Sweets Eating

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我是靠谱客的博主俭朴寒风，最近开发中收集的这篇文章主要介绍Sweets Eating，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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.

这个题自己倒是会算但是每想到这种的递推公式，哇太惨了，还是自己没找到他们之间的联系唉。。。
可一想想那些是联系的
k=1和K=2联系和K+m联系
把 K+M的联系先写出来
当i<m sum[i]=a[i];
当i>=m sum[i]=sum[i-1]+a[i];
而那一天你真正的惩罚还要加上前一个的值，这样才是这个的惩罚。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iomanip>
long long a[1000010], sum[100100];
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + 1 + n);
for (int i = 1; i <= n; i++)
{
if (i < m) sum[i] = a[i];
else
sum[i] = sum[i - m] + a[i];
}
long long s = 0;
for (int i = 1; i <= n; i++)
{
s = s + sum[i];
cout << s << " ";
}
}