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    $0xfd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L72:
  movl   $0x0,-0xfb4(%rbp)
  lea    -0xfb0(%rbp),%rax
  mov    $0xfa0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0xfc4(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0xfc4(%rbp),%eax
  test   %eax,%eax
  je     L25
  movl   $0x1,-0xfb8(%rbp)
  jmp    L74
L67:
  lea    -0xfbc(%rbp),%rdx
  lea    -0xfc0(%rbp),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0xfbc(%rbp),%edx
  mov    -0xfb8(%rbp),%eax
  add    %edx,%eax
  test   %eax,%eax
  js     L83
  mov    -0xfbc(%rbp),%edx
  mov    -0xfb8(%rbp),%eax
  add    %eax,%edx
  mov    -0xfc4(%rbp),%eax
  cmp    %eax,%edx
  jle    L30
L83:
  movl   $0x1,-0xfb4(%rbp)
  jmp    L1
L30:
  mov    -0xfbc(%rbp),%edx
  mov    -0xfb8(%rbp),%eax
  add    %edx,%eax
  mov    -0xfc0(%rbp),%edx
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
L1:
  addl   $0x1,-0xfb8(%rbp)
L74:
  mov    -0xfc4(%rbp),%eax
  cmp    %eax,-0xfb8(%rbp)
  jle    L67
  cmpl   $0x0,-0xfb4(%rbp)
  jne    L117
  movl   $0x1,-0xfb8(%rbp)
  jmp    L19
L31:
  mov    -0xfb8(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L109
  movl   $0x1,-0xfb4(%rbp)
  jmp    L117
L109:
  addl   $0x1,-0xfb8(%rbp)
L19:
  mov    -0xfc4(%rbp),%eax
  cmp    %eax,-0xfb8(%rbp)
  jle    L31
L117:
  cmpl   $0x1,-0xfb4(%rbp)
  jne    L151
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L72
L151:
  movl   $0x1,-0xfb8(%rbp)
  jmp    L86
L125:
  mov    -0xfb8(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xfb8(%rbp)
L86:
  mov    -0xfc4(%rbp),%eax
  cmp    %eax,-0xfb8(%rbp)
  jl     L125
  mov    -0xfc4(%rbp),%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    L72
L25:
  nop
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L153
  call   <__stack_chk_fail@plt>
L153:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D19
data_type: string
name: retrieve_data
arguments: data_label: D40
data_type: string
name: retrieve_data
arguments: data_label: D0
data_type: string
name: retrieve_data
arguments: data_label: D47
data_type: string
name: retrieve_data
arguments: data_label: D42
data_type: string

tool: D19: %d

tool: D40: %d %d

tool: D0: -1

tool: D47: %d 

tool: D42: %d


assistant: ```c
int main() {
  int n, i, c, p, posicao[1000], erro;
  while (1) {
    erro = 0;
    memset(posicao, 0, sizeof(posicao));
    scanf("%d", &n);
    if (n == 0)
      break;
    else {
      for (i = 1; i <= n; i++) {
        scanf("%d %d", &c, &p);
        if ((i + p) < 0 || (i + p) > n)
          erro = 1;
        else
          posicao[i + p] = c;
      }
      if (erro == 0) {
        for (i = 1; i <= n; i++) {
          if (posicao[i] == 0) {
            erro = 1;
            break;
          }
        }
      }
      if (erro == 1) {
        printf("-1\n");
      } else {
        for (i = 1; i <= n - 1; i++) {
          printf("%d ", posicao[i]);
        }
        printf("%d\n", posicao[n]);
      }
    }
  }
  return 0;
}
```