ACM-ICPC 2018 徐州赛区网络预赛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

 (L is the length of [ l, r ] that equals to r - l + 1).

Now Ryuji has qq questions, you should answer him:

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

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

Input

First line contains two integers n and .

The next line contains n integers represent .

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, cc represents [ l , r ]. 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

 

 

真菜，这次，做题做的一脸懵，三个人心态都炸了。。

 

 

很简单的推理

进一步化成

然后维护前缀和a[i]和i*a[i]就好了，这里用的是树状数组

 

#include
using namespace std;
typedef long long ll;
ll a[100005],b[100005],
n,q;
void add(ll a[],int x,ll val)
{
    while(x<=n)
    {
        a[x]+=val;
        x += x&-x;
    }
}
ll query(ll a[],int x)
{
    ll ans = 0;
    while(x)
    {
        ans += a[x];
        x -= x&-x;
    }
    return ans;
}
int main()
{
    ll t,l,r;
    scanf("%d%d",&n,&q);
    for(int i=0;i