이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
// Judges with GCC >= 12 only needs Ofast
// #pragma GCC optimize("O3,no-stack-protector,fast-math,unroll-loops,tree-vectorize")
// MLE optimization
// #pragma GCC optimize("conserve-stack")
// Old judges
// #pragma GCC target("sse4.2,popcnt,lzcnt,abm,mmx,fma,bmi,bmi2")
// New judges. Test with assert(__builtin_cpu_supports("avx2"));
// #pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,bmi2,fma,tune=native")
// Atcoder
// #pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,bmi2,fma")
#include<bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
uniform_real_distribution<> pp(0.0,1.0);
#define ld long double
#define pii pair<int,int>
#define piii pair<int,pii>
#define fi first
#define se second
const int inf=1e9;
const int mod=998244353;
const int mod2=1e9+7;
const int maxn=100005;
const int maxm=200005;
const int maxq=500005;
const int maxl=20;
const int maxa=1000005;
int power(int a,int n){
int res=1;
while(n){
if(n&1) res=res*a%mod;
a=a*a%mod;n>>=1;
}
return res;
}
int t[maxn];
void solve(){
int n,k;cin >> n >> k;
for(int i=1;i<=n;i++) cin >> t[i];
int res=t[n]-t[1]+1;
vector<int> s;
for(int i=2;i<=n;i++) s.push_back(t[i]-t[i-1]-1);
sort(s.begin(),s.end(),greater<int>());
for(int i=0;i<k-1;i++) res-=s[i];
cout << res << '\n';
}
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);cout.tie(NULL);
int test=1;//cin >> test;
while(test--) solve();
}
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