Dados dos enteros A y B. La tarea es contar cuántos números en el intervalo [ A, B ] tienen un número impar de divisores.
Ejemplos:
Input : A = 1, B = 10 Output : 3 Input : A = 5, B = 15 Output : 1
Enfoque ingenuo:
el enfoque simple sería iterar a través de todos los números entre el rango [A, B] y verificar si su número de divisores es impar.
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program to find count of numbers having
// odd number of divisors in given range
#include <bits/stdc++.h>
using namespace std;
// Function to count numbers having odd
// number of divisors in range [A, B]
int OddDivCount(int a, int b)
{
// variable to odd divisor count
int res = 0;
// iterate from a to b and count their
// number of divisors
for (int i = a; i <= b; ++i) {
// variable to divisor count
int divCount = 0;
for (int j = 1; j <= i; ++j) {
if (i % j == 0) {
++divCount;
}
}
// if count of divisor is odd
// then increase res by 1
if (divCount % 2) {
++res;
}
}
return res;
}
// Driver code
int main()
{
int a = 1, b = 10;
cout << OddDivCount(a, b) << endl;
return 0;
}
Java
// Java program to find count of numbers having
// odd number of divisors in given range
import java.io.*;
class GFG {
// Function to count numbers having odd
// number of divisors in range [A, B]
static int OddDivCount(int a, int b)
{
// variable to odd divisor count
int res = 0;
// iterate from a to b and count their
// number of divisors
for (int i = a; i <= b; ++i) {
// variable to divisor count
int divCount = 0;
for (int j = 1; j <= i; ++j) {
if (i % j == 0) {
++divCount;
}
}
// if count of divisor is odd
// then increase res by 1
if ((divCount % 2) != 0) {
++res;
}
}
return res;
}
// Driver code
public static void main(String[] args)
{
int a = 1, b = 10;
System.out.println(OddDivCount(a, b));
}
// This code is contributed by ajit.
}
Python3
# Python3 program to find count # of numbers having odd number # of divisors in given range # Function to count numbers # having odd number of divisors # in range [A, B] def OddDivCount(a, b): # variable to odd divisor count res = 0 # iterate from a to b and count # their number of divisors for i in range(a, b + 1) : # variable to divisor count divCount = 0 for j in range(1, i + 1) : if (i % j == 0) : divCount += 1 # if count of divisor is odd # then increase res by 1 if (divCount % 2) : res += 1 return res # Driver code if __name__ == "__main__": a = 1 b = 10 print(OddDivCount(a, b)) # This code is contributed # by ChitraNayal
C#
// C# program to find count of numbers having
// odd number of divisors in given range
using System;
class Geeks {
// Function to count numbers having odd
// number of divisors in range [A, B]
static int OddDivCount(int a, int b)
{
// variable to odd divisor count
int res = 0;
// iterate from a to b and count their
// number of divisors
for (int i = a; i <= b; ++i) {
// variable to divisor count
int divCount = 0;
for (int j = 1; j <= i; ++j) {
if (i % j == 0) {
++divCount;
}
}
// if count of divisor is odd
// then increase res by 1
if ((divCount % 2) != 0) {
++res;
}
}
return res;
}
// Driver code
public static void Main(String[] args)
{
int a = 1, b = 10;
Console.WriteLine(OddDivCount(a, b));
}
}
PHP
<?php
// PHP program to find count of
// numbers having odd number of
// divisors in given range
// Function to count numbers having odd
// number of divisors in range [A, B]
function OddDivCount($a, $b)
{
// variable to odd divisor count
$res = 0;
// iterate from a to b and count
// their number of divisors
for ($i = $a; $i <= $b; ++$i)
{
// variable to divisor count
$divCount = 0;
for ($j = 1; $j <= $i; ++$j)
{
if ($i % $j == 0)
{
++$divCount;
}
}
// if count of divisor is odd
// then increase res by 1
if ($divCount % 2)
{
++$res;
}
}
return $res;
}
// Driver code
$a = 1;
$b = 10;
echo OddDivCount($a, $b) ;
// This code is contributed
// by Shivi_Aggarwal
?>
Javascript
<script>
// Javascript program to find count of
// numbers having odd number of divisors
// in given range
// Function to count numbers having odd
// number of divisors in range [A, B]
function OddDivCount(a, b)
{
// Variable to odd divisor count
let res = 0;
// Iterate from a to b and count their
// number of divisors
for(let i = a; i <= b; ++i)
{
// Variable to divisor count
let divCount = 0;
for(let j = 1; j <= i; ++j)
{
if (i % j == 0)
{
++divCount;
}
}
// If count of divisor is odd
// then increase res by 1
if ((divCount % 2) != 0)
{
++res;
}
}
return res;
}
// Driver code
let a = 1, b = 10;
document.write(OddDivCount(a, b));
// This code is contributed by suresh07
</script>
Producción:
3
Complejidad temporal: O(n 2 )
Espacio Auxiliar: O(1)
Mejor enfoque:
un número puede representarse mediante el producto de sus factores primos con las potencias adecuadas. Esas potencias se pueden usar para obtener la cantidad de factores que tiene un número entero. Si el número es num y se puede representar como (a p1 ) * (b p2 ) * (c p3 )
Entonces la cuenta de factores de num es (p1 + 1) * (p2 + 1) * (p3 + 1)
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count
// of divisors of a number
int divisor(int a)
{
int div = 1, count = 0;
for (int i = 2; i <= sqrt(a); i++) {
// Count the powers of the current
// prime i which divides a
while (a % i == 0) {
