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Maximum Number of Unique Prime Factors in C++
Given the task is to find the maximum number of unique prime factors a number can have in the range of [1, N] where N is given.
Let’s now understand what we have to do using an example −
Input − N=100
Output − 3
Explanation − Let us take 30 in the range of [1, 100]
30 = 3 * 2 * 5 = unique prime factors. Therefore, in the range of [1, 100] a maximum of 3 unique factors can be found.
Input − N=300
Output − 4
Approach used in the below program as follows
In function MaxPrime() we will first check if (N < 2). If so then simply return zero, otherwise proceed.
Then we will use the sieve of Eratosthenes to find out all the prime numbers before the given number N.
Initialize two variables pro=1 and max=0 of type int to store the product and the final answer respectively.
Inside the sieve of Eratosthenes we will multiply the first set of prime numbers until the product remains smaller than N by writing − pro=*p;
Check if (pro > N). If so then return max and add 1 to max.
Outside the sieve, also return max.
Example
#include <bits/stdc++.h>
using namespace std;
int MaxPrime(int N){
if (N < 2)
return 0;
// Using Sieve of Eratosthenes
bool Arr[N+1];
memset(Arr, true, sizeof(Arr));
int pro = 1, max = 0;
for (int p=2; p*p<=N; p++){
if (Arr[p] == true){
for (int i=p*2; i<=N; i += p)
Arr[i] = false;
/*Multiply first set of prime numbers while product remains smaller than N*/
pro *= p;
if (pro > N)
return max;
max++;
}
}
return max;
}
//Main function
int main(){
int N = 300;
cout << MaxPrime(N);
return 0;
}
Output
If we run the above code we will get the following output −
4
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