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Count the nodes whose sum with X is a Fibonacci number

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  • Last Updated : 24 Sep, 2021

Given a tree, and the weights of all the nodes and an integer X, the task is to count all the nodes i such that (weight[i] + X) is a Fibonacci Number.
First, few Fibonacci numbers are: 

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, … 

Examples: 

Input: 
 

X = 5 
Output:
Only the nodes 3 and 5 give a fibonacci number when 5 is added to them. 
i.e. (3 + 5) = 8 and (16 + 5) = 21 are both Fibonacci numbers. 

Approach: Perform dfs on the tree and count all the nodes sum of whose weight with x is a Fibonacci number.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int ans = 0, x;
 
vector<int> graph[100];
vector<int> weight(100);
 
// Function that returns true if
// x is a perfect square
bool isPerfectSquare(long double x)
{
    // Find floating point value of
    // square root of x
    long double sr = sqrt(x);
 
    // If square root is an integer
    return ((sr - floor(sr)) == 0);
}
 
// Function that returns true
// if n is a fibonacci number
bool isFibonacci(int n)
{
    return isPerfectSquare(5 * n * n + 4)
           || isPerfectSquare(5 * n * n - 4);
}
 
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weight of the current node
    // gives a fibonacci number
    // when x is added to it
    if (isFibonacci(weight[node] + x))
        ans += 1;
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
    x = 5;
 
    // Weights of the node
    weight[1] = 4;
    weight[2] = 5;
    weight[3] = 3;
    weight[4] = 25;
    weight[5] = 16;
    weight[6] = 34;
 
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
    graph[5].push_back(6);
 
    dfs(1, 1);
 
    cout << ans;
 
    return 0;
}

Java




// Java implementation of the
// above approach
import java.util.*;
 
@SuppressWarnings("unchecked")
 
class GFG{
     
static int ans = 0, x;
 
static ArrayList []graph = new ArrayList[100];
static ArrayList weight = new ArrayList();
   
// Function that returns true if
// x is a perfect square
static boolean isPerfectSquare(double x)
{
   
  // Find floating point value of
  // square root of x
  double sr = Math.sqrt(x);
   
  // If square root is an integer
  return ((sr - Math.floor(sr)) == 0);
}
   
// Function that returns true
// if n is a fibonacci number
static boolean isFibonacci(int n)
{
  return isPerfectSquare(5 * n * n + 4) ||
         isPerfectSquare(5 * n * n - 4);
}
   
// Function to perform dfs
static void dfs(int node, int parent)
{
   
  // If weight of the current node
  // gives a fibonacci number
  // when x is added to it
  if (isFibonacci((int)weight.get(node) + x))
    ans += 1;
  
  for(int to : (ArrayList<Integer>)graph[node])
  {
    if (to == parent)
      continue;
     
    dfs(to, node);
  }
}
   
// Driver Code
public static void main(String[] args)
{
  x = 5;
   
  for(int i = 0; i < 100; i++)
  {
    weight.add(0);
    graph[i] = new ArrayList();
  }
   
  // Weights of the node
  weight.add(1, 4);
  weight.add(2, 5);
  weight.add(3, 3);
  weight.add(4, 25);
  weight.add(5, 16);
  weight.add(6, 34);
   
  // Edges of the tree
  graph[1].add(2);
  graph[2].add(3);
  graph[2].add(4);
  graph[1].add(5);
  graph[5].add(6);
  
  dfs(1, 1);
   
  System.out.println(ans);
}
}
 
// This code is contributed by pratham76

Python3




# Python3 implementation of the approach
import math
ans, x = 0, 0
  
graph = [[] for i in range(100)]
weight = [0]*(100)
  
