B. Roma and Changing Signs
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Roma works in a company that sells TVs. Now he has to prepare a report for the last year.

Roma has got a list of the company's incomes. The list is a sequence that consists of n integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly k changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.

The operation of changing a number's sign is the operation of multiplying this number by -1.

Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly k changes.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 105), showing, how many numbers are in the sequence and how many swaps are to be made.

The second line contains a non-decreasing sequence, consisting of n integers ai (|ai| ≤ 104).

The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.

Output

In the single line print the answer to the problem — the maximum total income that we can obtain after exactly k changes.

Sample test(s)
input
3 2
-1 -1 1
output
3
input
3 1
-1 -1 1
output
1
Note

In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.

In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1.

 

先对数组排序，一开始用冒泡，结果超时了QAQ，改快排了。然后分两种情况，负数大于等于 change次数的 直接把负数从小到大依次change。负数比change次数小的，把负数都change，再排序，剩下change次数都对最小的数操作。

附代码：

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 
 4 int arr[100000]={0};
 5 
 6 int cmp(const void *a,const void *b)
 7 {
 8     return *(int *)a-*(int *)b;
 9 }
10 
11 int main(void)
12 {
13     int n, k, i, j, sum=0, count=0;
14     int temp;
15 
16     scanf("%d %d", &n, &k);
17     for( i=0; i<n; i++)
18     {
19         scanf("%d", &arr[i]);
20         if( arr[i]<0 )
21             count++;
22     }
23     qsort( arr, n, sizeof(int), cmp);
24     if( count>k)
25     {
26         for( i=0; i<k; i++)
27             arr[i]*=-1;
28     }
29     else if( count == k)
30     {
31         for( i=0; i<k; i++)
32             arr[i]*=-1;
33     }
34     else
35     {
36         for( i=0; i<count; i++)
37             arr[i]*=-1;
38         qsort( arr, n, sizeof(int), cmp);
39         for( i=0; i<(k-count); i++)
40             arr[0]*=-1;
41     }
42     for( i=0; i<n; i++)
43         sum+=arr[i];
44     printf("%d", sum);
45     return 0;
46 }