徐州赛区网络赛 Ryuji doesn't want to study 线段树维护区间和

题目链接：https://nanti.jisuanke.com/t/31460

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, cc represents [ 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

要计算题目里类似卷积一样的东西，可以把原数组变成a[i]*=(n-i+1)然后建线段树维护区间和，记为p1（l,r）

再保持数组原样不变，单独维护区间和，记为p2（l，r）

1操作时，直接输出p1(l,r)-p2(l,r)*(n-r+1)

2操作时，更新两个前缀和

#include<bits/stdc++.h>
#include<math.h>
#define MOD  1000000007
#define For(i,m,n) for(int i=m;i<=n;i++)
#define LL long long
#define inf 0x3f3f3f3f
#define lan(a,b) memset(a,b,sizeof(a))
#define sqr(x) (x*x)
using namespace std;

#define maxn 100010



struct node
{
    int l,r;
    LL k;//标记值
    LL m;//存储值
};

node tree[2][maxn*4];
LL a[2][maxn];

void build(int n ,int l,int r,int t)
{
    tree[t][n].l=l;
    tree[t][n].r=r;
    if(l==r)
    {

        tree[t][n].m=a[t][l];//////////////////////////视情况
        return;
    }
    int mid=(l+r)/2;
    build(n<<1,l,mid,t);
    build(n<<1|1,mid+1,r,t);
    tree[t][n].m=tree[t][n<<1].m+tree[t][n<<1|1].m;/////////////////////看线段树种类
}

void pushup(int n,int t)
{
    tree[t][n].m=tree[t][n<<1].m+tree[t][n<<1|1].m;////////////////////看线段树种类
}

void update(int n,int l,int r,LL c,int t)
{
    if(tree[t][n].l==l&&tree[t][n].r==r)
    {
        tree[t][n].m=c;
        return;
    }
    int mid=(tree[t][n].r+tree[t][n].l)/2;
    if(r<=mid)
        update(n<<1,l,r,c,t);
    else if(l>=mid+1)
        update(n<<1|1,l,r,c,t);
    else
    {
        update(n<<1,l,mid,c,t);
        update(n<<1|1,mid+1,r,c,t);
    }
    pushup(n,t);
}

LL query(int n,int l,int r,int t)
{
    if(tree[t][n].l==l&&tree[t][n].r==r)
    {
        return tree[t][n].m;
    }
    int mid=(tree[t][n].r+tree[t][n].l)/2;

    LL res=0;
    if(r<=mid)
        res+=query(n<<1,l,r,t);
    else if(l>=mid+1)
        res+=query(n<<1|1,l,r,t);
    else
    {
        res+=query(n<<1,l,mid,t);////////////////////看线段树种类
        res+=query(n<<1|1,mid+1,r,t);////////////////////看线段树种类
    }
    return res;
}

int main()
{
    LL n,q;
    while(~scanf("%lld%lld",&n,&q))
    {
        lan(tree,0);
        lan(a,0);
        For(i,1,n)
            scanf("%lld",&a[0][i]),a[1][i]=a[0][i];
        build(1,1,n,1);
        LL tem=n;
        For(i,1,n)
        {
            a[0][i]*=tem;
            tem--;
//            printf("a%d=%lld\n",i,a[0][i]);
        }
        build(1,1,n,0);
        while(q--)
        {
            LL op,l,r;
            scanf("%lld%lld%lld",&op,&l,&r);
            if(op==1)
            {
                LL ans=0;
                ans=query(1,l,r,0)-query(1,l,r,1)*(n-r);
//                printf("%lld %lld %lld\n",query(1,l,r,0),query(1,l,r,1),n-r);
                printf("%lld\n",ans);
            }
            else
            {
                update(1,l,l,r*(n-l+1),0);
                update(1,l,l,r,1);
            }
        }
    }
    return 0;
}