user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x34(%rbp)
  lea    -0x34(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L37
L12:
  movl   $0x1,-0x30(%rbp)
  movl   $0x1,-0x2c(%rbp)
  lea    -0x2c(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x30(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L99
L65:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x28(%rbp)
L99:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jl     L65
  mov    -0x30(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <mergesort@plt>
  movl   $0x0,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L107
L102:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x24(%rbp)
  addl   $0x1,-0x28(%rbp)
L107:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jl     L102
  movl   $0x0,-0x28(%rbp)
  jmp    L54
L13:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x20(%rbp)
  addl   $0x1,-0x28(%rbp)
L54:
  mov    -0x2c(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jl     L13
  mov    -0x30(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x28(%rbp)
  jmp    L11
L104:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x1c(%rbp)
  subl   $0x1,-0x28(%rbp)
L11:
  mov    -0x30(%rbp),%eax
  mov    -0x2c(%rbp),%edx
  sub    %edx,%eax
  cmp    %eax,-0x28(%rbp)
  jge    L104
  mov    -0x24(%rbp),%eax
  sub    -0x20(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x20(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  mov    -0x1c(%rbp),%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,%ecx
  neg    %ecx
  cmovns %ecx,%eax
  cmp    %eax,%edx
  jl     L68
  mov    -0x20(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,%esi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L16
L68:
  mov    -0x1c(%rbp),%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,%esi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L16:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L37:
  mov    -0x34(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x34(%rbp)
  test   %eax,%eax
  jne    L12
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type: string
name: unpack_data
arguments: data_tag: D46
type: string
name: unpack_data
arguments: data_tag: D30
type: string

tool: D8: %d

tool: D46: %d%d

tool: D30: %d


assistant: ```c
int main() {
  int tc = 0;
  scanf("%d", &tc);
  while (tc--) {
    int t = 1;
    int x = 1;
    int i, j, k, l;
    int *n;
    scanf("%d%d", &t, &x);
    n = (int *)malloc(t * sizeof(int));
    for (i = 0; i < t; i++) {
      scanf("%d", n + i);
    }
    mergesort(n, 0, t - 1);
    int sum, sum1, sum2;
    sum = sum1 = sum2 = 0;
    for (i = 0; i < t; i++)
      sum = sum + n[i];
    for (i = 0; i < x; i++)
      sum1 = sum1 + n[i];
    for (i = t - 1; i > t - x - 1; i--)
      sum2 = sum2 + n[i];
    int sum11 = sum - sum1;
    int sum22 = sum - sum2;
    if (abs(sum1 - sum11) >= abs(sum2 - sum22))
      printf("%d\n", abs(sum1 - sum11));
    else
      printf("%d\n", abs(sum2 - sum22));
    free(n);
  }
  return 0;
}
```