Codeforces Round #498 (Div. 3) B. Polycarp's Practice

题目链接：http://codeforces.com/contest/1006/problem/B

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 $\mathrm{nn problems\; with\; difficulties a1,a2,\dots ,ana1,a2,\dots ,an,\; respectively.\; His\; plan\; is\; to\; practice\; for\; exactly kk days.\; Each\; day\; he\; has\; to\; solve\; at\; least\; one\; problem\; from\; his\; list.\; Polycarp\; solves\; the\; problems\; in\; the\; order\; they\; are\; given\; in\; his\; list,\; he\; cannot\; skip\; any\; problem\; from\; his\; list.\; He\; has\; to\; solve\; all nn problems\; in\; exactly kk days.}$

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 $\mathrm{kk days\; he\; will\; solve\; all\; the nn problems.}$

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

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 $\mathrm{nn problems\; between kk days\; satisfying\; the\; conditions\; above\; in\; such\; a\; way,\; that\; the total\; profit is\; maximum.}$

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

Input

The first line of the input contains two integers $\mathrm{nn and kk (1\le k\le n\le 20001\le k\le n\le 2000)\; —\; the\; number\; of\; problems\; and\; the\; number\; of\; days,\; respectively.}$

The second line of the input contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 20001\le ai\le 2000)\; —\; difficulties\; of\; problems\; in\; Polycarp\text{'}s\; list,\; in\; the\; order\; they\; are\; placed\; in\; the\; list\; (i.e.\; in\; the\; order\; Polycarp\; will\; solve\; them).}$

Output

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

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

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: $[1,2000],[2000,2][1,2000],[2000,2].\; The total\; profit of\; this\; distribution\; is 2000+2000=40002000+2000=4000.$

$题意：给你n个问题，你要在k天内把他解决，每天至少解决一个问题（要按照顺序来解决问题，不能重复解决问题），一天中最困难的问题就是这一天的profit，问总的profit最大是多少。$

$题解：很容易知道maxsum就是前k个最困难的问题和。每天解决的问题就是每个最困难的问题和上一个最困难的问题之间的问题个数。$

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
    int a;
    int b;
}dian[2010];
bool cmp(node one,node two){
    return one.b<two.b;//恢复相对顺序
}
bool cmp1(node one,node two){
    return one.a>two.a;//为了求k项最大的和 
}
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&dian[i].a);
        dian[i].b=i;
    }
    sort(dian+1,dian+1+n,cmp1);
    long long sum=0;
    for (int i=1;i<=k;i++)
    sum+=dian[i].a;
    sort(dian+1,dian+1+k,cmp);//最大的k项恢复相对位置。******重点。 
    printf("%d
",sum);
    if(k==1)
    printf("%d
",n);
    else
    {
        for (int i=1;i<=k;i++)//每天解决的问题就是每个最困难的问题和上一个最困难的问题之间的问题个数。 
        {
            if(i==1)//第一个困难问题特判一下 
            printf("%d",dian[i].b); 
            else if(i==k)//最后一个困难问题特判一下 
            printf(" %d
",n-dian[k-1].b);
            else
            printf(" %d",dian[i].b-dian[i-1].b);
        }
    }
}