D. Ehab and the Expected XOR Problem
Given two integers n and x, construct an array that satisfies the following conditions:
- for any element ai in the array, 1≤ai<2^n
- there is no non-empty subsegment with bitwise XOR equal to 0 or x,
- its length l should be maximized.
A sequence b is a subsegment of a sequence a if b can be obtained from a by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The only line contains two integers n and x (1≤n≤18 1≤x<2^18).
Output
The first line should contain the length of the array l.
If l is positive, the second line should contain l space-separated integers a1, a2, ……, al (1≤ai<2^n) — the elements of the array a.
If there are multiple solutions, print any of them.
solution
这道题当时候在写的时候并不知道怎么写，题目大意就是要求子区间的异或和不能够0或者x，求最长的区间数字。
首先对于这道题我们可以得到一个显而易见的结论那就是，对于所有的数字进行前缀异或和，如果前缀数组中不存在相同的两个数字，那么这个序列的连续子区间的异或和肯定不会为零，那么对于x>=2^n的情况显然最长的长度就是2^n-1。而当x<2^n时，我们得到x的情况当且仅当sum[l]^sum[r]=x时，我们才会出现异或和为0的情况，那么显然我们可以要a^b=x的情况只能存在a,b中间的一个即可，所以我们这样就可以确定我们的前缀异或和数组了。那么求原数组a[i]=sum[i]^sum[i-1]即可，代码如下：
1 #include<bits/stdc++.h>
2 #include<vector>
3 using namespace std;
4 long long n,x;
5 bool pd[2000010];
6 int main(){
7 cin>>n>>x;
8 memset(pd,0,sizeof pd);
9 long long cnm=(1<<n);
10 for (int i=0;i<cnm;i++)
11 if (!pd[i])
12 pd[i^x]=1;
13 int ans=0;
14 for (int i=1;i<cnm;i++)
15 if (!pd[i])
16 ans++;
17 cout<<ans<<endl;
18 int l=0;
19 for (int i=1;i<cnm;i++)
20 if (!pd[i]) {
21 cout<<(i^l)<<‘ ‘;
22 l=i;
23 }
24 }
原文地址：https://www.cnblogs.com/beafreeman/p/11001328.html