CodeForces 621A-Wet Shark and Odd and Even

我不是女神ヾ 2022-08-21 02:10 200阅读 0赞

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.

Examples

Input

3
    1 2 3

Output

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.

解题思路：

找出能被2整除的最大和。偶数全要，奇数判断个数，奇数个要减去最小值。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define INF 0x3f3f3f3f
    #define LL __int64
    using namespace std;
    
    //LL ji[100100];
    //LL ou[100002];
    
    int main()
    {
    	int n;
    	while(scanf("%d",&n)!=EOF)
    	{
    		LL minx=INF;
    		int i,j=0;
    		LL ans=0;
    		for(i=0;i<n;i++)
    		{
    			LL x;
    			scanf("%I64d",&x);
    			if(x%2==0)
    			{
    				ans+=x;
    			}
    			else
    			{
    				j++;
    				minx=min(minx,x);
    				ans+=x;
    			}
    		}		
    		if(j%2==0)
    		{
    			printf("%I64d\n",ans);
    		}
    		else
    		{
    			LL wc=ans-minx;
    			printf("%I64d\n",wc);
    		}
    		
    	}
    	return 0;
    }