Codeforces 242E - XOR on Segment

二进制下的每一位用一个线段树维护，一共拆分成32棵线段树

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int N = 1e5 + 10;
int n, m;
int a[N];
struct node {
    int l, r;
    int cnt, tag;
}tr[32][N * 4];
void pushup(int u, int x)
{
    tr[x][u].cnt = tr[x][u << 1].cnt + tr[x][u << 1 | 1].cnt;
}
void pushdown(int u, int x)
{
    if (tr[x][u].tag) {
        node &U = tr[x][u];
        node &L = tr[x][u << 1];
        node &R = tr[x][u << 1 | 1];
        U.tag = !U.tag;
        L.cnt = L.r - L.l + 1 - L.cnt;
        L.tag = !L.tag;
        R.cnt = R.r - R.l + 1 - R.cnt;
        R.tag = !R.tag;
    }
}
void build(int u, int l, int r, int x)
{
    if (l == r) {
        tr[x][u] = {l, r, (a[l] >> x & 1)};
        return;
    }
    tr[x][u] = {l, r};
    int mid = l + r >> 1;
    build(u << 1, l, mid, x);
    build(u << 1 | 1, mid + 1, r, x);
    pushup(u, x);
}
LL query(int u, int l, int r, int x)
{
    if (tr[x][u].l >= l && tr[x][u].r <= r)   {
        return tr[x][u].cnt;
    }
    pushdown(u, x);
    int mid = tr[x][u].l + tr[x][u].r >> 1;
    LL res = 0;
    if (l <= mid)   res = query(u << 1, l, r, x);
    if (r > mid)    res += query(u << 1 | 1, l, r, x);
    return res;   
}
void modify(int u, int l, int r, int num, int x)
{
    if (tr[x][u].l >= l && tr[x][u].r <= r) {
        if (num == 1)   tr[x][u].cnt = tr[x][u].r - tr[x][u].l + 1 - tr[x][u].cnt, tr[x][u].tag = !tr[x][u].tag;
        return ;
    }
    pushdown(u, x);
    int mid = tr[x][u].l + tr[x][u].r >> 1;
    if (l <= mid)   modify(u << 1, l, r, num, x);
    if (r > mid)    modify(u << 1 | 1, l , r, num, x);
    pushup(u, x);
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ )  
        scanf("%d", &a[i]);
    for (int i = 0; i < 32; i ++ )
        build(1, 1, n, i);
    scanf("%d", &m);
    while (m -- ) {
        int type, l, r, d;
        scanf("%d%d%d", &type, &l, &r);
        if (type == 1) {
            LL res = 0, p = 1;
            for (int i = 0; i < 32; i ++ ) {
                res += query(1, l, r, i) * p;
                p *= 2;
            }
            printf("%lld\n", res);
        }
        else {
            scanf("%d", &d);
            for (int i = 0; i < 32; i ++ )
                modify(1, l, r, (d >> i & 1), i);
        }
    }
    return 0;
}