思路：水题，直接全部加起来，并且维护一个最小的奇数，因为如果最后结果是奇数的话那么减去那个最小的奇数就会变成最大的偶数啦
#include<bits\stdc++.h>
using namespace std;
#define LL long long
int main()
{
int n;
cin >> n;
LL ans = 0;
LL mins = 1e9;
while(n--)
{
LL temp;
scanf("%lld",&temp);
ans+=temp;
if(temp&1)
mins = min(temp,mins);
}
if(ans&1)
ans-=mins;
cout << ans << endl;
}
A. Wet Shark and Odd and Even
time limit per test
memory limit per test
input
output
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.
n integers, the sum is an even integer 0.
Input
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.
Examples
input
3 1 2 3
output
6
input
5
999999999 999999999 999999999 999999999 999999999
output
3999999996
Note
6.
999 999 999.
