Dado un número entero 
. La tarea es encontrar todos los factores de N imprimir el producto de cuatro factores de N tal que:
- La suma de los cuatro factores es igual a N.
- El producto de los cuatro factores es máximo.
Si no es posible encontrar 4 de esos factores, imprima «No es posible» .
Nota: Los cuatro factores pueden ser iguales entre sí para maximizar el producto.
Ejemplos :
Input : 24 Output : Product -> 1296 All factors are -> 1 2 3 4 6 8 12 24 Choose the factor 6 four times, Therefore, 6+6+6+6 = 24 and product is maximum. Input : 100 Output : Product -> 390625 All the factors are -> 1 2 4 5 10 10 20 25 50 100 Choose the factor 25 four times.
En el segundo ejemplo, el producto será máximo cuando los cuatro factores sean iguales a 25. La suma de los cuatro factores es igual a ‘N’. Aunque podemos elegir el mismo factor cuatro veces para maximizar el producto.
A continuación se muestra el algoritmo paso a paso para resolver este problema:
- Primero encuentre los factores de un número ‘N’ pasando de 1 a la raíz cuadrada de ‘N’ y compruebe si ‘i’ y ‘n/i’ dividen a N y almacénelos en un vector.
- Ordene el vector e imprima cada elemento.
- Encuentra tres números para maximizar el producto con el cuarto número, usando tres bucles.
- Reemplace el siguiente producto máximo con el producto anterior.
- Imprime el producto cuando encuentres los cuatro factores.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find four factors of N
// with maximum product and sum equal to N
#include <bits/stdc++.h>
using namespace std;
// Function to find factors
// and to print those four factors
void findfactors(int n)
{
vector<int> vec;
// inserting all the factors in a vector s
for (int i = 1; i * i <= n; i++) {
if (n % i == 0) {
vec.push_back(i);
vec.push_back(n / i);
}
}
// sort the vector
sort(vec.begin(), vec.end());
// print all the factors
cout << "All the factors are -> ";
for (int i = 0; i < vec.size(); i++)
cout << vec[i] << " ";
cout << endl;
// Any elements is divisible by 1
int maxProduct = 1;
bool flag = 1;
// using three loop we'll find
// the three maximum factors
for (int i = 0; i < vec.size(); i++) {
for (int j = i; j < vec.size(); j++) {
for (int k = j; k < vec.size(); k++) {
// storing the fourth factor in y
int y = n - vec[i] - vec[j] - vec[k];
// if the fourth factor become negative
// then break
if (y <= 0)
break;
// we will replace more optimum number
// than the previous one
if (n % y == 0) {
flag = 0;
maxProduct = max(vec[i] * vec[j] * vec[k] * y,
maxProduct);
}
}
}
}
// print the product if the numbers exist
if (flag == 0)
cout << "Product is -> " << maxProduct << endl;
else
cout << "Not possible" << endl;
}
// Driver code
int main()
{
int n;
n = 50;
findfactors(n);
return 0;
}
Java
import java.util.Collections;
import java.util.Vector;
// Java program to find four factors of N
// with maximum product and sum equal to N
class GFG
{
// Function to find factors
// and to print those four factors
static void findfactors(int n)
{
Vector<Integer> vec = new Vector<Integer>();
// inserting all the factors in a vector s
for (int i = 1; i * i <= n; i++)
{
if (n % i == 0)
{
vec.add(i);
vec.add(n / i);
}
}
// sort the vector
Collections.sort(vec);
// print all the factors
System.out.println("All the factors are -> ");
for (int i = 0; i < vec.size(); i++)
{
System.out.print(vec.get(i) + " ");
}
System.out.println();
// Any elements is divisible by 1
int maxProduct = 1;
boolean flag = true;
// using three loop we'll find
// the three maximum factors
for (int i = 0; i < vec.size(); i++)
{
for (int j = i; j < vec.size(); j++)
{
for (int k = j; k < vec.size(); k++)
{
// storing the fourth factor in y
int y = n - vec.get(i) - vec.get(j) -
vec.get(k);
// if the fourth factor become negative
// then break
if (y <= 0)
{
break;
}
// we will replace more optimum number
// than the previous one
if (n % y == 0)
{
flag = false;
maxProduct = Math.max(vec.get(i) * vec.get(j) *
vec.get(k) * y, maxProduct);
}
}
}
}
// print the product if the numbers exist
if (flag == false)
{
System.out.println("Product is -> " + maxProduct);
}
else
{
System.out.println("Not possible");
}
}
// Driver code
public static void main(String[] args)
{
int n;
n = 50;
findfactors(n);
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python 3 to find four factors
# of N with maximum product and
# sum equal to N
# import every thing from
# math library
from math import *
# Function to find factors
# and to print those four factors
def findfactors(n) :
vec = []
# inserting all the factors
# in a list vec
for i in range(1, int(sqrt(n)) + 1) :
if n % i == 0 :
vec.append(i)
vec.append(n // i)
# sort the list
vec.sort()
print("All the factors are -> ",
end = "")
# print all the factors
for i in range(len(vec)) :
print(vec[i], end = " ")
print()
# Any elements is divisible by 1
maxProduct = 1
flag = 1
# using three loop we'll find
# the three maximum factors
for i in range(0, len(vec)) :
