
arguments: label: D54
data_type: string
name: unpack_data
arguments: label: D61
data_type: string
name: unpack_data
arguments: label: D102
data_type: string

tool: D5: 

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

tool: D34: %d

tool: D97: ------------------------------------------------------------------------

tool: D20: Numero de elementos (N) :

tool: D22: Ingrese %d elementos en el arreglo :


tool: D52: element - %d : 

tool: D6: 
Arreglo original : 

tool: D54: % 5d

tool: D61: 

Los valores en orden inverso son :

tool: D102: 


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");
}
```