AtCoder Beginner Contest 128 D - equeue

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#include<vector> 
#include<queue> 
typedef long long ll;
using namespace std;
ll maxn=0;
int main()
{
    int sum,k;cin>>sum>>k;
    deque<ll> num;
    for(int i=1;i<=sum;i++) {
        ll temp;cin>>temp;
        num.push_back(temp);
    }
    for(int i=0;i<=k;i++){
        for(int j=0;j<=k;j++){
            if(i+j>k) continue;
            priority_queue<ll,vector<ll>,greater<ll>> op;//重点
            deque<ll> copy=num;
            for(int m=0;m<i&&!copy.empty();m++) {
                op.push(copy.front());copy.pop_front();
            }
            for(int m=0;m<j&&!copy.empty();m++) {
                op.push(copy.back());copy.pop_back();
            }
            ll res=k-i-j;
            op.push(0);
            while(op.top()<0&&res>0&&!op.empty()){
                res--;op.pop();
            }
            ll sam=0;
            while(!op.empty()){
                sam+=op.top();op.pop();
            }
            maxn=max(sam,maxn);
        }
    }
    cout<<maxn<<"\n";
}

思路：重点是只枚举一端的长度，好让res有操作空间的trick以及双端队列stl的使用