I understand that this problem might seem intimidating at first. Let's break it down step-by-step to make it clearer. I'll explain the problem, the solution approach, and then go through the code in detail.

Problem Breakdown:

1. We're dealing with a function F(N) defined as follows:
   • F(0) = 1
   • F(1) = 1
   • For N ≥ 2: F(N) = (F(N-1) + 7*F(N-2) + N/4) modulo (10^9 + 7)
2. We need to calculate F(N) for a given N.
3. The hint suggests using lists (vectors in C++) instead of recursion to avoid timeouts.

Solution Approach:

1. We'll use a bottom-up dynamic programming approach.
2. We'll store the calculated values in a vector to avoid recalculating them.
3. We'll iterate from 2 to N, calculating each F(i) using the previous values.

Now, let's go through the code in detail:

#include <iostream>
#include <vector>
using namespace std;

const int MOD = 1e9 + 7;

int main() {
    int N;
    cin >> N;
    
    vector<long long> F(N + 1);
    F[0] = 1;
    F[1] = 1;
    
    for (int i = 2; i <= N; i++) {
        F[i] = (F[i-1] + 7LL * F[i-2] + i / 4) % MOD;
    }
    
    cout << F[N] << endl;
    
    return 0;
}

Important Notes:

• We use integer division for i / 4, so 2/4 will indeed be 0, not 0.5.
• The modulo operation is applied at each step to prevent integer overflow.

This solution efficiently calculates F(N) for large N values without recursion, avoiding potential stack overflow or timeout issues. The time complexity is O(N), and the space complexity is also O(N) due to the vector storing all values from 0 to N.