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Author: MathProgrammer

Explanations/Explanations - 10/Serega and Coat Rack Explanation.txt

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+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.
+
+-----------------------------------------
+
+Sort all the hook prices.
+
+Count the number of unhappy guests = max(guests - hooks, 0)
+
+It it guests - hooks, if guests is greater and 0 if hooks is greater.
+
+All the unhappy guests collect a fine.
+
+--------------------------------------
+
+int main()
+{
+    int no_of_hooks, fine, no_of_guests;
+    scanf("%d %d", &no_of_hooks, &fine);
+
+    vector <int> hook_prices(no_of_hooks);
+    for(int i = 0; i < no_of_hooks; i++)
+        scanf("%d", &hook_prices[i]);
+
+
+    sort(all(hook_prices));
+
+    scanf("%d", &no_of_guests);
+
+    int unhappy_guests = max(no_of_guests - no_of_hooks, 0);
+    int total_fine = unhappy_guests*fine;
+
+    int no_of_satisfied_guests = no_of_guests - unhappy_guests;
+    int income = 0;
+    for(int i = 0; i < no_of_satisfied_guests; i++)
+        income += hook_prices[i];
+
+    printf("%d\n", income - total_fine);
+
+    return 0;
+}