xor or and 线段树

每一位维护一颗线段树

(-1)^1 =-2

(-2)^1=-1

#include <cstdio>
#include<iostream>
using namespace std;
//#define int long long
#define si signed
#define sc(x) scanf("%d", &x);
#define P pair<int, int>
int lazy[4][4000005];
int sum[4][4000005];
int A[1000005];
int n, m;
void pushdown(int id, int x, int l, int r)
{
    if (lazy[id][x] == 1)
    {
        lazy[id][x] = 0;
        int mid = l + r >> 1;
        lazy[id][x << 1] ^= 1;
        sum[id][x << 1] = (mid - l + 1) - sum[id][x << 1];
        lazy[id][x << 1 | 1] ^= 1;
        sum[id][x << 1 | 1] = (r - mid) - sum[id][x << 1 | 1];
    }
    else if (lazy[id][x] == -2)
    {
        lazy[id][x] = 0;
        int mid = l + r >> 1;
        lazy[id][x << 1] = -2;
        sum[id][x << 1] = (mid - l + 1);
        lazy[id][x << 1 | 1] = -2;
        sum[id][x << 1 | 1] = (r - mid);
    }
    else if (lazy[id][x] == -1)
    {
        lazy[id][x] = 0;
        int mid = l + r >> 1;
        lazy[id][x << 1] = -1;
        sum[id][x << 1] = 0;
        lazy[id][x << 1 | 1] = -1;
        sum[id][x << 1 | 1] = 0;
    }
}
void pushup(int id, int x)
{
    sum[id][x] = sum[id][x << 1] + sum[id][x << 1 | 1];
}
void build(int id, int l, int r, int x)
{
    lazy[id][x] = 0;
    if (l == r)
    {
        sum[id][x] = ((A[l] >> id) & 1);
        return;
    }
    int mid = (l + r) / 2;
    build(id, l, mid, x << 1);
    build(id, mid + 1, r, x << 1 | 1);
    pushup(id, x);
}
void update(int id, int l, int r, int x, int type, int L, int R)
{
    if (type == 2)
    {
        // cout<<type<<"type"<<endl;

        if (l >= L && r <= R)
        {
            lazy[id][x] ^= 1;
            sum[id][x] = (r - l + 1) - sum[id][x];
            //cout<<L<<' '<<R<<endl;
            return;
        }
        pushdown(id, x, l, r);
        int mid = (l + r) >> 1;
        if (L <= mid)
            update(id, l, mid, x << 1, type, L, R);
        if (R > mid)
            update(id, mid + 1, r, x << 1 | 1, type, L, R);
        pushup(id, x);
    }
    else if (type == 4)
    {

        if (l >= L && r <= R)
        {
            lazy[id][x] = -1;
            sum[id][x] = 0;
            return;
        }
        pushdown(id, x, l, r);
        int mid = (l + r) >> 1;
        if (L <= mid)
            update(id, l, mid, x << 1, type, L, R);
        if (R > mid)
            update(id, mid + 1, r, x << 1 | 1, type, L, R);
        pushup(id, x);
    }
    else
    {

        if (l >= L && r <= R)
        {
            lazy[id][x] = -2;
            sum[id][x] = r - l + 1;
            return;
        }
        pushdown(id, x, l, r);
        int mid = (l + r) >> 1;
        if (L <= mid)
            update(id, l, mid, x << 1, type, L, R);
        if (R > mid)
            update(id, mid + 1, r, x << 1 | 1, type, L, R);
        pushup(id, x);
    }
}
int query(int id, int l, int r, int x, int L, int R)
{
    int ans = 0;
    if (L <= l && r <= R)
    {
        return sum[id][x];
    }
    pushdown(id, x, l, r);
    int mid = l + r >> 1;
    if (L <= mid)
        ans += query(id, l, mid, x << 1, L, R);
    if (R > mid)
        ans += query(id, mid + 1, r, x << 1 | 1, L, R);
    return ans;
}
si main()
{
    int T;
    sc(T)
    string s;
    while (T--)
    {
        sc(n) sc(m);
        for (int i = 0; i <= n; i++)
            sc(A[i]) 
        for (int i = 0; i <4; i++)
        {
            build(i, 0, n-1, 1);
        }
        
        int l,r,x;
        while (m--)
        {
            cin >> s;
            if (s[0] == 'S')
            {
                sc(l) sc(r)
                int ans=0;
                for(int i=0;i<4;i++){
                    ans +=query(i,0,n-1,1,l,r)*(1<<i);
                }
                cout<<ans<<'\n';
            }
            else if (s[0] == 'X')
            {
                sc(x)
                sc(l) sc(r)
                
                for(int i=0;i<4;i++){
                    if((x>>i)&1){
                        update(i,0,n-1,1,2,l,r);
                    }
                }
            }
            else if (s[0] == 'O')
            {
                sc(x)
                sc(l) sc(r)
                
                for(int i=0;i<4;i++){
                    if((x>>i)&1){
                        update(i,0,n-1,1,3,l,r);
                    }
                }
            }
            else
            {  
                sc(x)
                sc(l) sc(r)
               
                for(int i=0;i<4;i++){
                    if(!((x>>i)&1)){
                        update(i,0,n-1,1,4,l,r);
                    }
                }
            }
        }
    }

    //system("pause");
}

 

来源：https://www.cnblogs.com/liulex/p/11807030.html