Dados tres enteros a, b, n. Su tarea es imprimir el número de números entre a y b, incluyéndolos también que tienen n-divisores. Un número se llama n-divisor si tiene un total de n divisores, incluido el 1 y él mismo.
Ejemplos:
Input : a = 1, b = 7, n = 2 Output : 4 There are four numbers with 2 divisors in range [1, 7]. The numbers are 2, 3, 5, and 7.
Enfoque ingenuo:
el enfoque ingenuo es verificar todos los números entre a y b, cuántos de ellos son números n-divisor para hacer esto , averigüe el número de cada divisor para cada número . Si es igual a n, entonces es un número n-divisor
. Enfoque eficiente:
cualquier número se puede escribir en la forma de su descomposición en factores primos, siendo el número x y p1, p2…pm son los números primos que dividen x, por lo que x = p1 e1 * p2 e2 ….pm em donde e1, e2…em son los exponentes de los números primos p1, p2….pm. Entonces el número de divisores de x será (e1+1)*(e2+1)…*(em+1).
Ahora, la segunda observación es para números primos mayores que sqrt(x) su exponente no puede exceder 1. Probemos esto por contradicción, supongamos que hay un número primo P mayor que sqrt(x) y su exponente E en factorización prima de x es mayor que uno (E >= 2) entonces P^E sqrt(x) entonces P^E > (sqrt(x)) E y E >= 2 entonces P E siempre será mayor que x
La tercera observación es que el número de números primos es mayor que sqrt(x) en la descomposición en factores primos de x siempre será menor que igual a 1. Esto también se puede probar de manera similar por contradicción como arriba.
Ahora, para resolver este problema,
paso 1: aplique la criba de eratóstenes y calcule los números primos hasta sqrt (b).
Paso 2:Recorra cada número de a a b y calcule los exponentes de cada número primo en ese número dividiendo repetidamente ese número por el número primo y use la fórmula número de divisores (x) = (e1+1)*(e2+1)….(em +1).
Paso 3: Si después de dividir por todos los números primos menores que igual a la raíz cuadrada de ese número si el número > 1 esto significa que hay un número primo mayor que su raíz cuadrada que divide y su exponente siempre será uno como se demostró anteriormente.
C++
// C++ program to count numbers with n divisors
#include<bits/stdc++.h>
using namespace std;
// applying sieve of eratosthenes
void sieve(bool primes[], int x)
{
primes[1] = false;
// if a number is prime mark all its multiples
// as non prime
for (int i=2; i*i <= x; i++)
{
if (primes[i] == true)
{
for (int j=2; j*j <= x; j++)
primes[i*j] = false;
}
}
}
// function that returns numbers of number that have
// n divisors in range from a to b. x is sqrt(b) + 1.
int nDivisors(bool primes[], int x, int a, int b, int n)
{
// result holds number of numbers having n divisors
int result = 0;
// vector to hold all the prime numbers between 1
// ans sqrt(b)
vector <int> v;
for (int i = 2; i <= x; i++)
if (primes[i] == true)
v.push_back (i);
// Traversing all numbers in given range
for (int i=a; i<=b; i++)
{
// initialising temp as i
int temp = i;
// total holds the number of divisors of i
int total = 1;
int j = 0;
// we need to use that prime numbers that
// are less than equal to sqrt(temp)
for (int k = v[j]; k*k <= temp; k = v[++j])
{
// holds the exponent of k in prime
// factorization of i
int count = 0;
// repeatedly divide temp by k till it is
// divisible and accordingly increase count
while (temp%k == 0)
{
count++;
temp = temp/k;
}
// using the formula no.of divisors =
// (e1+1)*(e2+1)....
total = total*(count+1);
}
// if temp is not equal to 1 then there is
// prime number in prime factorization of i
// greater than sqrt(i)
if (temp != 1)
total = total*2;
// if i is a ndvisor number increase result
if (total == n)
result++;
}
return result;
}
// Returns count of numbers in [a..b] having
// n divisors.
int countNDivisors(int a, int b, int n)
{
int x = sqrt(b) + 1;
// primes[i] = true if i is a prime number
bool primes[x];
// initialising each number as prime
memset(primes, true, sizeof(primes));
sieve(primes, x);
return nDivisors(primes, x, a, b, n);
}
// driver code
int main()
{
int a = 1, b = 7, n = 2;
cout << countNDivisors(a, b, n);
return 0;
}
Java
// Java program to count numbers with n divisors
import java.util.*;
class GFG{
// applying sieve of eratosthenes
static void sieve(boolean[] primes, int x)
{
primes[1] = true;
// if a number is prime mark all its multiples
// as non prime
for (int i=2; i*i <= x; i++)
{
if (primes[i] == false)
{
for (int j=2; j*i <= x; j++)
primes[i*j] = true;
}
}
}
// function that returns numbers of number that have
// n divisors in range from a to b. x is sqrt(b) + 1.
