Xenia and Bit Operations
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：输入一个数n，然后有m个操作，然后接下来有2^n个数d的序列a[1<
#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const double pi=acos(-1.0);
const int maxn=(1<<17)+1;
int M[maxn*2];
bool lazy[maxn*2];//标记
int n,m;
void get_up(int rt){//向上更新
if (lazy[rt*2]==0){//如过其儿子是0，说明合并的时候是用或运算
M[rt]=M[rt*2]|M[rt*2+1];
}
else M[rt]=M[rt*2]^M[rt*2+1];//不然就是异或运算
lazy[rt]=lazy[rt*2]^1;
}
void build (int rt,int l,int r){
if (r==l){
scanf("%d",&M[rt]);//输入
lazy[rt]=0;//标记如果是0其父节点更新的时候就是或（|），1就是亦或（^）
return ;
}
int mid=(l+r)/2;
build(rt*2,l,mid);
build (rt*2+1,mid+1,r);
get_up(rt);
}
void update(int rt,int l,int r,int p,int v){//将第p个点的值改为v
if (l==r){
M[rt]=v;//找到就修改
}
else {
int mid=(l+r)/2;
if (p<=mid){//p在左区间
update(rt*2,l,mid,p,v);
}
else update(rt*2+1,mid+1,r,p,v);//点在右区间
if (lazy[rt*2]==0){//更新节点
M[rt]=M[2*rt]|M[2*rt+1];
}
else M[rt]=M[rt*2]^M[2*rt+1];
}
}
int main()
{ int p,b;
while (scanf ("%d%d",&n,&m)!=EOF){
build(1,1,1<<(n));
for (int i=0;i<m;i++){
scanf ("%d%d",&p,&b);
update(1,1,1<<(n),p,b);
printf ("%d\n",M[1]);
}
}
return 0;
}