【CF】CodeForces 301A Yaroslav and Sequence

最新推荐文章于 2019-05-29 18:37:00 发布

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最新推荐文章于 2019-05-29 18:37:00 发布

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本文介绍了一个关于通过改变序列中元素的符号来最大化序列和的问题。Yaroslav可以每次选择序列中的n个元素并将其符号反转。文章探讨了如何找到最佳方案以达到最大可能的总和，并提供了一个具体的实现算法。

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Yaroslav has an array, consisting of (2·n - 1) integers. In a single operation Yaroslav can change the sign of exactly n elements in the array. In other words, in one operation Yaroslav can select exactly n array elements, and multiply each of them by -1.

Yaroslav is now wondering: what maximum sum of array elements can be obtained if it is allowed to perform any number of described operations?

Help Yaroslav.

Input

The first line contains an integer n (2 ≤ n ≤ 100). The second line contains (2·n - 1) integers — the array elements. The array elements do not exceed 1000 in their absolute value.

Output

In a single line print the answer to the problem — the maximum sum that Yaroslav can get.

Sample test(s)

input
2
50 50 50

output
150

input
2
-1 -100 -1

output
100

Note

In the first sample you do not need to change anything. The sum of elements equals 150.

In the second sample you need to change the sign of the first two elements. Then we get the sum of the elements equal to 100.

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <queue>
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
int n;
int a[250];
int main()
{
	//freopen("input.txt","r",stdin);
	cin>>n;

	int w=0;
	int ok=0;
	int sum=0;
	for (int i = 0; i < 2*n-1; ++i)
	{
		scanf("%d",&a[i]);
		if(a[i]==0) ok=1;
		if(a[i]<0) w++,a[i]=-a[i];

		sum+=a[i];
	}
	if(ok) cout<<sum<<endl;
	else
	{
		int minnn=INF;
		for(int i=0;i<2*n-1;i++) minnn=min(minnn,a[i]);
		if(w==n||w%2==0||(w>n&&(w-n)%2==0)||(n>w&&(n-w)%2==0)) //重点在这里
			cout<<sum<<endl;
		else 
			cout<<sum-2*minnn<<endl;
	}
	return 0;
}