Codeforces 460d little Victor and set (construction, enumeration)

Almost none of the recent CF attacks, and I feel quite confused. However, I seem to be in a bad state. I can only know how to solve the problem.

InputL, Bytes,R, Bytes,K(1 digit ≤ DigitLLimit ≤ limitRLimit ≤ limit 10^{12; 1 ≤ bytesKLimit ≤ limitMin(106. Explain,RAccept-Encoding-LRows + rows 1 )).}

Selecting at most K numbers from [L, R] minimizes the exclusive or value of the selected number, and outputs the smallest exclusive OR value and the scheme.

 

Classification: first, if R-l + 1 <= 4, the enumeration set is solved.

First, we will discuss the situation of R-l + 1> = 5:

At this time, you can select at least five numbers. Therefore, at least four consecutive numbers meet the requirements of 2x, 2x + 1, 2x + 2, and 2x + 3.

When K = 1, it is clear that the solution is {L }. When K = 2, it is clear that the solution is {2x, 2x + 1 }. K> = 4, obviously the solution is {2x, 2x + 1, 2x + 2, 2x + 3 }.

When K = 3, consider the following:

First, the exclusive or value is 1 (refer to k = 2)

We are looking for whether the difference or value can be 0. If yes, it is obvious that three numbers are selected. Set x> Y> Z.

111... 1111

111... 1110

000... 0001

Obviously, the first half of X, Y, and Z must be like this, but since we want to make X, Y, and Z as close as possible, the first half of X, Y, and Z must be as follows:

11

10

01

After that, each time you add a digit, it may be Yi = Zi = 1, xi = 0 or xi = Zi = 1, YI = 0 or xi = YI = 1, Zi = 0.

Because X, Y, and Z are as close as possible, YI = Zi = 1, Zi = 0 is obviously adopted.

Therefore, the binary format of X, Y, and Z is as follows:

110... 0

101... 1

011... 1

At this point, the problem is roughly solved. The remaining issues are some details, but the problem is not serious.

 

#include <cstdio>#include <cstring>#include <iostream>#include <vector>using namespace std;#define ll long longint cnt(int i){    int ret=0;    while(i) i-=i&(-i), ++ret;    return ret;}int main(){    ll l,r;    int k;    while(~scanf("%I64d%I64d%d",&l,&r,&k)){        if(r-l+1<5){            int n=r-l+1;            ll ansxor=1ll<<60;            vector<ll>val;            for(int i=1;i<(1<<n);++i){                ll xx=0;                for(int j=0;j<n;++j)                    if(i&(1<<j)) xx^=l+j;                if(xx<ansxor && cnt(i)<=k){                    ansxor=xx;                    val.clear();                    for(int j=0;j<n;++j) if(i&(1<<j)) val.push_back(l+j);                }            }            printf("%I64d\n",ansxor);            printf("%d\n",val.size());            for(int i=0;i<val.size();++i) printf("%I64d%c",val[i],i==val.size()-1?‘\n‘:‘ ‘);        }        else if(r-l+1>=5){            if(k==1){printf("%I64d\n1\n%I64d\n",l,l);continue;}            if(k==2){                if(l&1) l++;                puts("1");                puts("2");                printf("%I64d %I64d\n",l,l+1);            }            else if(k>=4){                if(l&1) l++;                puts("0");                puts("4");                printf("%I64d %I64d %I64d %I64d\n",l,l+1,l+2,l+3);            }            else if(k==3){                ll x=-1,y,z;                for(ll i=3;i<=r;i=i<<1){                    if((i^(i-1))>=l){                        x=i;                        y=i - 1;                        z=i^(i-1);                        break;                    }                }                if(x!=-1){                    puts("0");                    puts("3");                    printf("%I64d %I64d %I64d\n",x,y,z);                }                else {                    if(l&1) l++;                    puts("1");                    puts("2");                    printf("%I64d %I64d\n",l,l+1);                }            }        }    }    return 0;}

 

Codeforces 460d little Victor and set (construction, enumeration)