Given two numbers N and K. Find the number of ways to represent N as the sum of K Fibonacci numbers.
Examples:
Input : n = 12, k = 1 Output : 0 Input : n = 13, k = 3 Output : 2 Explanation : 2 + 3 + 8, 3 + 5 + 5.
Approach: The Fibonacci series is f(0)=1, f(1)=2 and f(i)=f(i-1)+f(i-2) for i>1. Let’s suppose F(x, k, n) be the number of ways to form the sum x using exactly k numbers from f(0), f(1), …f(n-1). To find a recurrence for F(x, k, n), notice that there are two cases: whether f(n-1) in the sum or not.
- If f(n-1) is not in the sum, then x is formed as a sum using exactly k numbers from f(0), f(1), …, f(n-2).
- If f(n-1) is in the sum, then the remaining x-f(n-1) is formed using exactly k-1 numbers from f(0), f(1), …, f(n-1). (Notice that f(n-1) is still included because duplicate numbers are allowed.).
So the recurrence relation will be:
F(x, k, n)= F(x, k, n-1)+F(x-f(n-1), k-1, n)
Base cases:
- If k=0, then there are zero numbers from the series, so the sum can only be 0. Hence, F(0, 0, n)=1.
- F(x, 0, n)=0, if x is not equals to 0.
Also, there are other cases that make F(x, k, n)=0, like the following:
- If k>0 and x=0 because having at least one positive number must result in a positive sum.
- If k>0 and n=0 because there’s no possible choice of numbers left.
- If x<0 because there’s no way to form a negative sum using a finite number of nonnegative numbers.
Below is the implementation of above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// to store fibonacci numbers// 42 second number in fibonacci series// largest possible integerint fib[43] = { 0 };// Function to generate fibonacci seriesvoid fibonacci(){ fib[0] = 1; fib[1] = 2; for (int i = 2; i < 43; i++) fib[i] = fib[i - 1] + fib[i - 2];}// Recursive function to return the // number of ways int rec(int x, int y, int last){ // base condition if (y == 0) { if (x == 0) return 1; return 0; } int sum = 0; // for recursive function call for (int i = last; i >= 0 and (float)fib[i] * (float)y >= (float)x; i--) { if (fib[i] > x) continue; sum += rec(x - fib[i], y - 1, i); } return sum;}// Driver codeint main(){ fibonacci(); int n = 13, k = 3; cout << "Possible ways are: " << rec(n, k, 42); return 0;} |
C
// C implementation of above approach#include <stdio.h>// to store fibonacci numbers// 42 second number in fibonacci series// largest possible integerint fib[43] = { 0 };// Function to generate fibonacci seriesvoid fibonacci(){ fib[0] = 1; fib[1] = 2; for (int i = 2; i < 43; i++) fib[i] = fib[i - 1] + fib[i - 2];}// Recursive function to return the // number of ways int rec(int x, int y, int last){ // base condition if (y == 0) { if (x == 0) return 1; return 0; } int sum = 0; // for recursive function call for (int i = last; i >= 0 && (float)fib[i] * (float)y >= (float)x; i--) { if (fib[i] > x) continue; sum += rec(x - fib[i], y - 1, i); } return sum;}// Driver codeint main(){ fibonacci(); int n = 13, k = 3; printf("Possible ways are: %d",rec(n, k, 42)); return 0;}// This code is contributed by kothavvsaakash. |
Java
//Java implementation of above approachpublic class AQW { //to store fibonacci numbers //42 second number in fibonacci series //largest possible integer static int fib[] = new int[43]; //Function to generate fibonacci series static void fibonacci() { fib[0] = 1; fib[1] = 2; for (int i = 2; i < 43; i++) fib[i] = fib[i - 1] + fib[i - 2]; } //Recursive function to return the //number of ways static int rec(int x, int y, int last) { // base condition if (y == 0) { if (x == 0) return 1; return 0; } int sum = 0; // for recursive function call for (int i = last; i >= 0 && (float)fib[i] * (float)y >= (float)x; i--) { if (fib[i] > x) continue; sum += rec(x - fib[i], y - 1, i); } return sum; } //Driver code public static void main(String[] args) { fibonacci(); int n = 13, k = 3; System.out.println("Possible ways are: "+ rec(n, k, 42)); }} |
