user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  lea    D31(%rip),%rdi
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    $0x1f,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x138,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  lea    0x18(%rsp),%rax
  mov    %rax,0x8(%rsp)
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jg     L15
  jmp    L142
  nopl   0x0(%rax)
L90:
  add    $0x1,%r14d
  cmp    %r14d,0x14(%rsp)
  jle    L142
L15:
  mov    0x8(%rsp),%rsi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x18(%rsp),%rax
  cmp    $0x1,%rax
  je     L57
L34:
  test   %rax,%rax
  je     L108
L72:
  cmp    $0x1,%rax
  jbe    L90
  lea    0x118(%rsp),%rcx
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L71:
  mov    %rax,%rsi
  add    $0x1,%edi
  sub    $0x8,%rcx
  shr    $0x3f,%rsi
  lea    (%rax,%rsi,1),%rdx
  add    %rsi,%rax
  and    $0x1,%edx
  sar    %rax
  sub    %rsi,%rdx
  mov    %rdx,0x8(%rcx)
  cmp    $0x1,%rax
  jne    L71
  mov    %r13d,%ecx
  lea    0x20(%rsp),%r12
  movq   $0x1,0x18(%rsp)
  sub    %edi,%ecx
  movslq %ecx,%rcx
  movq   $0x1,0x20(%rsp,%rcx,8)
  cmp    $0x1e,%edi
  jg     L45
  and    $0x1fffffff,%ecx
  xor    %eax,%eax
  mov    %r12,%rdi
  rep stos %rax,%es:(%rdi)
L45:
  mov    %r12,%rax
  lea    0x120(%rsp),%rcx
  jmp    L91
  nopw   0x0(%rax,%rax,1)
L96:
  movq   $0x1,(%rax)
L74:
  add    $0x8,%rax
  cmp    %rcx,%rax
  je     L86
L91:
  mov    (%rax),%rdx
  test   %rdx,%rdx
  je     L96
  cmp    $0x1,%rdx
  jne    L74
  movq   $0x0,(%rax)
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L91
  xchg   %ax,%ax
L86:
  xor    %r15d,%r15d
  xor    %ebp,%ebp
  nopl   (%rax)
L93:
  mov    %r13d,%esi
  mov    (%r12,%r15,8),%rbx
  mov    $0x2,%edi
  sub    %r15d,%esi
  add    $0x1,%r15
  call   <pow@plt>
  imul   %rax,%rbx
  add    %rbx,%rbp
  cmp    $0x20,%r15
  jne    L93
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  add    $0x1,%r14d
  call   <printf@plt>
  cmp    %r14d,0x14(%rsp)
  jg     L15
L142:
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L66
  add    $0x138,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L108:
  mov    $0xffffffff,%esi
  lea    D20(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  jmp    L72
  nopl   0x0(%rax,%rax,1)
L57:
  mov    $0xfffffffe,%esi
  lea    D20(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  jmp    L34
L66:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D31
type_infered: string
name: read_data
arguments: tag: D39
type_infered: string
name: read_data
arguments: tag: D20
type_infered: string

tool: D31: %d

tool: D39: %ld

tool: D20: %ld


assistant: ```c
int main() {
  int test, i;
  scanf("%d", &test);
  for (i = 0; i < test; i++) {
    long int n;
    scanf("%ld", &n);
    int j = 0;
    long int c = 0;
    long int arr[32];
    if (n == 1) {
      long int s = 4294967294;
      printf("%ld\n", s);
    }
    if (n == 0) {
      long int g = 4294967295;
      printf("%ld\n", g);
    }
    if (n != 1 && n != 0) {
      while (c != 1) {
        arr[31 - j] = n % 2;
        c = n / 2;
        n = n / 2;
        j = j + 1;
      }
      arr[31 - j] = 1;
      int k;
      for (k = 0; k <= 31 - j - 1; k++) {
        arr[k] = 0;
      }
      int u;
      for (u = 0; u < 32; u++) {
        if (arr[u] == 0) {
          arr[u] = 1;
        } else if (arr[u] == 1) {
          arr[u] = 0;
        }
      }
      int h;
      long int count = 0;
      for (h = 0; h < 32; h++)
        count = count + arr[h] * (pow(2, 31 - h));
      printf("%ld\n", count);
    }
  }
  return 0;
}
```