Hex-a-bonacci Lightoj 1006

This problem is pretty easy, it only requires  memoization to solve.For this purpose we will use an array and save previous solved subproblems rather than call it again.

Codes:

#include <stdio.h>
#include <iostream>
using namespace std;
long long t;
long long a;
long long b;
long long c;
long long d;
long long e;
long long f;
long long n;
long long x[10005];
int fn( int n ) {
        for (int i = 0; i <= n; i++) {
            if( i == 0 ) {
                x[i] = a;
            continue;
            }
            if( i == 1 ) {
            x[i] = b;
            continue;
            }
            if( i == 2 ) {
            x[i] = c;
            continue;}
            if( i == 3 ) {
            x[i] = d;
            continue;}
            if( i == 4 ) {
            x[i] = e;
            continue;}
            if( i == 5 ) {
                x[i] = f;
            continue;

        }

        x[i] = x[i-1] + x[i-2] + x[i-3] + x[i-4] + x[i-5] + x[i-6];

        x[i] = x[i] % 10000007;

}

return x[n];

}
int main()
{
    long long ans;
    cin >> t;
    for (long long i = 1; i <= t; i++) {
        cin >> a;
        cin >> b;
        cin >> c;
        cin >> d;
        cin >> e;
        cin >> f;
        cin >> n;
        ans = fn(n);
        ans = ans % 10000007;
        cout << "Case "<< i <<": "<< ans << endl;
}}