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] imes L + a[l+1] imes (L-1) + cdots + a[r-1] imes 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
5 3 1 2 3 4 5 1 1 3 2 5 0 1 4 5
样例输出
10 8
题目来源
题意：有n个数，你可以对这n个数进行m次操作，可以进行的操作有两种，第一种将b位置的数置为c，第二种求a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r]
分析：注意表达式：a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r]可以化为：
化成这个式子后，我们每次只要维护a[i]*(n-i+1)和a[i]就可以了，更新的时候加上c-a[i]就可以将a[i]更新为c
做题目做少了，打网络赛的时候都没有想到这样化简式子
参考博客：https://blog.csdn.net/qq_39826163/article/details/82586489
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5+10;
const ll mod = 2e9+7;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll n, m, q, a[maxn], t1[maxn], t2[maxn];
ll lowbit( ll x ) {
return x&(-x);
}
void update1( ll x, ll v ) {
while( x <= n ) {
t1[x] += v;
x += lowbit(x);
}
}
void update2( ll x, ll v ) {
while( x <= n ) {
t2[x] += v;
x += lowbit(x);
}
}
ll query1( ll x ) {
ll sum = 0;
while(x) {
sum += t1[x];
x -= lowbit(x);
}
return sum;
}
ll query2( ll x ) {
ll sum = 0;
while(x) {
sum += t2[x];
x -= lowbit(x);
}
return sum;
}
int main() {
scanf("%lld%lld",&n,&m);
for( ll i = 1; i <= n; i ++ ) {
scanf("%lld",&a[i]);
update1(i,a[i]);
update2(i,a[i]*(n-i+1));
}
while( m -- ) {
ll k, b, c;
scanf("%lld%lld%lld",&k,&b,&c);
if( k == 1 ) {
ll sum1 = query1(c) - query1(b-1);
ll sum2 = query2(c) - query2(b-1);
ll ans = sum2-(n-c)*sum1;
printf("%lld
",ans);
} else {
update1(b,c-a[b]);
update2(b,(c-a[b])*(n-b+1));
a[b] = c;
}
}
return 0;
}