# 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:

`// 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

2

## Recommended Posts:

- Count of cells in a matrix which give a Fibonacci number when the count of adjacent cells is added
- Given a n-ary tree, count number of nodes which have more number of children than parents
- Count the number of non-reachable nodes
- Count all pairs of adjacent nodes whose XOR is an odd number
- Count the number of nodes at given level in a tree using BFS.
- Count the number of nodes at a given level in a tree using DFS
- Check if a M-th fibonacci number divides N-th fibonacci number
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Print levels with odd number of nodes and even number of nodes
- Count Fibonacci numbers in given range in O(Log n) time and O(1) space
- Number of ways to represent a number as sum of k fibonacci numbers
- Finding number of digits in n'th Fibonacci number
- Count BST nodes that lie in a given range
- Nth Even Fibonacci Number
- Count the nodes in the given tree whose weight is even

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.