计算给定树的权重以X为因子的节点

给定一棵树，以及所有节点的权重，任务是计算权重可被x整除的节点。

例子：

Input:

x = 5
Output: 2
Only the nodes 1 and 2 have weights divisible by 5.

方法：在树上执行dfs，对于每个节点，检查其权重是否可被x整除。如果是，则增加计数。

执行：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
long ans = 0;
int x;
vector graph[100];
vector weight(100);
  
// Function to perform dfs
void dfs(int node, int parent)
{
  
    // If weight of the current node
    // is divisible by x
    if (weight[node] % x == 0)
        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] = 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;
}

Java

// Java implementation of the approach 
import java.util.*;
  
class GFG
{
      
static long ans = 0; 
static int x; 
static Vector> graph=new Vector>(); 
static Vector weight=new Vector(); 
  
// Function to perform dfs 
static void dfs(int node, int parent) 
{ 
  
    // If weight of the current node 
    // is divisible by x 
    if (weight.get(node) % x == 0) 
        ans += 1; 
  
    for (int i = 0; i < graph.get(node).size(); i++) 
    { 
        if (graph.get(node).get(i) == parent) 
            continue; 
        dfs(graph.get(node).get(i), node); 
    } 
} 
  
// Driver code 
public static void main(String args[])
{ 
    x = 5; 
  
    // Weights of the node 
    weight.add(0); 
    weight.add(5); 
    weight.add(10);; 
    weight.add(11);; 
    weight.add(8); 
    weight.add(6); 
      
    for(int i = 0; i < 100; i++)
    graph.add(new Vector());
  
    // Edges of the tree 
    graph.get(1).add(2); 
    graph.get(2).add(3); 
    graph.get(2).add(4); 
    graph.get(1).add(5); 
  
    dfs(1, 1); 
  
    System.out.println(ans); 
}
} 
  
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach 
ans = 0
  
graph = [[] for i in range(100)]
weight = [0] * 100
  
# Function to perform dfs 
def dfs(node, parent):
    global ans,x
      
    # If weight of the current node 
    # is divisible by x 
    if (weight[node] % x == 0):
        ans += 1
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
  
# Driver code 
x = 5
  
# 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].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
  
dfs(1, 1)
print(ans)
  
# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG
{
      
static long ans = 0; 
static int x; 
static List> graph = new List>(); 
static List weight = new List(); 
  
// Function to perform dfs 
static void dfs(int node, int parent) 
{ 
  
    // If weight of the current node 
    // is divisible by x 
    if (weight[node] % x == 0) 
        ans += 1; 
  
    for (int i = 0; i < graph[node].Count; i++) 
    { 
        if (graph[node][i] == parent) 
            continue; 
        dfs(graph[node][i], node); 
    } 
} 
  
// Driver code 
public static void Main(String []args)
{ 
    x = 5; 
  
    // Weights of the node 
    weight.Add(0); 
    weight.Add(5); 
    weight.Add(10);; 
    weight.Add(11);; 
    weight.Add(8); 
    weight.Add(6); 
      
    for(int i = 0; i < 100; i++)
    graph.Add(new List());
  
    // Edges of the tree 
    graph[1].Add(2); 
    graph[2].Add(3); 
    graph[2].Add(4); 
    graph[1].Add(5); 
  
    dfs(1, 1); 
  
    Console.WriteLine(ans); 
}
}
  
// This code contributed by Rajput-Ji

输出：

2

复杂度分析：

• 时间复杂度： O(N)。
  在DFS中，树的每个节点都处理一次，因此，当树中总共有N个节点时，由于DFS导致的复杂度为O(N)。因此，时间复杂度为O(N)。
• 辅助空间： O(1)。
  不需要任何额外的空间，因此空间复杂度是恒定的。