First, sort \(A\) so that \(A_1 \le A_2 \le \dots \le A_N\).

When \(A_1 \le A_2 \le \dots \le A_{k} \le X_i \le A_{k+1} \le \dots \le A_N\), the answer is expressed as

\((X_i-A_1) + (X_i-A_2) + \dots + (X_i-A_k) + (A_{k+1}-X_i) + \dots +(A_N-X_i) = \sum^{k}_{j=1} (X_i-A_j) + \sum^{N}_{j=k+1} (A_j-X_i).\)

\(\sum^{k}_{j=1} (X_i-A_j) = kX_i - \sum^{k}_{j=1} A_j\), whose \(kX_i\) can obviously be calculated in an \(O(1)\) time, and \(\sum^{k}_{j=1} A_j\) can be calculated in an \(O(1)\) time for each query by precalculating the cumulative sum in an \(O(N)\) time. Also, \(\sum^{N}_{j=k+1} (A_j-X_i)\) can be calculated in the similar way.

Additionally, we can find \(k\) such that \(A_k \le X_i \le A_{k+1}\) with binary search in \(O(\log N)\) time per query.

Therefore, the problem has been solved in a total of \(O(N + Q \log N)\) time.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  int n,q;
  cin >> n >> q;
  vector<long long> a(n);
  for(auto &nx : a){cin >> nx;}
  sort(a.begin(),a.end());

  vector<long long> rw(n+1,0);
  for(long long i=0;i<n;i++){rw[i+1]=rw[i]+a[i];}

  for(int i=0;i<q;i++){
    long long x;
    cin >> x;
    int st=0,fi=n-1;
    while(st<=fi){
      int te=(st+fi)/2;
      if(a[te]<x){st=te+1;}
      else{fi=te-1;}
    }
    long long res=x*st;
    res-=rw[fi+1];
    res+=(rw[n]-rw[st]);
    res-=x*(n-st);
    cout << res << '\n';
  }
  return 0;
}