开了两颗树，一颗求和一颗求各个数的平方的和
WA #1 的数据中我的程序输出有较大偏差，不像精度问题，个人码风还比较整洁，思路应该没问题，请问是还有什么特殊情况没考虑到吗？

#include<bits/stdc++.h>
#define mmt(a,b) memset(a,b,sizeof(a))
#define test(x) cout<<#x<<" = "<<x<<endl
#define file(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
#define endl '\n'
#define enter putchar(endl)
#define ls i<<1
#define rs i<<1|1
// #define double long double
//#define int long long
using namespace std
const int inf=0x3f3f3f3f,N=1e5+10;
int n,m;
double a[N];
int pow2(int x)
{
	return x*x;
}
double _pow2(double x)
{
	return x*x;
}
struct Tree
{
	struct node{
		int l,r;
		double val,lazy;
	}tree[N<<2];
	int len(int i)
	{
		return tree[i].r-tree[i].l+1;
	}
	void build(int l,int r,int i)
	{
		tree[i].l=l; tree[i].r=r; tree[i].lazy=0;
		if(l==r)
		{
			tree[i].val=a[l];
			return;
		}
		int mid=l+r>>1;
		build(l,mid,ls);
		build(mid+1,r,rs);
		tree[i].val=tree[ls].val+tree[rs].val;
	}
	void pushdown(int i)
	{
		tree[ls].lazy+=tree[i].lazy; tree[rs].lazy+=tree[i].lazy;
		tree[ls].val+=tree[i].lazy* (tree[ls].r-tree[ls].l+1); 
		tree[rs].val+=tree[i].lazy* (tree[rs].r-tree[rs].l+1);
		tree[i].lazy=0;
	}
	void modify(double x,int l,int r,int i)
	{
		if(l<=tree[i].l&&tree[i].r<=r)
		{
			tree[i].val+=x*(tree[i].r-tree[i].l+1);
			tree[i].lazy+=x;
			return;
		}
		pushdown(i);
		if(tree[ls].r>=l)modify(x,l,r,ls);
		if(tree[rs].l<=r)modify(x,l,r,rs);
		tree[i].val=tree[ls].val+tree[rs].val;
	}
	double query(int l,int r,int i)
	{
		if(l<=tree[i].l&&tree[i].r<=r)
			return tree[i].val;
		pushdown(i);
		double res=0;
		if(l<=tree[ls].r)res+=query(l,r,ls);
		if(tree[rs].l<=r)res+=query(l,r,rs);
		return res;
	}
}t1;
struct Tree2
{
	struct node{
		int l,r;
		double val,lazy;
	}tree[N<<2];
	int len(int i)
	{
		return tree[i].r-tree[i].l+1;
	}
	void build(int l,int r,int i)
	{
		tree[i].l=l; tree[i].r=r; tree[i].lazy=0;
		if(l==r)
		{
			tree[i].val= a[l]*a[l];
			return;
		}
		int mid=l+r>>1;
		build(l,mid,ls);
		build(mid+1,r,rs);
		tree[i].val=tree[ls].val+tree[rs].val;
	}
	void pushdown(int i)
	{
		tree[ls].lazy+=tree[i].lazy; tree[rs].lazy+=tree[i].lazy;
		tree[ls].val+=pow2(tree[i].lazy) * len(ls) +
					tree[i].lazy*t1.query(tree[ls].l,tree[ls].r,1)*2; 
		tree[rs].val+=pow2(tree[i].lazy) * len(rs) +
					tree[i].lazy*t1.query(tree[rs].l,tree[rs].r,1)*2; 
		tree[i].lazy=0;
	}
	void modify(double x,int l,int r,int i)
	{
		if(l<=tree[i].l&&tree[i].r<=r)
		{
			tree[i].val+=pow2(x)*len(i) + t1.query(tree[i].l,tree[i].r,1)*2*x;
			tree[i].lazy+=x;
			return;
		}
		pushdown(i);
		if(tree[ls].r>=l)modify(x,l,r,ls);
		if(tree[rs].l<=r)modify(x,l,r,rs);
		tree[i].val=tree[ls].val+tree[rs].val;
	}
	double query(int l,int r,int i)
	{
		if(l<=tree[i].l&&tree[i].r<=r)
			return tree[i].val;
		pushdown(i);
		double res=0;
		if(l<=tree[ls].r)res+=query(l,r,ls);
		if(tree[rs].l<=r)res+=query(l,r,rs);
		return res;
	}
}t2;
signed main()
{
	int i,j;
	cin>>n>>m;
	for(i=1;i<=n;i++)
		cin>>a[i];
	t1.build(1,n,1);
	t2.build(1,n,1);
	while(m--)
	{
		int op,l,r;
		double x;
		cin>>op>>l>>r;
		if(op==1)
		{
			cin>>x;
			t2.modify(x,l,r,1);
			t1.modify(x,l,r,1);
		}
		if(op==2)
			printf("%.4lf\n",t1.query(l,r,1)*1.0/(r-l+1));
		if(op==3)
			printf("%.4lf\n",1.0*t2.query(l,r,1) / (r-l+1) - _pow2( t1.query(l,r,1)*1.0 / (r-l+1) )  );
	}
	return 0;
	
}