static int nDivisors(boolean[] primes, int x, int a, int b, int n)
{
// result holds number of numbers having n divisors
int result = 0;
// vector to hold all the prime numbers between 1
// ans sqrt(b)
ArrayList<Integer> v=new ArrayList<Integer>();
for (int i = 2; i <= x; i++)
if (primes[i] == false)
v.add(i);
// Traversing all numbers in given range
for (int i=a; i<=b; i++)
{
// initialising temp as i
int temp = i;
// total holds the number of divisors of i
int total = 1;
int j = 0;
// we need to use that prime numbers that
// are less than equal to sqrt(temp)
for (int k = v.get(j); k*k <= temp; k = v.get(++j))
{
// holds the exponent of k in prime
// factorization of i
int count = 0;
// repeatedly divide temp by k till it is
// divisible and accordingly increase count
while (temp%k == 0)
{
count++;
temp = temp/k;
}
// using the formula no.of divisors =
// (e1+1)*(e2+1)....
total = total*(count+1);
}
// if temp is not equal to 1 then there is
// prime number in prime factorization of i
// greater than sqrt(i)
if (temp != 1)
total = total*2;
// if i is a ndvisor number increase result
if (total == n)
result++;
}
return result;
}
// Returns count of numbers in [a..b] having
// n divisors.
static int countNDivisors(int a, int b, int n)
{
int x = (int)Math.sqrt(b) + 1;
// primes[i] = true if i is a prime number
boolean[] primes=new boolean[x+1];
// initialising each number as prime
sieve(primes, x);
return nDivisors(primes, x, a, b, n);
}
// driver code
public static void main(String[] args)
{
int a = 1, b = 7, n = 2;
System.out.println(countNDivisors(a, b, n));
}
}
// This code is contributed by mits
Python3
# Python3 program to count numbers # with n divisors import math; # applying sieve of eratosthenes def sieve(primes, x): primes[1] = False; # if a number is prime mark all # its multiples as non prime i = 2; while (i * i <= x): if (primes[i] == True): j = 2; while (j * i <= x): primes[i * j] = False; j += 1; i += 1; # function that returns numbers of number # that have n divisors in range from a to b. # x is sqrt(b) + 1. def nDivisors(primes, x, a, b, n): # result holds number of numbers # having n divisors result = 0; # vector to hold all the prime # numbers between 1 and sqrt(b) v = []; for i in range(2, x + 1): if (primes[i]): v.append(i); # Traversing all numbers in given range for i in range(a, b + 1): # initialising temp as i temp = i; # total holds the number of # divisors of i total = 1; j = 0; # we need to use that prime numbers that # are less than equal to sqrt(temp) k = v[j]; while (k * k <= temp): # holds the exponent of k in prime # factorization of i count = 0; # repeatedly divide temp by k till it is # divisible and accordingly increase count while (temp % k == 0): count += 1; temp = int(temp / k); # using the formula no.of divisors = # (e1+1)*(e2+1).... total = total * (count + 1); j += 1; k = v[j]; # if temp is not equal to 1 then there is # prime number in prime factorization of i # greater than sqrt(i) if (temp != 1): total = total * 2; # if i is a ndivisor number # increase result if (total == n): result += 1; return result; # Returns count of numbers in [a..b] # having n divisors. def countNDivisors(a, b, n): x = int(math.sqrt(b) + 1); # primes[i] = true if i is a prime number # initialising each number as prime primes = [True] * (x + 1); sieve(primes, x); return nDivisors(primes, x, a, b, n); # Driver code a = 1; b = 7; n = 2; print(countNDivisors(a, b, n)); # This code is contributed by mits
C#
// C# program to count numbers with n divisors
using System.Collections;
using System;
class GFG{
// applying sieve of eratosthenes
static void sieve(bool[] primes, int x)
{
primes[1] = true;
// if a number is prime mark all its multiples
// as non prime
for (int i=2; i*i <= x; i++)
{
if (primes[i] == false)
{
for (int j=2; j*i <= x; j++)
primes[i*j] = true;
}
}
}
// function that returns numbers of number that have
// n divisors in range from a to b. x is sqrt(b) + 1.
static int nDivisors(bool[] primes, int x, int a, int b, int n)
{
// result holds number of numbers having n divisors
int result = 0;
// vector to hold all the prime numbers between 1
// ans sqrt(b)
ArrayList v=new ArrayList();
for (int i = 2; i <= x; i++)
if (primes[i] == false)
v.Add(i);
// Traversing all numbers in given range
for (int i=a; i<=b; i++)
{
// initialising temp as i
int temp = i;
// total holds the number of divisors of i
int total = 1;
int j = 0;
// we need to use that prime numbers that
// are less than equal to sqrt(temp)
for (int k = (int)v[j]; k*k <= temp; k = (int)v[++j])
{
// holds the exponent of k in prime
// factorization of i
int count = 0;
// repeatedly divide temp by k till it is
// divisible and accordingly increase count
while (temp%k == 0)
{
count++;
temp = temp/k;
}
// using the formula no.of divisors =
// (e1+1)*(e2+1)....
total = total*(count+1);
}
// if temp is not equal to 1 then there is
// prime number in prime factorization of i
// greater than sqrt(i)
if (temp != 1)
total = total*2;
// if i is a ndivisor number increase result
if (total == n)
result++;
}
return result;
}
// Returns count of numbers in [a..b] having
// n divisors.
