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++
#include <bits/stdc++.h>
using namespace std;
int fib[43] = { 0 };
void fibonacci()
{
fib[0] = 1;
fib[1] = 2;
for (int i = 2; i < 43; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
int rec(int x, int y, int last)
{
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
int sum = 0;
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;
}
int main()
{
fibonacci();
int n = 13, k = 3;
cout << "Possible ways are: "
<< rec(n, k, 42);
return 0;
}
|
C
#include <stdio.h>
int fib[43] = { 0 };
void fibonacci()
{
fib[0] = 1;
fib[1] = 2;
for (int i = 2; i < 43; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
int rec(int x, int y, int last)
{
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
int sum = 0;
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;
}
int main()
{
fibonacci();
int n = 13, k = 3;
printf("Possible ways are: %d",rec(n, k, 42));
return 0;
}
|
Java
public class AQW {
static int fib[] = new int[43];
static void fibonacci()
{
fib[0] = 1;
fib[1] = 2;
for (int i = 2; i < 43; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
static int rec(int x, int y, int last)
{
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
int sum = 0;
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;
}
public static void main(String[] args) {
fibonacci();
int n = 13, k = 3;
System.out.println("Possible ways are: "+ rec(n, k, 42));
}
}
|
Python3
fib = [0] * 43
def fibonacci():
fib[0] = 1
fib[1] = 2
for i in range(2, 43):
fib[i] = fib[i - 1] + fib[i - 2]
def rec(x, y, last):
if y == 0:
if x == 0:
return 1
return 0
Sum, i = 0, last
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
if __name__ == "__main__":
fibonacci()
n, k = 13, 3
print("Possible ways are:", rec(n, k, 42))
|
C#
using System;
class GFG
{
static int[] fib = new int[43];
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];
}
public static int rec(int x, int y, int last)
{
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
int sum = 0;
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;
}
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));
}
}
|
PHP
<?php
$fib = array_fill(0, 43, 0);
function fibonacci()
{
global $fib;
$fib[0] = 1;
$fib[1] = 2;
for ($i = 2; $i < 43; $i++)
$fib[$i] = $fib[$i - 1] +
$fib[$i - 2];
}
function rec($x, $y, $last)
{
global $fib;
if ($y == 0)
{
if ($x == 0)
return 1;
return 0;
}
$sum = 0;
for ($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;
}
fibonacci();
$n = 13;
$k = 3;
echo "Possible ways are: " .
rec($n, $k, 42);
?>
|
Javascript
<script>
let fib=new Array(43);
function fibonacci()
{
fib[0] = 1;
fib[1] = 2;
for (let i = 2; i < 43; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
function rec(x,y,last)
{
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
let sum = 0;
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;
}
fibonacci();
let n = 13, k = 3;
document.write("Possible ways are: "+ rec(n, k, 42));
</script>
|
Output: Possible ways are: 2