Codeforces Round #341 (Div. 2) --A. Wet Shark and Odd and Even

A. Wet Shark and Odd and Even

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.

Note, that if Wet Shark uses no integers from the n integers, the sum is an even integer 0.

Input

The first line of the input contains one integer, n (1 ≤ n ≤ 100 000). The next line contains n space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.

Output

Print the maximum possible even sum that can be obtained if we use some of the given integers.

Sample test(s)

input

3
1 2 3

output

6

input

5
999999999 999999999 999999999 999999999 999999999

output

3999999996

Note

果真水的不行，只会做第一题：（慢慢补吧T T）

第一题大体题意：

给你n个数，取一些数，得到最大的偶数。

思路很简单

先把总和加起来，在判断sum奇偶，如果是奇数，减去最小的奇数就行了！

#include<cstdio>
using namespace std;
typedef long long ll;
const ll INF = 1e10;
int main()
{
    int n;
    while(~scanf("%d",&n)){
        ll sum = 0,min_a=INF;
        for(int i = 0; i < n; ++i){
            ll k;
            scanf("%I64d",&k);
            if (k%2)
                if (k < min_a)min_a=k;
            sum+=k;
        }
        if (sum % 2)sum-=min_a;
        printf("%I64d\n",sum);
    }
    return 0;
}