codeforces#1157D. Ehab and the Expected XOR Problem（构造）

Topic links:

http://codeforces.com/contest/1174/problem/D

Meaning of the questions:

A sequence configuration satisfies the following conditions

• His XOR value of all the sub-segments is not equal to $ x $
• $1 \le a_i<2^n$

A maximum output of such a sequence

data range:

$ 1 \ n \ the $ 18
$ 1 \ the x <2 ^ {18} $

analysis: 

When the game confused $ subsegment $ and $ subsequence $, former sub-paragraph must be continuous, the latter sequence may be discontinuous

After the match official to see the solution to a problem

Sequence hypotheses is configured $ a_i $, it is a prefix and XOR $ $ B_i

即：$b_i=a_1\bigoplus a_2 \bigoplus a_3\bigoplus a_4.....\bigoplus a_i$

$ B_i $ must meet the following criteria

• No repeated elements namely $ b_i \ neq b_j $
• Not one of the elements of the exclusive OR value of $ X $
• There is no $ x $

On the second, we can know that if $ g $ array added $ b $, then $ g \ bigoplus x $ is not $ b $ array, so choose which one of these two numbers on the line

After obtaining the array b $ a_i = b_i \ oplus b_ {i-1} $

ac Code:

#include<bits/stdc++.h>
#define ll long long
#define pa pair<int,int>
using namespace std;
const int maxn=1e5+10;
const int maxm=1e6+10;
const ll mod=998244353;
int ans[maxn];
inline ll cal(int st,int len)
{
    return (ll)len*(2*st+len-1)/2;
}
int main()
{
    int n,k;
    while(scanf("%d %d",&n,&k)==2)
    {
        if(cal(1,k)>n)
        {
            printf("NO\n");
            continue;
        }
        if(n==4&&k==2)
        {
            printf("NO\n");
            continue;
        }
        else if(n==8&&k==3)
        {
            printf("NO\n");
            continue;
        }
        int st=1,en=n;
        while(st!=en)
        {
            int md=(st+en)/2;
            if(cal(md+1,k)<=n)st=md+1;
            else en=md;
        }
        for(int i=1;i<=k;i++)
            ans[i]=st+i-1;
        int now=n-cal(st,k),inde=k;
        while(now)
        {
            if(ans[inde]+1<=2*ans[inde-1])ans[inde]++,inde--,now--;
            else inde=k;
        }
        printf("YES\n");
        for(int i=1;i<=k;i++)
        {
            printf("%d",ans[i]);
            if(i==k)printf("\n");
            else printf(" ");
        }
    }
    return 0;
}