样例过了，下载第一组数据第二个询问方差有问题。 求助，不知道错在哪里。

下面是第一组数据：

in

8 15
8.46 6.03 3.73 0.32 7.43 3.71 8.04 8.22 
3 1 8
1 2 8 -2.7566713364794850E+0000
1 2 8  2.1308819339610636E+0000
1 1 6  1.5912831262685359E+0000
1 1 8 -2.7779214559122920E+0000
1 1 8 -6.5134523715823889E-0001
3 2 8
1 1 6 -8.5440817382186651E-0001
3 2 8
2 2 8
3 2 7
1 3 8 -1.8737916438840330E+0000
1 1 7  2.5193137815222144E+0000
1 3 8  1.2835426828823984E+0000
3 3 8

out

7.4145
5.2609
6.2101
1.8256
6.1810
5.8854

code

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;
double a[N];
int n, m;

struct node {
	int l, r;
	double sum = 0, sqm = 0;
	double lazy = 0;
}t[N<<2];

void pushup(int p) {
	t[p].sum = t[p<<1].sum + t[p<<1|1].sum;
	t[p].sqm = t[p<<1].sqm + t[p<<1|1].sqm;
}

void build(int p, int l, int r) {
	t[p].l = l, t[p].r = r;
	if (l == r) {
		t[p].sum = a[l];
		t[p].sqm = a[l] * a[l];
		return;
	}
	int mid = l+r>>1;
	build(p<<1, l, mid);
	build(p<<1|1, mid+1, r);
	pushup(p);
}

void change(int p) {
	double k = t[p].lazy;
	t[p].sqm += 2*k*t[p].sum + (t[p].r-t[p].l+1)*k*k;
	t[p].sum += k * (t[p].r-t[p].l+1);
}

void pushdown(int p) {
	if (t[p].lazy) {
		t[p<<1].lazy += t[p].lazy;
		t[p<<1|1].lazy += t[p].lazy;
		change(p<<1);
		change(p<<1|1);
		t[p].lazy = 0;
	}
}

void updata(int p, int L, int R, double k) {
	cout << " " << p << " " << L << " " << R << " " << k << endl;
	if (L <= t[p].l && t[p].r <= R) {
		t[p].lazy += k;
		change(p);
		return;
	}
	pushdown(p);
	int mid = t[p].l+t[p].r>>1;
	if (L <= mid) updata(p<<1, L, R, k);
	if (mid < R) updata(p<<1|1, L, R, k);
	pushup(p);
}

pair<double, double> query(int p, int L, int R) {
	if (L <= t[p].l && t[p].r <= R) {
		return {t[p].sum, t[p].sqm};
	}
	pushdown(p);
	int mid = t[p].l+t[p].r>>1;
	double sum = 0, sqm = 0;
	if (L <= mid) {
		pair<double, double> q = query(p<<1, L, R);
		sum += q.first;
		sqm += q.second;
	}
	if (mid < R) {
		pair<double, double> q = query(p<<1|1, L, R);
		sum += q.first;
		sqm += q.second;
	}
	return {sum, sqm};
}

int main() {
//	ios::sync_with_stdio(0);
//	cin.tie(0);
	cin >> n >> m;
	for (int i=1; i<=n; i++) cin >> a[i];
	build(1, 1, n);
	while (m--) {
		int o, x, y;
		double k;
		cin >> o >> x >> y;
		if (o == 1) {
			cin >> k;
			updata(1, x, y, k);
		}
		if (o == 2) {
			printf("%.4lf\n", query(1, x, y).first/(y-x+1));
//			cout << fixed << setprecision(4) << query(1, x, y).first/(y-x+1) << endl;
		}
		if (o == 3) {
			pair<double, double> q = query(1, x, y);
			double ave = q.first/(y-x+1);
			printf("%.4lf\n", q.second/(y-x+1)-ave*ave);
//			cout << fixed << setprecision(4) << q.second/(y-x+1)-ave*ave<< endl;
		}
	}
	return 0;
}