标签：-- eve clu 变化 nta center operator and bit
Maxim and Array
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
5 3 1
5 4 3 5 2
5 4 3 5 -1
5 3 1
5 4 3 5 5
5 4 0 5 5
5 3 1
5 4 4 5 5
5 1 4 5 5
3 2 7
5 4 2
5 11 -5
题意：一个序列长为n，有k个x让你去给序列任意一个数加||减x，然后要求序列各项乘积最大，输出变化后的序列
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+9;
struct num {
ll abs,val,idx;
num(ll val,ll abs,ll idx):val(val),abs(abs),idx(idx) {}
bool operator < (const num& rhs)const {
return abs>rhs.abs;
}
};
ll a[N],n,k,x,zero,neg;
int main() {
//freopen("f.txt","r",stdin);
priority_queue<num>q;
scanf("%I64d%I64d%I64d",&n,&k,&x);
zero=neg=0;
for(int i=1; i<=n; i++) {
scanf("%I64d",&a[i]);
if(a[i]==0)zero++;
if(a[i]<0)neg++;
q.push(num(a[i],abs(a[i]),i));
}
if(zero>k) {
for(int i=1; i<=n; i++)printf("%I64d ",a[i]);
return 0;
}
if(neg%2==0) {
if(zero) {
neg++;
zero--,k--;
num t=q.top();
q.pop();
a[t.idx]-=x;
t.val-=x;
t.abs+=x;
q.push(t);
}
}
while(zero--) {
k--;
num t=q.top();
q.pop();
a[t.idx]+=x;
t.val+=x;
t.abs+=x;
q.push(t);
}
if(neg%2==0) {
num t=q.top();
q.pop();
if(t.val<0) {
while(t.val<=0&&k)t.val+=x,k--;
} else
while(t.val>=0&&k)t.val-=x,k--;
a[t.idx]=t.val;
t.abs=abs(t.val);
q.push(t);
}
while(k--) {
num t=q.top();
q.pop();
if(t.val<0)a[t.idx]-=x;
else a[t.idx]+=x;
t.abs+=x;
q.push(t);
}
for(int i=1; i<=n; i++)printf("%I64d ",a[i]);
return 0;
}
标签：-- eve clu 变化 nta center operator and bit
原文地址：https://www.cnblogs.com/geraldg/p/11247812.html
