东北赛H. Skyscraper（树状数组求前缀和）（想法）（前缀和的应用）

H. Skyscraper

time limit per test

4.0 s

memory limit per test

512 MB

input

standard input

output

standard output

At the main street of Byteland, there will be built nn skyscrapers, standing sequentially one next to other. If look leftside right, sequence of their height will be a1,a2,…,ana1,a2,…,an.

Initially the street is empty, every skyscraper's height is 00. Hamster is the leader of the construction team. In each stage, Hamster can select a range [l,r][l,r], then the team will work on this range. Specifically, assume the height sequence is h1,h2,…,hnh1,h2,…,hn, then hl,hl+1,…,hrhl,hl+1,…,hr will increase by 11during this stage. When hi=aihi=ai holds for all i∈[1,n]i∈[1,n], the project will be closed.

The plan may be changed for many times. There will be mm events of 22 kinds below:

• 1 l r k (1≤l≤r≤n,1≤k≤1051≤l≤r≤n,1≤k≤105), for all x∈[l,r]x∈[l,r], change axax to ax+kax+k.
• 2 l r (1≤l≤r≤n1≤l≤r≤n), assume a1,a2,…,al−1,ar+1,ar+2,…,an=0a1,a2,…,al−1,ar+1,ar+2,…,an=0, ask for the minimum number of required stages to close the project.

Input

The first line of the input contains an integer T(1≤T≤1000)T(1≤T≤1000), denoting the number of test cases.

In each test case, there are two integers n,m(1≤n,m≤100000)n,m(1≤n,m≤100000) in the first line, denoting the number of skyscrapers and events.

In the second line, there are nn integers a1,a2,...,an(1≤ai≤100000)a1,a2,...,an(1≤ai≤100000).

For the next mm lines, each line describes an event.

It is guaranteed that ∑n≤106∑n≤106 and ∑m≤106∑m≤106.

Output

For each query event, print a single line containing an integer, denoting the answer.

Example

input

Copy

1
5 4
1 3 1 4 5
2 1 5
1 3 4 2
2 2 4
2 1 5

output

Copy

7
6
6

仔细读了官方题解之后突然明白

原来本题并不需要区间更新

本题巧妙地运用了前缀和

在每次更新是我们把l+x r+1-x

这样我们其实也是妙的进行了一次区间更新

#include<bits/stdc++.h>
#define N 100005
using namespace std;
long long n;
long long bsum[N];
long long csum[N];
long long a[N];
long long b[N];
long long c[N];
long long lowbit(long long x)
{
    return x&(-x);
}

void updateb(long long k,long long x)
{
    for(long long i=k; i<=n; i+=lowbit(i))
        bsum[i]+=x;
}
long long getsumb(long long x)
{
    long long ans=0;
    for(long long i=x; i; i-=lowbit(i)) //i要大于0
        ans+=bsum[i];
    return ans;
}

void updatec(long long k,long long x)
{
    for(long long i=k; i<=n; i+=lowbit(i))
        csum[i]+=x;
}
long long getsumc(long long x)
{
    long long ans=0;
    for(long long i=x; i; i-=lowbit(i)) //i要大于0
        ans+=csum[i];
    return ans;
}


int main()
{
    long long T;
    cin>>T;
    while(T--)
    {
        long long m;
        n=2;
        memset(bsum,0,sizeof(bsum));
        memset(csum,0,sizeof(csum));
        a[0]=0;
        cin>>n>>m;
        for(long long i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
        }
        for(long long i=1;i<=n;i++)//预处理b
        {
            b[i]=a[i]-a[i-1];
            updateb(i,b[i]);
        }
        for(long long i=1;i<=n;i++)//预处理c
        {
            if(b[i]>0) c[i]=b[i];
            else c[i]=0;
            updatec(i,c[i]);
        }
        while(m--)
        {
            long long op;
            scanf("%lld",&op);
            if(op==2)
            {
                long long l,r;
                scanf("%lld%lld",&l,&r);
                long long ans=0;
                ans=getsumb(l)+getsumc(r)-getsumc(l);
                //cout<<getsumb(l)<<" "<<getsumc(r)<<" "<<getsumc(l)<<endl;
                printf("%lld
",ans);
            }
            else
            {
                long long l,r,temp3;
                long long add1,add2;
                scanf("%lld%lld%lld",&l,&r,&temp3);
                if(b[l]<0)
                {
                    add1=temp3+b[l];
                    add1=max(1LL*0,add1);
                }
                else add1=temp3;
                if(b[r+1]>0)
                {
                    add2=min(b[r+1],temp3);
                }
                else add2=0;
                b[l]+=temp3;
                b[r+1]-=temp3;
                updateb(l,temp3);
                updateb(r+1,-temp3);
                updatec(l,add1);
                updatec(r+1,-add2);
            }
        }

    }
    return 0;
}