暴力+构造

If r?-?l?≤?4 we can all subsets of size not greater than k. Else, if k?=?1, obviously that answer is l. If k?=?2, answer is 1, because xor of numbers 2x and 2x?+?1 equls 1. If k?≥?4 answer is 0 because xor of to pairs with xor 1 is 0.

If k?=?3, we can choose numbers 2x and 2x?+?1 with xor 1. So we need to know, if we can get xor equals 0. Suppose that there are 3 such numbers x, y and z (r?≥?x?>?y?>?z?≥?l) with xor equals 0. Consider the most non-zero bit of numberx. At the same bit of y it‘s also 1, because xor equls 0, and y?>?z. Consider the next bit of numbers. If z have 0 there, we have to do next: set the previous bit of numbers x and y equals 0, and set current bit equals 1. Obviously xor still equals 0, z hadn‘t changed and numbers x and y stood closer to z, so they are still at [l,?r].And x?>?y.Consider the next bit of numbers. If z has zero here than we will change most bits of x и y at the same way and so on. z?>?0, so somewhen we will get to bit in which z has 1. Since xorequals 0, the same bit of x would be 1 because x?>?y, and y would have 0 there. At the next bits we will change bit in x to 0, and in numbers y and z to 1.Finally z would be greater or equal than before, and x would be less or greater than before, and x?>?y?>?z would be correct. So, we have the next: if such numbers x, y and z exist than also exist numbers:

1100…000

1011…111

0111…111

with xor equals 0. There are not much such triples, so we can check them.

8 15 3

1
2
10 11

8 30 7

0
5
14 9 28 11 16

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

LL L,R,K;
LL ans=0x3f3f3f3f3f3f3f3f;

int main()
{
    cin>>L>>R>>K;
    if(R-L+1<=4)
    {
        LL m=R-L+1;
        LL sig=0;
        for(LL i=1;i<(1LL<<m);i++)
        {
            LL temp=0;
            LL wei=0;
            LL si=i;
            while(si)
            {
                wei++;
                si=si&(si-1);
            }
            if(wei>K) continue;
            for(LL j=0;j<m;j++)
            {
                if(i&(1LL<<j))
                {
                    temp^=L+j;
                }
            }
            if(temp<ans)
            {
                ans=temp;
                sig=i;
            }
        }
        cout<<ans<<endl;
        LL wei=0;
        LL tsig=sig;
        while(tsig)
        {
            wei++;
            tsig=tsig&(tsig-1);
        }
        cout<<wei<<endl;
        for(LL i=0;i<m;i++)
        {
            if(sig&(1<<i))
            {
                cout<<L+i<<" ";
            }
        }
        cout<<endl;
    }
    else
    {
        if(K==1)
        {
            cout<<L<<endl;
            cout<<1<<endl;
            cout<<L<<endl;
        }
        else if(K==2)
        {
            if(L%2) L++;
            cout<<1<<endl;
            cout<<2<<endl;
            cout<<L<<" "<<L+1<<endl;
        }
        else if(K==3)
        {
            bool flag=false;

            LL mx=3,mi=1;
            while(mx<=R)
            {
                if(mi>=L)
                {
                    flag=true;
                    cout<<0<<endl<<3<<endl;
                    cout<<mx<<" "<<mx-1<<" "<<mi<<endl;
                    break;
                }

                mx<<=1LL;
                mi<<=1LL; mi++;
            }

            if(flag==false)
            {
                if(L%2) L++;
                cout<<1<<endl;
                cout<<2<<endl;
                cout<<L<<" "<<L+1<<endl;
            }
        }
        else
        {
            cout<<0<<endl;
            cout<<4<<endl;
            if(L%2) L++;
            cout<<L<<" "<<L+1<<" "<<L+2<<" "<<L+3<<endl;
        }
    }
    return 0;
}

Codeforces 460 D. Little Victor and Set