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Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

# C. Array Splitting

You are given a sorted array 𝑎1,𝑎2,…,𝑎𝑛 (for each index 𝑖>1 condition 𝑎𝑖≥𝑎𝑖−1 holds) and an integer 𝑘.

You are asked to divide this array into 𝑘 non-empty consecutive subarrays. Every element in the array should be included in exactly one subarray.

Let 𝑚𝑎𝑥(𝑖) be equal to the maximum in the 𝑖-th subarray, and 𝑚𝑖𝑛(𝑖) be equal to the minimum in the 𝑖-th subarray. The cost of division is equal to ∑𝑖=1𝑘(𝑚𝑎𝑥(𝑖)−𝑚𝑖𝑛(𝑖)). For example, if 𝑎=[2,4,5,5,8,11,19] and we divide it into 3 subarrays in the following way: [2,4],[5,5],[8,11,19], then the cost of division is equal to (4−2)+(5−5)+(19−8)=13.

Calculate the minimum cost you can obtain by dividing the array 𝑎 into 𝑘 non-empty consecutive subarrays.

## Input

The first line contains two integers 𝑛 and 𝑘 (1≤𝑘≤𝑛≤3⋅105).

The second line contains 𝑛 integers 𝑎1,𝑎2,…,𝑎𝑛 (1≤𝑎𝑖≤109, 𝑎𝑖≥𝑎𝑖−1).

## Output

Print the minimum cost you can obtain by dividing the array 𝑎 into 𝑘 nonempty consecutive subarrays.

## Examples

input
6 3
4 8 15 16 23 42
output
12
input
4 4
1 3 3 7
output
0
input
8 1
1 1 2 3 5 8 13 21
output
20

## Note

In the first test we can divide array 𝑎 in the following way: [4,8,15,16],[23],[42].

## 代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 300000 + 5;
int n, k;
int a[maxn],b[maxn];
int main() {
scanf("%d%d",&n,&k);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
for(int i=0;i<n-1;i++){
b[i]=a[i+1]-a[i];
}
int sum=0;
sort(b,b+n-1);
for(int i=0;i<n-k;i++){
sum=sum+b[i];
}
printf("%d\n",sum);
return 0;
}
posted @ 2019-07-27 14:50  qscqesze  阅读(176)  评论(0编辑  收藏