E. Negatives and Positives

E. Negatives and Positives

Given an array $a$ consisting of $n$ elements, find the maximum possible sum the array can have after performing the following operation any number of times:

• Choose $2$ adjacent elements and flip both of their signs. In other words choose an index $i$ such that $1 \leq i \leq n - 1$ and assign $a_i = -a_i$ and $a_{i+1} = -a_{i+1}$.

Input

The input consists of multiple test cases. The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2\cdot10^5$) — the length of the array.

The following line contains $n$ space-separated integers $a_1,a_2,\dots,a_n$ ($-10^9 \leq a_i \leq 10^9$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot10^5$.

Output

For each test case, output the maximum possible sum the array can have after performing the described operation any number of times.

Example

input

5
3
-1 -1 -1
5
1 5 -5 0 2
3
1 2 3
6
-1 10 9 8 7 6
2
-1 -1

output

1
13
6
39
2

Note

For the first test case, by performing the operation on the first two elements, we can change the array from $[-1, -1, -1]$ to $[1, 1, -1]$, and it can be proven this array obtains the maximum possible sum which is $1 + 1 + (-1) = 1$.

For the second test case, by performing the operation on $-5$ and $0$, we change the array from $[1, 5, -5, 0, 2]$ to $[1, 5, -(-5), -0, 2] = [1, 5, 5, 0, 2]$, which has the maximum sum since all elements are non-negative. So, the answer is $1 + 5 + 5 + 0 + 2 = 13$.

For the third test case, the array already contains only positive numbers, so performing operations is unnecessary. The answer is just the sum of the whole array, which is $1 + 2 + 3 = 6$.

 

解题思路

　　比赛的时候一直想着贪心怎么写，然后一直在那wa，最后还是老老实实写dp过了，回来看题解的时候发现还真是贪心解法，就是一直没发现题目的某个性质。

　　注意到不管我们执行了多少次的操作，数组中负数个数的奇偶性总是保持不变（在一次取反后，要么是增加或者减少两个负数，要么是增加一个负数和减少一个负数）。

　　因此如果一开始数组中有偶数个负数，那么我们总是可以通过若干次的操作使得负数的个数变成$0$，因此答案就是所有数的绝对值之和。

　　如果一开始数组中有奇数个负数，那么最终至少要有一个负数，所有我们可以让数组中绝对值最小的那个数为负数（如果有$0$的话则选择$0$），那么答案就是其他数的绝对值之和再减去这个最小数的绝对值。

　　AC代码如下：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

void solve() {
    int n;
    scanf("%d", &n);
    int cnt = 0, minv = 2e9;
    LL s = 0;
    while (n--) {
        int x;
        scanf("%d", &x);
        if (x < 0) cnt++, x *= -1;    // 统计负数
        s += x;
        minv = min(minv, x);    // 找到绝对值最小的那个数
    }
    if (cnt & 1) s -= 2 * minv;    // 有奇数个负数
    printf("%lld\n", s);
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

　　再给出dp的做法。

　　定义状态$f(i,0)$表示前$i$个数中，且第$i$个数没有改变正负性的所有方案的最大值。$f(i,1)$表示前$i$个数中，且第$i$个数有改变正负性的所有方案的最大值。根据前一个数$a_{i-1}$是否有改变正负性来划分集合，状态转移方程为

\begin{cases}
f(i,0) = \max\{ f(i-1,0), f(i-1,1) \} + a_i \\
f(i,1) = \max \{ f(i-1,0) -2a_{i-1}, f(i-1,1) + 2a_{i-1} \} - a_i
\end{cases}

　　AC代码如下：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

int a[N];
LL f[N][2];

void solve() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", a + i);
    }
    f[2][0] = a[1] + a[2], f[2][1] = -a[1] + -a[2];
    for (int i = 3; i <= n; i++) {
        f[i][0] = max(f[i - 1][0], f[i - 1][1]) + a[i];
        f[i][1] = max(f[i - 1][0] - 2 * a[i - 1], f[i - 1][1] + 2 * a[i - 1]) - a[i];
    }
    printf("%lld\n", max(f[n][0], f[n][1]));
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

　　Codeforces Round #849 (Div. 4) Editorial：https://codeforces.com/blog/entry/112282