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.

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:

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// 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;
}

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Output:

2


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