ACM-ICPC 2018 徐州赛区网络赛 H. Ryuji doesn't want to study（树状数组）

本文介绍了一种使用两个树状数组解决特定类型查询问题的方法。通过对序列进行操作，包括查询特定区间的加权和及更新序列元素值，有效地解决了问题。

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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

样例输出复制

10
8

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

题意：

有长度为N的一个序列和M次操作。

操作1是计算a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r]这个值，L是r-l+1这个长度。

操作2是把a[i]修改成另一个值。

思路：

用开两个树状数组，一个存a[i],另一个存(n-i+1)*a[i]。

当计算操作1时，用存(n-i+1)*a[i]的树状数组得到一个值减去存a[i]的树状数组的值，可以消去多余的系数。

当遇到操作2时，两个树状数组的值都要修改。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
ll tr[maxn][2];
ll a[maxn];
int lowbit(int i)
{
    return i&(-i);
}
void add(int x,ll c,int op)
{
    while(x<maxn)
    {
        tr[x][op]+=c;
        x+=lowbit(x);
    }
}
ll query(int x,int op)
{
    ll ans=0;
    while(x)
    {
        ans+=tr[x][op];
        x-=lowbit(x);
    }
    return ans;
}
int main()
{
    int n,q,op,l,r;
    while(~scanf("%d%d",&n,&q))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
            add(i,a[i]*(n-i+1),1);
            add(i,a[i],2);
        }
        for(int i=1;i<=q;i++)
        {
            scanf("%d%d%d",&op,&l,&r);
            if(op&1)
            {
                printf("%lld\n",query(r,1)-query(l-1,1)-(query(r,2)-query(l-1,2))*(n-r));
            }
            else
            {
                add(l,-a[l]*(n-l+1),1);
                add(l,-a[l],2);
                a[l]=r;
                add(l,a[l]*(n-l+1),1);
                add(l,a[l],2);
            }
        }
    }
    return 0;
}