CodeForces-1006B-Polycarp's Practice

B. Polycarp's Practice

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp is practicing his problem solving skill. He has a list of nn problems with difficulties a1,a2,…,ana1,a2,…,an, respectively. His plan is to practice for exactly kk days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all nn problems in exactly kk days.

Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in kk days he will solve all the nn problems.

The profit of the jj-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the jj-th day (i.e. if he solves problems with indices from ll to rr during a day, then the profit of the day is maxl≤i≤raimaxl≤i≤rai). The total profit of his practice is the sum of the profits over all kk days of his practice.

You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all nnproblems between kk days satisfying the conditions above in such a way, that the total profit is maximum.

For example, if n=8,k=3n=8,k=3 and a=[5,4,2,6,5,1,9,2]a=[5,4,2,6,5,1,9,2], one of the possible distributions with maximum total profit is: [5,4,2],[6,5],[1,9,2][5,4,2],[6,5],[1,9,2]. Here the total profit equals 5+6+9=205+6+9=20.

Input

The first line of the input contains two integers nn and kk (1≤k≤n≤20001≤k≤n≤2000) — the number of problems and the number of days, respectively.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤20001≤ai≤2000) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).

Output

In the first line of the output print the maximum possible total profit.

In the second line print exactly kk positive integers t1,t2,…,tkt1,t2,…,tk (t1+t2+⋯+tkt1+t2+⋯+tk must equal nn), where tjtj means the number of problems Polycarp will solve during the jj-th day in order to achieve the maximum possible total profit of his practice.

If there are many possible answers, you may print any of them.

Examples

input

Copy

8 3
5 4 2 6 5 1 9 2

output

Copy

20
3 2 3

input

Copy

5 1
1 1 1 1 1

output

Copy

1
5

input

Copy

4 2
1 2000 2000 2

output

Copy

4000
2 2

Note

The first example is described in the problem statement.

In the second example there is only one possible distribution.

In the third example the best answer is to distribute problems in the following way: [1,2000],[2000,2][1,2000],[2000,2]. The total profit of this distribution is 2000+2000=40002000+2000=4000.

Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

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题解：就是找前K大的几个数，输出值几个大数之间的距离；找到前K大的数的位置，然后往两侧扩展即可。

AC代码为：

#include<bits/stdc++.h>

using namespace std;

int a[2010],vis[2010],ans[2010];

struct Node{

    int num,id;

} node[2010];

bool cmp(Node a,Node b)

{

    return a.num>b.num;

}

bool cmp1(Node a, Node b)

{

    return a.id<b.id;

}

int main()

{

    ios::sync_with_stdio(false);

    cin.tie(0);

    int n,k;

    cin>>n>>k;

    memset(vis,0,sizeof vis);

    for(int i=1;i<=n;i++)

    {

        cin>>a[i];

        node[i].num=a[i];

        node[i].id=i;

    }

    sort(node+1,node+1+n,cmp);

    int sum=0;

    for(int i=1;i<=k;i++) sum+=node[i].num,vis[node[i].id]=1;

    cout<<sum<<endl;

    int temp=0,flag1,flag2;

    for(int i=1;i<=n;i++)

    {

        if(vis[i])

        {

            flag1=i-1;flag2=i+1;

            while(!vis[flag1] && flag1>0&&flag1<=n) vis[flag1]=1,flag1--;

            while(!vis[flag2] && flag2>0&&flag2<=n) vis[flag2]=1,flag2++;

            ans[temp++]=flag2-flag1-1; 

            i=flag2-1;

        }

    }

    for(int i=0;i<temp;i++) i==temp-1 ? cout<<ans[i]<<endl : cout<<ans[i]<<" ";

    return 0;

}