Educational Codeforces Round 69 (Rated for Div. 2)
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].
题意：题意大概是 定义一个子数组代价为 数组里的最大值减去最小值，然后 给我们一个有序数组和一个k值，
问 把这个数组分成k个子数组的最小的总价值。
思路：先拿样例数组 【4 8 15 16 23 42】举例，
首先当k等于1时，当前代价就是原数组的代价为 最大值减最小值 （42 - 4） = 38;
当k等于2时，当前代价可以为 （8 - 4）+（42 - 15）= 31； 或者 （15 - 4）+（42 - 16）= 37 ...
求 最小代价 列举之后 发现应该是 （23 - 4）+（42 - 42）= 19。
这时候你可能发现了，当数组分成子数组时，有一段相邻元素的差值将不被算入代价，
例如上面的例子最小值 19 就是 相邻元素 （23，42）的差值 19 未被算入 ，然后就是 38-19=19。
然后拓展，k变大时，只需要去除k-1个相邻元素差值大的，就可以计算出最小代价，由此得解。。。
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
6 #include<map>
7 #include<set>
8 #include<vector>
9 #include<queue>
10 using namespace std;
11 #define ll long long
12 const int mod=1e9+7;
13 const int inf=1e9+7;
14
15 const int maxn=3e5+5;
16 int num[maxn];
17
18 int main()
19 {
20 ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
21 int n,k;
22
23 while(cin>>n>>k)
24 {
25 vector<int>v;
26 ll int ans=0;
27 int x;
28 for(int i=0;i<n;i++)
29 {
30 cin>>num[i];
31
32 if(i>0)
33 {
34 x=num[i]-num[i-1];
35 ans+=x;//预处理
36 v.push_back(x);
37 }
38 }
39
40 sort(v.begin(),v.end(),greater<int>());//由大到小排序
41 k--;//去除k-1个相邻差值
42
43 for(int i=0;i<k;i++)
44 ans-=v[i];
45
46 cout<<ans<<endl;
47 }
48
49 return 0;
50 }