873A - Chores.cpp

/*
Luba has to do n chores today. i-th chore takes ai units of time to complete. It is guaranteed that for every  the condition ai?=?ai?-?1 is met, so the sequence is sorted.

Also Luba can work really hard on some chores. She can choose not more than k any chores and do each of them in x units of time instead of ai ().

Luba is very responsible, so she has to do all n chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously.

Input
The first line contains three integers n,?k,?x (1?=?k?=?n?=?100,?1?=?x?=?99) — the number of chores Luba has to do, the number of chores she can do in x units of time, and the number x itself.

The second line contains n integer numbers ai (2?=?ai?=?100) — the time Luba has to spend to do i-th chore.

It is guaranteed that , and for each  ai?=?ai?-?1.

Output
Print one number — minimum time Luba needs to do all n chores.

Examples
inputCopy
4 2 2
3 6 7 10
outputCopy
13
inputCopy
5 2 1
100 100 100 100 100
outputCopy
302
Note
In the first example the best option would be to do the third and the fourth chore, spending x?=?2 time on each instead of a3 and a4, respectively. Then the answer is 3?+?6?+?2?+?2?=?13.

In the second example Luba can choose any two chores to spend x time on them instead of ai. So the answer is 100·3?+?2·1?=?302
*/




/*
#include<bits/stdc++.h>
using namespace std;
 
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int n,k, x;
    cin >> n >> k >> x;
   
    vector <int> v(n);
   
    for (auto &x: v) cin >> x;
   
    int sum = 0;
   
    for (int i  = 0; i < n - k; i++) sum += v[i];
    for (int i = n - k; i < n; i++) sum += x;
   
    cout << sum;
    
    return 0;
}
*/







#include<bits/stdc++.h>
using namespace std;
 
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int n,k, x;
    cin >> n >> k >> x;
   
    vector <int> v(n);
   
    for (auto &x: v) cin >> x;
   
    int sum = 0;
   
    for (int i  = 0; i < n - k; i++) sum += v[i];
    
    sum += x * k;
   
    cout << sum;
    
    return 0;
}