[Codeforces 339D] Xenia and Bit Operations

[题目链接]

        https://codeforces.com/problemset/problem/339/D

[算法]

        线段树模拟即可

        时间复杂度 ：O(MN)

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 18
const int MAXS = 1 << MAXN;

int n,m;
int a[MAXS];

struct SegmentTree
{
        struct Node
        {
                int l,r;
                int value,d;        
        }    Tree[MAXS << 2];
        inline void update(int index)
        {
                if (Tree[index].d == 1) Tree[index].value = Tree[index << 1].value | Tree[index << 1 | 1].value;
                else Tree[index].value = Tree[index << 1].value ^ Tree[index << 1 | 1].value;
        }
        inline void build(int index,int l,int r)
        {
                Tree[index].l = l;
                Tree[index].r = r;
                if (l == r)
                {
                        Tree[index].value = a[l];
                        Tree[index].d = 0;
                        return;
                }
                int mid = (l + r) >> 1;
                build(index << 1,l,mid);
                build(index << 1 | 1,mid + 1,r);
                Tree[index].d = (Tree[index << 1].d + 1) % 2;
                update(index);
        }
        inline void modify(int index,int pos,int val)
        {
                if (Tree[index].l == Tree[index].r)
                {
                        Tree[index].value = val;
                        return;
                }
                int mid = (Tree[index].l + Tree[index].r) >> 1;
                if (mid >= pos) modify(index << 1,pos,val);
                else modify(index << 1 | 1,val);
                update(index);
        }
        inline int query()
        {
                return Tree[1].value;
        }
} T;

template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
    x *= f;
}

int main()
{
        
        read(n); read(m);
        for (int i = 1; i <= (1 << n); i++) read(a[i]);
        T.build(1,1,(1 << n));
        while (m--)
        {
                int x,y;
                read(x); read(y);
                T.modify(1,x,y);
                printf("%d\n",T.query());
        }
        
        return 0;
    
}