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B. Yet Another Array Partitioning Task time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

An array bb is called to be a subarray of aa if it forms a continuous subsequence of aa , that is, if it is equal to alal , al+1al+1 , …… , ar

ar for some l,rl,r .

Suppose mm is some known constant. For any array, having mm or more elements, let‘s define it‘s beauty as the sum of mm largest elements of that array. For example:

• For array x=[4,3,1,5,2]x=[4,3,1,5,2] and m=3m=3 , the 33 largest elements of xx are 55 , 44 and 33 , so the beauty of
  xx is 5+4+3=125+4+3=12 .
• For array x=[10,10,10]x=[10,10,10] and m=2m=2 , the beauty of xx is 10+10=2010+10=20 .

You are given an array a1,a2,…,ana1,a2,…,an , the value of the said constant mm and an integer kk . Your need to split the array aa into exactly kk subarrays such that:

• Each element from aa belongs to exactly one subarray.
• Each subarray has at least mm elements.
• The sum of all beauties of kk subarrays is maximum possible.

Input

The first line contains three integers nn , mm and kk (2≤n≤2?1052≤n≤2?105 , 1≤m1≤m , 2≤k2≤k , m?k≤nm?k≤n ) — the number of elements in aa , the constant mm in the definition of beauty and the number of subarrays to split to.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (?109≤ai≤109?109≤ai≤109 ).

Output

In the first line, print the maximum possible sum of the beauties of the subarrays in the optimal partition.

In the second line, print k?1k?1 integers p1,p2,…,pk?1p1,p2,…,pk?1 (1≤p1<p2<…<pk?1<n1≤p1<p2<…<pk?1<n ) representing the partition of the array, in which:

• All elements with indices from 11 to p1p1 belong to the first subarray.
• All elements with indices from p1+1p1+1 to p2p2 belong to the second subarray.
• …… .
• All elements with indices from pk?1+1pk?1+1 to nn belong to the last, kk -th subarray.

If there are several optimal partitions, print any of them.

Examples Input Copy

9 2 3
5 2 5 2 4 1 1 3 2

Output Copy

21
3 5 

Input Copy

6 1 4
4 1 3 2 2 3

Output Copy

12
1 3 5 

Input Copy

2 1 2
-1000000000 1000000000

Output Copy

0
1 

Note

In the first example, one of the optimal partitions is [5,2,5][5,2,5] , [2,4][2,4] , [1,1,3,2][1,1,3,2] .

• The beauty of the subarray [5,2,5][5,2,5] is 5+5=105+5=10 .
• The beauty of the subarray [2,4][2,4] is 2+4=62+4=6 .
• The beauty of the subarray [1,1,3,2][1,1,3,2] is 3+2=53+2=5 .

The sum of their beauties is 10+6+5=2110+6+5=21 .

In the second example, one optimal partition is [4][4] , [1,3][1,3] , [2,2][2,2] , [3][3] .

大意：
這個題目是給你n個數據，讓你分成k段，每一段至少m個數，並且把每一段的前m大的數求和
輸出求和的結果，和分段的位置、
思路：
是求這每一段的前m大的數，其實可以轉化成，求這一組數前m*k大的數之和。
至於求分段的位置，這個就有點麻煩了，可以把每一個位置初始標記為0，
然後在求sum的同時，將求過的數位置標記成1，之後求分段位置的時候，
就可以進行累加，一旦累加之和為m，說明這m個數組成了一段，也就是在這個位置將數組分段。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
struct node
{
    int x,id;
}a[maxn];
bool vis[maxn];
bool cmp(node a,node b)
{
    return a.x>b.x;
}
int main()
{
    int n,m,k;
    memset(vis,0,sizeof(vis));
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].x);
        a[i].id=i;
    }
    sort(a+1,a+1+n,cmp);
    ll sum=0;
    for(int i=1;i<=m*k;i++)
    {
        sum+=a[i].x;
        vis[a[i].id]=1;
        //printf("a.id=%d\n",a[i])
    }
    printf("%I64d\n",sum);
    int s=0;;
    for(int i=1;i<=n;i++)
    {
        s=s+vis[i];
      //  printf("%d %d\n",i,s);
        if(s==m&&k!=1)
        {
        	k--;
        	s=0;
            printf("%d ",i);
        }
    }
    printf("\n");
    return 0;
}

　　

B. Yet Another Array Partitioning Task ——cf