Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is a Powerful Number.
A number n is said to be Powerful Number if, for every prime factor p of it, p2 also divides it.
Example:
Input:
Output: 3
Explanation:
4, 16 and 25 are powerful weights in the tree.
Approach: To solve the problem mentioned above we have to perform Depth First Search(DFS) on the tree and for every node, check if it’s weight is a powerful number or not. If yes then increment the count.
Below is the implementation of the above approach:
C++
// C++ implementation to Count the nodes in the // given tree whose weight is a powerful number #include <bits/stdc++.h> using namespace std; int ans = 0; vector<int> graph[100]; vector<int> weight(100); // Function to check if the number is powerful bool isPowerful(int n) { // First divide the number repeatedly by 2 while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } // Check if only 2^1 divides n, // then return false if (power == 1) return false; } // Check if n is not a power of 2 // then this loop will execute for (int factor = 3; factor <= sqrt(n); factor += 2) { // Find highest power of "factor" // that divides n int power = 0; while (n % factor == 0) { n = n / factor; power++; } // Check if only factor^1 divides n, // then return false if (power == 1) return false; } // n must be 1 now // if it is not a prime number. // Since prime numbers are not powerful, // we return false if n is not 1. return (n == 1); } // Function to perform dfs void dfs(int node, int parent) { // Check if weight of the current node // is a powerful number if (isPowerful(weight[node])) ans += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code int main() { // Weights of the node weight[1] = 5; weight[2] = 10; weight[3] = 11; weight[4] = 8; weight[5] = 6; // Edges of the tree graph[1].push_back(2); graph[2].push_back(3); graph[2].push_back(4); graph[1].push_back(5); dfs(1, 1); cout << ans; return 0; } |
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Java
//Java implementation to Count the nodes in the //given tree whose weight is a powerful number import java.util.*; class GFG { static int ans = 0; static Vector<Integer>[] graph = new Vector[100]; static int[] weight = new int[100]; // Function to check if the number is powerful static boolean isPowerful(int n) { // First divide the number repeatedly by 2 while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } // Check if only 2^1 divides n, // then return false if (power == 1) return false; } // Check if n is not a power of 2 // then this loop will execute for (int factor = 3; factor <= Math.sqrt(n); factor += 2) { // Find highest power of "factor" // that divides n int power = 0; while (n % factor == 0) { n = n / factor; power++; } // Check if only factor^1 divides n, // then return false if (power == 1) return false; } // n must be 1 now // if it is not a prime number. // Since prime numbers are not powerful, // we return false if n is not 1. return (n == 1); } // Function to perform dfs static void dfs(int node, int parent) { // Check if weight of the current node // is a powerful number if (isPowerful(weight[node])) ans += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code public static void main(String[] args) { for (int i = 0; i < graph.length; i++) graph[i] = new Vector<Integer>(); // Weights of the node weight[1] = 5; weight[2] = 10; weight[3] = 11; weight[4] = 8; weight[5] = 6; // Edges of the tree graph[1].add(2); graph[2].add(3); graph[2].add(4); graph[1].add(5); dfs(1, 1); System.out.print(ans); } } // This code is contributed by Princi Singh |
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C#
// C# implementation to count the // nodes in thegiven tree whose weight // is a powerful number using System; using System.Collections.Generic; class GFG{ static int ans = 0; static List<int>[] graph = new List<int>[100]; static int[] weight = new int[100]; // Function to check if the number // is powerful static bool isPowerful(int n) { // First divide the number // repeatedly by 2 while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } // Check if only 2^1 divides n, // then return false if (power == 1) return false; } // Check if n is not a power of 2 // then this loop will execute for(int factor = 3; factor <= Math.Sqrt(n); factor += 2) { // Find highest power of "factor" // that divides n int power = 0; while (n % factor == 0) { n = n / factor; power++; } // Check if only factor^1 divides n, // then return false if (power == 1) return false; } // n must be 1 now // if it is not a prime number. // Since prime numbers are not powerful, // we return false if n is not 1. return (n == 1); } // Function to perform dfs static void dfs(int node, int parent) { // Check if weight of the current node // is a powerful number if (isPowerful(weight[node])) ans += 1; foreach (int to in graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code public static void Main(String[] args) { for(int i = 0; i < graph.Length; i++) graph[i] = new List<int>(); // Weights of the node weight[1] = 5; weight[2] = 10; weight[3] = 11; weight[4] = 8; weight[5] = 6; // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); dfs(1, 1); Console.Write(ans); } } // This code is contributed by amal kumar choubey |
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Output:
1
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