给定加权树,任务是计算权重数字总和为质数的节点数。
例子:
Input:
Output: 2
Explanation:
Node 1: digitSum(144) = 1 + 4 + 4 = 9
Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10
Node 3: digitSum(21) = 2 + 1 = 3
Node 4: digitSum(5) = 5
Node 5: digitSum(77) = 7 + 7 = 14
Only the sum of digits of the weights of nodes 3 and 4 are prime.
方法:为解决上述问题,我们必须对树和每个节点执行DFS,并检查其权重的数字总和是否为质数。如果是,则增加计数。要检查数字总和是否为质数,我们将使用Eratosthenes筛。创建一个筛子,这将有助于我们确定度数是否在O(1)时间内为素数。
下面是上述方法的实现:
C++
// C++ program to Count Nodes
// which has Prime Digit
// Sum Weight in a Tree
#include
using namespace std;
int MAX = 1000000;
int ans = 0;
vector graph[100];
vector weight(100);
// Function to create Sieve
// to check primes
void SieveOfEratosthenes(
bool prime[], int p_size)
{
// false here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for (int i = p * 2;
i <= p_size;
i += p)
prime[i] = false;
}
}
}
// Function to return the
// sum of the digits of n
int digitSum(int n)
{
int sum = 0;
while (n) {
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to perform dfs
void dfs(int node,
int parent,
bool prime[])
{
// If sum of the digits
// of current node's weight
// is prime then increment ans
int sum = digitSum(weight[node]);
if (prime[sum])
ans += 1;
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node, prime);
}
}
// Driver code
int main()
{
// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
// Edges of the tree
graph[1].push_back(2);
graph[2].push_back(3);
graph[2].push_back(4);
graph[1].push_back(5);
bool prime[MAX];
memset(prime, true, sizeof(prime));
SieveOfEratosthenes(prime, MAX);
dfs(1, 1, prime);
cout << ans;
return 0;
}
Java
// Java program to Count Nodes
// which has Prime Digit
// Sum Weight in a Tree
import java.util.*;
class GFG{
static int MAX = 1000000;
static int ans = 0;
static Vector []graph =
new Vector[100];
static int []weight = new int[100];
// Function to create Sieve
// to check primes
static void SieveOfEratosthenes(boolean prime[],
int p_size)
{
// false here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// Update all multiples of p,
// set them to non-prime
for (int i = p * 2;
i < p_size; i += p)
prime[i] = false;
}
}
}
// Function to return the
// sum of the digits of n
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to perform dfs
static void dfs(int node,
int parent,
boolean prime[])
{
// If sum of the digits
// of current node's weight
// is prime then increment ans
int sum = digitSum(weight[node]);
if (prime[sum])
ans += 1;
for (int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node, prime);
}
}
// Driver code
public static void main(String[] args)
{
// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
for (int i = 0; i < graph.length; i++)
graph[i] = new Vector();
// Edges of the tree
graph[1].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
boolean []prime = new boolean[MAX];
Arrays.fill(prime, true);
SieveOfEratosthenes(prime, MAX);
dfs(1, 1, prime);
System.out.print(ans);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python program to Count Nodes
# which has Prime Digit
# Sum Weight in a Tree
from typing import List
MAX = 1000000
ans = 0
graph = [[] for _ in range(100)]
weight = [0 for _ in range(100)]
# Function to create Sieve
# to check primes
def SieveOfEratosthenes(prime: List[bool], p_size: int) -> None:
# false here indicates
# that it is not prime
prime[0] = False
prime[1] = False
p = 2
while p * p <= p_size:
# If prime[p] is not changed,
# then it is a prime
if (prime[p]):
# Update all multiples of p,
# set them to non-prime
for i in range(p * 2, p_size + 1, p):
prime[i] = False
p += 1
# Function to return the
# sum of the digits of n
def digitSum(n: int) -> int:
sum = 0
while (n):
sum += n % 10
n = n // 10
return sum
# Function to perform dfs
def dfs(node: int, parent: int, prime: List[bool]) -> None:
global ans
# If sum of the digits
# of current node's weight
# is prime then increment ans
sum = digitSum(weight[node])
if (prime[sum]):
ans += 1
for to in graph[node]:
if (to == parent):
continue
dfs(to, node, prime)
# Driver code
if __name__ == "__main__":
# Weights of the node
weight[1] = 144
weight[2] = 1234
weight[3] = 21
weight[4] = 5
weight[5] = 77
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
prime = [True for _ in range(MAX + 1)]
SieveOfEratosthenes(prime, MAX)
dfs(1, 1, prime)
print(ans)
# This code is contributed by sanjeev2552
C#
// C# program to Count Nodes
// which has Prime Digit
// Sum Weight in a Tree
using System;
using System.Collections.Generic;
class GFG{
static int MAX = 1000000;
static int ans = 0;
static List []graph =
new List[100];
static int []weight = new int[100];
// Function to create Sieve
// to check primes
static void SieveOfEratosthenes(bool []prime,
int p_size)
{
// false here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// Update all multiples of p,
// set them to non-prime
for (int i = p * 2;
i < p_size; i += p)
prime[i] = false;
}
}
}
// Function to return the
// sum of the digits of n
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to perform dfs
static void dfs(int node,
int parent,
bool []prime)
{
// If sum of the digits
// of current node's weight
// is prime then increment ans
int sum = digitSum(weight[node]);
if (prime[sum])
ans += 1;
foreach (int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node, prime);
}
}
// Driver code
public static void Main(String[] args)
{
// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
for (int i = 0; i < graph.Length; i++)
graph[i] = new List();
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
bool []prime = new bool[MAX];
for (int i = 0; i < prime.Length; i++)
prime[i] = true;
SieveOfEratosthenes(prime, MAX);
dfs(1, 1, prime);
Console.Write(ans);
}
}
// This code is contributed by Rajput-Ji
Javascript
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,如果树中总共有N个节点,则由于DFS导致的复杂度为O(N)。同样,为了处理每个节点,使用了SieveOfEratosthenes()函数,该函数的复杂度也为O(sqrt(N)),但由于此函数仅执行一次,因此不会影响整体时间复杂度。因此,时间复杂度为O(N)。 - 辅助空间: O(N)。
素数数组使用了额外的空间,因此空间复杂度为O(N)。
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