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    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L75
  lea    0x14(%rsp),%r15
  lea    D33(%rip),%r14
  lea    D21(%rip),%r13
  jmp    L169
L89:
  mov    $0x0,%r8d
  jmp    L101
L126:
  lea    -0x1(%rsi),%eax
  sub    %edi,%esi
  cmp    %esi,%eax
  jge    L130
  mov    $0x0,%ecx
  mov    $0x0,%edi
  jmp    L44
L130:
  mov    $0x0,%ecx
  jmp    L66
L56:
  lea    D14(%rip),%rdi
  call   <printf@plt>
L137:
  mov    %r12,%rdi
  call   <free@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L75
L169:
  movl   $0x1,0x10(%rsp)
  movl   $0x1,0x14(%rsp)
  lea    0x10(%rsp),%rsi
  mov    %r15,%rdx
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    0x10(%rsp),%edx
  test   %edx,%edx
  jle    L155
  mov    %rax,%rbp
  mov    $0x0,%ebx
L100:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0x10(%rsp),%edx
  add    $0x4,%rbp
  cmp    %ebx,%edx
  jg     L100
L155:
  sub    $0x1,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <mergesort@plt>
  mov    0x10(%rsp),%esi
  test   %esi,%esi
  jle    L89
  mov    %r12,%rax
  mov    %esi,%edx
  lea    (%r12,%rdx,4),%rdx
  mov    $0x0,%r8d
L28:
  add    (%rax),%r8d
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L28
L101:
  mov    0x14(%rsp),%edi
  test   %edi,%edi
  jle    L126
  mov    %r12,%rax
  mov    %edi,%edx
  lea    (%r12,%rdx,4),%rdx
  mov    $0x0,%ecx
L124:
  add    (%rax),%ecx
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L124
  lea    -0x1(%rsi),%eax
  sub    %edi,%esi
L66:
  movslq %eax,%rdx
  lea    (%r12,%rdx,4),%rdx
  mov    $0x0,%edi
L159:
  add    (%rdx),%edi
  sub    $0x1,%eax
  sub    $0x4,%rdx
  cmp    %eax,%esi
  jle    L159
L44:
  mov    %r8d,%eax
  sub    %ecx,%eax
  sub    %eax,%ecx
  mov    %ecx,%esi
  neg    %esi
  cmovns %esi,%ecx
  sub    %edi,%r8d
  sub    %r8d,%edi
  mov    %edi,%esi
  neg    %esi
  cmovs  %edi,%esi
  cmp    %esi,%ecx
  jl     L56
  mov    %ecx,%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L137
L75:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L110
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L110:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D21
datatype: string
name: parse_data
arguments: data_label: D33
datatype: string
name: parse_data
arguments: data_label: D14
datatype: string

tool: D21: %d

tool: D33: %d%d

tool: D14: %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;
}
```