本文主要是介绍CodeForces - 1006B. Polycarp's Practice，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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 nn problems 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
8 3 5 4 2 6 5 1 9 2
Output
20 3 2 3
Input
5 1 1 1 1 1 1
Output
1 5
Input
4 2 1 2000 2000 2
Output
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 .
思路：定义一个结构体，一个存储 数据一个存位置下标，先 按照数据的降序排序，将最大的前K个数的下标存放在一个数组里，
再将数组按照升序进行排列，通过循环输出，具体细节代码注释。
此外，此题也可以有pair动态 数组（也属于结构体），原理相同
#include<bits/stdc++.h>
using namespace std;
struct A
{int shu;int loca;
}a[2008];
int maxn[20008],location[20008];
int cmp(A c,A d)
{return c.shu>d.shu;
}
int main()
{int n,k,ans=0;scanf("%d%d",&n,&k);for(int i=1;i<=n;i++){scanf("%d",&a[i].shu);a[i].loca=i;}sort(a+1,a+n+1,cmp);for(int i=1;i<=k;i++){maxn[i]=a[i].shu;ans+=a[i].shu;}printf("%d\n",ans);for(int i=1;i<=k;i++){location[i]=a[i].loca;}sort(location+1,location+1+k);int temp=0;for(int i=1;i<=k-1;i++)//为了避免最后一个最大值的坐标不是最后他一位，通过此出进行处理{printf("%d ",(location[i]-location[i-1]));temp+=(location[i]-location[i-1]);//将前面的每组数据的个数加起来}printf("%d\n",n-temp);//总的数据减去前面 的就是最后一组的数据个数return 0;}
这篇关于CodeForces - 1006B. Polycarp's Practice的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
