2018 徐州网络赛 H

Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i].

Unfortunately, the longer he learns, the fewer he gets.

That means, if he reads books from ll to rr, he will get a[l] \times L + a[l+1] \times (L-1) + \cdots + a[r-1] \times 2 + a[r]a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r] (LL is the length of [ ll, rr ] that equals to r - l + 1r−l+1).

Now Ryuji has qq questions, you should answer him:

11. If the question type is 11, you should answer how much knowledge he will get after he reads books [ ll, rr ].

22. If the question type is 22, Ryuji will change the ith book's knowledge to a new value.

Input

First line contains two integers nn and qq (nn, q \le 100000q≤100000).

The next line contains n integers represent a[i]( a[i] \le 1e9)a[i](a[i]≤1e9) .

Then in next qq line each line contains three integers aa, bb, cc, if a = 1a=1, it means question type is 11, and bb, ccrepresents [ ll , rr ]. if a = 2a=2 , it means question type is 22 , and bb, cc means Ryuji changes the bth book' knowledge to cc

Output

For each question, output one line with one integer represent the answer.

样例输入复制

5 3
1 2 3 4 5
1 1 3
2 5 0
1 4 5

样例输出复制

10
8

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

 

思路：

维护两个树状数组。

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
typedef long long ll;
ll t1[maxn],t2[maxn],a[maxn];
inline int lowbit(int x){return x&-x;}
void update(int x,ll val,ll *tree){
    for (int i=x; i<maxn; i+=lowbit(i)) tree[i]+=val;
}
ll Query(int x,ll *tree){
    ll res=0;
    for (int i=x; i; i-=lowbit(i)) res+=tree[i];
    return res;
}
int n,q;
int op,l,r;
int main(){
    scanf("%d%d",&n,&q);
    for (int i=1; i<=n; i++) {
        scanf("%lld",&a[i]);
        update(i,a[i]*(n-i+1),t1);
        update(i,a[i],t2);
    }

    while(q--){
        scanf("%d",&op);
        if(op==1){
            scanf("%d%d",&l,&r);
            ll tmp=Query(r,t1)-Query(l-1,t1);
            tmp-=1ll*(Query(r,t2)-Query(l-1,t2))*(n-r);
            cout<<tmp<<endl;
        } else {
            scanf("%d%d",&l,&r);
            update(l,(r-a[l])*(n-l+1),t1);
            update(l,(r-a[l]),t2);
            a[l]=r;
        }
    }
    return 0;
}