[codeforces 339]D. Xenia and Bit Operations

[codeforces 339]D. Xenia and Bit Operations

试题描述

Xenia the beginner programmer has a sequence a, consisting of 2ⁿ non-negative integers: a₁, a₂, ..., a_2ⁿ. 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, ..., a_2^n - 1 or a_2ⁿ, 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 a_p = b. After each query, you need to print the new value v for the new sequence a.

输入

The first line contains two integers n and m (1 ≤ n ≤ 17, 1 ≤ m ≤ 10⁵). The next line contains 2ⁿ integers a₁, a₂, ..., a_2ⁿ (0 ≤ a_i < 2³⁰). Each of the next m lines contains queries. The i-th line contains integers p_i, b_i (1 ≤ p_i ≤ 2ⁿ, 0 ≤ b_i < 2³⁰) — the i-th query.

输出

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

输入示例

2 4
1 6 3 5
1 4
3 4
1 2
1 2

输出示例

1
3
3
3

数据规模及约定

见“输入”

题解

我们发现“运算之后得到新序列”这一过程只会进行 n 次，事实上，这整个过程就像一个倒过来的完全二叉树，我们不妨将初始序列得到的这颗完全二叉树建出来；以后每改动一个数字即改变了一个叶节点的权值，受到影响需要改动的节点为该叶节点到根节点（最终答案）的路径，于是每次修改时间复杂度也是 O(n) 的。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
    if(Head == Tail) {
        int l = fread(buffer, 1, BufferSize, stdin);
        Tail = (Head = buffer) + l;
    }
    return *Head++;
}
int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

#define maxn 20
#define maxm 1048576
int n, q, ToT, A[maxm], fa[maxm];

int main() {
	n = read(); q = read();
	for(int i = 0; i < (1 << n); i++) A[i] = read();
	
	memset(fa, -1, sizeof(fa));
	ToT = 1 << n; bool cur = 0;
	int nl = 0, nr = (1 << n) - 1;
	for(int i = 1; i <= n; i++, cur ^= 1) {
		int nnl = -1, nnr = -1;
		for(int j = nl; j < nr; j += 2) {
			if(cur) A[ToT++] = A[j] ^ A[j^1], fa[j] = fa[j^1] = ToT - 1;
			else A[ToT++] = A[j] | A[j^1], fa[j] = fa[j^1] = ToT - 1;
			if(nnl < 0) nnl = ToT;
			nnr = ToT;
		}
		nl = nnl; nr = nnr;
	}
//	printf("%d
", ToT);
	
	while(q--) {
		int p = read() - 1; A[p] = read();
		cur = 0;
		while(fa[p] >= 0) {
			if(cur) A[fa[p]] = A[p] ^ A[p^1];
			else A[fa[p]] = A[p] | A[p^1];
			p = fa[p]; cur ^= 1;
		}
		printf("%d
", A[ToT-1]);
	}
	
	return 0;
}