Problem:
Score : 200 points

Problem Statement

In some other world, today is the day before Christmas Eve.

Mr. Takaha is buying N items at a department store. The regular price of the i-th item (1 \leq i \leq N) is p_i yen (the currency of Japan).

He has a discount coupon, and can buy one item with the highest price for half the regular price. The remaining N-1 items cost their regular prices. What is the total amount he will pay?

Constraints

2 \leq N \leq 10

100 \leq p_i \leq 10000

p_i is an even number.

Input

Input is given from Standard Input in the following format:

N
p_1
p_2
:
p_N

Output

Print the total amount Mr. Takaha will pay.

Sample Input 1

3
4980
7980
6980

Sample Output 1

15950

The 7980-yen item gets the discount and the total is 4980 + 7980 / 2 + 6980 = 15950 yen.

Note that outputs such as 15950.0 will be judged as Wrong Answer.

Sample Input 2

4
4320
4320
4320
4320

Sample Output 2

15120

Only one of the four items gets the discount and the total is 4320 / 2 + 4320 + 4320 + 4320 = 15120 yen.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

int main(){
	int N;
	int *p;
	int max = 0;
	int index = -1;

	scanf("%d", &N);
    p = (int*)malloc(sizeof(int)*N);
    for (int i = 0; i < N; i++){
    	scanf("%d", p+i);
    }

    for (int i = 0; i < N; i++){
    	if (p[i] > max){
    		max = p[i];
    		index = i;
    	}
    }

    int sum = 0;
    for (int i = 0; i < N; i++){
    	sum += p[i];
    }
    sum = (sum - p[index]) + p[index] / 2;
    printf("%d", sum);

}