自己写的时候码力不足，比较痛苦，但是没找到写的简单清晰的代码可以参考
于是憋了一下午写了一份，供他人参考
#include<iostream>
#include<vector>
#include<queue>
#include<tuple>
#include<map>
#include<array>
#include<algorithm>
#include<stack>
#include<bitset>
#include<set>
using namespace std;
#define ffp(x,y,z) for(int (x)=(y);(x)<=(z);(x++))
#define ll long long int
#define q_ read()
#define pii pair<int,int>
const ll MOD = 998244353;
const ll lINF = 0x3f3f3f3f3f3f3f3f;
const int iINF = 0x3f3f3f3f;
inline ll read()
{ll x = 0, f = 1;char ch = getchar();while (ch < '0' || ch>'9'){if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9')x = x * 10 + ch - '0', ch = getchar();return x * f;
}void solve()
{ll l = q_;ll r = q_;//递归求解//从大到小来填数字ll base = l-1;vector<ll>ans(r - base +1, 0);ffp(i, 1, r - base){ans[i] = i + base;}auto find = [&](auto&& find, ll L, ll R, ll bit)->void{if (R <= L) { return; }//查看bit是否在L和R之内 数字为连续，即判断，当前位在L到R是否符合一半一半的特征/* ll va = 1ll <<bit;while(check(L,R,bit) != 1){bit--;}va += 1ll << bit;*/ll va = 0;for (ll i = 32; i >= 0; i--){ll b = 1ll << i;if (((b & L) == 0) && ((b & R) != 0)){va |= b;break;}if (((b & L) != 0) && ((b & R) != 0)){va |= b;}}//查看左边和右边的数量ll cnt = min(R - va + 1,va - L);ffp(i, 1, cnt){swap(ans[va + i - 1 - base], ans[va - i - base]);}find(find, L, va - cnt - 1, bit - 1);find(find, va + cnt, R, bit - 1);};find(find, l, r, 33);ll res = 0;ffp(i, 1, r - l + 1){res += (ans[i] | (i + base));} cout << res << endl;ffp(i, 1, r - l + 1){cout << ans[i] << ' ';}cout << endl;return;
}int main()
{int t = 1;t = q_;while (t--){solve();}return 0;
}