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. 
) 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 (
) — 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, 
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