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Count total divisors of A or B in a given range in C++
We are given four integers L, R, A and B. The goal is to find the count of numbers in range [L,R] that fully divide either A or B or both.
We will do this by traversing from L to R and for each number if number%A==0 or number%B==0 then increment count of divisors.
Let’s understand with examples.
Input − L=10, R=15, A=4, B=3
Output − Count of divisors of A or B − 2
Explanation −
Number 12 is fully divisible by 3 and 4. Number 15 is fully divisible by 3 only. Total divisors=2
Input − L=20, R=30, A=17, B=19
Output − Count of divisors of A or B − 0
Explanation − No number between 20 and 30 fully divisible by A or B or both.
Approach used in the below program is as follows
We have taken four variables A, B, L and R.
Function countDivisors(int l, int r, int a, int b) takes all as input and returns the divisors of A or B or both that lie in range [L, R].
Take the initial count as 0.
Starting from i=L to i=R, if i%a==0 or i%b==0 increment count.
At the end of loop count as divisors of A or B.
Return count as result.
Example
#include <bits/stdc++.h>
using namespace std;
int countDivisors(int l, int r, int a,int b){
int count = 0;
for (int i = l; i <= r; i++){
if(i%a==0 || i%b==0)
{ count++ ; }
}
return count;
}
int main(){
int L=5;
int R=15;
int A=2;
int B=5;
cout <<endl<< "Total divisors of A and B : "<<countDivisors(L,R,A,B);
return 0;
}Output
If we run the above code it will generate the following output −
Total divisors of A and B : 7
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