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📜  计算与 X 之和为斐波那契数的节点

📅  最后修改于: 2022-05-13 01:57:16.146000             🧑  作者: Mango

计算与 X 之和为斐波那契数的节点

给定一棵树,所有节点的权重和一个整数X ,任务是计算所有节点i使得(weight[i] + X)是一个斐波那契数。
首先,很少有斐波那契数是:

例子:

方法:在树上执行dfs,计算所有节点的权重与x为斐波那契数的总和。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
int ans = 0, x;
 
vector graph[100];
vector 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;
}


Java
// Java implementation of the
// above approach
import java.util.*;
 
@SuppressWarnings("unchecked")
 
class GFG{
     
static int ans = 0, x;
 
static ArrayList []graph = new ArrayList[100];
static ArrayList weight = new ArrayList();
   
// Function that returns true if
// x is a perfect square
static boolean isPerfectSquare(double x)
{
   
  // Find floating point value of
  // square root of x
  double sr = Math.sqrt(x);
   
  // If square root is an integer
  return ((sr - Math.floor(sr)) == 0);
}
   
// Function that returns true
// if n is a fibonacci number
static boolean isFibonacci(int n)
{
  return isPerfectSquare(5 * n * n + 4) ||
         isPerfectSquare(5 * n * n - 4);
}
   
// Function to perform dfs
static void dfs(int node, int parent)
{
   
  // If weight of the current node
  // gives a fibonacci number
  // when x is added to it
  if (isFibonacci((int)weight.get(node) + x))
    ans += 1;
  
  for(int to : (ArrayList)graph[node])
  {
    if (to == parent)
      continue;
     
    dfs(to, node);
  }
}
   
// Driver Code
public static void main(String[] args)
{
  x = 5;
   
  for(int i = 0; i < 100; i++)
  {
    weight.add(0);
    graph[i] = new ArrayList();
  }
   
  // Weights of the node
  weight.add(1, 4);
  weight.add(2, 5);
  weight.add(3, 3);
  weight.add(4, 25);
  weight.add(5, 16);
  weight.add(6, 34);
   
  // Edges of the tree
  graph[1].add(2);
  graph[2].add(3);
  graph[2].add(4);
  graph[1].add(5);
  graph[5].add(6);
  
  dfs(1, 1);
   
  System.out.println(ans);
}
}
 
// This code is contributed by pratham76


Python3
# Python3 implementation of the approach
import math
ans, x = 0, 0
  
graph = [[] for i in range(100)]
weight = [0]*(100)
  
# Function that returns true if
# x is a perfect square
def isPerfectSquare(x):
   
    # Find floating point value of
    # square root of x
    sr = math.sqrt(x);
  
    # If square root is an integer
    return ((sr - math.floor(sr)) == 0)
  
# Function that returns true
# if n is a fibonacci number
def isFibonacci(n):
    return isPerfectSquare(5 * n * n + 4) or isPerfectSquare(5 * n * n - 4)
  
# Function to perform dfs
def dfs(node, parent):
    global ans
    # If weight of the current node
    # gives a fibonacci number
    # when x is added to it
    if (isFibonacci(weight[node] + x)):
        ans += 1
  
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
 
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].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
graph[5].append(6)
 
dfs(1, 1)
 
print(ans)
 
# This code is contributed by divyesh072019.


C#
// C# implementation of the
// above approach
using System;
using System.Collections;
class GFG{
      
static int ans = 0, x;
static ArrayList []graph = new ArrayList[100];
static ArrayList weight = new ArrayList();
  
// Function that returns true if
// x is a perfect square
static bool isPerfectSquare(double x)
{
  // Find floating point value of
  // square root of x
  double sr = Math.Sqrt(x);
 
  // If square root is an integer
  return ((sr - Math.Floor(sr)) == 0);
}
  
// Function that returns true
// if n is a fibonacci number
static bool isFibonacci(int n)
{
    return isPerfectSquare(5 * n * n + 4) ||
           isPerfectSquare(5 * n * n - 4);
}
  
// Function to perform dfs
static void dfs(int node, int parent)
{
  // If weight of the current node
  // gives a fibonacci number
  // when x is added to it
  if (isFibonacci((int)weight[node] + x))
    ans += 1;
 
  foreach (int to in graph[node])
  {
    if (to == parent)
      continue;
    dfs(to, node);
  }
}
  
// Driver Code
public static void Main(string[] args)
{
  x = 5;
 
  for(int i = 0; i < 100; i++)
  {
    weight.Add(0);
    graph[i] = new ArrayList();
  }
 
  // 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].Add(2);
  graph[2].Add(3);
  graph[2].Add(4);
  graph[1].Add(5);
  graph[5].Add(6);
 
  dfs(1, 1);
  Console.Write(ans);
}
}
 
// This code is contributed by rutvik_56


Javascript


输出:
2