Find original numbers from gcd() every pair in C++

Concept

With respect of a given array array[] containing GCD of every possible pair of elements of another array, our task is to determine the original numbers which are used to compute the GCD array.

Input

array[] = {6, 1, 1, 13}

Output

13 6
gcd(13, 13) = 13
gcd(13, 6) = 1
gcd(6, 13) = 1
gcd(6, 6) = 6

Input

arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 6, 6, 8, 11, 13, 3, 3}

Output

13 11 8 6 6

Method

At first, sort the array in decreasing order.
Largest element will always be one of the original numbers. Maintain that number and delete it from the array.
Calculate GCD of the element taken in the previous step with the current element beginning from the largest and discard the GCD value from the given array.

Example

Live Demo

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Shows utility function to print
// the contents of an array
void printArr(int array[], int n1){
   for (int i = 0; i < n1; i++)
      cout << array[i] << " ";
}
// Shows function to determine the required numbers
void findNumbers(int array[], int n1){
   // Used to sort array in decreasing order
   sort(array, array + n1, greater<int>());
   int freq1[array[0] + 1] = { 0 };
   // Here, count frequency of each element
   for (int i = 0; i < n1; i++)
      freq1[array[i]]++;
   // Shows size of the resultant array
   int size1 = sqrt(n1);
   int brr1[size1] = { 0 }, x1, l1 = 0;
   for (int i = 0; i < n1; i++) {
      if (freq1[array[i]] > 0) {
         // Here, store the highest element in
         // the resultant array
         brr1[l1] = array[i];
         //Used to decrement the frequency of that element
         freq1[brr1[l1]]--;
         l1++;
         for (int j = 0; j < l1; j++) {
            if (i != j) {
               // Calculate GCD
               x1 = __gcd(array[i], brr1[j]);
               // Decrement GCD value by 2
               freq1[x1] -= 2;
            }
         }
      }
   }
   printArr(brr1, size1);
}
// Driver code
int main(){
   /* int array[] = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,1, 1, 1, 6, 6, 6, 8, 11, 13, 3, 3}; */
   int array[] = { 6, 1, 1, 13};
   int n1 = sizeof(array) / sizeof(array[0]);
   findNumbers(array, n1);
   return 0;
}

Output

13 6

Arnab Chakraborty

Published on 25-Jul-2020 10:53:55

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