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
  lea    -0x61000(%rsp),%r11
L160:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L160
  sub    $0xaa8,%rsp
  lea    D26(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x61a98(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  nopl   (%rax)
L70:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    $0xffffffff,%eax
  je     L62
  test   %eax,%eax
  jle    L57
  xor    %ebx,%ebx
  xor    %r14d,%r14d
  lea    0x10(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L131:
  lea    0x0(%rbp,%rbx,4),%rsi
  mov    %r12,%rdi
  mov    %rbp,%r15
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  add    0x0(%rbp,%rbx,4),%r14d
  add    $0x1,%rbx
  cmp    %ebx,%edi
  jg     L131
  pxor   %xmm0,%xmm0
  pxor   %xmm2,%xmm2
  pxor   %xmm1,%xmm1
  mov    %r14d,%eax
  cltd
  cvtsi2ss %r14d,%xmm0
  idiv   %edi
  cvtsi2ss %edi,%xmm2
  divss  %xmm2,%xmm0
  cvtsi2ss %eax,%xmm1
  ucomiss %xmm0,%xmm1
  jp     L16
  jne    L16
  test   %edi,%edi
  jle    L163
  lea    -0x1(%rdi),%edx
  cmp    $0x2,%edx
  jbe    L151
  mov    %edi,%edx
  movd   %eax,%xmm4
  pxor   %xmm1,%xmm1
  shr    $0x2,%edx
  pshufd $0x0,%xmm4,%xmm3
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x20(%rsp,%rdx,1),%rdx
  nopl   (%rax)
L10:
  movdqa (%r15),%xmm0
  add    $0x10,%r15
  psubd  %xmm3,%xmm0
  movdqa %xmm0,%xmm2
  psrad  $0x1f,%xmm2
  pxor   %xmm2,%xmm0
  psubd  %xmm2,%xmm0
  paddd  %xmm0,%xmm1
  cmp    %rdx,%r15
  jne    L10
  movdqa %xmm1,%xmm0
  mov    %edi,%edx
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%edx
  paddd  %xmm0,%xmm1
  movdqa %xmm1,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm1
  movd   %xmm1,%esi
  test   $0x3,%dil
  je     L68
L89:
  movslq %edx,%rcx
  mov    0x10(%rsp,%rcx,4),%ecx
  sub    %eax,%ecx
  mov    %ecx,%r8d
  neg    %r8d
  cmovns %r8d,%ecx
  add    %ecx,%esi
  lea    0x1(%rdx),%ecx
  cmp    %ecx,%edi
  jle    L68
  movslq %ecx,%rcx
  mov    0x10(%rsp,%rcx,4),%ecx
  sub    %eax,%ecx
  mov    %ecx,%r8d
  neg    %r8d
  cmovns %r8d,%ecx
  add    $0x2,%edx
  add    %ecx,%esi
  cmp    %edx,%edi
  jle    L68
  movslq %edx,%rdx
  mov    0x10(%rsp,%rdx,4),%edx
  sub    %eax,%edx
  mov    %edx,%eax
  neg    %eax
  cmovs  %edx,%eax
  add    %eax,%esi
L68:
  sar    %esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L70
  nopw   0x0(%rax,%rax,1)
L57:
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  pxor   %xmm5,%xmm5
  cvtsi2ss %eax,%xmm1
  divss  %xmm1,%xmm0
  ucomiss %xmm5,%xmm0
  jp     L16
  je     L163
  nopl   0x0(%rax,%rax,1)
L16:
  lea    D7(%rip),%rdi
  call   <puts@plt>
  jmp    L70
  nopl   0x0(%rax)
L62:
  mov    0x61a98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  add    $0x61aa8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L163:
  xor    %esi,%esi
  jmp    L68
L151:
  xor    %esi,%esi
  xor    %edx,%edx
  jmp    L89
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D26
type: string
name: extract_symbolic_data
arguments: data_tag: D15
type: string
name: extract_symbolic_data
arguments: data_tag: D7
type: string

tool: D26: %d

tool: D15: %d


tool: D7: -1

assistant: ```c
int main() {
  int i, j, c, len, num, temp, t, res;
  int arr[100001];
  while (1) {
    c = len = 0;
    scanf("%d", &num);
    if (num == -1)
      break;
    for (i = 0; i < num; i++) {
      scanf("%d", &arr[i]);
      c += arr[i];
    }
    if (c / num != (float)c / num) {
      printf("-1\n");
      continue;
    }
    temp = c / num;
    for (i = 0; i < num; i++)
      len += ((arr[i] - temp) >= 0 ? (arr[i] - temp) : -(arr[i] - temp));
    printf("%d\n", len / 2);
  }
  return 0;
}
```