user: 这段汇编代码用C语言表示是什么样的:

```asm
<print_multi_set_permutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L9
  movl   $0x0,-0x10(%rbp)
  jmp    L143
L6:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L143:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L6
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L126
L9:
  movl   $0x0,-0x10(%rbp)
  jmp    L141
L136:
  cmpl   $0x0,-0x10(%rbp)
  je     L78
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L55
L78:
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L4
L20:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L21
  addl   $0x1,-0x8(%rbp)
L21:
  addl   $0x1,-0xc(%rbp)
L4:
  mov    -0xc(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L20
  movl   $0x0,-0xc(%rbp)
  jmp    L98
L18:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L53
  addl   $0x1,-0x4(%rbp)
L53:
  addl   $0x1,-0xc(%rbp)
L98:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L18
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L55
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <print_multi_set_permutation>
L55:
  addl   $0x1,-0x10(%rbp)
L141:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L136
L126:
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
datatype: string

tool: D0: %3d

assistant: ```c
void print_multi_set_permutation(int n, int b[], int a[], int cur) {
  int i, j;
  if (cur == n) {
    for (i = 0; i < n; ++i)
      printf("%3d", a[i]);
    putchar('\n');
  } else
    for (i = 0; i < n; ++i) {
      if (!i || b[i] != b[i - 1]) {
        int c1, c2;
        c1 = c2 = 0;
        for (j = 0; j < cur; ++j)
          if (a[j] == b[i])
            c1++;
        for (j = 0; j < n; ++j)
          if (b[j] == b[i])
            c2++;
        if (c1 < c2) {
          a[cur] = b[i];
          print_multi_set_permutation(n, b, a, cur + 1);
        }
      }
    }
}
```