codeforces 721D D. Maxim and Array (STL||优先队列)

Recently Maxim has found an array of n integers, needed by no one. He immediately come up with idea of changing it: he invented positive integer
x and decided to add or subtract it from arbitrary array elements. Formally, by applying single operation Maxim chooses integer
i (1 ≤ i ≤ n) and replaces the
i-th element of array
ai either with
ai + x or with
ai - x. Please note that the operation may be applied more than once to the same position.

Maxim is a curious minimalis, thus he wants to know what is the minimum value that the product of all array elements (i.e.


) can reach, if Maxim would apply no more than
k operations to it. Please help him in that.

Input
The first line of the input contains three integers n, k and
x (1 ≤ n, k ≤ 200 000, 1 ≤ x ≤ 109) — the number of elements in the array, the maximum number of operations and the number invented
by Maxim, respectively.

The second line contains n integers
a1, a2, ..., an (

) —
the elements of the array found by Maxim.

Output
Print n integers
b1, b2, ..., bn in the only line — the array elements after applying no more than
k operations to the array. In particular,


should stay true for every 1 ≤ i ≤ n, but the product of all array elements should be
minimum possible.

If there are multiple answers, print any of them.

Examples

Input

5 3 1
5 4 3 5 2

Output

5 4 3 5 -1

Input

5 3 1
5 4 3 5 5

Output

5 4 0 5 5

Input

5 3 1
5 4 4 5 5

Output

5 1 4 5 5

Input

3 2 7
5 4 2

Output

5 11 -5

题意：

给你n个数，k次操作，每次操作可以将第i个数+x或者-x，输出k次操作后这n个数乘积最小时的数列

解题：每次处理绝对值最小的数，如果当前乘积是负数，该数尽量远离0。如果当前成绩是正数，该数靠近0。。

我是用优先队列来处理，重载运算符

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int N = 2e5+100;
typedef long long ll;
struct node
{
ll x;
ll index;
}a
;

bool operator<(node a , node b){
return abs(a.x)>abs(b.x);

}
priority_queue<node>que;
int main()
{
ios::sync_with_stdio(false);
ll n,k,x,i,j;
int flag=1;
node temp;
cin>>n>>k>>x;
for(i=1;i<=n;i++) {
cin>>a[i].x;
a[i].index=i;
que.push(a[i]);
if(a[i].x<0) flag=-flag;
}
while(k--) {
temp=que.top();
que.pop();
if(temp.x<0) {
if(flag==-1) a[temp.index].x-=x;
else a[temp.index].x +=x;
if(a[temp.index].x>=0) flag=-flag;
}
else {
if(flag==-1) a[temp.index].x+=x;
else a[temp.index].x-=x;
if(a[temp.index].x<0) flag=-flag;
}
que.push(a[temp.index]);
}
for(i=1;i<=n;i++) cout<<a[i].x<<" ";
cout<<endl;
return 0;
}