CF 621A Wet Shark and Odd and Even

最新推荐文章于 2021-07-11 17:05:34 发布

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最新推荐文章于 2021-07-11 17:05:34 发布

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分类专栏： Codeforce

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本文介绍了一个算法问题，即如何从一组整数中选择部分整数以获得最大的偶数和。文章提供了详细的解题思路及AC代码实现，通过判断总和的奇偶性和移除最小奇数值来确保结果为最大偶数。

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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

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.

其实我也只在比赛的时候写了个A题而已，还WA了一次，

题意：求和，要求出最大的偶数和，

可以这样理解，如果加完是奇数，则要减去最小的奇数；如果加完是偶数，则就是最大值

AC代码如下：

#include <iostream>
#include <algorithm>
#include <stdio.h> 
using namespace std ;
long long a[100000];
int main()
{
	int n ;
	while(cin>>n)
	{		
		long long  sum = 0 ;
		for(int i = 0 ; i < n ;i++)
		{
			cin>>a[i];
		}
		long long sum2=0;
		for(int i = 0 ; i < n ; i++)
		{
			sum+=a[i];
		}
		sort(a,a+n);
		for(int i = 0 ; i < n ; i++)
		{
			int j ;
			if(sum%2!=0)
			{
				for(j=0;j<n;j++)
				{
					if(a[j]%2!=0)
					{
						sum-=a[j];
						break;
					}
				}
			}	
		}
		cout<<sum<<endl;
	}
	return 0 ;
}