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 $a[i]$.
Unfortunately, the longer he learns, the fewer he gets.
That means, if he reads books from ll to rr, he will get $a[l] imes L + a[l+1] imes (L-1) + cdots + a[r-1] imes 2 + a[r]$ ($L$ 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 1, you should answer how much knowledge he will get after he reads books [ $l$, $r$ ].
2. If the question type is 2, Ryuji will change the ith book's knowledge to a new value.
Input
First line contains two integers $n$ and $q$ ($n$, $q le 100000$).
The next line contains n integers represent $a[i]$($a[i] le 1e9$).
Then in next qq line each line contains three integers $a,b,c$, if $a = 1$, it means question type is $1$, and $b$, $c$ represents [ $l$ , $r$ ].
If $a = 2$, it means question type is $2$ , and $b$, $c$ means Ryuji changes the bth book' knowledge to $c$.
Output
For each question, output one line with one integer represent the answer.
样例输入
5 3 1 2 3 4 5 1 1 3 2 5 0 1 4 5
样例输出
10 8
题意：
给出 $n$ 本书编号 $1$ 到 $n$，每本书权值为 $w[i]$，给出 $q$ 个操作，
操作 $1$，给出区间 $[l,r]$，则区间长度为 $L = r - l + 1$，查询的答案应为 $a[l] imes L + a[l+1] imes (L-1) + cdots + a[r-1] imes 2 + a[r]$，
操作 $2$，把在编号为 $b$ 的书的权值改成 $c$。
题解：
线段树维护两个和：
一个是普通的区间和 $sumlimits_{i = l}^r {w[i]} = w[l] + cdots + w[r]$；
另一个是 $sumlimits_{i = l}^r {left[ {w[i] imes left( {n - i + 1} ight)} ight]} = w[l] imes left( {n - l + 1} ight) + cdots + w[r] imes left( {n - r + 1} ight)$。
那么，对于所有的查询：
$egin{array}{l}
Qleft( {l,r}
ight) \
= wleft[ l
ight] imes left( {r - l + 1}
ight) + wleft[ {l + 1}
ight] imes left( {r - l}
ight) + cdots + wleft[ r
ight] imes 1 \
= sumlimits_{i = l}^r {left[ {wleft[ i
ight] imes left( {r - i + 1}
ight)}
ight]} \
= sumlimits_{i = l}^r {left[ {wleft[ i
ight] imes left( {n - i + 1 - n + r}
ight)}
ight]} \
{
m{ = }}sumlimits_{i = l}^r {left[ {wleft[ i
ight] imes left( {n - i + 1}
ight) - wleft[ i
ight] imes left( {n - r}
ight)}
ight]} \
= sumlimits_{i = l}^r {left[ {wleft[ i
ight] imes left( {n - i + 1}
ight)}
ight]} - left( {n - r}
ight)sumlimits_{i = l}^r {wleft[ i
ight]} \
end{array}$
AC代码：
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=1e5+10; int n,q; ll a[maxn]; /********************************* Segment Tree - st *********************************/ struct Node{ int l,r; ll val,sum; }node[4*maxn]; void pushup(int root) { node[root].val=node[root*2].val+node[root*2+1].val; node[root].sum=node[root*2].sum+node[root*2+1].sum; } void build(int root,int l,int r) { if(l>r) return; node[root].l=l; node[root].r=r; node[root].val=0; node[root].sum=0; if(l==r) { node[root].val=a[l]; node[root].sum=a[l]*(n-l+1); } else { int mid=l+(r-l)/2; build(root*2,l,mid); build(root*2+1,mid+1,r); pushup(root); } } void update(int root,int pos,ll val) { if(node[root].l==node[root].r) { node[root].val=val; node[root].sum=val*(n-pos+1); return; } int mid=node[root].l+(node[root].r-node[root].l)/2; if(pos<=mid) update(root*2,pos,val); if(pos>mid) update(root*2+1,pos,val); pushup(root); } ll askval(int root,int st,int ed) { if(st>node[root].r || ed<node[root].l) return 0; if(st<=node[root].l && node[root].r<=ed) return node[root].val; else return askval(root*2,st,ed)+askval(root*2+1,st,ed); } ll asksum(int root,int st,int ed) { if(st>node[root].r || ed<node[root].l) return 0; if(st<=node[root].l && node[root].r<=ed) return node[root].sum; else return asksum(root*2,st,ed)+asksum(root*2+1,st,ed); } /********************************* Segment Tree - ed *********************************/ int main() { scanf("%d%d",&n,&q); for(int i=1;i<=n;i++) scanf("%d",&a[i]); build(1,1,n); for(int i=1;i<=q;i++) { int type; scanf("%d",&type); if(type==1) { int a,b; scanf("%d%d",&a,&b); ll A=asksum(1,a,b); ll B=askval(1,a,b); //cout<<A<<" "<<B<<endl; printf("%lld ",A-(n-b)*B); } else { int a; ll b; scanf("%d%lld",&a,&b); update(1,a,b); } } }