CF 873 A. Chores【sort】

A. Chores
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
input
4 2 2
3 6 7 10
output
13
input
5 2 1
100 100 100 100 100
output
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.

题意：
输入三个数，分别表示接下来数列的个数、可替换的数字数量、可替换成的数字。

对下面的数组进行排序，然后选择把前面n-m个累加求和，然后把后面m*k个加上。

得到正解！

#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<iomanip>#include<queue>#include<cmath>using namespace std;int a[1000];int main(){    int n, m, k;    while (~scanf("%d%d%d", &n, &m, &k))    {        for (int i = 0; i < n; i++)        {            cin >> a[i];        }        sort(a, a + n);        int ans = m*k;        for (int i = 0; i < n - m; i++)        {            ans += a[i];        }        printf("%d\n", ans);    }    return 0;}