D. Maxim and Array(贪心+优先队列模板)

D. Maxim and Array

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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

题目大意:给你n个数，然后给你K次改变机会，可以让n个数中的任意一个数增加或者减少x，求k次改变后的数使之乘积最小
解题思路:要让乘积最小，肯定是负数最小，所以判断负数的个数，然后负数个数奇数可以乘积最小，然后再去使绝对值最小的数字不断增加x，这道题用了优先队列，之前没有用过，所以当作模板去处理

#include<iostream>
#include<cstdio>
#include<stdio.h>
#include<cstring>
#include<cstdio>
#include<climits>
#include<cmath>
#include<vector>
#include <bitset>
#include<algorithm>
#include <queue>
#include<map>
using namespace std;

struct point
{
friend bool operator< (point n1, point n2)
{
return n1.sum> n2.sum;
}
long long int sum;
long long int i;
int flag;
}poin[200005], pp;
priority_queue<point> qua;
bool cmp(point x, point y)
{
return x.sum < y.sum;
}
bool cmp1(point x, point y)
{
return x.i < y.i;
}
long long int n, k, x, i, sum, ss;
long long int a[200005];
int main()
{
cin >> n >> k >> x;
sum = 0;
for (i = 1; i <= n; i++)
{
cin >> ss;
if (ss < 0)
{
poin[i].sum = -ss;
poin[i].flag = -1;
sum++;
}
else
{
poin[i].sum = ss;
poin[i].flag = 1;
}
poin[i].i = i;
qua.push(poin[i]);
}

if (sum % 2 == 0)
{
pp = qua.top();
qua.pop();
while (pp.sum>=0)
{
pp.sum -= x;
k--;
if (k == 0)
{
break;
}
}
if (pp.sum < 0)
{
pp.flag = -pp.flag;
pp.sum = -pp.sum;
}
qua.push(pp);
}
while (k--)
{
pp = qua.top();
qua.pop();
pp.sum += x;
qua.push(pp);
}
while (!qua.empty())
{
pp = qua.top();
qua.pop();
if (pp.flag == 1)
{
a[pp.i] = pp.sum;
}
else
{
a[pp.i] = -pp.sum;
}
}
for (i = 1; i < n; i++)
{
cout << a[i] << " ";
}
cout << a[i] << endl;
}