Count the nodes whose sum with X is a Fibonacci number

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: 2
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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ 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; 
} 

Output:

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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