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

题目链接：http://codeforces.com/problemset/problem/721/D

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 15 4 3 5 2

output

5 4 3 5 -1 

input

5 3 15 4 3 5 5

output

5 4 0 5 5 

input

5 3 15 4 4 5 5

output

5 1 4 5 5 

input

3 2 75 4 2

output

5 11 -5 

题解：

1.首先在输入时，统计负数的个数，目的是知道初始状态的乘积是正数（包括0）还是负数。

2.如果乘积为正数，那么就需要减小某个数的绝对值，使其改变符号， 而且步数越少越好，由此推出：减少绝对值最小的那个数的绝对值，可以最快改变符号，使得乘积变为负数。

3.如果乘积已经为负，那么就需要增加某个数的绝对值， 使得乘积的绝对值尽可能大， 根据基本不等式： a+b>=2*根号（a*b），若要a*b的值最大，则a==b。 可以得出结论：当a与b的差值越小时，a*b越大。或者可以自己手动推算一遍， 也可以得出这个结论。由此推出：增加绝对值最小的那个数的绝对值，使得乘积的绝对值尽可能大。

4.总的来说：就是需要对绝对值最小的数进行操作。当乘积为正或为0时， 减小其绝对值，直到符号改变； 当乘积为负时， 增加其绝对值， 使其乘积的绝对值尽可能大。 用优先队列维护。

学习之处： 

a+b=常数， 当a与b的差值越小时，a*b越大（a*b>=0）。

这里的a*b，可以假设a为绝对值最小的那个数， b为其他数的绝对值的乘积。所以这个结论也适用于多个数的乘积（将多个转为2个）。

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <string>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <sstream>#include <algorithm>using namespace std;#define ms(a, b)  memset((a), (b), sizeof(a))#define eps 0.0000001typedef long long LL;const int INF = 2e9;const LL LNF = 9e18;const int mod = 1e9+7;const int maxn = 2e5+10;struct node{    LL v, s, pos;    bool operator<(const node&x)const{        return v>x.v;    }};priority_queue<node>q;LL a[maxn], n, k, x, flag;void init(){    scanf("%lld%lld%lld",&n, &k, &x);    while(!q.empty()) q.pop();    flag = 0;    for(int i = 1; i<=n; i++)    {        node e;        scanf("%lld",&a[i]);        e.v = abs(a[i]);        e.s = (a[i]<0);        e.pos = i;        q.push(e);        if(a[i]<0) flag = !flag;    }}void solve(){    while(k--)    {        node e = q.top();        q.pop();        if(!flag)        {            e.v -= x;            if(e.v<0)            {                e.v = -e.v;                e.s = !e.s;                flag = !flag;            }        }        else        {            e.v += x;        }        a[e.pos] = e.v;        if(e.s) a[e.pos] = -a[e.pos];        q.push(e);    }    for(int i = 1; i<=n; i++)        printf("%lld ",a[i]);    putchar('\n');}int main(){//    int T;//    scanf("%d",&T);//    while(T--)    {        init();        solve();    }    return 0;}