样例过了，但是一直超内存

sum维护区间和，sqr维护区间平方和

所有标1的变量都对应sum,标2对应sqr

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, m;
double a[N], sum[N * 4], sqr[N * 4];
void build (int k, int l, int r) {
	if (l == r) {
		sum[k] = a[l];
		sqr[k] = a[l] * a[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(k * 2, l, mid);
	build(k * 2 + 1, mid + 1, r);
	sum[k] = sum[k * 2] + sum[k * 2 + 1];
	sqr[k] = sqr[k * 2] + sqr[k * 2 + 1]; 
}
double tag1[N * 4], tag2[N * 4];
void pushdown(int k, int l, int r);
double ask1(int k, int l, int r, int x, int y) {
	if (r < x || y < l) return 0;
	if (x <= l && r <= y) return sum[k];
	pushdown(k, l, r);
	int mid = (l + r) >> 1;
	return ask1(k * 2, l, mid, x, y) + ask1(k * 2 + 1, mid + 1, r, x, y);
}
double ask2(int k, int l, int r, int x, int y) {
	if (r < x || y < l) return 0;
	if (x <= l && r <= y) {
		return sqr[k];
	}
	pushdown(k, l, r);
	int mid = (l + r) >> 1;
	return ask2(k * 2, l, mid, x, y) + ask2(k * 2 + 1, mid + 1, r, x, y);
}
void pushdown(int k, int l, int r) {
	int mid = (l + r) >> 1;
	if (tag1[k]) {
		sum[k * 2] += tag1[k] * (mid - l + 1);
		sum[k * 2 + 1] += tag1[k] * (r - mid);
		tag1[k * 2] += tag1[k],
		tag1[k * 2 + 1] += tag1[k];
		tag1[k] = 0;
	}
	if (tag2[k]) {
		sqr[k * 2] += 2 * tag2[k] * ask1(1, 1, n, l, mid) + tag2[k] * tag2[k] * (mid - l + 1);
		sqr[k * 2 + 1] += 2 * tag2[k] * ask1(1, 1, n, mid + 1, r) + tag2[k] * tag2[k] * (r - mid);
		tag2[k * 2] += tag2[k],
		tag2[k * 2 + 1] += tag2[k];
		tag2[k] = 0;
	}
}
void update1(int k, int l, int r, int x, int y, double v) {
	if (r < x || y < l) return;
	if (x <= l && r <= y) {
		sum[k] += v * (r - l + 1);
		tag1[k] += v;
		return;
	}
	pushdown(k, l, r);
	int mid = (l + r) >> 1;
	update1(k * 2, l, mid, x, y, v);
	update1(k * 2 + 1, mid + 1, r, x, y, v);
	sum[k] = sum[k * 2] + sum[k * 2 + 1];
}
void update2(int k, int l, int r, int x, int y, double v) {
	if (r < x || y < l) return;
	if (x <= l && r <= y) {
		sqr[k] += 2 * v * ask1(1, 1, n, l, r) + v * v * (r - l + 1);
		tag2[k] += v;
		return;
	}
	pushdown(k, l, r);
	int mid = (l + r) >> 1;
	update2(k * 2, l, mid, x, y, v);
	update2(k * 2 + 1, mid + 1, r, x, y, v);
	sqr[k] = sqr[k * 2] + sqr[k * 2 + 1];
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) 
		scanf("%lf", &a[i]);
	build(1, 1, n);
	int op, x, y;
	double k;
	for (int i = 1; i <= m; i++) {
		scanf("%d", &op);
		if (op == 1) {
			scanf("%d%d%lf", &x, &y, &k);
			update2(1, 1, n, x, y, k);
			update1(1, 1, n, x, y, k);
		} else if (op == 2) {
			scanf("%d%d", &x, &y);
			printf("%.4lf\n", ask1(1, 1, n, x, y) / (y - x + 1));
		} else if (op == 3) {
			scanf("%d%d", &x, &y);
			double s = ask1(1, 1, n, x, y);
			//cout << ask2(1, 1, n, x, y) << " ";
			double tmp = ask2(1, 1, n, x, y) - s * (s / (y - x + 1));
			printf("%.4f\n", tmp / (y - x + 1));
		}
		fflush(stdin); 
	}
	return 0;
}