codeforces 721D Maxim and Array

  给定一个序列，每次可以选一个ai，ai+x或者ai-x。操作次数不超过k次。问结果序列之积最小是多少?

  联想到和一定的序列在每一个元素最相近的时候积最小，所以用一个优先队列，每次取出绝对值最小的元素，正＋x，负－x就可以了。

  虽然如此但是contest的时候还是wa了，因为有比较多的分类需要讨论。首先就是0怎么处理，如果有零，不能粗暴的决定变正还是负，要根据负数的个数的奇偶性判断。处理完0之后如果负数的个数是偶数个，仍然要处理。

  

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+'0';tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn 300000
struct node
{
    ll v;
    int idx;
    bool operator<(const node&rhs)const
    {
        return abs(v)>abs(rhs.v);
    }
}b[maxn];
priority_queue<node>q;
int n,k;
ll x;
ll a[maxn];
int cntf,cntz;
void read()
{
    FOR(i,0,n-1) {
        cin>>a[i];
        b[i].idx=i;
        b[i].v=a[i];
        if(a[i]<0) cntf++;
        else if(a[i]==0) cntz++;
    }
}
void solve()
{
    FOR(i,0,n-1) q.push(b[i]);
    int flg=0;
    while(k&&cntz)
    {
        k--;
        cntz--;
        node tmp=q.top();
        q.pop();
        if(cntf%2==1)
        {
            a[tmp.idx]+=x;
            tmp.v+=x;
        }
        else
        {
            a[tmp.idx]-=x;
            tmp.v-=x;
            cntf++;
        }
        q.push(tmp);
    }
    if(cntf%2==0)
    {
        node tmp=q.top();
        q.pop();
        if(tmp.v>0)
        {
            while(k&&tmp.v>=0)
            {
                k--;
                tmp.v-=x;
                a[tmp.idx]-=x;
            }
        }
        else
        {
            while(k&&tmp.v<=0)
            {
                k--;
                tmp.v+=x;
                a[tmp.idx]+=x;
            }
        }
        q.push(tmp);
    }
    while(k)
    {
        k--;
        node tmp=q.top();
        q.pop();
        if(tmp.v>0||(tmp.v==0&&flg==2)){
            tmp.v+=x;
            a[tmp.idx]+=x;
        }
        else if(tmp.v<0||(tmp.v==0&&flg==1))
        {
            tmp.v-=x;
            a[tmp.idx]-=x;
        }
        q.push(tmp);
    }
    FOR(i,0,n-1)
    {
        if(i!=n-1)
            cout<<a[i]<<" ";
        else cout<<a[i]<<endl;
    }
}
int main()
{
    cin>>n>>k>>x;
    read();
    solve();
    return 0;
}