user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1d0,%rsp
  mov    %edi,-0x1c4(%rbp)
  mov    %rsi,-0x1d0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x2,-0x1c4(%rbp)
  je     L23
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L23:
  addq   $0x8,-0x1d0(%rbp)
  mov    -0x1d0(%rbp),%rax
  mov    (%rax),%rax
  lea    -0x1a0(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <Load_Mem@plt>
  mov    %eax,-0x1a8(%rbp)
  cmpl   $0x64,-0x1a8(%rbp)
  je     L46
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L46:
  movl   $0x0,-0x1b0(%rbp)
  jmp    L53
L2:
  mov    -0x1b0(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x1ac(%rbp)
  jmp    L44
L31:
  mov    -0x1b0(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jle    L95
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,-0x1a4(%rbp)
  mov    -0x1b0(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  mov    -0x1b0(%rbp),%eax
  cltq
  mov    -0x1a4(%rbp),%edx
  mov    %edx,-0x1a0(%rbp,%rax,4)
L95:
  addl   $0x1,-0x1ac(%rbp)
L44:
  mov    -0x1ac(%rbp),%eax
  cmp    -0x1a8(%rbp),%eax
  jl     L31
  addl   $0x1,-0x1b0(%rbp)
L53:
  mov    -0x1b0(%rbp),%eax
  cmp    -0x1a8(%rbp),%eax
  jl     L2
  mov    -0x1a8(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L9
  mov    -0x1a8(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1a8(%rbp),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  sub    $0x1,%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x1b4(%rbp)
  jmp    L40
L9:
  mov    -0x1a8(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,-0x1b4(%rbp)
L40:
  mov    -0x1b4(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L87
  call   <__stack_chk_fail@plt>
L87:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D8
datatype: string
name: parse_data
arguments: label: D10
datatype: string
name: parse_data
arguments: label: D22
datatype: string

tool: D8: usage: ./P1-2 valuefile

tool: D10: valuefiles must contain 100 entries

tool: D22: The median of the array is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int Nums[100];
  int NumNums, Median;
  int Load_Mem(char *, int *);
  if (argc != 2) {
    printf("usage: ./P1-2 valuefile\n");
    exit(1);
  }
  NumNums = Load_Mem(*++argv, Nums);
  if (NumNums != 100) {
    printf("valuefiles must contain 100 entries\n");
    exit(1);
  }
  int i, j, temp;
  for (i = 0; i < NumNums; i++) {
    for (j = i + 1; j < NumNums; j++) {
      if (Nums[i] > Nums[j]) {
        temp = Nums[j];
        Nums[j] = Nums[i];
        Nums[i] = temp;
      }
    }
  }
  if (NumNums % 2 == 0) {
    Median = (Nums[NumNums / 2] + Nums[NumNums / 2 - 1]) / 2;
  } else {
    Median = (Nums[NumNums / 2]);
  }
  printf("The median of the array is %d\n", Median);
  exit(0);
}
```