D2. Max Sum OR (Hard Version) （应该写的比较清晰）

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<map>
#include<algorithm>
#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
    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)->void
        {
            if (R <= L) { return; }
            //查看va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 = 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 - base] , ans[va + i - 1 - base]);
            }

            find(find, L, va - cnt - 1);
            find(find, va + cnt, R);
        };

    find(find, l, r);
    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;
}