Nth Element Of Modified Fibonacci Series

Posted: 23 Oct, 2020
Difficulty: Easy

PROBLEM STATEMENT

Note:

The series is 1-based indexed.
Input Format:
The first line contains an integer T denoting the number of test cases. Then each test case follows.

The first line of each test case contains three space-separated integers ‘X’, 'Y', and ‘N’, respectively where ‘X’ and ‘Y’ represent the first and second element of the series while N represents which number of the sequence we have to find out.
Output Format:
For each test case, print a single line containing a single integer denoting the Nth number of the series.

The output of each test case will be printed in separate lines.
Note
You are not required to print the expected output; it has already been taken care of. Just implement the given function.
Constraints:
1 <= T <= 100
1 <= N <= 10 ^ 18
-10 ^ 6 <= X, Y <= 10 ^ 6

Time limit: 1 sec. Approach 1

Let’s define a dp array of size N, where dp[i] represents the i-th Fibonacci number. For each block, let’s compute the answer in a top-down fashion, starting with the leftmost blocks (dp and dp). We will iterate through the array for i = 2 to N and then we can fill the array in a top-down manner like:

dp[i] = dp[i-1] + dp[i-2].

Since, during this method, we only need the previous two states, we can store the last two states in two variables instead of using a whole array.

Here is the algorithm:

1. Initialise variable sum = 0 that stores the sum of the previous two values.
2. Now, run a loop from i = 2 to N-1 and for each index update value of sum = X + Y  and X = Y, Y = sum.
3. Finally, return the sum, which is the required Nth element.