First, we consider how can we determine if it can grab \(N\) treasures while satisfying the conditions in the problem statement when its head is at \(x=k_0\).

Consider a sequence \((Y_1,Y_2, \ldots, Y_N)\) consisting of the distances from the head to each treasure, \(\lvert X_i-k_0\rvert\) \((1\leq i\leq N)\), sorted in ascending order.
If \(Y_i\leq L_i\) holds for all \(1\leq i\leq N\), then it can grab them while satisfying the conditions, specifically by using the \(i\)-th leg to grab the treasure distant by \(Y_i\) from the head.
On the other hand, if \(Y_i>L_i\), then there is no conforming way to grab the treasure, because at least \((N-i+1)\) treasures are distant from the head by at least \((L_i+1)\), but there are at most \((N-i)\) legs that can grab them.

Naively, one can sort \((Y_1,Y_2,\ldots,Y_N)\) in an \(O(N\log N)\) time using algorithms like quick sort, or one can even inspect the nearest treasures to \(x=k_0\) on both sides in a total of \(O(N)\) time, so the decision problem can be solved in a total of \(O(N\log N)\) or \(O(N)\) time.

However, the range of the candidates of the head is \(-2\times 10^{18}\leq x\leq 2\times 10^{18}\), so it is almost impossible to consider it for all such candidates within the time limit. Instead, we consider when does the answer change from \(x=k_0\) to \(x=k_0+1\).

• If \(x=k_0\) satisfies the condition, but \(x=k_0+1\) does not

In this case, a valid correspondence between a leg and a treasure in \(x=k_0\) should become invalid. That is, there should be \(1\leq i,j\leq N\) such that \(k_0-L_j\leq X_i\leq k_0+L_j\) and (\(X_i<k_0+1-L_j\) or \(k_0+1+L_j<X_i\)). Such \(k_0\), which should be an integer, is limited to those with \(k_0=X_i+L_j\), so there are at most \(N^2\) such candidates.

• If \(x=k_0\) does not satisfy the condition, but \(x=k_0+1\) does

In this case, an invalid correspondence between a leg and a treasure in \(x=k_0\) should become valid. That is, there should be \(1\leq i,j\leq N\) such that \(k_0+1-L_j\leq X_i\leq k_0+1+L_j\) and (\(X_i<k_0-L_j\) or \(k_0+L_j<X_i\)). Such \(k_0\), which should be an integer, is limited to those with \(k_0=X_i-L_j-1\), so there are at most \(N^2\) such candidates too.

Therefore, the set \(S=\{X_i+L_j|1\leq i,j\leq N \}\bigcup \{X_i-L_j-1|1\leq i,j\leq N \}\) has at most \(2N^2\) elements. When its elements are sorted in ascending order as \(S_1,S_2,\ldots,S_{|S|}\), then for each \(2\leq i\leq |S|\), all \(x\) with \(S_{i-1}+1\leq x\leq S_i\) satisfy the condition if and only if \(x=S_i\) does. Here, those within \(x\leq S_1\) and \(S_{|S|}+1\leq x\) never satisfies the condition because it is impossible to satisfy the condition when the head is far enough.

Therefore, it is sufficient to determine if \(x=S_i\) satisfies the condition for each \(2\leq i\leq |S|\).
Enumerating and sorting the elements of \(S\) costs \(O(N^2\log N)\), and checking each element costs a total of \(O(N^3)\) (or \(O(N^3\log N)\)) time, so the problem has been solved fast enough.

Sample code in C++:

#include <bits/stdc++.h>
using namespace std;

int n;
vector<long long>l,x;

bool judge(long long k){
  int le=-1,ri=0,nx;
  for(int i=0;i<n;i++){
    if(x[i]<k)le++,ri++;
    else break;
  }
  for(int i=0;i<n;i++){
    if(le<0)nx=ri,ri++;
    else if(ri>=n)nx=le,le--;
    else if((k-x[le])>(x[ri]-k))nx=ri,ri++;
    else nx=le,le--;
    if(abs(x[nx]-k)>l[i])return false;
  }
  return true;
}


int main(void) {
  long long val,ans=0;
  vector<long long>s;

	cin>>n;
  for(int i=0;i<n;i++){
    cin>>val;
    x.push_back(val);
  }
  for(int i=0;i<n;i++){
    cin>>val;
    l.push_back(val);
  }

  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      s.push_back(x[i]+l[j]);
      s.push_back(x[i]-l[j]-1);
    }
  }
  sort(s.begin(),s.end());

  int sz=s.size();
  for(int i=1;i<sz;i++){
    if(judge(s[i]))ans+=s[i]-s[i-1];
  }
  cout<<ans<<endl;
  return 0;
}