Codeforces Round #182 (Div. 2)---C. Yaroslav and Sequence(不错的题,分析找规律)

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.

2*n-1个数,每次可以选择n个数,改变它们的符号
可以发现,当n是奇数,我们可以令任意数目的数改变符号
当n是偶数,我们可以使得2*k对数改变符号,这里枚举下求个最大值就行

/*************************************************************************
    > File Name: cf-182-c.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月06日 星期三 20时22分21秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int arr[220];

int main() {
    int n;
    while (~scanf("%d", &n)) {
        int cnt1 = 0, cnt2 = 0;
        for (int i = 1; i <= 2 * n - 1; ++i) {
            scanf("%d", &arr[i]);
            if (arr[i] < 0) {
                ++cnt1;
            }
            else {
                ++cnt2;
            }
        }
        sort(arr + 1, arr + 2 * n);
        if (n & 1) {
            int sum = 0;
            for (int i = 1; i <= 2 * n - 1; ++i) {
                sum += abs(arr[i]);
            }
            printf("%d\n", sum);
        }
        else {
            int sum = 0;
            if (cnt1 & 1) {
                for (int k = 0; 2 * k <= 2 * n - 1; ++k) {
                    int tmp = 0;
                    for (int i = 1; i <= 2 * n - 1; ++i) {
                        if (i <= k * 2) {
                            tmp -= arr[i];
                        }
                        else {
                            tmp += arr[i];
                        }
                    }
                    sum = max(sum, tmp);
                }
            }
            else {
                for (int i = 1; i <= 2 * n - 1; ++i) {
                    sum += abs(arr[i]);
                }
            }
            printf("%d\n", sum);
        }
    }
    return 0;
}