Codeforces F. Maxim and Array(构造贪心)

题目描述:

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 *a**i* either with *a**i + x* or with *a**i - 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. img) 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, ..., *a**n* (img) — the elements of the array found by Maxim.

Output

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

Copy

5 3 1
5 4 3 5 2

Output

Copy

5 4 3 5 -1 

Input

Copy

5 3 1
5 4 3 5 5

Output

Copy

5 4 0 5 5 

Input

Copy

5 3 1
5 4 4 5 5

Output

Copy

5 1 4 5 5 

Input

Copy

3 2 7
5 4 2

Output

Copy

5 11 -5 

思路:

题目是要求给一个数列,k次操作在某个数上加或减x,让数列乘积最小。因为数有正负,要分情况讨论。

如果现在负数个数是偶数,就是乘积是个正数,应该相办法让它变成负,就让绝对值最小的变,因为它距离零最近。如果要变的是正数,就减x,如果是负数要变,就加x,即使不能让乘积编号也可以让乘积变小。

如果现在负数个数是奇数,乘积是个负数,我们要让乘积的绝对值更大来使乘积更小。就变绝对值最小的,是正数就加x,是负数就减x,因为当一堆数大小越接近,这堆数的乘积越大。

注意每次变化要更新负数的个数,要快速取得绝对值最小的元素,要维护一个结构体的优先级队列,需要在结构体的定义里重载<运算符。

刚开始我沙雕用了两个队列,一个存整数,一个存负数,每次选绝对值小的还要比较,\(if-else\)写了一大堆最后还错了。-_-||详见后面的代码。

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#define max_n 200005
using namespace std;
int n,k,x;
long long a[max_n];
struct node
{
    int id;
    long long val;
    bool operator<(const node& a) const
    {
        return abs(val-0)>abs(a.val-0);
    }
};
int cnt = 0;
priority_queue<node> que;
int main()
{
    cin >> n >> k >> x;
    for(int i = 0;i<n;i++)
    {
        cin >> a[i];
        node nw;
        nw.val = a[i];
        nw.id = i;
        if(a[i]<0)
        {
            cnt++;
        }
        que.push(nw);
    }
    while(k)
    {
        node nw = que.top();
        int id = nw.id;
        if(cnt%2==0)
        {
            if(a[id]<0)
            {
                a[id] += x;
                if(a[id]>=0)
                {
                    cnt--;
                }
            }
            else
            {
                a[id] -= x;
                if(a[id]<0)
                {
                    cnt++;
                }
            }
        }
        else
        {
            if(a[id]<0)
            {
                a[id] -= x;
            }
            else
            {
                a[id] += x;
            }
        }
        nw.val = a[id];
        que.pop();
        que.push(nw);
        k--;
    }
    for(int i = 0;i<n;i++)
    {
        cout << a[i] << " ";
    }
    cout << endl;
}

不明哪里写错的代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#define max_n 200005
using namespace std;
long long n,k,x;
long long a[max_n];
struct node
{
    int id;
    long long val;
    bool operator<(const node& a) const
    {
        return abs(val-0)>abs(a.val-0);
    }
};
priority_queue<node> fque;
priority_queue<node> zque;
int main()
{
    cin >> n >> k >> x;
    for(int i = 0;i<n;i++)
    {
        cin >> a[i];
        node nw;
        nw.val = a[i];
        nw.id = i;
        if(a[i]<0)
        {

            fque.push(nw);
        }
        else
        {
            zque.push(nw);
        }
    }
    //cout << "input " << endl;
    while(k)
    {
        /*for(int i = 0;i<n;i++)
        {
            cout << a[i] << " ";
        }
        cout << endl;*/
        if(fque.size()%2==0)
        {
            if(fque.size()==0)
            {
                int id = zque.top().id;
                a[id] -= x;
                node nw;
                nw.id = id;
                nw.val = a[id];
                if(a[id]<0)
                {
                    zque.pop();
                    fque.push(nw);
                }
                else
                {
                    zque.pop();
                    zque.push(nw);
                }
            }
            else if(zque.size()==0)
            {
                int id = fque.top().id;
                a[id] += x;
                node nw;
                nw.id = id;
                nw.val = a[id];
                if(a[id]>=0)
                {
                    fque.pop();
                    zque.push(nw);
                }
                else
                {
                    fque.pop();
                    fque.push(nw);
                }
            }
            else
            {
                int gapz = abs(zque.top().val-0);
                int idz = zque.top().id;
                int gapf = abs(fque.top().val-0);
                int idf = fque.top().id;
                if(gapz<gapf)
                {
                    a[idz] -= x;
                    node nw;
                    nw.id = idz;
                    nw.val = a[idz];
                    if(a[idz]<0)
                    {
                        zque.pop();
                        fque.push(nw);
                    }
                    else
                    {
                        zque.pop();
                        zque.push(nw);
                    }
                }
                else
                {
                    a[idf] += x;
                    node nw;
                    nw.id = idf;
                    nw.val = a[idf];
                    if(a[idf]>=0)
                    {
                        fque.pop();
                        zque.push(nw);
                    }
                    else
                    {
                        fque.pop();
                        fque.push(nw);
                    }
                }
            }
        }
        else
        {
            if(zque.size()==0)
            {
                int id = fque.top().id;
                a[id] -= x;
                node nw;
                nw.id = id;
                nw.val = a[id];
                fque.pop();
                fque.push(nw);
            }
            else
            {
                int gapz = abs(zque.top().val-0);
                int idz = zque.top().id;
                int gapf = abs(fque.top().val-0);
                int idf = fque.top().id;
                if(gapz<gapf)
                {
                    a[idz] += x;
                    zque.pop();
                    node nw;
                    nw.id = idz;
                    nw.val = a[idz];
                    zque.push(nw);
                }
                else
                {
                    a[idf] -= x;
                    fque.pop();
                    node nw;
                    nw.id = idf;
                    nw.val = a[idf];
                    fque.push(nw);
                }
            }
        }
        k--;
    }
    for(int i = 0;i<n;i++)
    {
        cout << a[i] << " ";
    }
    cout << endl;
    return 0;

}

参考文章:

木流牛马,D. Maxim and Array,https://www.cnblogs.com/thunder-110/p/9340279.html

posted @ 2019-08-16 19:09  小张人  阅读(...)  评论(...编辑  收藏
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