Educational Codeforces Round 69 (Rated for Div. 2) C. Array Splitting 水题

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].

题目大意

给你长度为n的从小到大排列的数组，你须要切为k个连续的子序列数组，每一个数组的切分代价是max(a[i])-min(a[j])，如今问你总代价最小是多少

题解

每一个子序列的代价其实就是最大值减去最小值，就是最后面的数减去最前面的数，再拆分细致一点，就是差值的和。

咱们令b[i]=a[i+1]-a[i]以后，他须要切成k个连续的子序列，实际上就是去掉k-1个最大的差值。

举个简单例子 4 8 15 16 23 42 这个序列须要切分红三份。

他们的差值分别为: 4, 7, 1, 7, 19。若是不须要切分的话，答案就是42-4，或者全部差值的和。

若是切分红三份，实际上就是去掉了最大的3-1个差值罢了。

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;
}