user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  lea    D32(%rip),%rdi
  lea    D24(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x4d8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4c8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%ebp
  test   %ebp,%ebp
  jle    L121
  lea    0x14(%rsp),%r12
  mov    $0x1,%ebx
  nopl   0x0(%rax)
L43:
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%r12
  call   <scanf@plt>
  mov    0xc(%rsp),%ebp
  cmp    %ebx,%ebp
  jge    L43
  mov    0x14(%rsp),%ebx
  test   %ebp,%ebp
  jle    L166
  lea    -0x1(%rbp),%edx
  lea    0x18(%rsp),%rax
  lea    (%rax,%rdx,4),%rcx
  mov    %ebx,%edx
  jmp    L46
  cs nopw 0x0(%rax,%rax,1)
L159:
  mov    (%rax),%edx
  add    $0x4,%rax
L46:
  cmp    %edx,%ebx
  cmovl  %edx,%ebx
  cmp    %rcx,%rax
  jne    L159
  test   %ebx,%ebx
  js     L25
  movslq %ebx,%rax
  lea    0x330(%rsp),%rdi
  xor    %esi,%esi
  lea    0x4(,%rax,4),%rdx
  call   <memset@plt>
L25:
  mov    $0x1,%eax
  lea    0x10(%rsp),%rdi
  nopl   0x0(%rax,%rax,1)
L128:
  movslq (%rdi,%rax,4),%rdx
  add    $0x1,%rax
  addl   $0x1,0x330(%rsp,%rdx,4)
  cmp    %eax,%ebp
  jge    L128
  test   %ebx,%ebx
  jle    L70
L76:
  lea    0x334(%rsp),%rax
  mov    $0x1,%edx
  nopl   (%rax)
L66:
  add    $0x1,%edx
  mov    -0x4(%rax),%ecx
  add    %ecx,(%rax)
  add    $0x4,%rax
  cmp    %edx,%ebx
  jge    L66
  lea    0x10(%rsp),%rdi
  test   %ebp,%ebp
  jle    L97
L70:
  movslq %ebp,%rdx
  nopl   0x0(%rax)
L143:
  movslq (%rdi,%rdx,4),%rcx
  sub    $0x1,%rdx
  movslq 0x330(%rsp,%rcx,4),%r8
  mov    %r8,%rax
  mov    %ecx,0x1a0(%rsp,%r8,4)
  sub    $0x1,%eax
  mov    %eax,0x330(%rsp,%rcx,4)
  test   %edx,%edx
  jg     L143
L97:
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  mov    $0x1,%ebx
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  lea    0x1a0(%rsp),%r12
  lea    D21(%rip),%rbp
  test   %eax,%eax
  jle    L16
  nopl   0x0(%rax,%rax,1)
L95:
  mov    (%r12,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jge    L95
L16:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4c8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L137
  add    $0x4d8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L121:
  mov    0x14(%rsp),%ebx
L166:
  test   %ebx,%ebx
  js     L97
  movslq %ebx,%rax
  xor    %esi,%esi
  lea    0x330(%rsp),%rdi
  lea    0x4(,%rax,4),%rdx
  call   <memset@plt>
  test   %ebx,%ebx
  jg     L76
  jmp    L97
L137:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D32
type_guessed: string
name: disassemble_data
arguments: label: D24
type_guessed: string
name: disassemble_data
arguments: label: D40
type_guessed: string
name: disassemble_data
arguments: label: D4
type_guessed: string
name: disassemble_data
arguments: label: D21
type_guessed: string

tool: D32: Enter the number of elements to be sorted: 

tool: D24: %d

tool: D40: Enter %d elements to be sorted: 

tool: D4: The sorted elements are: 

tool: D21: %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;
}
```