洛谷P2574 XOR的艺术

 

思路：线段树（lazy标记）

样例：
10 6
1011101001
0 2 4
1 1 5
0 3 7
1 1 10
0 1 4
1 2 6
输出：
3
6
1

#include<iostream>
#include<bits/stdc++.h>
#define ll long long
using namespace std;
#define endl '\n'
#define ls (o<<1)
#define rs (o<<1|1)
#define mid (l+r>>1)
inline ll read() {
    ll s = 0; bool w = true;
    char c = getchar();
    for (; c < '0' || c>'9'; c = getchar())if (c == '-')w = false;
    for (; c >= '0' && c <= '9'; c = getchar())s = (s << 1) + (s << 3) + (c ^ 48);
    return w ? s : -s;
}
const int N = 2e5 + 10;
ll s[N << 2], lz[N << 2];//s存放1的个数，lz为lazy标记（不用lazy会超时）
string ss;
inline void build(ll o, ll l, ll r) {//建树
    lz[o] = 0;
    if (l == r) {
        s[o] = ss[l] - '0';
        return;
    }
    build(ls, l, mid);
    build(rs, mid + 1, r);
    s[o] = (s[ls] + s[rs]);
}
inline void pd(ll o, ll l, ll r) {//懒标记下传

    s[ls] = mid - l + 1 - s[ls];
    //区间内取反，3个0，5个1，变成5个0，3个1，就是用长度减去之前1的个数就好了
    //eg:8-5=3；
    s[rs] = r - mid - s[rs];
    lz[ls] ^= lz[o];
    lz[rs] ^= lz[o];
    lz[o] = 0;
}
inline void ud(ll o, ll l, ll r, ll ql, ll qr) {//更新操作
    if ( ql <= l && r <= qr) {
        lz[o] ^= 1;//表示翻转一次，两次反转就又回去了
        s[o] = r - l + 1 - s[o];
        return;
    }
    if (lz[o]) { pd(o, l, r); }//记住要下传
    if (ql <= mid)ud(ls, l, mid, ql, qr);
    if (qr > mid)ud(rs, mid + 1, r, ql, qr);
    s[o] = s[ls] + s[rs];
}
inline ll query(ll o, ll l, ll r, ll ql, ll qr) {
    if (ql <= l && r <= qr) {
        return s[o];
    }
    ll sum = 0;
    if (lz[o])pd(o, l, r);//一定要下传，debug了半天!!!!!!
    if (ql <= mid)sum += query(ls, l, mid, ql, qr);
    if (mid < qr)sum += query(rs, mid + 1, r, ql, qr);
    return sum;
}
void solve() {
    ll n = read(), w = read();
    cin >> ss;
    ss = " " + ss;
    char c = getchar();
    build(1, 1, n);
    for (int i = 1; i <= w; ++i) {
        ll x = read(), y = read(), z = read();

        if (x == 0) {
            ud(1, 1, n, y, z);
        }
        else cout << query(1, 1, n, y, z) << endl;
    }

}
signed main() {
    ll T = 1;
    while (T--) {
        solve();
    }
    return 0;
}