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  • 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<<'
    ';
                }
                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");
    }
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  • 原文地址:https://www.cnblogs.com/liulex/p/11807030.html
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