Find the GCD of N Fibonacci Numbers with given Indices
Given indices of N Fibonacci numbers. The task is to find the GCD of the Fibonacci numbers present at the given indices.
The first few Fibonacci numbers are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89…
Note: The indices start from zero. That is, 0th Fibonacci number = 0.
Examples:
Input: Indices = {2, 3, 4, 5}
Output: GCD of the fibonacci numbers = 1
Input: Indices = {3, 6, 9}
Output: GCD of the fibonacci numbers = 2
Brute force Approach: The brute force solution is to find all the Fibonacci numbers present at the given indices and compute the GCD of all of them, and print the result.
Efficient Approach: An efficient approach is to use the property:
GCD(Fib(M), Fib(N)) = Fib(GCD(M, N))
The idea is to calculate the GCD of all the indices and then find the Fibonacci number at the index gcd_1( where gcd_1 is the GCD of the given indices).
Below is the implementation of the above approach:
C++
// C++ program to Find the GCD of N Fibonacci // Numbers with given Indices #include <bits/stdc++.h> using namespace std; // Function to return n'th // Fibonacci number int getFib(int n) { /* Declare an array to store Fibonacci numbers. */ int f[n + 2]; // 1 extra to handle case, n = 0 int i; // 0th and 1st number of the series // are 0 and 1 f[0] = 0; f[1] = 1; for (i = 2; i <= n; i++) { // Add the previous 2 numbers in the series // and store it f[i] = f[i - 1] + f[i - 2]; } return f[n]; } // Function to Find the GCD of N Fibonacci // Numbers with given Indices int find(int arr[], int n) { int gcd_1 = 0; // find the gcd of the indices for (int i = 0; i < n; i++) { gcd_1 = __gcd(gcd_1, arr[i]); } // find the fibonacci number at // index gcd_1 return getFib(gcd_1); } // Driver code int main() { int indices[] = { 3, 6, 9 }; int N = sizeof(indices) / sizeof(int); cout << find(indices, N); return 0; } |
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Java
// Java program to Find the GCD of N Fibonacci // Numbers with given Indices import java.io.*; // Function to return n'th // Fibonacci number public class GFG { // Recursive function to return gcd of a and b static int __gcd(int a, int b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a-b, b); return __gcd(a, b-a); } static int getFib(int n) { /* Declare an array to store Fibonacci numbers. */ int f[] = new int[n + 2]; // 1 extra to handle case, n = 0 int i; // 0th and 1st number of the series // are 0 and 1 f[0] = 0; f[1] = 1; for (i = 2; i <= n; i++) { // Add the previous 2 numbers in the series // and store it f[i] = f[i - 1] + f[i - 2]; } return f[n]; } // Function to Find the GCD of N Fibonacci // Numbers with given Indices static int find(int arr[], int n) { int gcd_1 = 0; // find the gcd of the indices for (int i = 0; i < n; i++) { gcd_1 = __gcd(gcd_1, arr[i]); } // find the fibonacci number at // index gcd_1 return getFib(gcd_1); } // Driver code public static void main (String[] args) { int indices[] = { 3, 6, 9 }; int N = indices.length; System.out.println( find(indices, N)); } } |
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Python 3
# Python program to Find the # GCD of N Fibonacci Numbers # with given Indices from math import * # Function to return n'th # Fibonacci number def getFib(n) : # Declare an array to store # Fibonacci numbers. f = [0] * (n + 2) # 1 extra to handle case, n = 0 # 0th and 1st number of the # series are 0 and 1 f[0], f[1] = 0, 1 # Add the previous 2 numbers # in the series and store it for i in range(2, n + 1) : f[i] = f[i - 1] + f[i - 2] return f[n] # Function to Find the GCD of N Fibonacci # Numbers with given Indices def find(arr, n) : gcd_1 = 0 # find the gcd of the indices for i in range(n) : gcd_1 = gcd(gcd_1, arr[i]) # find the fibonacci number # at index gcd_1 return getFib(gcd_1) # Driver code if __name__ == "__main__" : indices = [3, 6, 9] N = len(indices) print(find(indices, N)) # This code is contributed by ANKITRAI1 |
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C#
// C# program to Find the GCD // of N Fibonacci Numbers with // given Indices using System; // Function to return n'th // Fibonacci number class GFG { // Recursive function to // return gcd of a and b static int __gcd(int a, int b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } static int getFib(int n) { /* Declare an array to store Fibonacci numbers. */ int []f = new int[n + 2]; // 1 extra to handle case, n = 0 int i; // 0th and 1st number of // the series are 0 and 1 f[0] = 0; f[1] = 1; for (i = 2; i <= n; i++) { // Add the previous 2 numbers // in the series and store it f[i] = f[i - 1] + f[i - 2]; } return f[n]; } // Function to Find the GCD // of N Fibonacci Numbers // with given Indices static int find(int []arr, int n) { int gcd_1 = 0; // find the gcd of the indices for (int i = 0; i < n; i++) { gcd_1 = __gcd(gcd_1, arr[i]); } // find the fibonacci number // at index gcd_1 return getFib(gcd_1); } // Driver code public static void Main () { int []indices = { 3, 6, 9 }; int N = indices.Length; Console.WriteLine(find(indices, N)); } } // This code is contributed // by Shashank |
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PHP
<?php // PHP program to Find the GCD of // N Fibonacci Numbers with given // Indices // Function to return n'th // Fibonacci number function gcd($a, $b) { return $b ? gcd($b, $a % $b) : $a; } function getFib($n) { /* Declare an array to store Fibonacci numbers. */ // 1 extra to handle case, n = 0 $f = array_fill(0, ($n + 2), NULL); // 0th and 1st number of the // series are 0 and 1 $f[0] = 0; $f[1] = 1; for ($i = 2; $i <= $n; $i++) { // Add the previous 2 numbers // in the series and store it $f[$i] = $f[$i - 1] + $f[$i - 2]; } return $f[$n]; } // Function to Find the GCD of N Fibonacci // Numbers with given Indices function find(&$arr, $n) { $gcd_1 = 0; // find the gcd of the indices for ($i = 0; $i < $n; $i++) { $gcd_1 = gcd($gcd_1, $arr[$i]); } // find the fibonacci number // at index gcd_1 return getFib($gcd_1); } // Driver code $indices = array(3, 6, 9 ); $N = sizeof($indices); echo find($indices, $N); // This code is contributed // by ChitraNayal ?> |
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Output:
2
