Codeforces Round #430 D. Vitya and Strange Lesson
2017-08-30 20:09
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Today at the lesson Vitya learned a very interesting function — mex. Mex of a sequence of numbers is the minimum non-negative number that is not present in the sequence as element. For example, mex([4, 33, 0, 1, 1, 5]) = 2 and mex([1, 2, 3]) = 0.
Vitya quickly understood all tasks of the teacher, but can you do the same?
You are given an array consisting of n non-negative integers, and m queries. Each query is characterized by one number x and consists of the following consecutive steps:
Perform the bitwise addition operation modulo 2 (xor) of each array element with the number x.
Find mex of the resulting array.
Note that after each query the array changes.
Input
First line contains two integer numbers n and m (1 ≤ n, m ≤ 3·105) — number of elements in array and number of queries.
Next line contains n integer numbers ai (0 ≤ ai ≤ 3·105) — elements of then array.
Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·105).
Output
For each query print the answer on a separate line.
题目大意:
定义mex数为数组中第一个没有出现的非负整数.有m个操作,每个操作有一个x,将数组中所有的元素都异或x,然后询问当前的mex
解题报告:
考场上搞了一个小时,原来看错题了,其实只是简单的Trie树基本操作,
mex数:如果左子树没满直接走左子树,不然就走右子树.
异或操作:如果x的该位为1,交换该节点的左右子树,打上标记即可
Vitya quickly understood all tasks of the teacher, but can you do the same?
You are given an array consisting of n non-negative integers, and m queries. Each query is characterized by one number x and consists of the following consecutive steps:
Perform the bitwise addition operation modulo 2 (xor) of each array element with the number x.
Find mex of the resulting array.
Note that after each query the array changes.
Input
First line contains two integer numbers n and m (1 ≤ n, m ≤ 3·105) — number of elements in array and number of queries.
Next line contains n integer numbers ai (0 ≤ ai ≤ 3·105) — elements of then array.
Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·105).
Output
For each query print the answer on a separate line.
题目大意:
定义mex数为数组中第一个没有出现的非负整数.有m个操作,每个操作有一个x,将数组中所有的元素都异或x,然后询问当前的mex
解题报告:
考场上搞了一个小时,原来看错题了,其实只是简单的Trie树基本操作,
mex数:如果左子树没满直接走左子树,不然就走右子树.
异或操作:如果x的该位为1,交换该节点的左右子树,打上标记即可
#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #include <cmath> #define RG register #define il inline #define iter iterator #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) using namespace std; const int N=6e6+10,maxdep=21; int gi(){ int str=0;char ch=getchar(); while(ch>'9' || ch<'0')ch=getchar(); while(ch>='0' && ch<='9')str=(str<<1)+(str<<3)+ch-48,ch=getchar(); return str; } struct node{ int l,r,s,rev; }t ; int n,root=0,tot=0,w[30],m; void insert(int &rt,int x,int d){ if(!rt)rt=++tot; if(d==-1){ t[rt].s=1;return ; } if(x&w[d])insert(t[rt].r,x,d-1); else insert(t[rt].l,x,d-1); t[rt].s=t[t[rt].l].s&t[t[rt].r].s; } void pushdown(int rt,int d){ if(!t[rt].rev)return ; int k=t[rt].rev; t[t[rt].l].rev^=k;t[t[rt].r].rev^=k; if(d>=1 && (k&w[d-1])){ swap(t[t[rt].l].l,t[t[rt].l].r);swap(t[t[rt].r].l,t[t[rt].r].r); } t[rt].rev=0; } int query(int rt,int d){ if(d==-1)return 0; pushdown(rt,d); if(!t[t[rt].l].s)return query(t[rt].l,d-1); return query(t[rt].r,d-1)+w[d]; } void work() { int x; n=gi();m=gi(); w[0]=1;for(int i=1;i<=maxdep;i++)w[i]=w[i-1]<<1; for(int i=1;i<=n;i++){ x=gi();insert(root,x,maxdep); } while(m--){ scanf("%d",&x); t[root].rev^=x; printf("%d\n",query(root,maxdep)); } } int main() { work(); return 0; }
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