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Find four factors of N with maximum product and sum equal to N - Set-2 in C++
Concept
With respect of a given integer N, our task is to determine all factors of N and print the product of four factors of N so that −
- The sum of the four factors is equal to N.
- The product of the four factors is largest.
It has been seen that if it is impossible to determine 4 such factors then print “Not possible”. It should be noted thatall the four factors can be equal to each other to maximize the product.
Input
N = 60
Output
All the factors are -> 1 2 3 4 5 6 10 12 15 20 30 60 Product is -> 50625
Select the factor 15 four times,
Therefore, 15+15+15+15 = 60 and product is largest.
Method
Here a method which takes a complexity of O(P^3), where P is the number of factors of N has been explained.
So an efficient method of time complexity O(N^2) can be obtained with the help of following steps.
- We store all the factors of given number in a container.
- Now we iterate for all pairs and store their sum in a different container.
- We have to mark the index (element1 + element2) with pair(element1, element2) to obtain the elements by which the sum was obtained.
- Again we iterate for all the pair_sums, and verify if n-pair_sum exists in the same container, as a result both the pairs form the quadruple.
- Implement the pair hash array to obtain the elements by which the pair was formed.
- Finally, store the largest of all such quadruples, and print it at the end.
Example
// C++ program to find four factors of N
// with maximum product and sum equal to N
#include <bits/stdc++.h>
using namespace std;
// Shows function to find factors
// and to print those four factors
void findfactors1(int q){
vector<int> vec1;
// Now inserting all the factors in a vector s
for (int i = 1; i * i <= q; i++) {
if (q % i == 0) {
vec1.push_back(i);
vec1.push_back(q / i);
}
}
// Used to sort the vector
sort(vec1.begin(), vec1.end());
// Used to print all the factors
cout << "All the factors are -> ";
for (int i = 0; i < vec1.size(); i++)
cout << vec1[i] << " ";
cout << endl;
// So any elements is divisible by 1
int maxProduct1 = 1;
bool flag1 = 1;
// implementing three loop we'll find
// the three largest factors
for (int i = 0; i < vec1.size(); i++) {
for (int j = i; j < vec1.size(); j++) {
for (int k = j; k < vec1.size(); k++) {
// Now storing the fourth factor in y
int y = q - vec1[i] - vec1[j] - vec1[k];
// It has been seen that if the fouth factor become negative
// then break
if (y <= 0)
break;
// So we will replace more optimum number
// than the previous one
if (q % y == 0) {
flag1 = 0;
maxProduct1 = max(vec1[i] * vec1[j] * vec1[k] *y,maxProduct1);
}
}
}
}
// Used to print the product if the numbers exist
if (flag1 == 0)
cout << "Product is -> " << maxProduct1 << endl;
else
cout << "Not possible" << endl;
}
// Driver code
int main(){
int q;
q = 60;
findfactors1(q);
return 0;
}Output
All the factors are -> 1 2 3 4 5 6 10 12 15 20 30 60 Product is -> 50625
Updated on: 24-Jul-2020
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