user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf68,%rsp
  lea    D1(%rip),%rbp
  mov    %rbp,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x3f58(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rsi
  call   <scanf@plt>
  cmpq   $0x0,0x20(%rsp)
  jle    L43
  lea    0x28(%rsp),%rax
  lea    0x30(%rsp),%rbx
  movq   $0x1,0x8(%rsp)
  mov    %rax,0x18(%rsp)
  lea    0x1fc0(%rsp),%rax
  lea    D20(%rip),%r15
  mov    %rax,(%rsp)
  nopw   0x0(%rax,%rax,1)
L20:
  mov    $0x1f90,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x28(%rsp),%r14
  test   %r14,%r14
  jle    L134
  lea    0x38(%rsp),%r13
  mov    $0x1,%r12d
  nopl   0x0(%rax)
L69:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  add    $0x1,%r12
  add    $0x8,%r13
  call   <scanf@plt>
  mov    0x28(%rsp),%r14
  cmp    %r12,%r14
  jge    L69
L134:
  cmpq   $0x1,0x8(%rsp)
  jne    L25
L110:
  mov    (%rsp),%rdi
  mov    $0x1f90,%edx
  mov    %rbx,%rsi
  call   <memcpy@plt>
  addq   $0x1,0x8(%rsp)
  mov    0x8(%rsp),%rax
  cmp    %rax,0x20(%rsp)
  jl     L43
  mov    %r14,0x10(%rsp)
  jmp    L20
  nopw   0x0(%rax,%rax,1)
L25:
  mov    0x10(%rsp),%rax
  cmp    %rax,%r14
  mov    %rax,%r13
  cmovge %r14,%r13
  cmp    $0x1,%r13
  jle    L57
  mov    $0x1,%r12d
  nopl   0x0(%rax,%rax,1)
L128:
  mov    (%rsp),%rax
  mov    (%rbx,%r12,8),%rsi
  mov    %r15,%rdi
  add    (%rax,%r12,8),%rsi
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12,%r13
  jne    L128
L71:
  mov    0x30(%rsp,%r13,8),%rsi
  mov    %rbp,%rdi
  add    0x1fc0(%rsp,%r13,8),%rsi
  call   <printf@plt>
L97:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L110
L57:
  jne    L97
  jmp    L71
  nopl   (%rax)
L43:
  mov    0x28(%rsp),%rax
  cmp    $0x1,%rax
  jle    L106
  mov    $0x1,%r12d
  lea    0x30(%rsp),%rbx
  lea    D20(%rip),%r13
  nopl   (%rax)
L37:
  mov    (%rbx,%r12,8),%rsi
  mov    %r13,%rdi
  add    $0x1,%r12
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  cmp    %r12,%rax
  jg     L37
L106:
  test   %rax,%rax
  jle    L27
  mov    0x30(%rsp,%rax,8),%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
L27:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x3f58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x3f68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D1
type_guessed: string
name: unpack_data
arguments: label: D20
type_guessed: string

tool: D1: %lld

tool: D20: %lld 

assistant: ```c
main() {
  long long n, m, i, j, k, l, cur[1010], per[1010], pm, cm, maxline = 0;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) {
    memset(cur, 0, sizeof(cur));
    scanf("%lld", &m);
    for (j = 1; j <= m; j++)
      scanf("%lld", &cur[j]);
    cm = m;
    if (cm >= pm)
      maxline = cm;
    else
      maxline = pm;
    if (i != 1) {
      for (j = 1; j < maxline; j++)
        printf("%lld ", per[j] + cur[j]);
      if (maxline >= 1)
        printf("%lld", per[maxline] + cur[maxline]);
      putchar('\n');
      maxline = 0;
    }
    memcpy(per, cur, sizeof(cur));
    pm = cm;
  }
  for (i = 1; i < m; i++)
    printf("%lld ", cur[i]);
  if (m >= 1)
    printf("%lld", cur[m]);
  putchar('\n');
  return 0;
}
```