user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x1b0(%rbp),%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1b0(%rbp),%eax
  mov    %eax,%esi
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x1ac(%rbp)
  jmp    L65
L82:
  lea    -0x1a0(%rbp),%rdx
  mov    -0x1ac(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x1ac(%rbp)
L65:
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L82
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x1ac(%rbp)
  jmp    L33
L35:
  mov    -0x1ac(%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,-0x1ac(%rbp)
L33:
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L35
  movl   $0x1,-0x1ac(%rbp)
  jmp    L78
L50:
  movl   $0x0,-0x1a8(%rbp)
  jmp    L100
L30:
  mov    -0x1a8(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1a8(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jl     L67
  mov    -0x1a8(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,-0x1a4(%rbp)
  mov    -0x1a8(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1a8(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  mov    -0x1a8(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x1a4(%rbp),%edx
  mov    %edx,-0x1a0(%rbp,%rax,4)
L67:
  addl   $0x1,-0x1a8(%rbp)
L100:
  mov    -0x1b0(%rbp),%eax
  sub    -0x1ac(%rbp),%eax
  cmp    %eax,-0x1a8(%rbp)
  jl     L30
  addl   $0x1,-0x1ac(%rbp)
L78:
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L50
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x1ac(%rbp)
  jmp    L99
L106:
  mov    -0x1ac(%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,-0x1ac(%rbp)
L99:
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L106
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D23
type_infered: string
name: unpack_data
arguments: tag: D48
type_infered: string
name: unpack_data
arguments: tag: D64
type_infered: string
name: unpack_data
arguments: tag: D55
type_infered: string
name: unpack_data
arguments: tag: D56
type_infered: string
name: unpack_data
arguments: tag: D36
type_infered: string

tool: D23: Ingresar la longitud de la lista

tool: D48: %d

tool: D64: ingrese %d valores presionando ENTER despues de cada dato:  


tool: D55: Lista ingresada :

tool: D56: %5d

tool: D36: 

Lista ordenada :

assistant: ```c
void main() {
  int a[100], i, j, n, t;
  printf("Ingresar la longitud de la lista\n");
  scanf("%d", &n);
  printf("ingrese %d valores presionando ENTER despues de cada dato:  \n", n);
  for (i = 0; i < n; i++)
    scanf("%d", &a[i]);
  printf("Lista ingresada :\n");
  for (i = 0; i < n; i++)
    printf("%5d", a[i]);
  for (i = 1; i < n; i++)
    for (j = 0; j < n - i; j++)
      if (a[j] >= a[j + 1]) {
        t = a[j];
        a[j] = a[j + 1];
        a[j + 1] = t;
      }
  printf("\n\nLista ordenada :\n");
  for (i = 0; i < n; i++)
    printf("%5d", a[i]);
  printf("\n");
}
```