Python3
# Python3 implementation of the above approach # To store fibonacci numbers 42 second # number in fibonacci series largest# possible integer fib = [0] * 43# Function to generate fibonacci# series def fibonacci(): fib[0] = 1 fib[1] = 2 for i in range(2, 43): fib[i] = fib[i - 1] + fib[i - 2] # Recursive function to return the # number of ways def rec(x, y, last): # base condition if y == 0: if x == 0: return 1 return 0 Sum, i = 0, last # for recursive function call while i >= 0 and fib[i] * y >= x: if fib[i] > x: i -= 1 continue Sum += rec(x - fib[i], y - 1, i) i -= 1 return Sum# Driver code if __name__ == "__main__": fibonacci() n, k = 13, 3 print("Possible ways are:", rec(n, k, 42))# This code is contributed # by Rituraj Jain |
C#
// C# implementation of above approachusing System; class GFG { // to store fibonacci numbers // 42 second number in fibonacci series // largest possible integer static int[] fib = new int[43]; // Function to generate fibonacci series public static void fibonacci() { fib[0] = 1; fib[1] = 2; for (int i = 2; i < 43; i++) fib[i] = fib[i - 1] + fib[i - 2]; } // Recursive function to return the // number of ways public static int rec(int x, int y, int last) { // base condition if (y == 0) { if (x == 0) return 1; return 0; } int sum = 0; // for recursive function call for (int i = last; i >= 0 && (float)fib[i] * (float)y >= (float)x; i--) { if (fib[i] > x) continue; sum += rec(x - fib[i], y - 1, i); } return sum; } // Driver code static void Main() { for(int i = 0; i < 43; i++) fib[i] = 0; fibonacci(); int n = 13, k = 3; Console.Write("Possible ways are: " + rec(n, k, 42)); } //This code is contributed by DrRoot_} |
PHP
<?php// PHP implementation of above approach// To store fibonacci numbers// 42 second number in fibonacci series// largest possible integer$fib = array_fill(0, 43, 0);// Function to generate// fibonacci seriesfunction fibonacci(){ global $fib; $fib[0] = 1; $fib[1] = 2; for ($i = 2; $i < 43; $i++) $fib[$i] = $fib[$i - 1] + $fib[$i - 2];}// Recursive function to return // the number of ways function rec($x, $y, $last){ global $fib; // base condition if ($y == 0) { if ($x == 0) return 1; return 0; } $sum = 0; // for recursive function call for ($i = $last; $i >= 0 and $fib[$i] * $y >= $x; $i--) { if ($fib[$i] > $x) continue; $sum += rec($x - $fib[$i], $y - 1, $i); } return $sum;}// Driver codefibonacci();$n = 13;$k = 3;echo "Possible ways are: " . rec($n, $k, 42);// This code is contributed by mits?> |
Javascript
<script>//Javascript implementation of above approach //to store fibonacci numbers //42 second number in fibonacci series //largest possible integer let fib=new Array(43); //Function to generate fibonacci series function fibonacci() { fib[0] = 1; fib[1] = 2; for (let i = 2; i < 43; i++) fib[i] = fib[i - 1] + fib[i - 2]; } //Recursive function to return the //number of ways function rec(x,y,last) { // base condition if (y == 0) { if (x == 0) return 1; return 0; } let sum = 0; // for recursive function call for (let i = last; i >= 0 && fib[i] * y >= x; i--) { if (fib[i] > x) continue; sum += rec(x - fib[i], y - 1, i); } return sum; } //Driver code fibonacci(); let n = 13, k = 3; document.write("Possible ways are: "+ rec(n, k, 42)); // This code is contributed by rag2127</script> |
Output:
Possible ways are: 2
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