ACM-ICPC 2018 徐州赛区网络预赛 H. Ryuji doesn't want to study

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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] * L + a[l+1] *(L-1) + …… + a[r-1] *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 nn and qq (n, q≤100000).

The next line contains n integers represent a[i](a[i]≤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, ccrepresents [ l, r ]. if a =2 , it means question type is 2 , 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

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

 #define  ull unsigned  long  long
 #define ll  long  long
 #define  N  100009
 #define  lowbit(x) x&(-x)
 ull c1[N],c2[N];
 int n,q;
 void   update1(int x,ull num)
 {
     while(x<=n)
     {
         c1[x]+=num;
         x+=lowbit(x);
     }
 }
 ull getsum1(int  x)
 {
     ull sum=;
     while(x>)
     {
         sum+=c1[x];
         x-=lowbit(x);
     }
     return  sum;
 }
 void   update2(int x,ull num)
 {
     while(x<=n)
     {
         c2[x]+=num;
         x+=lowbit(x);
     }
 }
 ull getsum2(int  x)
 {
     ull sum=;
     while(x>)
     {
         sum+=c2[x];
         x-=lowbit(x);
     }
     return  sum;
 }
 int  main()
 {
     scanf("%d%d",&n,&q);
     ull x;
     for(int i=;i<=n;i++)
     {
         scanf("%lld",&x);
         update1(i,x);
         update2(i,1ull*(n-i+)*x);//在前面加个 ull  保证整体是ull 的
     }
     int op,l,r;
     while(q--)
     {
         scanf("%d%d%d",&op,&l,&r);
         if(op==)
         {
             ull ans1=1ull*(n-r)*(getsum1(r)-getsum1(l-));
             ull ans2=getsum2(r)-getsum2(l-);
             printf("%lld\n",ans2-ans1);//"u"  是unsigned   int
         }
         else{
             ull temp=getsum1(l)-getsum1(l-);
             update1(l,1ull*(r-temp) );
             update2(l,1ull*(r-temp)*(n-l+)) ;
         }
     }
     return   ;
 }