D - Xenia and Bit Operations

Posted 2020-12-26 dreamkill

D. Xenia and Bit Operations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia the beginner programmer has a sequence a, consisting of 2n non-negative integers: a1,?a2,?...,?a2n. Xenia is currently studying bit operations. To better understand how they work, Xenia decided to calculate some value v for a.

Namely, it takes several iterations to calculate value v. At the first iteration, Xenia writes a new sequence a1 or a2,?a3 or a4,?...,?a2n?-?1 or a2n, consisting of 2n?-?1 elements. In other words, she writes down the bit-wise OR of adjacent elements of sequence a. At the second iteration, Xenia writes the bitwise exclusive OR of adjacent elements of the sequence obtained after the first iteration. At the third iteration Xenia writes the bitwise OR of the adjacent elements of the sequence obtained after the second iteration. And so on; the operations of bitwise exclusive OR and bitwise OR alternate. In the end, she obtains a sequence consisting of one element, and that element is v.

Let‘s consider an example. Suppose that sequence a?=?(1,?2,?3,?4). Then let‘s write down all the transformations (1,?2,?3,?4) ?→?(1 or 2?=?3,?3 or 4?=?7) ?→? (3 xor 7?=?4). The result is v?=?4.

You are given Xenia‘s initial sequence. But to calculate value v for a given sequence would be too easy, so you are given additional mqueries. Each query is a pair of integers p,?b. Query p,?b means that you need to perform the assignment ap?=?b. After each query, you need to print the new value v for the new sequence a.

Input

The first line contains two integers n and m (1?≤?n?≤?17,?1?≤?m?≤?105). The next line contains 2n integers a1,?a2,?...,?a2n (0?≤?ai?<?230). Each of the next m lines contains queries. The i-th line contains integers pi,?bi (1?≤?pi?≤?2n,?0?≤?bi?<?230) — the i-th query.

Output

Print m integers — the i-th integer denotes value v for sequence a after the i-th quer

 

 

线段树（刚学线段树拿过来练练手

 

 

#include <algorithm>

#include <stack>

#include <istream>

#include <stdio.h>

#include <map>

#include <math.h>

#include <vector>

#include <iostream>

#include <queue>

#include <string.h>

#include <set>

#include <cstdio>

#define FR(i,n) for(int i=0;i<n;i++)

#define MAX 2005

#define mkp pair <int,int>

using namespace std;

const int maxn = 1000000;

typedef long long ll;

const int  inf = 0x3fffff;

void read(ll &x) {

char ch; bool flag = 0;

for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == ‘-‘)) || 1); ch = getchar());

for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar());

x *= 1 - 2 * flag;

}

ll n;

int m;

struct inof{

  // int l,r;

   int flag;

   int val;

}Tree[maxn<<2];

int arr[maxn];

void build(int l,int r,int id)

{

    if(l==r)

    {

       Tree[id].flag=1;

       Tree[id].val=arr[l];

    }

    else

    {

    int mid = (l+r)/2;

    build(mid+1,r,2*id+1);

    build(l,mid,2*id);

    Tree[id].flag=Tree[id*2].flag^1;

    if(Tree[id].flag)

    {

        Tree[id].val=Tree[id*2].val^Tree[id*2+1].val;

    }

    else

    {

        Tree[id].val=Tree[id*2].val|Tree[id*2+1].val;

    }

    }

    //pushdown(l,r);

}

void update(int l,int r,int pos,int cnt,int id)

{

    if(l==r)

    {

        Tree[id].val=cnt;

    }

    else

    {

        int mid = (l+r)/2;

        if(pos<=mid)update(l,mid,pos,cnt,id*2);

        else update(mid+1,r,pos,cnt,id*2+1);

        if(Tree[id].flag)Tree[id].val=Tree[id*2+1].val^Tree[id*2].val;

        else Tree[id].val=Tree[id*2+1].val|Tree[id*2].val;

    }

}

int main() {

    read(n);

    cin>>m;

    n =(1<<n);

    for(int i=1;i<=n;i++){

        scanf("%d",&arr[i]);

    }//cout<<1<<endl;

    build(1,n,1);

    while(m--){

        int pos;int x;

        scanf("%d%d",&pos,&x);

        update(1,n,pos,x,1);

        printf("%d ",Tree[1].val);

   }

    return 0;

}