Prompt Title: Elegant Card Sequences.

To solve this problem using recursion, we can use the following approach:

Let's define a function countWays(cards, index, lastColor) that takes as input the following parameters:

• cards: an array of size N containing the colors of the cards, where cards[i] represents the color of the ith card.
• index: the index of the current card we are considering.
• lastColor: the color of the previous card we considered.

The function will return the number of possible elegant sequences that can be formed starting from the current card at index index.

Here's the recursive logic:

1. Base case: If we have considered all N cards, then we have found a valid sequence. Return 1.
2. Recursive case: For the current card at index index, we have two options: a. If the color of the current card is different from the color of the last card, then we can place it next to the last card. In this case, we can recursively count the number of ways to form an elegant sequence starting from the next card (i.e., call countWays(cards, index+1, cards[index])) and add it to our answer. b. If the color of the current card is the same as the color of the last card, then we cannot place it next to the last card. In this case, we cannot form any more elegant sequences starting from this point, so we simply return 0.

Here's the C++ code implementing the above approach:

c++Copy code
#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int MOD = 998244353;

int countWays(vector<int>& cards, int index, int lastColor) {
    // Base case: we have considered all cards
    if (index == cards.size()) {
        return 1;
    }

    // Recursive case
    int ans = 0;

    // Option 1: place the current card next to the last card
    if (cards[index] != lastColor) {
        ans = countWays(cards, index+1, cards[index]);
    }

    // Option 2: skip the current card
    ans = (ans + countWays(cards, index+1, lastColor)) % MOD;

    return ans;
}

int main() {
    int n;
    cin >> n;

    vector<int> cards(n);
    for (int i = 0; i < n; i++) {
        int a, b;
        cin >> a >> b;
        cards[i] = a;
    }

    // Count the number of elegant sequences starting from each card
    int ans = 0;
    for (int i = 0; i < n; i++) {
        ans = (ans + countWays(cards, i+1, cards[i])) % MOD;
    }

    cout << ans << endl;

    return 0;
}

The time complexity of this approach is O(2^N), which is exponential. However, we can optimize it by using dynamic programming to avoid recomputing the same subproblems multiple times.