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, 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
5 3
1 2 3 4 5
1 1 3
2 5 0
1 4 5样例输出复制
10
8题意：1.求
2 。 更新a[l] 为 w
题解：对于任意一项 i，它对区间和的贡献为 a[i]*(r-i+1) (r为区间右端点），因此上式可化为
可以进一步化简为 ,那么只要求a[i]的前缀和以及i*a[i]的前缀和就行了。
树状数组
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<vector>
using namespace std;
#define inf 0x3f3f3f
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a));
#define lowbit(x) x&-x;
typedef long long ll;
const int maxn = 1e5+5;
int n,q;
//更新两个前缀和
ll c1[maxn],c2[maxn];
ll a[maxn];
void update1(int x,ll val){
while(x <= n){
c1[x] += val;
x += lowbit(x);
}
}
ll sum1(int x){
ll res = 0;
while(x > 0){
res += c1[x];
x -= lowbit(x);
}
return res;
}
void update2(int x,ll val){
while(x <= n){
c2[x] += val;
x += lowbit(x);
}
}
ll sum2(int x){
ll res = 0;
while(x > 0){
res += c2[x];
x -= lowbit(x);
}
return res;
}
int main(){
while(~scanf("%d%d",&n,&q)){
mem(c1,0);
mem(c2,0);
int op,l,r;
for(int i = 1; i <= n; i++){
scanf("%lld",&a[i]);
}
for(int i = 1; i <= n; i++){
update1(i,a[i]);
update2(i,i*a[i]);
}
ll ans = 0;
for(int i = 1; i <= q; i++){
scanf("%d%d%d",&op,&l,&r);
if(op == 1){
ans = (r+1)*(sum1(r) - sum1(l-1));
ans -= (sum2(r) - sum2(l-1));
printf("%lld\n",ans);
}else{
update1(l,r-a[l]);
update2(l,l*(r-a[l]));
a[l] = r;
}
}
}
}