# Function that returns true if
# x is a perfect square
def isPerfectSquare(x):
   
    # Find floating point value of
    # square root of x
    sr = math.sqrt(x);
  
    # If square root is an integer
    return ((sr - math.floor(sr)) == 0)
  
# Function that returns true
# if n is a fibonacci number
def isFibonacci(n):
    return isPerfectSquare(5 * n * n + 4) or isPerfectSquare(5 * n * n - 4)
  
# Function to perform dfs
def dfs(node, parent):
    global ans
    # If weight of the current node
    # gives a fibonacci number
    # when x is added to it
    if (isFibonacci(weight[node] + x)):
        ans += 1
  
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
 
x = 5
  
# Weights of the node
weight[1] = 4
weight[2] = 5
weight[3] = 3
weight[4] = 25
weight[5] = 16
weight[6] = 34
 
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
graph[5].append(6)
 
dfs(1, 1)
 
print(ans)
 
# This code is contributed by divyesh072019.

C#




// C# implementation of the
// above approach
using System;
using System.Collections;
class GFG{
      
static int ans = 0, x;
static ArrayList []graph = new ArrayList[100];
static ArrayList weight = new ArrayList();
  
// Function that returns true if
// x is a perfect square
static bool isPerfectSquare(double x)
{
  // Find floating point value of
  // square root of x
  double sr = Math.Sqrt(x);
 
  // If square root is an integer
  return ((sr - Math.Floor(sr)) == 0);
}
  
// Function that returns true
// if n is a fibonacci number
static bool isFibonacci(int n)
{
    return isPerfectSquare(5 * n * n + 4) ||
           isPerfectSquare(5 * n * n - 4);
}
  
// Function to perform dfs
static void dfs(int node, int parent)
{
  // If weight of the current node
  // gives a fibonacci number
  // when x is added to it
  if (isFibonacci((int)weight[node] + x))
    ans += 1;
 
  foreach (int to in graph[node])
  {
    if (to == parent)
      continue;
    dfs(to, node);
  }
}
  
// Driver Code
public static void Main(string[] args)
{
  x = 5;
 
  for(int i = 0; i < 100; i++)
  {
    weight.Add(0);
    graph[i] = new ArrayList();
  }
 
  // Weights of the node
  weight[1] = 4;
  weight[2] = 5;
  weight[3] = 3;
  weight[4] = 25;
  weight[5] = 16;
  weight[6] = 34;
 
  // Edges of the tree
  graph[1].Add(2);
  graph[2].Add(3);
  graph[2].Add(4);
  graph[1].Add(5);
  graph[5].Add(6);
 
  dfs(1, 1);
  Console.Write(ans);
}
}
 
// This code is contributed by rutvik_56

Javascript




<script>
 
// JavaScript implementation of the
// above approach
 
var ans = 0, x;
var graph = Array.from(Array(100), ()=>Array());
var weight = [];
  
// Function that returns true if
// x is a perfect square
function isPerfectSquare(x)
{
  // Find floating point value of
  // square root of x
  var sr = Math.sqrt(x);
 
  // If square root is an integer
  return ((sr - Math.floor(sr)) == 0);
}
  
// Function that returns true
// if n is a fibonacci number
function isFibonacci(n)
{
    return isPerfectSquare(5 * n * n + 4) ||
           isPerfectSquare(5 * n * n - 4);
}
  
// Function to perform dfs
function dfs(node, parent)
{
  // If weight of the current node
  // gives a fibonacci number
  // when x is added to it
  if (isFibonacci(weight[node] + x))
    ans += 1;
 
  for(var to of graph[node])
  {
    if (to == parent)
      continue;
    dfs(to, node);
  }
}
  
// Driver Code
x = 5;
for(var i = 0; i < 100; i++)
{
  weight.push(0);
  graph[i] = [];
}
// Weights of the node
weight[1] = 4;
weight[2] = 5;
weight[3] = 3;
weight[4] = 25;
weight[5] = 16;
weight[6] = 34;
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
graph[5].push(6);
dfs(1, 1);
document.write(ans);
 
</script>

Output: 

2

 


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