// This C++ program finds the minimum removals
// so that the calculated gcd of remaining numbers will be more
// than the initial gcd of N numbers
#include <bits/stdc++.h>
using namespace std;
#define MAXN 100001
// storing smallest prime factor for every number
int spf1[MAXN];
// Calculates SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
void sieve1()
{
spf1[1] = 1;
for (int i = 2; i < MAXN; i++)
// marks smallest prime factor for every
// number to be itself.
spf1[i] = i;
// separately marks spf for every even
// number as 2
for (int i = 4; i < MAXN; i += 2)
spf1[i] = 2;
for (int i = 3; i * i < MAXN; i++) {
// checks if i is prime
if (spf1[i] == i) {
// marks SPF for all numbers divisible by i
for (int j = i * i; j < MAXN; j += i)
// marks spf1[j] if it is not
// previously marked
if (spf1[j] == j)
spf1[j] = i;
}
}
}
// Now a O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
vector<int> getFactorization1(int x)
{
vector<int> ret;
while (x != 1) {
ret.push_back(spf1[x]);
x = x / spf1[x];
}
return ret;
}
// So function which returns the minimal
// removals required to make gcd
// greater than previous
int minimumRemovals1(int a1[], int n)
{
int g = 0;
// finding initial gcd
for (int i = 0; i < n; i++)
g = __gcd(a1[i], g);
unordered_map<int, int> mpp;
// divides all number by initial gcd
for (int i = 0; i < n; i++)
a1[i] = a1[i] / g;
// iterating for all numbers
for (int i = 0; i < n; i++) {
// primt factorisation to get the prime
// factors of i-th element in the array
vector<int> p = getFactorization1(a1[i]);
set<int> s1;
// insert all the prime factors in
// set to remove duplicates
for (int j = 0; j < p.size(); j++) {
s1.insert(p[j]);
}
/// increase the count of prime
// factor in map for every element
for (auto it = s1.begin(); it != s1.end(); it++) {
int el = *it;
mpp[el] += 1;
}
}
int mini = INT_MAX;
int mini1 = INT_MAX;
// iterate in map and check for every factor
// and its count
for (auto it = mpp.begin(); it != mpp.end(); it++) {
int fir1 = it->first;
int sec1 = it->second;
// checking largest appearing factor
// which does not appears in any one or more
if ((n - sec1) <= mini) {
mini = n - sec1;
}
}
if (mini != INT_MAX)
return mini;
else
return -1;
}
// Driver code
int main()
{
int a1[] = { 6, 9, 15, 30 };
int n = sizeof(a1) / sizeof(a1[0]);
sieve1();
cout << minimumRemovals1(a1, n);
return 0;
}
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