cf-A. Wet Shark and Odd and Even(水)

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

3 1 2 3

output

6

input

5 999999999 999999999 999999999 999999999 999999999

output

3999999996
题解：第一次打cf，写了两道水题，剩下的就完全没思路啊。。。
这个题让找给的数中可以加到的最大可以被2整除的数；水；
代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%I64d",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%I64d",x)
#define P_ printf(" ")
const int MAXN=1e5+100;
int m[MAXN];
int main(){
	int n;
	while(~SI(n)){
		__int64 ans=0,temp,top=0;
		for(int i=0;i<n;i++){
			SL(temp);
			if(temp&1)m[++top]=temp;
			else ans+=temp;
		}
		sort(m+1,m+top+1);
		int t=1;
		if(top&1)t++;
		for(int i=t;i<=top;i++)ans+=m[i];
		printf("%I64d
",ans);
	}
	return 0;
}