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>
long long int max(long long int n, long long int tmp[], long long int *num, long long int *M) {
	long long int max_memo = tmp[0];
	long long int max_num = 0;
	for (int i = 1; i < n; i++) {
		if (max_memo < tmp[i]) {
			max_memo = tmp[i];
			max_num = i;
		}
	}
	*num = max_num;
	*M = max_memo;
	return 0;
}
int main() {
	long long int n;
	long long int num, m;
	int result = 0;
	long long int s[10];
	scanf("%lld", &n);
	for (int i = 0; i < n; i++){
		scanf("%lld", &s[i]);
		result += s[i];
	}
	max(n, s, &num, &m);
	printf("%d",result - (m / 2));
}