B. Nick and Array 简单简洁

因为在解决过程中没有看到比较简洁易懂的代码，所以在解决后发个博客给大家康康

利用公式ai = -ai -1处理数组，使数组积最大，我的思路是在输入时全部转成负值，同时找最小值，如果个数为奇数，转换最小值为正值，如果个数为偶数则不进行更多处理，最后输出

原链接：Codeforces - Problem Bhttps://codeforces.ml/contest/1180/problem/B

Nick had received an awesome array of integers a=[a1,a2,…,an]a=[a1,a2,…,an] as a gift for his 55 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⋅…ana1⋅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 ii (1≤i≤n1≤i≤n) and do ai:=−ai−1ai:=−ai−1.

For example, he can change array [3,−1,−4,1][3,−1,−4,1] to an array [−4,−1,3,1][−4,−1,3,1] after applying this operation to elements with indices i=1i=1 and i=3i=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⋅…ana1⋅a2⋅…an which can be received using only this operation in some order.If there are multiple answers, print any of them.

The first line contains integer nn (1≤n≤1051≤n≤105) — number of integers in the array.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (−106≤ai≤106−106≤ai≤106) — elements of the array

Output

Print nn 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.

样例：

in 4 2 2 2 2
out -3 -3 -3 -3
in 1 0
out 0
in 3 -3 -3 2
out -3 -3 2

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <iostream>
#include <algorithm>
using namespace std;int num[1000000];int main()
{int i;int N, nmin;scanf("%d", &N);nmin = 0;for (i = 0; i < N; i++){scanf("%d", &num[i]);if (num[i] >= 0) num[i] = num[i] * (-1) - 1; // 全部转换为负数后 存到数组里if (num[nmin] > num[i]) nmin = i; // 找到最小数}if (N%2) num[nmin] = num[nmin] * (-1) - 1; // 如果个数是奇数，转换最小数为正数for (i = 0; i < N; i++)printf("%d ", num[i]);return 0;
}

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B. Nick and Array 简单简洁

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