Count of all possible values of X such that A % X = B in C++

Given two integers A and B and a number X. The goal is to find the count of values that X can have so that A%X=B. For the above equation if, A==B then infinite values of X are possible, so return −1. If A < B then there would be no solution so return 0. If A>B then return count of divisors of (AB) as result.

For Example

Input

A=5, B=2

Output

Count of all possible values of X such that A % X = B are: 1

Explanation

5%3=2. So X is 3 here.

Input

A=10, B=10

Output

Count of all possible values of X such that A % X = B are: −1

Explanation

Here A==B so there are infinite solutions so −1 is returned.

Approach used in the below program is as follows −

In this approach we will calculate divisors of (A−B) using for loop from i=1 to i*i<=(A−B). If any i fully divides (A−B) then update count accordingly.

• Take integers A and B as input.

• If A<B then print 0 as result.

• If A==B then print −1 as result.

• For A>B, function possible_values(int A, int B) takes A and B and returns count of all possible values of X such that A % X = B.

• Take initial count as 0 and X=A−B.

• Traverse using for loop from i=1 to i*i<(A−B), for calculating divisors of X.

• If any i fully divides X then take temp=i, temp_2=B−1 and if i*i!=X then set temp_2 = X / i.

• If temp>B and temp_2 >B then increment count.

• At the end of loop return count as result.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int possible_values(int A, int B){
   int count = 0;
   int X = A − B;
   for (int i = 1; i * i <= A − B; i++){
      if(X % i == 0){
         int temp = i;
         int temp_2 = B − 1;
         if(i * i != X){
            temp_2 = X / i;
         }
         if(temp > B){
            count++;
         }
         if(temp_2 > B){
            count++;
         }
      }
   }
   return count;
}
int main(){
   int A = 15, B = 5;
   if(A < B){
      cout<<"Count of all possible values of X such that A % X = B are: "<<0;
   }
   else if(A == B){
      cout<<"Count of all possible values of X such that A % X = B are: "<<−1;
   }
   else{
      cout<<"Count of all possible values of X such that A % X = B are: "<<possible_values(A, B);
   }
   return 0;
}

Output

If we run the above code it will generate the following output −

Count of all possible values of X such that A % X = B are: 1