You are given a sorted array ????1,????2,…,???????? (for each index ????>1 condition ????????≥????????−1 holds) and an integer ????.
You are asked to divide this array into ???? non-empty consecutive subarrays. Every element in the array should be included in exactly one subarray.
Let ????????????(????) be equal to the maximum in the ????-th subarray, and ????????????(????) be equal to the minimum in the ????-th subarray. The cost of division is equal to ∑????=1????(????????????(????)−????????????(????)). For example, if ????=[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 ???? into ???? non-empty consecutive subarrays.
Input
The first line contains two integers ???? and ???? (1≤????≤????≤3⋅105).
The second line contains ???? integers ????1,????2,…,???????? (1≤????????≤109, ????????≥????????−1).
Output
Print the minimum cost you can obtain by dividing the array ???? into ???? 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 ???? 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，你需要切为k个连续的子序列数组，每个数组的切分代价是max(a[i])-min(a[j])，现在问你总代价最小是多少
题解
每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。
我们令b[i]=a[i+1]-a[i]之后，他需要切成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 这个序列需要切分成三份。
他们的差值分别为: 4, 7, 1, 7, 19。如果不需要切分的话，答案就是42-4，或者所有差值的和。
如果切分成三份，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;
}