for j in range(i, len(vec)) :
for k in range(j, len(vec)) :
# storing the fourth factor in y
y = n - vec[i] - vec[j] - vec[k]
# if the fourth factor become
# negative then break
if y <= 0 :
break
# we will replace more optimum
# number than the previous one
if n % y == 0 :
flag = 0
maxProduct = max(vec[i] * vec[j] *
vec[k] * y , maxProduct)
# print the product if the numbers exist
if flag == 0 :
print("Product is - >", maxProduct)
else :
print("Not possible")
# Driver Code
if __name__ == "__main__" :
n = 50
# function calling
findfactors(n)
# This code is contributed by ANKITRAI1
C#
// C# program to find four factors of N
// with maximum product and sum equal to N
using System;
using System.Collections.Generic;
class GFG
{
// Function to find factors
// and to print those four factors
static void findfactors(int n)
{
List<int> vec = new List<int>();
// inserting all the factors in a vector s
for (int i = 1; i * i <= n; i++)
{
if (n % i == 0)
{
vec.Add(i);
vec.Add(n / i);
}
}
// sort the vector
vec.Sort();
// print all the factors
Console.WriteLine("All the factors are -> ");
for (int i = 0; i < vec.Count; i++)
{
Console.Write(vec[i] + " ");
}
Console.WriteLine();
// Any elements is divisible by 1
int maxProduct = 1;
Boolean flag = true;
// using three loop we'll find
// the three maximum factors
for (int i = 0; i < vec.Count; i++)
{
for (int j = i; j < vec.Count; j++)
{
for (int k = j; k < vec.Count; k++)
{
// storing the fourth factor in y
int y = n - vec[i] - vec[j] -
vec[k];
// if the fourth factor become negative
// then break
if (y <= 0)
{
break;
}
// we will replace more optimum number
// than the previous one
if (n % y == 0)
{
flag = false;
maxProduct = Math.Max(vec[i] * vec[j] *
vec[k] * y, maxProduct);
}
}
}
}
// print the product if the numbers exist
if (flag == false)
{
Console.WriteLine("Product is -> " +
maxProduct);
}
else
{
Console.WriteLine("Not possible");
}
}
// Driver code
public static void Main(String[] args)
{
int n;
n = 50;
findfactors(n);
}
}
// This code is contributed by Rajput-Ji
PHP
<?php
// PHP program to find four factors
// of N with maximum product and
// sum equal to N
// Function to find factors
// and to print those four factors
function findfactors($n)
{
$vec = array();
// inserting all the
// factors in a vector s
for ($i = 1; $i * $i <= $n; $i++)
{
if ($n % $i == 0)
{
array_push($vec, $i);
array_push($vec, ($n / $i));
}
}
// sort the vector
sort($vec);
// print all the factors
echo "All the factors are -> ";
for ($i = 0; $i < sizeof($vec); $i++)
echo $vec[$i] . " ";
echo "\n";
// Any elements is divisible by 1
$maxProduct = 1;
$flag = 1;
// using three loop we'll find
// the three maximum factors
for ($i = 0; $i < sizeof($vec); $i++)
{
for ($j = $i;
$j < sizeof($vec); $j++)
{
for ($k = $j;
$k < sizeof($vec); $k++)
{
// storing the fourth factor in y
$y = $n - $vec[$i] -
$vec[$j] - $vec[$k];
// if the fourth factor become
// negative then break
if ($y <= 0)
break;
// we will replace more optimum
// number than the previous one
if ($n % $y == 0)
{
$flag = 0;
$maxProduct = max($vec[$i] * $vec[$j] *
$vec[$k] * $y, $maxProduct);
}
}
}
}
// print the product if
// the numbers exist
if ($flag == 0)
echo "Product is -> " .
$maxProduct ."\n";
else
echo "Not possible" ."\n";
}
// Driver code
$n = 50;
findfactors($n);
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// Javascript program to find four factors of N
// with maximum product and sum equal to N
// Function to find factors
// and to print those four factors
function findfactors(n)
{
let vec = [];
// inserting all the factors in a vector s
for (let i = 1; i * i <= n; i++)
{
if (n % i == 0)
{
vec.push(i);
vec.push(Math.floor(n / i));
}
}
// sort the vector
vec.sort(function(a,b){return a-b;});
// print all the factors
document.write("All the factors are -> ");
for (let i = 0; i < vec.length; i++)
{
document.write(vec[i] + " ");
}
document.write("<br>");
// Any elements is divisible by 1
let maxProduct = 1;
let flag = true;
// using three loop we'll find
// the three maximum factors
for (let i = 0; i < vec.length; i++)
{
for (let j = i; j < vec.length; j++)
{
for (let k = j; k < vec.length; k++)
{
// storing the fourth factor in y
let y = n - vec[i] - vec[j] -
vec[k];
// if the fourth factor
// become negative
// then break
if (y <= 0)
{
break;
}
// we will replace more
// optimum number
// than the previous one
if (n % y == 0)
{
flag = false;
maxProduct = Math.max(vec[i] *
vec[j] * vec[k] * y, maxProduct);
}
}
}
}
// print the product if the numbers exist
if (flag == false)
{
document.write("Product is -> " + maxProduct);
}
else
{
document.write("Not possible");
}
}
// Driver code
let n = 50;
findfactors(n);
// This code is contributed by avanitrachhadiya2155
</script>
Producción:
All the factors are -> 1 2 5 10 25 50 Product is -> 12500
Publicación traducida automáticamente
Artículo escrito por AmanSrivastava1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA