user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x4d0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x4d0(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x4d0(%rbp),%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x4cc(%rbp)
  jmp    L20
L39:
  lea    -0x4c0(%rbp),%rdx
  mov    -0x4cc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x4cc(%rbp)
L20:
  mov    -0x4d0(%rbp),%eax
  cmp    %eax,-0x4cc(%rbp)
  jle    L39
  mov    -0x4bc(%rbp),%eax
  mov    %eax,-0x4c8(%rbp)
  movl   $0x1,-0x4cc(%rbp)
  jmp    L61
L106:
  mov    -0x4cc(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  cmp    %eax,-0x4c8(%rbp)
  jge    L101
  mov    -0x4cc(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  mov    %eax,-0x4c8(%rbp)
L101:
  addl   $0x1,-0x4cc(%rbp)
L61:
  mov    -0x4d0(%rbp),%eax
  cmp    %eax,-0x4cc(%rbp)
  jle    L106
  movl   $0x0,-0x4cc(%rbp)
  jmp    L93
L14:
  mov    -0x4cc(%rbp),%eax
  cltq
  movl   $0x0,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x4cc(%rbp)
L93:
  mov    -0x4cc(%rbp),%eax
  cmp    -0x4c8(%rbp),%eax
  jle    L14
  movl   $0x1,-0x4c4(%rbp)
  jmp    L96
L56:
  mov    -0x4c4(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x4c4(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  add    $0x1,%edx
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x4c4(%rbp)
L96:
  mov    -0x4d0(%rbp),%eax
  cmp    %eax,-0x4c4(%rbp)
  jle    L56
  movl   $0x1,-0x4cc(%rbp)
  jmp    L16
L33:
  mov    -0x4cc(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x4cc(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  add    %eax,%edx
  mov    -0x4cc(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x4cc(%rbp)
L16:
  mov    -0x4cc(%rbp),%eax
  cmp    -0x4c8(%rbp),%eax
  jle    L33
  mov    -0x4d0(%rbp),%eax
  mov    %eax,-0x4c4(%rbp)
  jmp    L36
L30:
  mov    -0x4c4(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%ecx
  mov    -0x4c4(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%edx
  movslq %ecx,%rax
  mov    %edx,-0x330(%rbp,%rax,4)
  mov    -0x4c4(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x4c4(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  sub    $0x1,%edx
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  subl   $0x1,-0x4c4(%rbp)
L36:
  cmpl   $0x0,-0x4c4(%rbp)
  jg     L30
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x4cc(%rbp)
  jmp    L99
L146:
  mov    -0x4cc(%rbp),%eax
  cltq
  mov    -0x330(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4cc(%rbp)
L99:
  mov    -0x4d0(%rbp),%eax
  cmp    %eax,-0x4cc(%rbp)
  jle    L146
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D49
datatype: string
name: extract_symbolic_data
arguments: label: D18
datatype: string
name: extract_symbolic_data
arguments: label: D42
datatype: string
name: extract_symbolic_data
arguments: label: D59
datatype: string
name: extract_symbolic_data
arguments: label: D28
datatype: string

tool: D49: Enter the number of elements to be sorted: 

tool: D18: %d

tool: D42: Enter %d elements to be sorted: 

tool: D59: The sorted elements are: 

tool: D28: %d 

assistant: ```c
int main() {
  int A[100], B[100], C[100], i, n, k, j;
  printf("Enter the number of elements to be sorted: ");
  scanf("%d", &n);
  printf("Enter %d elements to be sorted: ", n);
  for (i = 1; i <= n; i++)
    scanf("%d", &A[i]);
  k = A[1];
  for (i = 1; i <= n; i++) {
    if (A[i] > k)
      k = A[i];
  }
  for (i = 0; i <= k; i++)
    C[i] = 0;
  for (j = 1; j <= n; j++)
    C[A[j]] = C[A[j]] + 1;
  for (i = 1; i <= k; i++)
    C[i] = C[i] + C[i - 1];
  for (j = n; j >= 1; j--) {
    B[C[A[j]]] = A[j];
    C[A[j]] = C[A[j]] - 1;
  }
  printf("The sorted elements are: ");
  for (i = 1; i <= n; i++)
    printf("%d ", B[i]);
  printf("\n");
  return 0;
}
```