Minimize the error

Problem - B - Codeforceshttps://codeforces.com/contest/960/problem/B The title means to give you   The length is n , a Array k1 operations ,b Array k2 operations

Each operation can +1 or -1

Their thinking

Put every a and b The pairing value of is calculated first , The question should be the square of the absolute value , Because we need to make the answer smaller , Reducing the large absolute value can make the final total result smaller , Then priority queue can be used to solve this problem

#include"bits/stdc++.h"
#define ll long long
#define pi pair<int,int>
#define inf 0x3f3f3f3f
#define  _for(i,a,b) for(int i=a;i<=b;i++)
#define  for_(i,a,b) for(int i=a;i<b;i++)
#define _fr(i,a,b) for(int i=a;i>=b;i--)
#define fr_(i,a,b) for(int i=a;i>b;i--)
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
const int N = 1e3+5;
const ll mod = 1e9+7;
const double lp=1.000000011;
map<char,int>mp;
int n,m,d,t;
priority_queue<ll,vector<ll>,less<ll>>q;
void solve(){
    int k1,k2;
    cin >> n >> k1 >> k2;
    vector<int>a(n),b(n),c(n);
    for_(i,0,n) cin >> a[i];
    for_(i,0,n){
        cin >> b[i];
        c[i]= abs(a[i]-b[i]);
        q.push(c[i]);
    }
    int ans = k1+k2;
    while (ans--){
        int lo  = q.top();
        q.pop();
        q.push(abs(lo-1));
    }
    ll sum = 0;
    while (!q.empty()){
        sum+=q.top()*q.top();
        q.pop();
    }
    cout << sum << endl;
}

int main()
{
    IOS;
    solve();
    return 0;
}