count++;
a = a / i;
}
// Update the count of divisors
div = div * (count + 1);
// Reset the count
count = 0;
}
// If the remaining a is prime then a^1
// will be one of its prime factors
if (a > 1) {
div = div * (2);
}
return div;
}
// Function to count numbers having odd
// number of divisors in range [A, B]
int OddDivCount(int a, int b)
{
// To store the count of elements
// having odd number of divisors
int res = 0;
// Iterate from a to b and find the
// count of their divisors
for (int i = a; i <= b; ++i) {
// To store the count of divisors of i
int divCount = divisor(i);
// If the divisor count of i is odd
if (divCount % 2) {
++res;
}
}
return res;
}
// Driver code
int main()
{
int a = 1, b = 10;
cout << OddDivCount(a, b);
return 0;
}
// This code is contributed by jrolofmeister
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the count
// of divisors of a number
static int divisor(int a)
{
int div = 1, count = 0;
for (int i = 2; i <= Math.sqrt(a); i++)
{
// Count the powers of the current
// prime i which divides a
while (a % i == 0)
{
count++;
a = a / i;
}
// Update the count of divisors
div = div * (count + 1);
// Reset the count
count = 0;
}
// If the remaining a is prime then a^1
// will be one of its prime factors
if (a > 1)
{
div = div * (2);
}
return div;
}
// Function to count numbers having odd
// number of divisors in range [A, B]
static int OddDivCount(int a, int b)
{
// To store the count of elements
// having odd number of divisors
int res = 0;
// Iterate from a to b and find the
// count of their divisors
for (int i = a; i <= b; ++i)
{
// To store the count of divisors of i
int divCount = divisor(i);
// If the divisor count of i is odd
if (divCount % 2 == 1)
{
++res;
}
}
return res;
}
// Driver code
public static void main(String[] args)
{
int a = 1, b = 10;
System.out.println(OddDivCount(a, b));
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 implementation of the approach # Function to return the count # of divisors of a number def divisor(a): div = 1; count = 0; for i in range(2, int(pow(a, 1 / 2)) + 1): # Count the powers of the current # prime i which divides a while (a % i == 0): count += 1; a = a / i; # Update the count of divisors div = div * (count + 1); # Reset the count count = 0; # If the remaining a is prime then a^1 # will be one of its prime factors if (a > 1): div = div * (2); return div; # Function to count numbers having odd # number of divisors in range [A, B] def OddDivCount(a, b): # To store the count of elements # having odd number of divisors res = 0; # Iterate from a to b and find the # count of their divisors for i in range(a, b + 1): # To store the count of divisors of i divCount = divisor(i); # If the divisor count of i is odd if (divCount % 2): res += 1; return res; # Driver code if __name__ == '__main__': a, b = 1, 10; print(OddDivCount(a, b)); # This code is contributed by PrinciRaj1992
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the count
// of divisors of a number
static int divisor(int a)
{
int div = 1, count = 0;
for (int i = 2;
i <= Math.Sqrt(a); i++)
{
// Count the powers of the current
// prime i which divides a
while (a % i == 0)
{
count++;
a = a / i;
}
// Update the count of divisors
div = div * (count + 1);
// Reset the count
count = 0;
}
// If the remaining a is prime then a^1
// will be one of its prime factors
if (a > 1)
{
div = div * (2);
}
return div;
}
// Function to count numbers having odd
// number of divisors in range [A, B]
static int OddDivCount(int a, int b)
{
// To store the count of elements
// having odd number of divisors
int res = 0;
// Iterate from a to b and find the
// count of their divisors
for (int i = a; i <= b; ++i)
{
// To store the count of divisors of i
int divCount = divisor(i);
// If the divisor count of i is odd
if (divCount % 2 == 1)
{
++res;
}
}
return res;
}
// Driver code
public static void Main(String[] args)
{
int a = 1, b = 10;
Console.WriteLine(OddDivCount(a, b));
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript implementation of the approach
// Function to return the count
// of divisors of a number
function divisor(a)
{
let div = 1, count = 0;
for (let i = 2; i <= Math.sqrt(a); i++)
{
// Count the powers of the current
// prime i which divides a
while (a % i == 0)
{
count++;
a = parseInt(a / i, 10);
}
// Update the count of divisors
div = div * (count + 1);
// Reset the count
count = 0;
}
// If the remaining a is prime then a^1
// will be one of its prime factors
if (a > 1)
{
div = div * (2);
}
return div;
}
// Function to count numbers having odd
// number of divisors in range [A, B]
function OddDivCount(a, b)
{
// To store the count of elements
// having odd number of divisors
let res = 0;
// Iterate from a to b and find the
// count of their divisors
for (let i = a; i <= b; ++i)
{
// To store the count of divisors of i
let divCount = divisor(i);
// If the divisor count of i is odd
if (divCount % 2 == 1)
{
++res;
}
}
return res;
}
let a = 1, b = 10;
document.write(OddDivCount(a, b));
</script>
Producción:
3
Complejidad de tiempo: O(n * logn)
Consulte este artículo para conocer un enfoque O(1).
Publicación traducida automáticamente
Artículo escrito por saurabh_shukla y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA