Codeforces Round #215 (Div. 2) A. Sereja and Coat Rack

A. Sereja and Coat Rack
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Sereja owns a restaurant for n people. The restaurant hall has a coat rack with n hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using the i-th hook costs ai rubles. Only one person can hang clothes on one hook.
Tonight Sereja expects m guests in the restaurant. Naturally, each guest wants to hang his clothes on an available hook with minimum price (if there are multiple such hooks, he chooses any of them). However if the moment a guest arrives the rack has no available hooks, Sereja must pay a d ruble fine to the guest.
Help Sereja find out the profit in rubles (possibly negative) that he will get tonight. You can assume that before the guests arrive, all hooks on the rack are available, all guests come at different time, nobody besides the m guests is visiting Sereja's restaurant tonight.
Input
The first line contains two integers n and d (1 ≤ n, d ≤ 100). The next line contains integers a1, a2, ...,an (1 ≤ ai ≤ 100). The third line contains integer m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the answer to the problem.
Sample test(s)
input

2 1

2 1

2

output

3

input

2 1

2 1

10

output

-5

Note
In the first test both hooks will be used, so Sereja gets 1 + 2 = 3 rubles.
In the second test both hooks will be used but Sereja pays a fine 8 times, so the answer is 3 - 8 = - 5.
오랫동안 문제를 풀지 않았는데, 간단한 정렬 문제 하나.
 

#include <iostream>

#include <cstdio>

#include <algorithm>

using namespace std;



const int maxn=110;

int a[maxn],n,m,d;



int main(){

    while(scanf("%d%d",&n,&d)!=EOF){

         int sum=0,ans=0;

         for(int i=0;i<n;i++){

            scanf("%d",&a[i]);

            sum+=a[i];

         }

         scanf("%d",&m);

         if(m>=n){

             ans=sum-(m-n)*d;

         }else{

             sort(a,a+n);

             for(int i=0;i<m;i++) ans+=a[i];

         }

         cout<<ans<<endl;

    }

    return 0;

}