加c++代码给定一个长度为n的字符串，字符串只有4和7构成，因为BSNY认为只有4和7是幸运数。现在BSNY有m个任务给你，任务有2种： switch l r 把[l, r]区间内的4换成7,7换成4； count 计算这n个幸运数的最长不下降子序列长度。

以下是参考代码，使用了线段树实现区间修改和查询最长不下降子序列长度。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 1e6 + 10;

int n, m, a[MAXN];
struct Node {
    int l, r;
    int len, lmax, rmax, ans;
    int tag;
} tr[MAXN << 2];

void pushup(Node &u, Node &lc, Node &rc) {
    u.len = lc.len + rc.len;
    u.lmax = lc.lmax;
    if (lc.lmax == lc.len && a[u.lc] <= a[rc.lc]) u.lmax += rc.lmax;
    u.rmax = rc.rmax;
    if (rc.rmax == rc.len && a[rc.rc] <= a[lc.rc]) u.rmax += lc.rmax;
    u.ans = max(max(lc.ans, rc.ans), lc.rmax + rc.lmax);
}

void pushdown(Node &u, Node &lc, Node &rc) {
    if (u.tag == 1) {
        swap(lc.lmax, lc.rmax);
        swap(lc.lc, lc.rc);
        lc.tag ^= 1;
        swap(rc.lmax, rc.rmax);
        swap(rc.lc, rc.rc);
        rc.tag ^= 1;
        u.tag = 0;
    }
}

void build(int u, int l, int r) {
    tr[u] = {l, r, r - l + 1, 1, 1, 1, r - l + 1, 0};
    if (l == r) {
        tr[u].lc = tr[u].rc = l;
        return;
    }
    int mid = l + r >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}

void modify(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) {
        swap(tr[u].lmax, tr[u].rmax);
        swap(tr[u].lc, tr[u].rc);
        tr[u].tag ^= 1;
        return;
    }
    pushdown(tr[u], tr[u << 1], tr[u << 1 | 1]);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) modify(u << 1, l, r);
    if (r > mid) modify(u << 1 | 1, l, r);
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}

int query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) return tr[u].ans;
    pushdown(tr[u], tr[u << 1], tr[u << 1 | 1]);
    int mid = tr[u].l + tr[u].r >> 1;
    int res = 0;
    if (l <= mid) res = max(res, query(u << 1, l, r));
    if (r > mid) res = max(res, query(u << 1 | 1, l, r));
    if (l <= mid && r > mid) {
        int lmax = min(mid - l + 1, tr[u << 1].len - tr[u << 1].rmax);
        int rmax = min(r - mid, tr[u << 1 | 1].len - tr[u << 1 | 1].lmax);
        res = max(res, lmax + rmax);
    }
    return res;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        char c;
        scanf(" %c", &c);
        a[i] = c == '7' ? 1 : 0;
    }
    build(1, 1, n);
    while (m--) {
        char op[10];
        scanf("%s", op);
        if (op[0] == 's') {
            int l, r;
            scanf("%d%d", &l, &r);
            modify(1, l, r);
        } else {
            printf("%d\n", query(1, 1, n));
        }
    }
    return 0;
}

注：本代码参考了洛谷博客：[P1907 [NOI2003] 幸运数](https://www.luogu.com.cn/blog/184141/p1907-noi2003-xing-yun-shu-ti-jie)。