题目传送
题意:
给你一个n大小的数组(n一定为奇数),现在你可以改变其中元素的符号,现在要求你在操作后,使得有至少一半的ai <= ai+1,至少有一半ai >= ai+1,现在让你构造出这个数组
思路:
我们直接正负正负的一直下去,由于n为奇数,那么肯定就满足条件了
AC代码
#include <bits/stdc++.h>
inline long long read(){char c = getchar();long long x = 0,s = 1;
while(c < '0' || c > '9') {if(c == '-') s = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x*10 + c -'0';c = getchar();}
return x*s;}
using namespace std;
#define NewNode (TreeNode *)malloc(sizeof(TreeNode))
#define Mem(a,b) memset(a,b,sizeof(a))
#define lowbit(x) (x)&(-x)
const int N = 1e5 + 10;
const long long INFINF = 0x7f7f7f7f7f7f7f;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
const double EEE = exp(1);
const int mod = 1e9+7;
const double II = acos(-1);
const double PP = (II*1.0)/(180.00);
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> piil;
signed main()
{
std::ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
// freopen("input.txt","r",stdin);
// freopen("output.txt","w",stdout);
int t;
cin >> t;
while(t--)
{
ll n;
cin >> n;
ll arr[n+5] = {0};
for(int i = 1;i <= n;i++)
{
cin >> arr[i];
if(i % 2 != 0 && arr[i] < 0)
arr[i] = arr[i]*-1;
else if(i % 2 == 0 && arr[i] > 0)
arr[i] = arr[i]*-1;
}
for(int i = 1;i <= n;i++)
i != n ? cout << arr[i] << " " : cout << arr[i] << endl;
}
}