Wet Shark and Odd and Even


[align=center]Wet Shark and Odd and Even[/align]
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Time Limit:2000MS    
Memory Limit:262144KB    
64bit IO Format:%I64d & %I64u
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Status
Practice
CodeForces 621A

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

6

Input

5
999999999 999999999 999999999 999999999 999999999

Output

3999999996

Sample Output

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.

思路：

用sort先从小到大排序，把所有的数加起来，先判断总和是不是偶数，若是偶数直接输出，若不是则减去经排序后的最小的奇数，减去第一个奇数时要用for循环先找到第一个奇数的位置。

其次注意数据类型为long long或者__int64。

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
long long a[100005];
int main()
{
int n;
scanf("%d",&n);
long long sum=0;
for(int i=0;i<n;i++)
{
scanf("%lld",&a[i]);
sum+=a[i];
}
sort(a,a+n);
for(int i=0;i<n;i++)
{
if(sum%2==0)
break;
else
{
if(a[i]%2!=0)
{
sum-=a[i];
}
}
}
printf("%lld\n",sum);
return 0;