给定一个整数
。任务是找到N的所有因子，打印N的四个因子的乘积，使得：
- 四个因子的总和等于N。
- 这四个因素的乘积最大。
如果无法找到4个这样的因素，则打印“不可能” 。
注意：为了使乘积最大化，这四个因素可以彼此相等。
例子：
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.
在第二个示例中，当四个因子等于25时，乘积将最大。四个因子的总和等于“ N”。尽管我们可以选择四次相同的因子来最大化产品。
以下是解决此问题的分步算法：
- 首先通过从1遍历到’N’的平方根来找到数字’N’的因数，然后检查’i’和’n / i’是否将N除以并将它们存储在向量中。
- 排序向量并打印每个元素。
- 使用三个循环，找到三个数字以使第四个数字的乘积最大化。
- 用上一个产品替换下一个最大产品。
- 找到四个因素后，请打印产品。
下面是上述方法的实现：
C++
// C++ program to find four factors of N
// with maximum product and sum equal to N
#include
using namespace std;
// Function to find factors
// and to print those four factors
void findfactors(int n)
{
vector 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 fouth 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 vec = new Vector();
// 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 fouth 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 Python 3
# 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 fouth 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 ANKITRAI1C#
// 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 vec = new List();
// 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 fouth 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
";
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 fouth 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
?>输出：
All the factors are -> 1 2 5 10 25 50
Product is -> 12500