单点更新线段树 RMQ

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

线段树：

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int N=200006;

int n, m, A[N];

struct Segment_Tree {
	struct Node {
		int s, t, dep, val;
		void init(int L, int R, int a) {
			s=L, t=R, dep=a, val=A[L];
		}
	} T[N<<2];

	void pushUp(Node &fa, Node &L, Node &R) {
		if(fa.dep&1) fa.val=L.val|R.val;
		else fa.val=L.val^R.val;
	}

	void build(int id, int dep, int L, int R) {
		T[id].init(L, R, dep);
		if(L==R) return;
		int mid=(L+R)>>1;
		build(id<<1, dep-1, L, mid);
		build(id<<1|1, dep-1, mid+1, R);
		pushUp(T[id], T[id<<1], T[id<<1|1]);
	}

	void update(int id, int pos, int val) {
		if(T[id].s==T[id].t) {
			T[id].val=val;
			return;
		}
		int mid=(T[id].s+T[id].t)>>1;
		if(pos<=mid) update(id<<1, pos, val);
		else update(id<<1|1, pos, val);
		pushUp(T[id], T[id<<1], T[id<<1|1]);
	}
} tree;

int main() {
//	freopen("in", "r", stdin);
	scanf("%d%d", &n, &m);
	int old=n;
	n=(1<<n);
	for(int i=1; i<=n; i++) scanf("%d", &A[i]);
	tree.build(1, old, 1, n);
	for(int i=0, p, q; i<m; i++) {
		scanf("%d%d", &p, &q);
		tree.update(1, p, q);
		printf("%d\n", tree.T[1].val);
	}
	return 0;
}

单点更新二叉树 模板

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#include <queue>
#include <deque>
#include <stack>
#include <complex>
#include <cassert>

using namespace std;

#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define sz(s) int((s).size())
#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define LLD "%lld"
#else
#define eprintf(...) 42
#define LLD "%I64d"
#endif
#define next _next
#define prev _prev
#define rank _rank
#define link _link
#define hash _hash

typedef long long ll;
typedef long long llong;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef complex <double> tc;

const int inf = int(1e9);
const double eps = 1e-9;
const double pi = 4 * atan(double(1));
const int N = 17;

int n;
int rmq[(1 << (N + 2))];

inline void update(int pos, int val){
	pos += (1 << n);
	rmq[pos] = val;
	int lev = 0;
	while(pos > 1){
		pos /= 2;
		if(lev == 0){
			rmq[pos] = (rmq[pos * 2] | rmq[pos * 2 + 1]);
		}
		else{
			rmq[pos] = (rmq[pos * 2] ^ rmq[pos * 2 + 1]);
		}
		lev = 1 - lev;
	}
}

int main(){
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	int m;
	scanf("%d %d", &n, &m);
	for(int i = 0; i < (1 << n); i++){
		int cur;
		scanf("%d", &cur);
		update(i, cur);
	}
	for(int i = 0; i < m; i++){
		int pos, val;
		scanf("%d %d", &pos, &val);
		pos--;
		update(pos, val);
		printf("%d\n", rmq[1]);
	}
	return 0;
}