Codeforces Round #374 (Div. 2) D. Maxim and Array 线段树+贪心

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 

题意：n个数，可以修改k次，每次可以+x或者-x，使得成绩最小；

思路：每次寻找绝对值最小的那个数，判断负数的个数，进行+x或者-x；

　　　ps：优先队列也可做；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=2e5+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
struct is
{
    ll num;
    int pos;
}tree[N<<];
ll ans[N];
void pushup(int pos)
{
    tree[pos].num=min(tree[pos<<].num,tree[pos<<|].num);
}
void buildtree(int l,int r,int pos)
{
    if(l==r)
    {
        tree[pos].num=abs(ans[l]);
        tree[pos].pos=l;
        return;
    }
    int mid=(l+r)>>;
    buildtree(l,mid,pos<<);
    buildtree(mid+,r,pos<<|);
    pushup(pos);
}
void update(int p,ll c,int l,int r,int pos)
{
    if(p==r&&p==l)
    {
        tree[pos].num=abs(c);
        return;
    }
    int mid=(l+r)>>;
    if(p<=mid)
    update(p,c,l,mid,pos<<);
    else
    update(p,c,mid+,r,pos<<|);
    pushup(pos);
}
int query(ll x,int l,int r,int pos)
{
    if(l==r&&tree[pos].num==x)
        return tree[pos].pos;
    int mid=(l+r)>>;
    if(tree[pos<<].num==x)
    return query(x,l,mid,pos<<);
    else
    return query(x,mid+,r,pos<<|);
}
int main()
{
    int n,m,k;
    int flag=;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=;i<=n;i++)
    {
        scanf("%lld",&ans[i]);
        if(ans[i]<)flag++;
    }
    buildtree(,n,);
    while(m--)
    {
        ll x=tree[].num;
        int pos=query(x,,n,);
        if(flag&)
        {
            if(ans[pos]>=)
            ans[pos]+=k;
            else
            ans[pos]-=k;
            update(pos,ans[pos],,n,);
        }
        else
        {
            if(ans[pos]>=)
            {
                ans[pos]=ans[pos]-k;
                if(ans[pos]<)
                flag++;
            }
            else
            {
                ans[pos]=ans[pos]+k;
                if(ans[pos]>=)
                flag--;
            }
            update(pos,ans[pos],,n,);
        }
    }
    for(int i=;i<=n;i++)
    printf("%lld ",ans[i]);
    return ;
}