user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D100(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1a8(%rbp),%rax
  mov    %rax,%rsi
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1a8(%rbp),%eax
  mov    %eax,%esi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x1a4(%rbp)
  jmp    L14
L47:
  mov    -0x1a4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1a0(%rbp),%rdx
  mov    -0x1a4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x1a4(%rbp)
L14:
  mov    -0x1a8(%rbp),%eax
  cmp    %eax,-0x1a4(%rbp)
  jl     L47
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x1a4(%rbp)
  jmp    L1
L8:
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x1a4(%rbp)
L1:
  mov    -0x1a8(%rbp),%eax
  cmp    %eax,-0x1a4(%rbp)
  jl     L8
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x1a8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x1a4(%rbp)
  jmp    L41
L22:
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  subl   $0x1,-0x1a4(%rbp)
L41:
  cmpl   $0x0,-0x1a4(%rbp)
  jns    L22
  lea    D86(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L58
  call   <__stack_chk_fail@plt>
L58:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D100
type_infered: string
name: read_data
arguments: tag: D57
type_infered: string
name: read_data
arguments: tag: D43
type_infered: string
name: read_data
arguments: tag: D58
type_infered: string
name: read_data
arguments: tag: D59
type_infered: string
name: read_data
arguments: tag: D1
type_infered: string
name: read_data
arguments: tag: D17
type_infered: string
name: read_data
arguments: tag: D56
type_infered: string
name: read_data
arguments: tag: D30
type_infered: string
name: read_data
arguments: tag: D86
type_infered: string

tool: D100: 

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

tool: D57: ------------------------------------------------------------------------

tool: D43: Numero de elementos (N) :

tool: D58: %d

tool: D59: Ingrese %d elementos en el arreglo :


tool: D1: element - %d : 

tool: D17: 
Arreglo original : 

tool: D56: % 5d

tool: D30: 

Los valores en orden inverso son :

tool: D86: 


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");
}
```