CF1174D Ehab and the Expected XOR Problem - 构造

Posted 2020-11-16 lrj124

题面

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.

题意

给两个数 (n) 和 (x)，构造一个满足以下条件的序列：
·对任何序列中的元素 (a_i)，1leq a_i<2^n1≤a_i<2^n$
·序列中没有非空连续子序列异或和为 (0) 或 (x)
·序列长度 (l) 应该最大

思路

思路比较巧妙，因为元素可重复不太好搞，就考虑构造一个答案序列 (a) 的异或前缀和 (b)，且 (b) 满足任意 (b_i xor b_j ot= x) 或 (0)。
因为若 (a xor b = c)，则 (a xor c = b)，所以从 (1) 枚举到 (2^n-1) ，每次用可行的 (i) 数加入答案并排除 (i xor x) 这个数。

代码

/************************************************
*Author        :  lrj124
*Created Time  :  2019.10.15.19:28
*Mail          :  [e-mail]
*Problem       :  cf1174d
************************************************/
#include <cstdio>
const int maxn = 1<<18;
int n,x,ans[maxn];
bool vis[maxn];
int main() {
    //freopen("cf1174d.in","r",stdin);
    //freopen("cf1174d.out","w",stdout);
    scanf("%d%d",&n,&x);
    vis[0] = vis[x] = true;
    for (int i = 1;i < 1<<n;i++)
        if (!vis[i]) {
            vis[i^x] = true;
            ans[++ans[0]] = i;
        }
    printf("%d
",ans[0]);
    for (int i = 1;i <= ans[0];i++) printf("%d ",ans[i]^(i ^ 1 ? ans[i-1] : 0));
    return 0;
}