Art of Los Valley P2574 XOR

topic

In fact, with almost 1,2 tree line, the only difference is the need lazy array and mark the operating time pass.

The number of segment tree left son of the right should be larger than the number of sons. So we need time intervals and then XOR, the element must be clear how many of each interval, then the number of elements in the interval 0,1 reversed just fine

#include <bits/stdc++.h>            
#define ll long long                
#define ls l, mid, root << 1        
#define rs mid + 1, r, root << 1 | 1
using namespace std;                
int n, m, ans[800100], a[800100], lazy[800100];//ans表示一个根节点下1的个数 
inline void pushup(int root)
{
    ans[root] = ans[root << 1] + ans[root << 1 | 1];
}
inline void build(int l, int r, int root)
{
    if (l == r)
    {
        ans[root] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(ls), build(rs);
    pushup(root);
}
inline void pushdown(int l, int r, int root)
{
    int mid = (l + r) >> 1;
    int len = r - l + 1;   
    if (lazy[root])       //如果该节点还没有异或1，那就异或 
    {
        lazy[root << 1] ^= 1;
        lazy[root << 1 | 1] ^= 1;
        ans[root << 1] = len - (len >> 1) - ans[root << 1];//当然还需要更新，原root<<1里有len-len>>1个数。 
        ans[root << 1 | 1] = (len >> 1) - ans[root << 1 | 1];
        lazy[root] = 0;
    }
}
void update(int l, int r, int root, int ql, int qr)
{
    int len = r - l + 1;
    if (ql <= l && r <=  qr)
    {
        lazy[root] ^= 1;
        ans[root] = len  - ans[root];
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(l, r, root);
    if (ql <= mid)
    update(ls, ql, qr);
    if (qr >= mid + 1)
    update(rs, ql, qr);
    pushup(root);
}
int query(int l, int r, int root, int ql, int qr)
{
    int res = 0;
    if (ql <= l && r <= qr)
        return ans[root];
    int mid = (l + r) >> 1;
    pushdown(l, r, root);
    if (ql <= mid)
    res += query(ls, ql, qr);
    if (qr >= mid + 1)
    res += query(rs, ql, qr);
    return res;
}
int main()
{
    cout << (5 >> 1);
    return 0; 
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        char c;
        cin >> c;
        a[i] = c - '0';
    }
    build(1, n, 1);
    while (m--)
    {
        int p, l, r;
        cin >> p >> l >> r;
        if (p)
            printf("%d\n", query(1, n, 1, l, r));
        else
            update(1, n, 1, l, r);
    }
    return 0;
}