static int countNDivisors(int a, int b, int n)
{
int x = (int)Math.Sqrt(b) + 1;
// primes[i] = true if i is a prime number
bool[] primes=new bool[x+1];
// initialising each number as prime
sieve(primes, x);
return nDivisors(primes, x, a, b, n);
}
// driver code
public static void Main()
{
int a = 1, b = 7, n = 2;
Console.WriteLine(countNDivisors(a, b, n));
}
}
// This code is contributed by mits
PHP
<?php
// PHP program to count numbers with n divisors
// applying sieve of eratosthenes
function sieve(&$primes, $x)
{
$primes[1] = false;
// if a number is prime mark all
// its multiples as non prime
for ($i = 2; $i * $i <= $x; $i++)
{
if ($primes[$i] == true)
{
for ($j = 2; $j * $i <= $x; $j++)
$primes[$i * $j] = false;
}
}
}
// function that returns numbers of number
// that have n divisors in range from a to
// b. x is sqrt(b) + 1.
function nDivisors($primes, $x, $a, $b, $n)
{
// result holds number of numbers
// having n divisors
$result = 0;
// vector to hold all the prime numbers
// between 1 ans sqrt(b)
$v = array();
for ($i = 2; $i <= $x; $i++)
if ($primes[$i] == true)
array_push($v, $i);
// Traversing all numbers in given range
for ($i = $a; $i <= $b; $i++)
{
// initialising temp as i
$temp = $i;
// total holds the number of
// divisors of i
$total = 1;
$j = 0;
// we need to use that prime numbers that
// are less than equal to sqrt(temp)
for ($k = $v[$j];
$k * $k <= $temp; $k = $v[++$j])
{
// holds the exponent of k in
// prime factorization of i
$count = 0;
// repeatedly divide temp by k till
// it is divisible and accordingly
// increase count
while ($temp % $k == 0)
{
$count++;
$temp = (int)($temp / $k);
}
// using the formula no.of divisors =
// (e1+1)*(e2+1)....
$total = $total * ($count + 1);
}
// if temp is not equal to 1 then there is
// prime number in prime factorization of i
// greater than sqrt(i)
if ($temp != 1)
$total = $total * 2;
// if i is a n divisor number increase result
if ($total == $n)
$result++;
}
return $result;
}
// Returns count of numbers in [a..b]
// having n divisors.
function countNDivisors($a, $b, $n)
{
$x = (int)(sqrt($b) + 1);
// primes[i] = true if i is a prime number
// initialising each number as prime
$primes = array_fill(0, $x + 1, true);
sieve($primes, $x);
return nDivisors($primes, $x, $a, $b, $n);
}
// Driver code
$a = 1;
$b = 7;
$n = 2;
print(countNDivisors($a, $b, $n));
// This code is contributed by mits
?>
Javascript
<script>
// Javascript program to count numbers with n divisors
// applying sieve of eratosthenes
function sieve(primes, x)
{
primes[1] = true;
// if a number is prime mark all its multiples
// as non prime
for (var i=2; i*i <= x; i++)
{
if (primes[i] == false)
{
for (var j=2; j*i <= x; j++)
primes[i*j] = true;
}
}
}
// function that returns numbers of number that have
// n divisors in range from a to b. x is sqrt(b) + 1.
function nDivisors(primes, x, a, b, n)
{
// result holds number of numbers having n divisors
var result = 0;
// vector to hold all the prime numbers between 1
// ans sqrt(b)
var v = [];
for (var i = 2; i <= x; i++)
if (primes[i] == false)
v.push(i);
// Traversing all numbers in given range
for (var i=a; i<=b; i++)
{
// initialising temp as i
var temp = i;
// total holds the number of divisors of i
var total = 1;
var j = 0;
// we need to use that prime numbers that
// are less than equal to sqrt(temp)
for (var k = v[j]; k*k <= temp; k = v[++j])
{
// holds the exponent of k in prime
// factorization of i
var count = 0;
// repeatedly divide temp by k till it is
// divisible and accordingly increase count
while (temp%k == 0)
{
count++;
temp = parseInt(temp/k);
}
// using the formula no.of divisors =
// (e1+1)*(e2+1)....
total = total*(count+1);
}
// if temp is not equal to 1 then there is
// prime number in prime factorization of i
// greater than sqrt(i)
if (temp != 1)
total = total*2;
// if i is a ndivisor number increase result
if (total == n)
result++;
}
return result;
}
// Returns count of numbers in [a..b] having
// n divisors.
function countNDivisors(a, b, n)
{
var x = parseInt(Math.sqrt(b)) + 1;
// primes[i] = true if i is a prime number
var primes = Array(x+1).fill(false);
// initialising each number as prime
sieve(primes, x);
return nDivisors(primes, x, a, b, n);
}
// driver code
var a = 1, b = 7, n = 2;
document.write(countNDivisors(a, b, n));
// This code is contributed by rutvik_56.
</script>
Producción:
4
Complejidad temporal: O(n)
Espacio auxiliar: O(n)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA