cf-A. Wet Shark and Odd and Even(水)

最新推荐文章于 2025-08-24 09:29:29 发布

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最新推荐文章于 2025-08-24 09:29:29 发布

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文章标签：数据结构与算法

WetShark面临一项任务：从一组整数中找到最大的偶数之和。本文将深入探讨如何解决这个问题，包括输入解析、筛选偶数和奇数，以及如何计算最大可能的偶数总和。

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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
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\n",ans);
	}
	return 0;
}