•  24.51%
•  1000ms
•  262144K

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.

樣例輸入複製

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

樣例輸出複製

10
8

題目來源

雖然可以用樹狀陣列做,但覺得線段樹的方法易於理解(吧)....

其實這個可以分解為len個字首和。設sum（l,r）表示l到r的和，則上式等於：

sum(l,l)+sum(l,l+1)+sum(l,l+2)+...+sum(l,r)

=sum(1,l)+sum(1,l+1)+sum(1,l+2)+...+sum(1,r)-（r-l+1）*sum(1,l-1)

則轉化為求字首和的區間和。然後對單點修改，相當於對要修改的位置i到n所有的字首和都加增量。

也就轉化為線段樹查詢區間和，區間修改。

區間修改用lazy標誌即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define lcvalue(rt) (lz[rt<<1]*(m-l+1)+tree[rt<<1])
#define rcvalue(rt) (lz[rt<<1|1]*(r-m)+tree[rt<<1|1])
#define PushUP(rt) tree[rt] = lcvalue(rt)+ rcvalue(rt)
const int maxn = 1e5+5;
ll tree[maxn<<2];
ll lz[maxn<<2];
int ip;
int n,q;
ll fo[maxn];
void build(int l,int r,int rt) {
    if (l == r) {
       tree[rt]=fo[ip++];
       return ;
    }
    ll m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUP(rt);
}
ll query(int L,int R,int l,int r,int rt,ll z){
    lz[rt]+=z;
    if(L>r || R<l)
        return 0;
    if (L <= l && r <= R) {
        return tree[rt]+lz[rt]*(r-l+1);
    }
    ll m = (l + r) >> 1;
    ll ret = 0;
       ret += query(L , R , lson,lz[rt]);
    ret += query(L , R , rson,lz[rt]);
    tree[rt]+=lz[rt]*(r-l+1);
    lz[rt]=0;
    return ret;
}
void update(int L,int R,ll add,int l,int r,int rt) {
    if (l == r) {
        tree[rt] += add;
        return ;
    }
    if (L <= l && r <= R) {
        lz[rt]+=add;
        return;
    }
    ll m = (l + r) >> 1;
    if (L <= m) update(L,R, add , lson);
    if(R>m)    update(L,R, add , rson);
    PushUP(rt);
}
int co[maxn];
int main(){
    cin>>n>>q;
    for(int i=1;i<=n;i++){
        scanf("%d",&co[i]);
        fo[i]=co[i]+fo[i-1];
    }
    ip=1;
    build(1,n,1);
    int l,r,len,v,d,flag;
    while(q--)
    {
        scanf("%d",&flag);
        if(flag==1){
            scanf("%d%d",&l,&r);
            len=r-l+1;
            printf("%lld\n",query(l,r,1,n,1,0)-len*query(l-1,l-1,1,n,1,0));
        }
        else{
            scanf("%d%d",&d,&v);
            update(d,n,v-co[d],1,n,1);
            co[d]=v;
        }
    }
}