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.
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).
In a single line print a single integer — the answer to the problem.
2 1 2 1 2
3
2 1 2 1 10
-5
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.
题意：旅店一共有n个衣架，有m个客人，每个客人来会选最便宜的衣架挂衣，旅店获得相应的收益，但是衣架不够的情况下要给每个客人赔款p
思路：水题，排序后分情况讨论即可
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
int n,d,m,i,j;
int a[10000];
while(~scanf("%d%d",&n,&d))
{
int s = 0;
memset(a,0,sizeof(a));
int sum = 0;
for(i = 0; i<n; i++)
{
scanf("%d",&a[i]);
s+=a[i];
}
sort(a,a+n);
scanf("%d",&m);
if(m==n)
printf("%d\n",s);
else if(m>n)
{
s-=(m-n)*d;
printf("%d\n",s);
}
else if(m<n)
{
sum = 0;
for(i = 0;i<m;i++)
sum+=a[i];
printf("%d\n",sum);
}
}
return 0;
}