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
  push   %r15
  push   %r14
  lea    -0x3c(%rbp),%rsi
  push   %r13
  lea    D28(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3c(%rbp)
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rax
  mov    %rsp,%rsi
  lea    0xf(,%rax,4),%rdx
  mov    %rax,%rcx
  mov    %rdx,%rax
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rsi
  and    $0xfffffffffffffff0,%rax
  cmp    %rsi,%rsp
  je     L139
L169:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L169
L139:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L103
L70:
  mov    %rsp,%r8
  test   %ecx,%ecx
  jle    L50
  mov    %r8,%r14
  mov    %r8,%r12
  xor    %ebx,%ebx
L76:
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%r12
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  cmp    %ebx,%eax
  jg     L76
  test   %eax,%eax
  jle    L50
  movl   $0x0,-0x44(%rbp)
  xor    %r15d,%r15d
L108:
  mov    (%r14),%edi
  mov    $0x4,%esi
  call   <calloc@plt>
  xor    %esi,%esi
  mov    %rax,%r12
  mov    (%r14),%eax
  mov    %r12,%rdi
  lea    0x0(,%rax,4),%edx
  call   <memset@plt>
  mov    (%r14),%eax
  cmp    $0x2,%eax
  jle    L182
  lea    0xc(%r12),%r9
  mov    $0x3,%edi
  cmp    $0x3,%eax
  je     L182
  nopl   (%rax)
L143:
  mov    %edi,%esi
  mov    $0x2,%ecx
  nop
L146:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L129
  cltd
  mov    %eax,%esi
  idiv   %ecx
  test   %edx,%edx
  jne    L129
  mov    %edi,(%r9)
  add    $0x1,%r15d
L129:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L146
  add    $0x1,%edi
  add    $0x4,%r9
  cmp    %edi,(%r14)
  jg     L143
L182:
  lea    (%r15,%r15,1),%esi
  lea    D12(%rip),%rdi
  mov    %r12,%r13
  xor    %ebx,%ebx
  call   <printf@plt>
  mov    (%r14),%eax
  test   %eax,%eax
  jle    L67
  nopl   0x0(%rax,%rax,1)
L161:
  mov    0x0(%r13),%esi
  test   %esi,%esi
  jne    L95
  add    $0x1,%ebx
  add    $0x4,%r13
  cmp    %ebx,%eax
  jg     L161
L67:
  mov    %r12,%rdi
  add    $0x4,%r14
  call   <free@plt>
  addl   $0x1,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  cmp    %eax,-0x3c(%rbp)
  jg     L108
L50:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L90
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nop
L95:
  lea    D4(%rip),%rdi
  add    $0x1,%ebx
  add    $0x4,%r13
  call   <printf@plt>
  mov    (%r14),%eax
  cmp    %ebx,%eax
  jg     L161
  jmp    L67
L103:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L70
L90:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D28
type_infered: string
name: unpack_data
arguments: data_label: D12
type_infered: string
name: unpack_data
arguments: data_label: D4
type_infered: string

tool: D28: %d

tool: D12: %d


tool: D4: %d 

assistant: ```c
int main() {
  int T = 0;
  scanf("%d", &T);
  int *r;
  char buffer[20];
  int t[T];
  for (int i = 0; i < T; i++) {
    scanf("%d", &t[i]);
  }
  int count = 0;
  int factor_count = 0;
  for (int i = 0; i < T; i++) {
    r = (int *)calloc(t[i], sizeof(int));
    memset(r, 0, t[i] * sizeof(int));
    for (int j = 2; j < t[i]; j++) {
      int num = j;
      for (int k = 2; k < j; k++) {
        int sub_count = 0;
        while (num % k == 0) {
          if (sub_count > 0) {
            count++;
            *(r + j) = j;
            break;
          }
          num /= k;
          sub_count++;
        }
      }
    }
    printf("%d\n", (count * 2));
    for (int k = 0; k < t[i]; k++) {
      if (*(r + k) != 0) {
        printf("%d ", *(r + k));
      }
    }
    free(r);
  }
  return 0;
}
```