# Maximum GCD of N integers with given product in C++

Suppose we two integers N and P. The P is the product of N unknown integers. We have to find the maximum possible GCD of those integers. Suppose N = 3, and P = 24, then different groups will be like {1, 1, 24}, {1, 2, 12}, {1, 3, 8}, {1, 4, 6}, {2, 2, 6}, {2, 3, 4}. The GCDs are: 1, 1, 1, 1, 2, 1. So answer is 2 here.

We will find all prime factors of P, and store them into hashmap. The N integers will have max GCD when the prime factor will be common in all the integers. So if P = p1k1 * p2k2 * … * pnkn. Here pi is the prime factor. Then max GCD will be res = p1k1/N * p2k2/N * … * pnkn/N.

## Example

Live Demo

#include <iostream>
#include <cmath>
#include <unordered_map>
using namespace std;
long getMaxGCD(long N, long p) {
int gcd = 1;
unordered_map<int, int> prime_factors;
for (int i = 2; i * i <= p; i++) {
while (p % i == 0) {
prime_factors[i]++;
p /= i;
}
}
if (p != 1)
prime_factors[p]++;
for (auto v : prime_factors)
gcd = gcd * pow(v.first, v.second / N);
return gcd;
}
int main() {
long n = 3;
long p = 24;
cout << "MAX GCD: " << getMaxGCD(n, p);
}

## Output

MAX GCD: 2

Updated on: 21-Oct-2019

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