Philosopher (set tree line consolidation)

• Direct maintenance product is certainly not feasible, accuracy will explode, so we have to maintain and logarithmic, and finally to the highest level you can calculate
• Then converted into the interval sum, Interval Ranking
• The combined properties of splitting section sort of way may be used to maintain the maximum increment block segment tree to resolve, the outer layer zone set to maintain the segment tree, then use the segment tree can be operated

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#include<set>
#include<cmath>
#define ll long long
#define M (1 << 18)
#define N 20000010
#define double long double
const double eps = 1e-8;
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
double c[M], ver[M];
int num[M], n, m;
int lowbit(int x) {
    return x & -x;
}
void add(int x, double v) {
    for(int i = x; i <= n; i += lowbit(i)) c[i] += v;
}

double query(int x) {
    double ans = 0;
    for(int i = x; i; i -= lowbit(i)) ans += c[i];
    return ans;
}

struct Note {
    int l, r, rt, op;
    Note(int ln = 0, int rn = 0, int rtn = 0, int opn = 0) {
        l = ln, r = rn, rt = rtn, op = opn;
    }
    bool operator < (const Note &b) const {
        return this->l < b.l;
    }
};
#define S set<Note>::iterator
set<Note> st;
int ls[N], rs[N], cnt[N], f;
double sum[N];

void pushup(int now) {
    cnt[now] = cnt[ls[now]] + cnt[rs[now]];
    sum[now] = sum[ls[now]] + sum[rs[now]];
}

int merge(int x, int y) {
    if(!x || !y) return x + y;
    ls[x] = merge(ls[x], ls[y]);
    rs[x] = merge(rs[x], rs[y]);
    cnt[x] = cnt[x] + cnt[y];
    sum[x] = sum[x] + sum[y];
    return x;
}

S insert(Note x) {
    add(x.l, sum[x.rt]);
    return st.insert(x).first;
}

void Del(S it) {
    add(it->l, -sum[it->rt]);
    st.erase(it);
}

void split(int x, int &rt1, int &rt2, int l, int r, int k) {
    rt1 = ++f;
    rt2 = ++f;
    if(l == r) {
        cnt[rt1] = k;
        sum[rt1] = ver[l] * k;
        cnt[rt2] = cnt[x] - cnt[rt1];
        sum[rt2] = sum[x] - sum[rt1];
        return;
    }
    int mid = (l + r) >> 1;
    if(cnt[ls[x]] >= k) {
        rs[rt2] = rs[x];
        split(ls[x], ls[rt1], ls[rt2], l, mid, k);
    } else {
        ls[rt1] = ls[x];
        split(rs[x], rs[rt1], rs[rt2], mid + 1, r, k - cnt[ls[x]]);
    }
    pushup(rt1);
    pushup(rt2);
}

S split(int x) {
    if(x > n) return st.end();
    S it = st.upper_bound(Note(x, 0, 0, 0));
    it--;
    Note hh = *it;
    if(hh.l == x) return it;
    int rt1, rt2;
    if(!hh.op) split(hh.rt, rt1, rt2, 1, n, x - hh.l);
    else split(hh.rt, rt2, rt1, 1, n, hh.r - x + 1);
    Del(it);
    insert(Note(hh.l, x - 1, rt1, hh.op));
    return insert(Note(x, hh.r, rt2, hh.op));
}

void build(int &x, int l, int r, int k) {
    x = ++f;
    cnt[x]++;
    sum[x] += ver[k];
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(k <= mid) build(ls[x], l, mid, k);
    else build(rs[x], mid + 1, r, k);
}


int calc(int l, int r) {
    S L = split(l), R = split(r + 1);
    R--;
    double ans = query(R->r) - query(L->l - 1);
    double out = pow(10, ans - floorl(ans) + eps);
    return floorl(out);
}

void updata(int l, int r, int op) {
    S L = split(l);
    split(r + 1);
    int rt = 0;
    for(S it = L; it != st.end() && (it->l) <= r; Del(it++)) {
        rt = merge(rt, it->rt);
    }
    insert(Note(l, r, rt, op));
}

int main() {
    n = read(), m = read();
    for(int i = 1; i <= n; i++) num[i] = read(), ver[i] = log10(i);
    for(int i = 1; i <= n; i++) {
        int now;
        build(now, 1, n, num[i]);
        insert(Note(i, i, now, 0));
    }
    while(m--) {
        int op = read(), l = read(), r = read();
        if(op == 2) cout << calc(l, r) << "\n";
        else {
            op = read() ^ 1;
            updata(l, r, op);
        }
    }
    return 0;
}