codeforces 339D 简单的线段树操作

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 m queries. 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 query.

Example
Input
2 4
1 6 3 5
1 4
3 4
1 2
1 2
Output
1
3
3
3
Note
For more information on the bit operations, you can follow this link: http://en.wikipedia.org/wiki/Bitwise_operation

题意：
输入n，m
以及2^n个数字组成的数组
接下来是m行
每行输入p，b，将a[p]的值改成b，然后计算数值，计算规则如下：

第一次计算b1 = a[1] | a[2] , b2 = a[ 3 ] | a[ 4 ] , b3 = a[ 5 ] | a[ 6 ] , b4 = a[ 7 ] | a[ 8 ]
第二次计算c1 = b1^b2 , c2 = b3^b4
第四次计算v = c1 | c2
输出v

简单来说 就是 | ^ 这两种运算符的交替运算，然后输出最终的数值

#include <bits/stdc++.h>using namespace std;const int MAXN = (1<<17)+10;int a[MAXN];int ff[MAXN<<2];int sum[MAXN<<2];void pb(int rt){    if(ff[rt]%2==0)    {        sum[rt]=sum[rt<<1|1]|sum[rt<<1];    }    else sum[rt]=sum[rt<<1|1]^sum[rt<<1];}void build(int l,int r,int rt){    sum[rt]=0;    ff[rt<<1]=ff[rt<<1|1]=0;    if(l==r)    {        sum[rt]=a[l];        ff[rt]=-1;        return ;    }    int mid=(l+r)>>1;    build(l,mid,rt<<1);    build(mid+1,r,rt<<1|1);    ff[rt]=ff[rt<<1]+1;    pb(rt);}void update(int l,int r,int p,int d,int rt){    if(l==r)    {        sum[rt]=d;        return ;    }    int mid=(l+r)>>1;    if(p<=mid) update(l,mid,p,d,rt<<1);    if(p>mid) update(mid+1,r,p,d,rt<<1|1);    pb(rt);}int main(){    int n,m;    scanf("%d%d",&n,&m);    n=(1<<n);    for(int i=1;i<=n;i++)        scanf("%d",&a[i]);    ff[1]=1;    build(1,n,1);    for(int i=1;i<=m;i++)    {        int p,d;        scanf("%d%d",&p,&d);        update(1,n,p,d,1);        printf("%d\n",sum[1] );    }}