Codeforces Round #341 (Div. 2) CF621A. Wet Shark and Odd and Even

A. Wet Shark and Odd and Even

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

In the first sample, we can simply take all three integers for a total sum of 6.

In the second sample Wet Shark should take any four out of five integers 999 999 999.

题意：

题目非常简单，给出几个数字，在几个数中选任意个数的数字组成一个最大的偶数。标准水题一道。

思路：

输入过程中把总和tot求出来，并把最小的奇数记录下来。如果tot % 2 == 0则输出，反之减去minn再输出。比赛过程中忘了哪里的bug居然还被hack了一次233333_(:з」∠)_。。

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
while(cin>>n)
{
long long tot = 0;
int minn = 1000000001;
for(int i = 0;i < n;i++)
{
int x;  cin>>x;
tot += x;
if(x % 2 != 0)
minn = min(x,minn);
}
if(tot % 2 != 0)
tot -= minn;
cout<<tot<<endl;
}
return 0;
}