4号关注，线段树求助，内有极小的hack

#include <bits/stdc++.h>
using namespace std;
#define srand srand(time(NULL))
#define random(x) rand() % (x)
#define il inline
#define ptc putchar
#define reg register
#define mp make_pair
typedef __int128 LL;
typedef long long ll;
typedef pair<int, int> PII;
namespace cyyh {
    template <typename T>
    il void read(T &x) {
        x = 0; T f = 1; char ch;
        while (!isdigit(ch = getchar())) f -= (ch == '-') << 1;
        while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch & 15), ch = getchar(); x *= f;
    }
    template <typename T>
    il void write(T x) {
        if (x < 0) ptc('-'), x = -x;
        if (x > 9) write(x / 10);
        ptc(x % 10 + '0');
    }
    template <typename T>
    il T lowbit(const T &x) {
        return x & -x;
    }
}
using namespace cyyh; 
const int N = 4e5 + 5;
int n, m, a[N];
struct node {
    int sum0, sum1, mx0, mx1, zq0, zq1, yq0, yq1, tag;
    // tag1表示区间更改的数，初始值为-1
    // tag2表示区间是否取反
    // sum表示区间内该数字的个数
    // mx表示区间内该数字的最大的连续一段
    // zq表示区间左端点开始该数最长的连续个数
    // yq表示区间右端点开始该数最长的连续个数
} tr[N];
void pushup(int id, int l, int r) {
    int ls = id << 1, rs = id << 1 | 1;
    tr[id].sum0 = tr[ls].sum0 + tr[rs].sum0;
    tr[id].sum1 = tr[ls].sum1 + tr[rs].sum1;
    tr[id].mx0 = max(tr[ls].yq0 + tr[rs].zq0, max(tr[ls].mx0, tr[rs].mx0));
    tr[id].mx1 = max(tr[ls].yq1 + tr[rs].zq1, max(tr[ls].mx1, tr[rs].mx1));
    tr[id].zq0 = tr[ls].zq0, tr[id].yq0 = tr[rs].yq0;
    tr[id].zq1 = tr[ls].zq1, tr[id].yq1 = tr[rs].yq1;
    int mid = l + r >> 1;
    if (tr[ls].zq0 == mid - l + 1) tr[id].zq0 += tr[rs].zq0;
    if (tr[ls].zq1 == mid - l + 1) tr[id].zq1 += tr[rs].zq1;
    if (tr[ls].yq0 == r - mid) tr[id].yq0 += tr[rs].yq0;
    if (tr[ls].yq1 == r - mid) tr[id].yq1 += tr[rs].yq1;
}
void pushdown(int id, int l, int r) {
    int mid = l + r >> 1, ls = id << 1, rs = id << 1 | 1;
    if (tr[id].tag == -1) return ;
    if (!tr[id].tag) {
        tr[ls].tag = tr[ls].mx1 = tr[ls].sum1 = tr[ls].yq1 = tr[ls].zq1 = 0;
        tr[ls].mx0 = tr[ls].sum0 = tr[ls].yq0 = tr[ls].zq0 = mid - l + 1;
        tr[rs].tag = tr[rs].mx1 = tr[rs].sum1 = tr[rs].yq1 = tr[rs].zq1 = 0;
        tr[rs].mx0 = tr[rs].sum0 = tr[rs].yq0 = tr[rs].zq0 = r - mid;
    } else if (tr[id].tag == 1) {
        tr[ls].tag = 1;
        tr[ls].mx0 = tr[ls].sum0 = tr[ls].yq0 = tr[ls].zq0 = 0;
        tr[ls].mx1 = tr[ls].sum1 = tr[ls].yq1 = tr[ls].zq1 = mid - l + 1;
        tr[rs].tag = 1;
        tr[rs].mx0 = tr[rs].sum0 = tr[rs].yq0 = tr[rs].zq0 = 0;
        tr[rs].mx1 = tr[rs].sum1 = tr[rs].yq1 = tr[rs].zq1 = r - mid;
    } else {
        swap(tr[ls].mx0, tr[ls].mx1);
        swap(tr[ls].sum0, tr[ls].sum1);
        swap(tr[ls].yq0, tr[ls].yq1);
        swap(tr[ls].zq0, tr[ls].zq1);
        if (tr[ls].tag == 1 || tr[ls].tag == 0) tr[ls].tag = !tr[ls].tag;
        else if (tr[ls].tag == -1) tr[ls].tag = 2;
        else tr[ls].tag = -1;
        swap(tr[rs].mx0, tr[rs].mx1);
        swap(tr[rs].sum0, tr[rs].sum1);
        swap(tr[rs].yq0, tr[rs].yq1);
        swap(tr[rs].zq0, tr[rs].zq1);
        if (tr[rs].tag == 1 || tr[rs].tag == 0) tr[rs].tag = !tr[rs].tag;
        else if (tr[rs].tag == -1) tr[rs].tag = 2;
        else tr[rs].tag = -1;
    }
    tr[id].tag = -1;
