codeforces 242-E section XOR

Topic link: https: //codeforces.com/problemset/problem/242/E
subject to the effect:
you n and a [i] ... a [n ].
m operations:
. 1 LR Query A [L] + ... + A [R & lt]
2 LRX



idea: We use multiple trees each maintenance on it.

#include <bits/stdc++.h>
#define LL long long
using namespace std;
#define mid (l+r)/2

int a[100005];
int sum[21][400005], add[21][400005];

void BT(int rt, int l, int r, int x){

    if(l==r){
        sum[x][rt]=(a[l]&(1<<x))?1:0;
        return ;
    }
    BT(rt<<1, l, mid, x);
    BT((rt<<1)+1, mid+1, r, x);
    sum[x][rt]=sum[x][rt<<1]+sum[x][(rt<<1)+1];
}

void GXdown(int rt, int x, int Le, int Re){
    add[x][rt<<1]+=add[x][rt];
    add[x][(rt<<1)+1]+=add[x][rt];
    if(add[x][rt]%2){
        sum[x][rt<<1]=Le-sum[x][rt<<1];
        sum[x][(rt<<1)+1]=Re-sum[x][(rt<<1)+1];
    }
    add[x][rt]=0;
}

void GX(int rt, int l, int r, int L, int R, int x){
    if(l==L&&r==R){
        add[x][rt]+=1;
        sum[x][rt]=(R-L+1)-sum[x][rt];
        return ;
    }
    GXdown(rt, x, mid-l+1, r-mid);

    if(R<=mid){
        GX(rt<<1, l, mid, L, R, x);
    }
    else if(L>mid){
        GX((rt<<1)+1, mid+1, r, L, R, x);
    }
    else{
        GX(rt<<1, l, mid, L, mid, x);
        GX((rt<<1)+1, mid+1, r, mid+1, R, x);
    }
    sum[x][rt]=sum[x][rt<<1]+sum[x][(rt<<1)+1];
}


int Q(int rt, int l, int r, int L, int R, int x){
    if(l==L&&R==r){
        return sum[x][rt];
    }
    GXdown(rt, x, mid-l+1, r-mid);

    if(R<=mid){
        return Q(rt<<1, l, mid, L, R, x);
    }
    else if(L>mid){
        return Q((rt<<1)+1, mid+1, r, L, R, x);
    }
    else{
        return Q(rt<<1, l, mid, L, mid, x)+Q((rt<<1)+1, mid+1, r, mid+1, R, x);
    }
}

int main(){

    int n, m;
    scanf("%d", &n);
    for(int i=1; i<=n; i++){
        scanf("%d", &a[i]);
    }
    for(int i=0; i<=20; i++){
        BT(1, 1, n, i);
    }
    scanf("%d", &m);
    while(m--){
        int k, l, r, x;
        scanf("%d", &k);
        if(k==1){
            scanf("%d%d", &l, &r);
            LL ans=0;
            for(int i=0; i<=20; i++){
                ans+=Q(1, 1, n, l, r, i)*(1ll<<i);
            }
            printf("%lld\n", ans);
        }
        else{
            scanf("%d%d%d", &l, &r, &x);
            for(int i=0; i<=20; i++){
                if(x&(1<<i)){
                    GX(1, 1, n, l, r, i);
                }
            }
        }
    }

    return 0;
}