Un número altamente totiente k es un entero que tiene más soluciones a la ecuación φ(x) = k, donde φ es la función totiente de Euler
La secuencia:
1, 2, 4, 8, 12, 24, 48, 72, 144, 240, 432, 480, 576, 720, 1152, 1440, 2880, 4320, 5760, 8640
Explicación :
- 1 tiene 2 soluciones
- 2 tiene 3 soluciones
- 4 tiene 4 soluciones
- 8 tiene 5 soluciones
- 12 tiene 6 soluciones
- 24 tiene 10 soluciones
Para un N dado , la tarea es imprimir los primeros N números altamente útiles.
Ejemplos:
Entrada: N = 10
Salida: 1, 2, 4, 8, 12, 24, 48, 72, 144, 240
Entrada: N = 20
Salida: 1, 2, 4, 8, 12, 24, 48, 72, 144 , 240, 432, 480, 576, 720, 1152, 1440, 2880, 4320, 5760, 8640
Enfoque:
Un enfoque eficiente es almacenar todos los valores de φ(x) hasta 10 5 en un mapa junto con sus frecuencias y luego ejecutar un bucle desde 1 hasta que el recuento del número de pacientes sea menor que N . Para cada i , verificaremos si la frecuencia de i es mayor que el elemento anterior, si es así, imprima el número y aumente el conteo; de lo contrario, incremente el número.
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program to find highly totient numbers
#include <bits/stdc++.h>
using namespace std;
// Function to find euler totient number
int phi(int n)
{
int result = n; // Initialize result as n
// Consider all prime factors of n and
// subtract their multiples from result
for (int p = 2; p * p <= n; ++p) {
// Check if p is a prime factor.
if (n % p == 0) {
// If yes, then update n and result
while (n % p == 0)
n /= p;
result -= result / p;
}
}
// If n has a prime factor greater than sqrt(n)
// (There can be at-most one such prime factor)
if (n > 1)
result -= result / n;
return result;
}
// Function to find first n highly totient numbers
void Highly_Totient(int n)
{
// Count of Highly totient numbers
// and value of count of phi of previous numbers
int count = 0, p_count = -1, i = 1;
// Store all the values of phi(x) upto
// 10^5 with frequencies
map<int, int> mp;
for (int i = 1; i < 100000; i++)
mp[phi(i)]++;
while (count < n) {
// If count is greater than count of
// previous element
if (mp[i] > p_count)
{
// Display the number
cout << i;
if(count < n-1)
cout << ", ";
// Store the value of phi
p_count = mp[i];
count++;
}
i++;
}
}
// Driver code
int main()
{
int n = 20;
// Function call
Highly_Totient(n);
return 0;
}
Java
// Java program to find highly totient numbers
import java.util.*;
class GFG
{
// Function to find euler totient number
static int phi(int n)
{
int result = n; // Initialize result as n
// Consider all prime factors of n and
// subtract their multiples from result
for (int p = 2; p * p <= n; ++p)
{
// Check if p is a prime factor.
if (n % p == 0)
{
// If yes, then update n and result
while (n % p == 0)
n /= p;
result -= result / p;
}
}
// If n has a prime factor greater than sqrt(n)
// (There can be at-most one such prime factor)
if (n > 1)
result -= result / n;
return result;
}
// Function to find first n highly totient numbers
static void Highly_Totient(int n)
{
// Count of Highly totient numbers
// and value of count of phi of previous numbers
int count = 0, p_count = -1;
// Store all the values of phi(x) upto
// 10^5 with frequencies
HashMap<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
for (int i = 1; i < 100000; i++)
{
if(mp.containsKey(phi(i)))
{
mp.put(phi(i), mp.get(phi(i)) + 1);
}
else
{
mp.put(phi(i), 1);
}
}
int i = 1;
while (count < n)
{
// If count is greater than count of
// previous element
if (mp.containsKey(i)&&mp.get(i) > p_count)
{
// Display the number
System.out.print(i);
if(count < n - 1)
System.out.print(", ");
// Store the value of phi
p_count = mp.get(i);
count++;
}
i++;
}
}
// Driver code
public static void main(String[] args)
{
int n = 20;
// Function call
Highly_Totient(n);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program to find highly totient numbers
# Function to find euler totient number
def phi(n):
result = n; # Initialize result as n
# Consider all prime factors of n and
# subtract their multiples from result
p = 2
while(p*p <= n):
