The Fibonacci sequence begins with Fibonacci(0) = 0 and Fibonacci(1)=1 as its respective first and second terms. After these first two elements, each subsequent element is equal to the sum of the previous two elements.
Here is the basic information you need to calculate Fibonacci(n):
- Fibonacci(0) = 0
- Fibonacci(1) = 1
- Fibonacci(n) = Fibonacci(n-1) + Fibonacci(n-2)
Given n, complete the fibonacci function so it returns Fibonacci(n).
Consider the Fibonacci sequence:
We want to know the value of Fibonacci(3). If we look at the sequence above,Fibonacci(3) evaluates to 2. Thus, we print 2 as our answer.
public static int Fibonacci(int n)
if (n == 0) return 0;
else if (n == 1) return 1;
return Fibonacci(n - 1) + Fibonacci(n - 2);
A left rotation operation on an array of size n shifts each of the array’s elements 1 unit to the left. For example, if 2 left rotations are performed on array [1,2,3,4,5], then the array would become [3,4,5,1,2].
Given an array of n integers and a number, d , perform d left rotations on the array. Then print the updated array as a single line of space-separated integers.
The first line contains two space-separated integers denoting the respective values of (the number of integers) and (the number of left rotations you must perform).
The second line contains space-separated integers describing the respective elements of the array’s initial state.
Print a single line of space-separated integers denoting the final state of the array after performing left rotations.
static int LeftRotationByD(int a,int k)
int b = new int[a.Length];
int length = a.Length;
for (int i = 0; i < length; i++)
index = i - k;
place = length + index;
if (index >= 0)
b[index] = a[i];
b[place] = a[i];
We can see the results for 1 and 4 rotations as follows: