1006 - Hex-a-bonacci

Author: Ismail Hosen

Daffodil International University

Problem Link

//Please follow from Mian function

//Just calculate each value once and Store that value into array.

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;



ll a, b, c, d, e, f, x, t, n;

ll mod=10000007;

ll idx[10004];

ll fn( ll n ) {

    if( n == 0 ) return a;

    if( n == 1 ) return b;

    if( n == 2 ) return c;

    if( n == 3 ) return d;

    if( n == 4 ) return e;

    if( n == 5 ) return f;

    if(idx[n]!= -1) return idx[n];

    else{

        idx[n] = (fn(n-1) + fn(n-2) + fn(n-3) + fn(n-4) + fn(n-5) + fn(n-6))%mod;

        return idx[n];

    }

}

int main()

{

    x= 1;

    cin>>t;

    for(int i=1; i<=t; i++){

        memset(idx, -1, sizeof(idx));

        cin>>a>>b>>c>>d>>e>>f>>n;

        cout<<"Case "<<i<<": "<<fn(n)%mod<<endl;

    }

}