Dados dos números ‘num’ y ‘divisor’, encuentre el resto cuando ‘num’ se divide por ‘divisor’. No se permite el uso del operador módulo o %.
Ejemplos:
Input: num = 100, divisor = 7 Output: 2 Input: num = 30, divisor = 9 Output: 3
Método 1 :
C++
// C++ program to find remainder without using
// modulo operator
#include <iostream>
using namespace std;
// This function returns remainder of num/divisor
// without using % (modulo) operator
int getRemainder(int num, int divisor)
{
return (num - divisor * (num / divisor));
}
// Driver program to test above functions
int main()
{
// cout << 100 %0;
cout << getRemainder(100, 7);
return 0;
}
Java
// Java program to find remainder without
// using modulo operator
import java.io.*;
class GFG {
// This function returns remainder of
// num/divisor without using % (modulo)
// operator
static int getRemainder(int num, int divisor)
{
return (num - divisor * (num / divisor));
}
// Driver program to test above functions
public static void main(String[] args)
{
// print 100 % 0;
System.out.println(getRemainder(100, 7));
}
}
// This code is contributed by Sam007.
Python3
# Python program to find remainder # without using modulo operator # This function returns remainder of # num / divisor without using % (modulo) # operator def getRemainder(num, divisor): return (num - divisor * (num // divisor)) # Driver program to test above functions num = 100 divisor = 7 print(getRemainder(num, divisor)) # This code is contributed by Danish Raza
C#
// C# program to find remainder without using
// modulo operator
using System;
class GFG {
// This function returns remainder of
// num/divisor without using %
// (modulo) operator
static int getRemainder(int num, int divisor)
{
return (num - divisor * (num / divisor));
}
// Driver program to test above functions
public static void Main()
{
// print 100 % 0;
Console.Write(getRemainder(100, 7));
}
}
// This code is contributed by Sam007.
PHP
<?php
// PHP program to find remainder
// without using modulo operator
// This function returns remainder
// of num/divisor without using
// % (modulo) operator
function getRemainder($num, $divisor)
{
$t = ($num - $divisor *
(int)($num / $divisor));
return $t;
}
// Driver Code
echo getRemainder(100, 7);
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript program to find remainder
// without using modulo operator
// This function returns remainder
// of num/divisor without using
// % (modulo) operator
function getRemainder(num, divisor)
{
let t = (num - divisor *
parseInt(num / divisor));
return t;
}
// Driver Code
document.write(getRemainder(100, 7));
// This code is contributed by _saurabh_jaiswal
</script>
Producción :
2
Complejidad de tiempo: O(1)
Espacio auxiliar: O(1)
Este método es aportado por Bishal Kumar Dubey
Método 2
La idea es simple, ejecutamos un ciclo para encontrar el mayor múltiplo de ‘divisor’ que es menor o igual que ‘num’. Una vez que encontramos dicho múltiplo, restamos el múltiplo de ‘num’ para encontrar el divisor.
A continuación se muestra la implementación de la idea anterior. Gracias a eleventyone por sugerir esta solución en un comentario.
C++
// C++ program to find remainder without using modulo operator
#include <iostream>
using namespace std;
// This function returns remainder of num/divisor without
// using % (modulo) operator
int getRemainder(int num, int divisor)
{
// Handle divisor equals to 0 case
if (divisor == 0) {
cout << "Error: divisor can't be zero \n";
return -1;
}
// Handle negative values
if (divisor < 0)
divisor = -divisor;
if (num < 0)
num = -num;
// Find the largest product of 'divisor' that is smaller
// than or equal to 'num'
int i = 1;
int product = 0;
while (product <= num) {
product = divisor * i;
i++;
}
// return remainder
return num - (product - divisor);
}
// Driver program to test above functions
int main()
{
// cout << 100 %0;
cout << getRemainder(100, 7);
return 0;
}
Java
// Java program to find remainder without
// using modulo operator
import java.io.*;
class GFG {
// This function returns remainder
// of num/divisor without using %
// (modulo) operator
static int getRemainder(int num, int divisor)
{
// Handle divisor equals to 0 case
if (divisor == 0) {
System.out.println("Error: divisor "
+ "can't be zero \n");
return -1;
}
// Handle negative values
if (divisor < 0)
divisor = -divisor;
if (num < 0)
num = -num;
// Find the largest product of 'divisor'
// that is smaller than or equal to 'num'
int i = 1;
int product = 0;
while (product <= num) {
product = divisor * i;
i++;
}
// return remainder
return num - (product - divisor);
}
// Driver program to test above functions
public static void main(String[] args)
{
// print 100 % 0;
System.out.println(getRemainder(100, 7));
}
}
// This code is contributed by Sam007.
