见down函数

#include <bits/stdc++.h>

using namespace std;

using db = double;

const int maxN = 1e5 + 7;
const db eps = 1e-7;

struct Tree {
	int l, r;
	db x2, sum;
	db tg;
} t[maxN << 2];
#define ls (p << 1)
#define rs (p << 1 | 1)
void upd(int p) {
	t[p].sum = t[ls].sum + t[rs].sum;
	t[p].x2 = t[ls].x2 + t[rs].x2;
}
void build(int L, int R, int p, db B[]) {
	t[p].l = L; t[p].r = R;
	if (L == R) {
		t[p].x2 = B[L] * B[L];
		t[p].sum = B[L];
	} else {
		int mid = (L + R) >> 1;
		build(L, mid, ls, B);
		build(mid + 1, R, rs, B);
		upd(p);
	}
}
void make(Tree &tmp, db lz) {
	int len = tmp.r - tmp.l + 1;
	tmp.x2 += lz * lz * len + 2 * lz * tmp.sum;
	tmp.sum += lz * len;
	tmp.tg += lz;
}
void down(int p) {
	if (t[p].tg == 0) return;//这样能过
  //if (t[p].tg < eps) return; //这样不能过
	make(t[ls], t[p].tg);
	make(t[rs], t[p].tg);
	t[p].tg = 0;
}
void C(int L, int R, int p, db v) {
	if (L <= t[p].l && t[p].r <= R) make(t[p], v);
	else {
		down(p);
		int mid = (t[p].l + t[p].r) >> 1;
		if (L <= mid) C(L, R, ls, v);
		if (R > mid) C(L, R, rs, v);
		upd(p);
	}
}
db query(int L, int R, int p) {
	if (L <= t[p].l && t[p].r <= R) return t[p].sum;
	down(p);
	int mid = (t[p].l + t[p].r) >> 1;
	db res = 0;
	if (L <= mid) res += query(L, R, ls);
	if (R > mid) res += query(L, R, rs);
	return res;
}
db ask(int L, int R, int p) {
	if (L <= t[p].l && t[p].r <= R) return t[p].x2;
	down(p);
	int mid = (t[p].l + t[p].r) >> 1;
	db res = 0;
	if (L <= mid) res += ask(L, R, ls);
	if (R > mid) res += ask(L, R, rs);
	return res;
}

int n, m;
db a[maxN];

int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	build(1, n, 1, a);
	
	while (m--) {
		int op, x, y;
		cin >> op >> x >> y;
		int len = y - x + 1;
		if (op == 1) {
			db k;
			cin >> k;
			C(x, y, 1, k);
		}
		if (op == 2) {
			db res = query(x, y, 1) / len;
			printf("%.4lf\n", res);
		}
		if (op == 3) {
			db res = 0;
			res += ask(x, y, 1);
			db tmp = query(x, y, 1) / len;
			res += tmp * tmp * len;
			res -= 2 * tmp * query(x, y, 1);
			res /= len;
			printf("%.4lf\n", res);
		}
	}
}