调到调不动了qwq~~~

此处附上我还算看的过去的 $code$,(看不懂的可以去看看 $Dpair$ 大佬的题解)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
const double eps = 1e-6;
inline void ac(){
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}
int n,m;
double a[N];
struct node{
    double sum,lazy;
}tr1[N << 3],tr2[N << 3];//tr1 : average_tree ; tr2 : fc_tree
inline void pushup(int pos){
    tr1[pos].sum = tr1[pos << 1].sum + tr1[pos << 1 | 1].sum;
    tr2[pos].sum = tr2[pos << 1].sum + tr2[pos << 1 | 1].sum;
}
inline void build(int pos,int l,int r){
    if(l == r){
        tr1[pos].sum = a[l];
        tr2[pos].sum = a[l] * a[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build(pos << 1,l,mid);
    build(pos << 1 | 1,mid + 1,r);
    pushup(pos);
}
inline void pushdown(int pos,int l,int r){
    int mid = (l + r) >> 1;
    tr2[pos << 1].sum += tr1[pos << 1].sum * (mid - l + 1) * 2 + (mid - l + 1) * tr2[pos].lazy *tr2[pos].lazy;
    tr2[pos << 1 | 1].sum += tr1[pos << 1 | 1].sum * (r - mid) + (r - mid) * tr2[pos].lazy * tr2[pos].lazy;
    tr2[pos << 1].lazy += tr2[pos].lazy;
    tr2[pos << 1 | 1].lazy +=tr2[pos].lazy;
    tr2[pos].lazy = 0; 
    tr1[pos << 1].sum += (mid - l + 1) * tr1[pos].lazy;
    tr1[pos << 1 | 1].sum += (r - mid) * tr1[pos].lazy;
    tr1[pos << 1].lazy += tr1[pos].lazy;
    tr1[pos << 1 | 1].lazy +=tr1[pos].lazy;
    tr1[pos].lazy = 0; 
}
inline void update(int pos,int l,int r,int L,int R,double k){
    if(L <= l && r <= R){
        tr2[pos].lazy += k;
        tr1[pos].lazy += k;
        tr2[pos].sum += tr1[pos].sum * 2 * k + (r - l + 1) * k * k;
        tr1[pos].sum += (r - l + 1) * k;
        return;
    }
    if(tr1[pos].lazy || tr2[pos].lazy)pushdown(pos,l,r);
    int mid = (l + r) >> 1;
    if(L <= mid)update(pos << 1,l,mid,L,R,k);
    if(R > mid)update(pos << 1 | 1,mid + 1,r,L,R,k);
    pushup(pos);
}
inline double query(node tr[],int pos,int l,int r,int L,int R){
    if(L <= l && r <= R)return tr[pos].sum;
    if(tr[pos].lazy)
        pushdown(pos,l,r);
    int mid = (l + r) >> 1;
    double ret = 0;
    if(L <= mid)ret += query(tr,pos << 1,l,mid,L,R);
    if(R > mid)ret += query(tr,pos << 1 | 1,mid + 1,r,L,R);
    return ret;
}
int main(){
    ac();
    cin >> n >> m;
    for(int i = 1;i <= n;i ++)cin >> a[i];
    build(1,1,n);
    while(m --){
        int opt,l,r;
        double k;
        cin >> opt >> l >> r;
        if(opt == 1){
            cin >> k;
            update(1,1,n,l,r,k);
        }
        else if(opt == 2){
            double ans = query(tr1,1,1,n,l,r) * 1.0 / ((r - l + 1) * 1.0);
            printf("%.4f\n",ans);
        }
        else {
            double ans = query(tr2,1,1,n,l,r) / (1.0 * (r - l + 1)) + query(tr1,1,1,n,l,r) * query(tr1,1,1,n,l,r) / ( (r - l + 1) * (r - l + 1) * 1.0 );
            printf("%.4f\n",ans);
        }
    }
    return 0;
}