Python3
# Python program to find remainder without # using modulo operator. This function # returns remainder of num / divisor without # using % (modulo) operator def getRemainder(num, divisor): # Handle divisor equals to 0 case if (divisor == 0): return False # Handle negative values if (divisor < 0): divisor = -divisor if (num < 0): num = -num # Find the largest product of 'divisor' # that is smaller than or equal to 'num' i = 1 product = 0 while (product <= num): product = divisor * i i += 1 # return remainder return num - (product - divisor) # Driver program to test above functions num = 100 divisor = 7 print(getRemainder(num, divisor)) # This code is contributed by Danish Raza
C#
// C# program to find remainder without
// using modulo operator
using System;
class GFG {
// This function returns remainder
// of num/divisor without using %
// (modulo) operator
static int getRemainder(int num, int divisor)
{
// Handle divisor equals to 0 case
if (divisor == 0) {
Console.WriteLine("Error: divisor "
+ "can't be zero \n");
return -1;
}
// Handle negative values
if (divisor < 0)
divisor = -divisor;
if (num < 0)
num = -num;
// Find the largest product of 'divisor'
// that is smaller than or equal to 'num'
int i = 1;
int product = 0;
while (product <= num) {
product = divisor * i;
i++;
}
// return remainder
return num - (product - divisor);
}
// Driver program to test above functions
public static void Main()
{
// print 100 %0;
Console.Write(getRemainder(100, 7));
}
}
// This code is contributed by Sam007.
PHP
<?php
// php program to find remainder without
// using modulo operator
// This function returns remainder of
// num/divisor without using % (modulo)
// operator
function getRemainder($num, $divisor)
{
// Handle divisor equals to 0 case
if ($divisor == 0)
{
echo "Error: divisor can't be zero \n";
return -1;
}
// Handle negative values
if ($divisor < 0) $divisor = -$divisor;
if ($num < 0) $num = -$num;
// Find the largest product of 'divisor'
// that is smaller than or equal to 'num'
$i = 1;
$product = 0;
while ($product <= $num)
{
$product = $divisor * $i;
$i++;
}
// return remainder
return $num - ($product - $divisor);
}
// Driver program to test above functions
echo getRemainder(100, 7);
// This code is contributed by ajit.
?>
Javascript
// Javascript program to find remainder without
// using modulo operator
// This function returns remainder of
// num/divisor without using % (modulo)
// operator
function getRemainder(num, divisor)
{
// Handle divisor equals to 0 case
if (divisor == 0)
{
document.write("Error: divisor can't be zero <br>");
return -1;
}
// Handle negative values
if (divisor < 0) divisor = -divisor;
if (num < 0) num = -num;
// Find the largest product of 'divisor'
// that is smaller than or equal to 'num'
let i = 1;
let product = 0;
while (product <= num)
{
product = divisor * i;
i++;
}
// return remainder
return num - (product - divisor);
}
// Driver program to test above functions
document.write(getRemainder(100, 7));
// This code is contributed by _saurabh_jaiswal
Producción :
2
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
Este artículo es una contribución de Chirag. Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Método 3
Sigue restando el denominador del numerador hasta que el numerador sea menor que el denominador.
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return num % divisor
// without using % (modulo) operator
int getRemainder(int num, int divisor)
{
// While divisor is smaller
// than n, keep subtracting
// it from num
while (num >= divisor)
num -= divisor;
return num;
}
// Driver code
int main()
{
int num = 100, divisor = 7;
cout << getRemainder(num, divisor);
return 0;
}
Java
// A Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return num % divisor
// without using % (modulo) operator
static int getRemainder(int num, int divisor)
{
// While divisor is smaller
// than n, keep subtracting
// it from num
while (num >= divisor)
num -= divisor;
return num;
}
// Driver code
public static void main(String[] args)
{
int num = 100, divisor = 7;
System.out.println(getRemainder(num, divisor));
}
}
// This code is contributed by Princi Singh
Python3
# Python3 implementation of the approach # Function to return num % divisor # without using % (modulo) operator def getRemainder(num, divisor): # While divisor is smaller # than n, keep subtracting # it from num while (num >= divisor): num -= divisor; return num; # Driver code if __name__ == '__main__': num = 100; divisor = 7; print(getRemainder(num, divisor)); # This code is contributed by Princi Singh
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return num % divisor
// without using % (modulo) operator
static int getRemainder(int num, int divisor)
{
// While divisor is smaller
// than n, keep subtracting
// it from num
while (num >= divisor)
num -= divisor;
return num;
}
// Driver code
public static void Main(String[] args)
{
int num = 100, divisor = 7;
Console.WriteLine(getRemainder(num, divisor));
}
}
// This code is contributed by PrinciRaj1992
Javascript
// Javascript implementation of the approach
// Function to return num % divisor
// without using % (modulo) operator
function getRemainder(num, divisor)
{
// While divisor is smaller
// than n, keep subtracting
// it from num
while (num >= divisor)
num -= divisor;
return num;
}
// Driver code
let num = 100, divisor = 7;
document.write(getRemainder(num, divisor));
// This code is contributed by _saurabh_jaiswal
Producción :
2
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA