CodeForces 339D Xenia and Bit Operations（线段树）

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 query.

Sample test(s)

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：

给你一对n，m。输入2^n个数，进行m次操作。

2^n个数进行先或再进行异或再进行或运算的方式最终求得一个结果并输出。

m次操作分别是输入p  和  b两个数，将p位置的数值改为b，在输出或和异或后的结果。

思路：

线段树单点更新就行了。

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn=1e6+5;
int flag=0;int a[maxn];
struct Seg{
    int l,r;
    int val,flag;
}seg[maxn<<2];

void build(int l,int r,int node,int flag)//构造线段树
{
    seg[node].l=l;seg[node].r=r;
    if(l==r){seg[node].val = a[l] ;return ;}
    int mid=(l+r)/2;
    build(l,mid,node<<1,1-flag);
    build(mid+1,r,node<<1|1,1-flag);
    if(flag==0)      //flag标记下，如果flag=0，那么这一层为或运算，否则为异或运算
    {seg[node].val=seg[node<<1].val|seg[node<<1|1].val;seg[node].flag=flag;}
    if(flag==1)
    {seg[node].val=seg[node<<1].val^seg[node<<1|1].val;seg[node].flag=flag;}
}
void query(int q,int k,int node)
{
    if(seg[node].l==q&&seg[node].r==q)
    {
        seg[node].val=k;return ;
    }
    int mid=(seg[node].l+seg[node].r)/2;
    if(q<=mid)
    query(q,k,node<<1);
    else
    query(q,k,node<<1|1);
    if(seg[node].flag==0)//更新节点，利用构造线段树时打下的标记
    seg[node].val=seg[node<<1].val|seg[node<<1|1].val;
    else
    seg[node].val=seg[node<<1].val^seg[node<<1|1].val;
}

void show(int node)//编译时检查的函数，可忽略
{

    printf("Node:%d\n",node);
    printf("Val:%d\n",seg[node].val);
    printf("Left:%d\n",seg[node].l);
    printf("Right:%d\n",seg[node].r);
    printf("Flag:%d\n",seg[node].flag);
    printf("\n");
    if(seg[node].l==seg[node].r)return ;
    show(node<<1);
    show(node<<1|1);
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        int nn=pow(2,n);
        for(int i=1;i<=nn;i++)
        {
            scanf("%d",&a[i]);
        }
        if(n%2==0)
        build(1,nn,1,1);
        else
        build(1,nn,1,0);
        //show(1);
        while(m--)
        {
            int c1,c2;
            scanf("%d%d",&c1,&c2);
            query(c1,c2,1);
            printf("%d\n",seg[1].val);//输出根节点的值就ok了
        }
    }


    return 0;
}