Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer ai.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1,2,3,…). Eating the sweet i at the d-th day will cause a sugar penalty of (d⋅ai), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1≤m≤n≤200 000).
The second line contains n integers a1,a2,…,an (1≤ai≤200 000).
Output
You have to output n integers x1,x2,…,xn on a single line, separed by spaces, where xk is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let’s analyze the answer for k=5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
Day 1: sweets 1 and 4
Day 2: sweets 5 and 3
Day 3 : sweet 6
Total penalty is 1⋅a1+1⋅a4+2⋅a5+2⋅a3+3⋅a6=6+4+8+6+6=30. We can prove that it’s the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x5=30.
思路：很好的一道题目。前缀和的巧妙应用。
排序之后，对于前面的糖果，肯定是放到后面吃的，贪心策略。但是如何计算就是一个问题了。我们先求出所有的前缀和来，对于前m个，就是单纯的前缀和了。但是对于m之后的，例如m=2，要吃6个糖果的时候，第5,6个糖果肯定是第一天吃，第3,4个糖果是第二天吃，第1,2个糖果是第一天吃。那么一共a[5]+a[6]+(a[3]+a[4])*2+(a[1]+a[2])*3个，转化一下等于(a[1]+a[2]+a[3]+a[4]+a[5]+a[6])+(a[3]+a[4])+(a[1+a[2])*2个，其中(a[3]+a[4])+(a[1]+a[2])*2正是第四天吃的，也就是sum[6]+ans[4]=sum[6]+ans[6-m]。这样就可以了。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=2e5+100;
int a[maxx];
ll val[maxx],ans[maxx];
int n,m;
int main()
{
scanf("%d%d",&n,&m);
val[0]=0;
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
ll sum=0;
sort(a+1,a+1+n);
for(int i=1;i<=n;i++) val[i]=val[i-1]+(ll)a[i];
for(int i=1;i<=m;i++) ans[i]=val[i];
for(int i=m+1;i<=n;i++) ans[i]=ans[i-m]+val[i];
for(int i=1;i<=n;i++) cout<<ans[i]<<" ";
cout<<endl;
return 0;
}
努力加油a啊，(o)/~
