http://codeforces.com/contest/1110/problem/B

题意：有一把尺子n处破了，总长m，可以用k张原无限长的胶布进行修补，问最少使用胶布多长；

思路：n个位置必补，先进性修补，再把间隔里取最短的n-k个修补；

#include<bits/stdc++.h>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=1e5+9;
const int mod=1e9+7;

template <class T>
inline void sc(T &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

int a[maxn],b[maxn];
int main()
{
    int n,m,k;
    cin>>n>>m>>k;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    if(k==n)
    {
        cout<<n<<endl;
        return 0;
    }
    int yu=n-k;
    for(int i=0;i<n-1;i++)
    {
        b[i]=a[i+1]-a[i]-1;
    }
    ll ans=0;
    sort(b,b+n-1);
    for(int i=0;i<n-k;i++) ans+=b[i];
    cout<<ans+n<<endl;
    return 0;
}