CodeForces--621A--Wet Shark and Odd and Even（数学水题）

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本文介绍了一种算法，用于解决从给定整数中选取若干数以获得最大偶数和的问题。输入为一系列整数，输出则是这些整数中能组成的最大偶数和。

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Wet Shark and Odd and Even

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

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 Input

Input

3
1 2 3

Output

Input

5
999999999 999999999 999999999 999999999 999999999

Output

3999999996

Hint

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.

Source

Codeforces Round #341 (Div. 2)

给出n个数，任取若干个数，得到最大偶数和

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long a[1000000];
int main()
{
	int n;
	scanf("%d",&n);
	long long sum=0,b;
	int l=0;
	for(int i=0;i<n;i++)
	{
		scanf("%lld",&b);
		if(b%2)
		{
			a[l++]=b;
			sum+=b;
		}
		else
		sum+=b;
	}
	sort(a,a+l);
	if(sum%2) printf("%lld\n",sum-a[0]);
	else printf("%lld\n",sum);
	return 0;
}