user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D35(%rip),%r12
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    -0x3c(%rbp),%rax
  mov    %rax,-0x58(%rbp)
  nopl   0x0(%rax,%rax,1)
L184:
  mov    -0x58(%rbp),%rsi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  mov    %rsp,-0x48(%rbp)
  call   <scanf@plt>
  mov    -0x3c(%rbp),%edx
  test   %edx,%edx
  je     L91
  movslq %edx,%rax
  mov    %rsp,%rsi
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdi
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rdi
  and    $0xfffffffffffffff0,%rcx
  sub    %rdi,%rsi
  cmp    %rsi,%rsp
  je     L213
L56:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L56
L213:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  je     L117
  orq    $0x0,-0x8(%rsp,%rcx,1)
L117:
  mov    %rax,%rcx
  mov    %rsp,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%r13
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L223
L198:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L198
L223:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L78
L121:
  lea    0x3(%rsp),%rbx
  mov    %rbx,%rax
  and    $0xfffffffffffffffc,%rbx
  shr    $0x2,%rax
  mov    %rbx,-0x68(%rbp)
  mov    %rax,-0x60(%rbp)
  test   %edx,%edx
  jle    L29
  mov    %rbx,%r14
  xor    %r15d,%r15d
  mov    %r13,%rbx
  nopl   0x0(%rax)
L235:
  mov    %r14,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  add    $0x1,%r15d
  add    $0x4,%rbx
  add    $0x4,%r14
  cmp    %r15d,%eax
  jg     L235
  movslq %eax,%r9
  mov    %rsp,%rsi
  lea    0xf(,%r9,4),%rax
  mov    %r9,%r14
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L14
L224:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L224
L14:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L137
  orq    $0x0,-0x8(%rsp,%rdx,1)
L137:
  mov    %rsp,%r15
  test   %r14d,%r14d
  jle    L29
  lea    0x0(,%r9,4),%rdx
  xor    %esi,%esi
  mov    %r15,%rdi
  mov    %r9,-0x50(%rbp)
  call   <memset@plt>
  mov    -0x60(%rbp),%rax
  mov    0x0(,%rax,4),%eax
  cmp    %eax,%r14d
  jle    L142
  test   %eax,%eax
  js     L142
  mov    -0x68(%rbp),%rsi
  mov    -0x50(%rbp),%r9
  mov    $0x1,%edx
  jmp    L199
L96:
  mov    -0x4(%r13,%rdx,4),%ecx
  mov    %ecx,(%r15,%rax,4)
  cmp    %rdx,%r9
  je     L208
  mov    (%rsi,%rdx,4),%eax
  add    %edx,%eax
  add    $0x1,%rdx
  test   %eax,%eax
  js     L142
  cmp    %eax,%r14d
  jle    L142
L199:
  cltq
  mov    (%r15,%rax,4),%ecx
  test   %ecx,%ecx
  je     L96
L142:
  lea    D3(%rip),%rdi
  call   <puts@plt>
L65:
  mov    -0x48(%rbp),%rsp
  jmp    L184
  nop
L208:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L175:
  mov    (%r15,%rbx,4),%esi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,-0x3c(%rbp)
  jg     L175
L29:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L65
  nop
L91:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L78:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L121
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D35
datatype: string
name: retrieve_data
arguments: label: D18
datatype: string
name: retrieve_data
arguments: label: D3
datatype: string
name: retrieve_data
arguments: label: D9
datatype: string

tool: D35: %d%d

tool: D18: %d

tool: D3: -1

tool: D9: %d 

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