Codeforces 797B - Odd sum

最新推荐文章于 2024-08-05 19:43:37 发布

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最新推荐文章于 2024-08-05 19:43:37 发布

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原文链接：http://www.cnblogs.com/widsom/p/6720400.html

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本文介绍了一道 CodeForces 上的题目 B.Oddsum 的解题思路及实现代码。题目要求找到一个整数序列中具有最大奇数和的子序列。文章详细解释了解决方案，并附上了 C++ 实现代码。

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You are given sequence a1, a2, ..., an of integer numbers of length n. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.

Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

You should write a program which finds sum of the best subsequence.

Input

The first line contains integer number n (1 ≤ n ≤ 105).

The second line contains n integer numbers a1, a2, ..., an ( - 104 ≤ ai ≤ 104). The sequence contains at least one subsequence with odd sum.

Output

Print sum of resulting subseqeuence.

Examples

input

4
-2 2 -3 1

output

input

3
2 -5 -3

output

-1
题目大意：求最大奇数和的子序列的和。
方法：先将正数加起来，得sum
　　　　如果sum为奇数，输出sum;
　　　　否则，sum1=sum-最小正奇数，sum2=sum+最大负奇数，输出max(sum1,sum2)。
代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<iostream>
#define ll long long
using namespace std;
const int N=1e5;
int a[N];
int temp1=-1e5,temp2=1e5; 
int main()
{
        int n;
        int sum=0;
        int index1=-1,index2=-1;
        int sum1=-0x7f7f7f7f,sum2=-0x7f7f7f7f;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            if(a[i]>0)
            {
                sum+=a[i];
                if(a[i]%2)
                {
                    temp2=min(temp2,a[i]);
                }
            }
            else if(a[i]<0)
            {
                if(a[i]%2)
                temp1=max(temp1,a[i]);
            }
        }
            if(sum%2)cout<<sum<<endl;
            else
            {
                    sum1=sum-temp2;
                    sum2=sum+temp1;
                cout<<max(sum1,sum2)<<endl;
            }
            return 0;
}