CodeForces 339D Xenia and Bit Operations 数据结构+线段树+点更新

CodeForces 339D Xenia and Bit Operations

Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

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Description

Xenia the beginner programmer has a sequence a, consisting of2ⁿ non-negative integers:a₁, a₂, ..., a_2ⁿ. Xenia is currently studying bit operations. To better understand how they work, Xenia decided to calculate some valuev 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, ..., a_2ⁿ - 1 or a_2ⁿ, consisting of 2^n - 1 elements. In other words, she writes down the bit-wise OR of adjacent elements of sequencea. At the second iteration, Xenia writes the bitwiseexclusive 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 isv.

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 isv = 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 additionalm queries. Each query is a pair of integersp, b. Queryp, b means that you need to perform the assignmenta_p = b. After each query, you need to print the new valuev for the new sequencea.

Input

The first line contains two integers n andm(1 ≤ n ≤ 17, 1 ≤ m ≤ 10⁵). The next line contains2ⁿ integersa₁, a₂, ..., a_2ⁿ(0 ≤ a_i < 2³⁰). Each of the nextm lines contains queries. Thei-th line contains integers p_i, b_i(1 ≤ p_i ≤ 2ⁿ, 0 ≤ b_i < 2³⁰) — thei-th query.

Output

Print m integers — the i-th integer denotes value v for sequencea after thei-th query.

Sample Input

Input

2 41 6 3 51 43 41 21 2

Output

1333

Hint

For more information on the bit operations, you can follow this link: http://en.wikipedia.org/wiki/Bitwise_operation

解题思路：

1，这题就是，首先给你2^n个数，第一层进行或运算，得到2^(n-1)个数，第二层进行异或运算，得到2^(n-2)个数，

第一层进行或运算，得到2^(n-3)个数直到只剩一个数，然后输出那个数。

2，或运算和异或运算是交替在每一层进行的。

3，有两个操作，一个是点赋值，另一个是查询

4，很明显就是线段树的点更新，核心的地方可能就是如何确定哪里进行异或/或操作吧，这个用一个数组标记也很好实现

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int maxn = 1<<17 ;int n,m;int sum[maxn<<2] ;int op[maxn<<2] ;int spow(int x){    int ret = 1 ;    for(int i=0;i<x;i++){        ret*=2 ;    }    return ret ;}void PushUp(int rt){    //printf("op:  %d = %d\n",rt,op[rt]);    if(op[rt]==0){        sum[rt] = (sum[rt<<1]^sum[rt<<1|1]) ;    }else{        sum[rt] = (sum[rt<<1]|sum[rt<<1|1]) ;        //printf("rwet:  %d = %d\n",rt,sum[rt]);    }}void build(int l,int r,int rt){    if(l==r){        scanf("%d",&sum[rt]);        //printf("rt:  %d = %d\n",rt,sum[rt]);        op[rt]==0 ;        return ;    }    int m = (l+r)>>1 ;    build(lson);    build(rson);    if(op[rt<<1]==0)op[rt]=1 ;    else op[rt] =0 ;    PushUp(rt) ;}void update(int p,int setv,int l,int r,int rt){    if(l==r){        sum[rt]=setv ;        return ;    }    int m = (l+r)>>1 ;    if(p<=m)update(p,setv,lson) ;    else update(p,setv,rson) ;    PushUp(rt) ;}int main(){    //freopen("in.txt","r",stdin);    while(~scanf("%d%d",&n,&m)){        n = spow(n) ;        //printf("n = %d\n",n);        build(1,n,1) ;        for(int i=0;i<m;i++){            int a ,b ;            scanf("%d%d",&a,&b);            update(a,b,1,n,1) ;            printf("%d\n",sum[1]);        }    }    return 0;}