Yaroslav and Sequence - CodeForces 301A 水题

A. Yaroslav and Sequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

思路：其实每次可以把任意两个的符号改变，那么就要考虑n是奇数还是偶数了，如果n是奇数(或n是偶数且其中有偶数个负数)，那么可以让所有数变成非负数，如果n是偶数且有奇数个负数，那么必须剩一个负数。另外考虑有0的情况。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
vector <int> vc1,vc2;
void solve()
{ int T,t,n,m,i,j,k,len1,len2,sum=0,ret,num0=0;
  scanf("%d",&n);
  for(i=1;i<=n*2-1;i++)
  { scanf("%d",&k);
    if(k>0)
     vc1.push_back(k);
    else if(k<0)
     vc2.push_back(k);
    else
     num0++;
  }
  len1=vc1.size();
  len2=vc2.size();
  for(i=0;i<len1;i++)
   sum+=vc1[i];
  for(i=0;i<len2;i++)
   sum-=vc2[i];
  if(num0>0 || len1==n*2-1 || n&1)
  { printf("%d\n",sum);
    return;
  }
  if(n%2==0 && len2%2==0)
  { printf("%d\n",sum);
    return;
  }
  ret=100000;
  for(i=0;i<len1;i++)
   ret=min(ret,vc1[i]);
  for(i=0;i<len2;i++)
   ret=min(ret,-vc2[i]);
  printf("%d\n",sum-2*ret);
}
int main()
{ solve();
}