An array b is called to be a subarray of a if it forms a continuous subsequence of a, that is, if it is equal to al, al+1, …, ar for some l,r.

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

For array x=[4,3,1,5,2] and m=3, the 3 largest elements of x are 5, 4 and 3, so the beauty of x is 5+4+3=12.
For array x=[10,10,10] and m=2, the beauty of x is 10+10=20.
You are given an array a1,a2,…,an, the value of the said constant m and an integer k. Your need to split the array a into exactly k subarrays such that:

Each element from a belongs to exactly one subarray.
Each subarray has at least m elements.
The sum of all beauties of k subarrays is maximum possible.
Input
The first line contains three integers n, m and k (2≤n≤2?105, 1≤m, 2≤k, m?k≤n) — the number of elements in a, the constant m in the definition of beauty and the number of subarrays to split to.

The second line contains n integers a1,a2,…,an (?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?1 integers p1,p2,…,pk?1 (1≤p1<p2<…<pk?1<n) representing the partition of the array, in which:

All elements with indices from 1 to p1 belong to the first subarray.
All elements with indices from p1+1 to p2 belong to the second subarray.
….
All elements with indices from pk?1+1 to n belong to the last, k-th subarray.
If there are several optimal partitions, print any of them.

Examples
Input
9 2 3
5 2 5 2 4 1 1 3 2
Output
21
3 5
Input
6 1 4
4 1 3 2 2 3
Output
12
1 3 5
Input
2 1 2
-1000000000 1000000000
Output
0
1
Note
In the first example, one of the optimal partitions is [5,2,5], [2,4], [1,1,3,2].

The beauty of the subarray [5,2,5] is 5+5=10.
The beauty of the subarray [2,4] is 2+4=6.
The beauty of the subarray [1,1,3,2] is 3+2=5.
The sum of their beauties is 10+6+5=21.

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

思路：首先我們可以知道的是，這個數(shù)組中最終要選取mk個數(shù)字，那么就是前mk大的數(shù)字和就是最大值。難點(diǎn)在于怎么去找分割的點(diǎn)。我的方法是map記錄，前k-1組中只要出現(xiàn)了m個（前m*k大的數(shù)字），就分成一組。最后一組，由后往前找，假設(shè)在位置j上湊夠了m個，就把k-1組的位置弄到j(luò)-1上，這樣就可以符合題目要求了。

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=2e5+100; int a[maxx]; int b[maxx]; int pos[maxx]; int n,m,k;inline bool cmp(int a,int b) {return a>b; } int main() {scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=n;i++) scanf("%d",&a[i]),b[i]=a[i];sort(b+1,b+1+n,cmp);map<int,int> mp;ll ans=0;for(int i=1,j=1;i<=n&&j<=m*k;i++,j++) ans+=b[i],mp[b[i]]++;int cnt=0,k1=0;int i;for(i=1;i<=n;i++){if(cnt<m){if(mp[a[i]]) mp[a[i]]--,cnt++;}if(cnt==m) pos[++k1]=i,cnt=0;if(k1==k-1) break;}cnt=0;for(i=n;;i--){if(mp[a[i]]) mp[a[i]]--,cnt++;if(cnt==m) break;}pos[k-1]=i-1;cout<<ans<<endl;for(int i=1;i<k;i++) cout<<pos[i]<<" ";return 0; }

努力加油a啊，(^o)/~

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