}
void modify(int l, int r, int x, int y, int id, int op) {
    // op为指定操作 
    if (l > y || r < x) return ;
    if (l >= x && r <= y) {
        if (!op) {
            tr[id].mx1 = tr[id].sum1 = tr[id].tag = tr[id].yq1 = tr[id].zq1 = 0;
            tr[id].sum0 = tr[id].mx0 = tr[id].yq0 = tr[id].zq0 = r - l + 1;
        } else if (op == 1) {
            tr[id].tag = 1;
            tr[id].mx1 = tr[id].sum1 = tr[id].tag = tr[id].yq1 = r - l + 1;
            tr[id].sum0 = tr[id].mx0 = tr[id].yq0 = tr[id].zq0 = 0;
        } else {
            swap(tr[id].mx0, tr[id].mx1);
            swap(tr[id].sum0, tr[id].sum1);
            swap(tr[id].yq0, tr[id].yq1);
            swap(tr[id].zq0, tr[id].yq1);
//          if(~tr[id].tag && tr[id].tag != 2) tr[id].tag ^= 1;
//          else tr[id].tag = (tr[id].tag == 2 ? -1 : 2);
            if (tr[id].tag == 1 || tr[id].tag == 0) tr[id].tag = !tr[id].tag;
            else if (tr[id].tag == -1) tr[id].tag = 2;
            else tr[id].tag = -1;
        } 
        return ;
    }
    pushdown(id, l, r);
    int mid = l + r >> 1;
    modify(l, mid, x, y, id << 1, op), modify(mid + 1, r, x, y, id << 1 | 1, op);
    pushup(id, l, r);
}
int query(int l, int r, int x, int y, int id) {
    if (l > y || r < x) return 0;
    if (l >= x && r <= y) return tr[id].sum1;
    int mid = l + r >> 1;
    pushdown(id << 1, l, r);
    return query(l, mid, x, y, id << 1) + query(mid + 1, r, x, y, id << 1 | 1);
}
node query2(int l, int r, int x, int y, int id) {
    if (l >= x && r <= y) return tr[id];
    int mid = l + r >> 1, ls = id << 1, rs = id << 1 | 1;
    pushdown(id, l, r);
    if (y <= mid) return query2(l, mid, x, y, ls);
    else if (x > mid) return query2(mid + 1, r, x, y, rs);
    node a = query2(l, mid, x, y, ls), b = query2(mid + 1, r, x, y, rs), res;
    // pushup
    res.mx1 = max(a.yq1 + b.zq1, max(a.mx1, b.mx1));
    res.zq1 = a.zq1, res.yq1 = b.yq1;
    if (a.zq1 == mid - l + 1) res.zq1 += b.zq1;
    if (b.yq1 == r - mid) res.yq1 += a.yq1;
    return res;
}
void build(int l, int r, int id) {
    tr[id].tag = -1;
    if (l == r) {
        tr[id].mx0 = tr[id].sum0 = tr[id].yq0 = tr[id].zq0 = !a[l];
        tr[id].mx1 = tr[id].sum1 = tr[id].yq1 = tr[id].zq1 = a[l];
        return ;
    }
    int mid = l + r >> 1;
    build(l, mid, id << 1), build(mid + 1, r, id << 1 | 1);
    pushup(id, l, r);
}
int main() {
    read(n), read(m);
    for (int i = 1; i <= n; ++i) read(a[i]);
    build(1, n, 1);
    //modify(1, n, 2, 3, 1, 2);
//  for (int i = 1; i <= n; ++i) {
//      for (int j = i; j <= n; ++j) {
//          cout << i << ' ' << j << ' ' << query(1, n, i, j, 1) << ' ' << endl;
//      }
//  }
    //write(query(1, n, 1, 3, 1)), ptc('\n');
//  return 0;
    while (m--) {
        int op, l, r;
        read(op), read(l), read(r);
        ++l, ++r;
        if (op == 0) modify(l, r, 1, n, 1, 0);
        else if (op == 1) modify(1, n, l, r, 1, 1);
        else if (op == 2) modify(1, n, l, r, 1, 2);
        else if (op == 3) write(query(1, n, l, r, 1)), ptc('\n');
        else if (op == 4) write(query2(1, n, l, r, 1).mx1), ptc('\n');
    }
    return 0;
}

萌新第4次写线段树，hack如下：

3 2
1 0 0
2 1 2
4 0 2

答案显然是3，而modify确实修改成功了的，查询的时候却总是出错，但是查询我死活看不出来问题在哪