Codeforces Round #341 (Div. 2)-A. Wet Shark and Odd and Even(奇数+奇数为偶数)

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

31 2 3

output

6

input

5999999999 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.

思路：

偶数加偶数肯定是偶数，而奇数加奇数就一定是偶数，所以只要将偶数个奇数相加，再加上偶数就是最大能被2整除的和。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<vector>#include<cstdio>#include<cmath>#include<set>using namespace std;#define CRL(a) memset(a,0,sizeof(a))typedef unsigned __int64 LL;typedef  __int64 ll;const int T = 103000;const int mod = 1000000007;ll v[T];int main(){#ifdef zsc    freopen("input.txt","r",stdin);#endifint n,m,i,j,k;while(~scanf("%d",&n)){ll sum = 0;for(i=0;i<n;++i){ scanf("%I64d",&v[i]); if(!(v[i]&1)){ sum += v[i]; i--,n--; }}sort(v,v+n);for(i=n-1;i>=n%2;--i){sum += v[i];}printf("%I64d\n",sum);}    return 0;}