Codeforces Round #374 (Div. 2) D. Maxim and Array

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：

给出  n 个数，限制 m 次操作，每次操作可以将 数组中的某个数加上或者减去 k 

然后让 得到的 n 个数乘积最小。

思路：

当 n 个数 都为正数的时候，他们绝对值越接近，那么他们的乘积越大。

那么如果他们之间有负数的话，并且负数的个数为奇数个，那么我们就可以让他们的绝对值越接近。

所以用多重map 对他们进行操作。

AC代码：

#include<bits/stdc++.h>
using namespace std;
multimap<long long int,long long int> mm;
long long int a[200005];
int main()
{
long long int n,m,k;
long long int num=0;
scanf("%I64d%I64d%I64d",&n,&m,&k);
for(int i=0;i<n;i++)
{
scanf("%I64d",&a[i]);
if(a[i]<0)
num++;
mm.insert(pair<long long int ,long long int>(fabs(a[i]),i));
}
int c;
for(int i=0;i<m;i++)
{
c=mm.begin()->first;
int j;
j=mm.begin()->second;
if(num&1)
{
if(a[j]>=0)
{
a[j]+=k;
}
else
{
a[j]-=k;
}
}
else
{
if(a[j]>=0)
{
a[j]=a[j]-k;
if(a[j]<0)
num++;
}
else
{
a[j]+=k;
if(a[j]>=0)
num--;
}
}
mm.erase(mm.begin());
mm.insert(pair<long long int,long long int>(fabs(a[j]),j));
}
for(int i=0;i<n;i++)
{
if(i==n-1)
cout<<a[i]<<endl;
else
cout<<a[i]<<" ";
}
}