Codeforces Round #569 (Div. 2) B. Nick and Array

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博客围绕Codeforces Round #569 (Div. 2)的B题展开，题目是通过特定操作使数组元素乘积最大。操作是选任意索引i，将ai变为 -ai - 1。给出了思路，即把非负数变负数排序，根据负数个数奇偶性处理，还给出了代码实现。

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B. Nick and Array

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Nick had received an awesome array of integers a=[a1,a2,…,an] as a gift for his 5 birthday from his mother. He was already going to explore its various properties but after unpacking he was disappointed a lot because the product a1⋅a2⋅…an of its elements seemed to him not large enough.

He was ready to throw out the array, but his mother reassured him. She told him, that array would not be spoiled after the following operation: choose any index i (1≤i≤n) and do ai:=−ai−1.

For example, he can change array [3,−1,−4,1] to an array [−4,−1,3,1] after applying this operation to elements with indices i=1 and i=3.

Kolya had immediately understood that sometimes it’s possible to increase the product of integers of the array a lot. Now he has decided that he wants to get an array with the maximal possible product of integers using only this operation with its elements (possibly zero, one or more times, as many as he wants), it is not forbidden to do this operation several times for the same index.

Help Kolya and print the array with the maximal possible product of elements a1⋅a2⋅…an which can be received using only this operation in some order.

If there are multiple answers, print any of them.

Input

The first line contains integer n (1≤n≤105) — number of integers in the array.

The second line contains n integers a1,a2,…,an (−106≤ai≤106) — elements of the array

Output

Print n numbers — elements of the array with the maximal possible product of elements which can be received using only this operation in some order from the given array.

If there are multiple answers, print any of them.

Examples

input

4
2 2 2 2

output

-3 -3 -3 -3

input

1
0

output

input

3
-3 -3 2

output

-3 -3 2

思路：

由于一个数只有变成符号相反的数减一和他本身两种形态，那么我的方法是：

全部大于等于0的数变成负数，排个序（升序降序都行），
1:如果是偶数个，直接sort回原来的位置，输出就可以了。
2:如果是奇数个，将最小的数变成正数，再sort变成原来的位置，如果最小的数是-1，那么说明全部数都是-1，那么就直接，随便一个变成0，直接输出就行

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<ctime>
#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cstring>
#include<string>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#define ll long long
#define dd double
#define mes(x,y) memset(x,y,sizeof(x))
using namespace std;

const int maxn=1e5+9;
const int mod=1e9+7;


ll GCD(ll a,ll b) {//最大公约数
    return b == 0 ? a : GCD(b, a % b);
}
struct node{
    ll number,flag;
}v[200030];
bool cmp1(node n1,node n2){
    return n1.number<n2.number;
}
bool cmp2(node n3,node n4){
    return n3.flag<n4.flag;
}
int main() {
    std::ios::sync_with_stdio(false);
    ll n, i, j, z, ero;
    while (cin >> n) {
        mes(v, 0);
        for (i = 0; i < n; i++) {
            cin >> v[i].number;
            v[i].flag = i;
            if (v[i].number >= 0)v[i].number = v[i].number * (-1) - 1;
        }
        if (n == 1) {
            if (v[0].number < 0)v[0].number = v[0].number * (-1) - 1;
            cout << v[0].number << endl;
            continue;
        }
        sort(v, v + n, cmp1);
        if (n % 2 != 0) {
            for (i = 0; i < n; i++) {
                if (v[i].number == -1)continue;
                v[i].number = v[i].number * (-1) - 1;
                break;
            }
            if (v[0].number == -1 && v[n - 1].number == -1)v[0].number = 0;
        }
        sort(v, v + n, cmp2);
        for (i = 0; i < n; i++) {
            cout << v[i].number << " ";
        }
        cout << endl;
    }
}