user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x150,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D121(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1150(%rbp),%rax
  mov    %rax,%rsi
  lea    D118(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x1148(%rbp)
  jmp    L13
L90:
  lea    -0xfb0(%rbp),%rdx
  mov    -0x1148(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D118(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x1148(%rbp)
L13:
  mov    -0x1150(%rbp),%eax
  cmp    %eax,-0x1148(%rbp)
  jl     L90
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x114c(%rbp),%rax
  mov    %rax,%rsi
  lea    D118(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D122(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x1148(%rbp)
  jmp    L94
L129:
  lea    -0x1140(%rbp),%rdx
  mov    -0x1148(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D118(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x1148(%rbp)
L94:
  mov    -0x114c(%rbp),%eax
  cmp    %eax,-0x1148(%rbp)
  jl     L129
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x1148(%rbp)
  jmp    L28
L80:
  movl   $0x0,-0x1144(%rbp)
  jmp    L146
L119:
  mov    -0x1148(%rbp),%eax
  cltq
  mov    -0x1140(%rbp,%rax,4),%edx
  mov    -0x1144(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jne    L41
  cmpl   $0x0,-0x1144(%rbp)
  jne    L138
  mov    -0xfac(%rbp),%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L35
L138:
  mov    -0x1150(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x1144(%rbp)
  jne    L159
  mov    -0x1150(%rbp),%eax
  sub    $0x2,%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L35
L159:
  mov    -0x1144(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%edx
  mov    -0x1144(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D102(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L35
L41:
  addl   $0x1,-0x1144(%rbp)
L146:
  mov    -0x1150(%rbp),%eax
  cmp    %eax,-0x1144(%rbp)
  jl     L119
L35:
  mov    -0x1150(%rbp),%eax
  cmp    %eax,-0x1144(%rbp)
  jne    L155
  lea    D126(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L155:
  addl   $0x1,-0x1148(%rbp)
L28:
  mov    -0x114c(%rbp),%eax
  cmp    %eax,-0x1148(%rbp)
  jl     L80
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D121
type: string
name: extract_symbolic_data
arguments: tag: D118
type: string
name: extract_symbolic_data
arguments: tag: D69
type: string
name: extract_symbolic_data
arguments: tag: D47
type: string
name: extract_symbolic_data
arguments: tag: D122
type: string
name: extract_symbolic_data
arguments: tag: D27
type: string
name: extract_symbolic_data
arguments: tag: D42
type: string
name: extract_symbolic_data
arguments: tag: D102
type: string
name: extract_symbolic_data
arguments: tag: D126
type: string

tool: D121: Please input the number of integers:

tool: D118: %d

tool: D69: Please input the integers:

tool: D47: Please input the times of query:

tool: D122: Please input the integers you want to query:

tool: D27: The result is:

tool: D42: %d


tool: D102: %d %d


tool: D126: NULL

assistant: ```c
int main() {
  int n, m, num[1000], query[100], i, j;
  printf("Please input the number of integers:");
  scanf("%d", &n);
  printf("Please input the integers:");
  for (i = 0; i < n; i++)
    scanf("%d", &num[i]);
  printf("Please input the times of query:");
  scanf("%d", &m);
  printf("Please input the integers you want to query:");
  for (i = 0; i < m; i++)
    scanf("%d", &query[i]);
  printf("The result is:\n");
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++)
      if (query[i] == num[j]) {
        if (j == 0)
          printf("%d\n", num[1]);
        else if (j == n - 1)
          printf("%d\n", num[n - 2]);
        else
          printf("%d %d\n", num[j - 1], num[j + 1]);
        break;
      }
    if (j == n)
      printf("NULL\n");
  }
  return 0;
}
```