Timus DP 1225-Flag

1225-I.cpp

#include <iostream>
using namespace std;
using ull = unsigned long long;

int main()
{
    int n;
    cin >> n;
    ull dp[n + 1][2]; // dp[i][0] i-th stripe ends with white 
    dp[1][0] = dp[1][1] = dp[2][0] = dp[2][1] = 1;
    for(int i = 3; i <= n; i++){
        dp[i][0] = dp[i - 1][1] + dp[i - 2][1];
        dp[i][1] = dp[i - 1][0] + dp[i - 2][0];
    }
    cout << dp[n][0] + dp[n][1] << '\n';

    return 0;
}

/* 
    dp[n][0] + dp[n][1] should return the answer for n;
    recursive relation:
            dp[n][0] = dp[n - 1][1] + dp[n - 2][1]
            dp[n][1] = dp[n - 1][0] + dp[n - 2][1] 

judgeID:249119DJ

*/

1225.cpp

#include <iostream>
using namespace std;
using ull = unsigned long long;

int main()
{
    int n;
    cin >> n;
    ull dp[n + 1];  
    dp[1] = dp[2] = 2;
    for(int i = 3; i <= n; i++){
        dp[i] = dp[i - 1] + dp[i - 2];    
    }
    cout << dp[n] << '\n';

    return 0;
}

First I solved with 2D array, latter I myself got to discover that my solution's recursive relation direct me to an easy approach using 1D array:
LooK:
answer for n = dp[n][0] + dp[n][1] = dp[n - 1][1] + dp[n - 2][1] + dp[n - 1][0] + dp[n - 2][1] never changes for different n.