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=0;
    while(x>0)
    {
        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=0;
    while(x>0)
    {
        sum+=c2[x];
        x-=lowbit(x);
    }
    return  sum;
}
int  main()
{
    scanf("%d%d",&n,&q);
    ull x;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&x);
        update1(i,x);
        update2(i,1ull*(n-i+1)*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==1)
        {
            ull ans1=1ull*(n-r)*(getsum1(r)-getsum1(l-1));
            ull ans2=getsum2(r)-getsum2(l-1);
            printf("%lld
",ans2-ans1);//"u"  是unsigned   int 
        }
        else{
            ull temp=getsum1(l)-getsum1(l-1);
            update1(l,1ull*(r-temp) );
            update2(l,1ull*(r-temp)*(n-l+1)) ;
        }
    }
    return   0;
}