$$\sum_{i=1}^k\sum_{j=1}^i a_j\\ =\sum_{i=1}^k (k-i+1)a_i\\ =(k+1)\sum_{i=1}^k a_i-\sum_{i=1}^k i\times a_i $$

分别维护 $\sum_{i=1}^k a_i$ 和 $\sum_{i=1}^k i\times a_i$ 即可。

#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#define int long long
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N=1e5+7;
int n,m,a[N],tr[N][2];
void add(int x,int k,int f){
	for(;x<=n;x+=lowbit(x)) tr[x][f]+=k;
}
int ask(int x,int f){
	int res=0;
	for(;x;x-=lowbit(x)) res+=tr[x][f];
	return res;
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		add(i,a[i],0);
		add(i,a[i]*i,1);
	}
	while(m--){
		string s;
		cin>>s;
		if(s=="Query"){
			int k;
			cin>>k;
			cout<<(k+1)*ask(k,0)-ask(k,1)<<'\n';
		}
		else{
			int x,y;
			cin>>x>>y;
			add(x,y-a[x],0);
			add(x,x*(y-a[x]),1);
			a[x]=y;
		}
	}
	return 0;
}