# 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 ` `using` `namespace` `std; ` ` `  `int` `ans = 0, x; ` ` `  `vector<``int``> graph; ` `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 = 4; ` `    ``weight = 5; ` `    ``weight = 3; ` `    ``weight = 25; ` `    ``weight = 16; ` `    ``weight = 34; ` ` `  `    ``// Edges of the tree ` `    ``graph.push_back(2); ` `    ``graph.push_back(3); ` `    ``graph.push_back(4); ` `    ``graph.push_back(5); ` `    ``graph.push_back(6); ` ` `  `    ``dfs(1, 1); ` ` `  `    ``cout << ans; ` ` `  `    ``return` `0; ` `} `

Output:

```2
```

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