# Check if p is a prime factor.
if (n % p == 0):
# If yes, then update n and result
while (n % p == 0):
n //= p;
result -= (result // p);
p += 1
# If n has a prime factor greater than sqrt(n)
# (There can be at-most one such prime factor)
if (n > 1):
result -= (result // n);
return result;
# Function to find first n highly totient numbers
def Highly_Totient(n):
# Count of Highly totient numbers
# and value of count of phi of previous numbers
count = 0
p_count = -1
# Store all the values of phi(x) upto
# 10^5 with frequencies
mp = dict()
i = 1
while i < 100000:
tmp = phi(i)
if tmp not in mp:
mp[tmp] = 0
mp[tmp] += 1;
i += 1
i = 1
while (count < n):
# If count is greater than count of
# previous element
if ((i in mp) and mp[i] > p_count):
# Display the number
print(i, end = '');
if(count < n - 1):
print(", ", end = '');
# Store the value of phi
p_count = mp[i];
count += 1
i += 1
# Driver code
if __name__=='__main__':
n = 20;
# Function call
Highly_Totient(n);
# This code is contributed by rutvik_56
C#
// C# program to find highly totient numbers
using System;
using System.Collections.Generic;
class GFG
{
// Function to find euler totient number
static int phi(int n)
{
int result = n; // Initialize result as n
// Consider all prime factors of n and
// subtract their multiples from result
for (int p = 2; p * p <= n; ++p)
{
// Check if p is a prime factor.
if (n % p == 0)
{
// If yes, then update n and result
while (n % p == 0)
n /= p;
result -= result / p;
}
}
// If n has a prime factor greater than sqrt(n)
// (There can be at-most one such prime factor)
if (n > 1)
result -= result / n;
return result;
}
// Function to find first n highly totient numbers
static void Highly_Totient(int n)
{
// Count of Highly totient numbers
// and value of count of phi of
// previous numbers
int count = 0, p_count = -1, i;
// Store all the values of phi(x) upto
// 10^5 with frequencies
Dictionary<int,
int> mp = new Dictionary<int,
int>();
for (i = 1; i < 100000; i++)
{
if(mp.ContainsKey(phi(i)))
{
mp[phi(i)] = mp[phi(i)] + 1;
}
else
{
mp.Add(phi(i), 1);
}
}
i = 1;
while (count < n)
{
// If count is greater than count of
// previous element
if (mp.ContainsKey(i)&&mp[i] > p_count)
{
// Display the number
Console.Write(i);
if(count < n - 1)
Console.Write(", ");
// Store the value of phi
p_count = mp[i];
count++;
}
i++;
}
}
// Driver code
public static void Main(String[] args)
{
int n = 20;
// Function call
Highly_Totient(n);
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// javascript program to find highly totient numbers
// Function to find euler totient number
function phi(n) {
var result = n; // Initialize result as n
// Consider all prime factors of n and
// subtract their multiples from result
for (var p = 2; p * p <= n; ++p) {
// Check if p is a prime factor.
if (n % p == 0) {
// If yes, then update n and result
while (n % p == 0)
n /= p;
result -= result / p;
}
}
// If n has a prime factor greater than sqrt(n)
// (There can be at-most one such prime factor)
if (n > 1)
result -= result / n;
return result;
}
// Function to find first n highly totient numbers
function Highly_Totient(n)
{
// Count of Highly totient numbers
// and value of count of phi of previous numbers
var count = 0, p_count = -1;
// Store all the values of phi(x) upto
// 10^5 with frequencies
var mp = new Map();
for (i = 1; i < 100000; i++) {
if (mp.has(phi(i))) {
mp.set(phi(i), mp.get(phi(i)) + 1);
} else {
mp.set(phi(i), 1);
}
}
var i = 1;
while (count < n) {
// If count is greater than count of
// previous element
if (mp.has(i) && mp.get(i) > p_count)
{
// Display the number
document.write(i);
if (count < n - 1)
document.write(", ");
// Store the value of phi
p_count = mp.get(i);
count++;
}
i++;
}
}
// Driver code
var n = 20;
// Function call
Highly_Totient(n);
// This code is contributed by gauravrajput1
</script>
Producción:
1, 2, 4, 8, 12, 24, 48, 72, 144, 240, 432, 480, 576, 720, 1152, 1440, 2880, 4320, 5760, 8640
Complejidad de tiempo: O (N), ya que estamos usando un bucle para atravesar N veces. [Como O(N) > O(sqrt(N)*logN), ya que usamos bucles anidados para atravesar sqrt(N)*logN veces]
Espacio auxiliar: O(100000), ya que estamos usando espacio extra para el mapa.
Este método no se puede utilizar para encontrar más de 1000 Número de pacientes altamente.
Publicación traducida automáticamente
Artículo escrito por andrew1234 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA