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  1. Practice
  2. Tutorials
  3. 30 Days of Code
  4. Day 19: Interfaces
  5. Discussions

Day 19: Interfaces

  • wy218032 + 1 comment
    //we could stop at sqrt(n)
//This is my Java solution. I think this may be better.

class Calculator implements AdvancedArithmetic{
    public int divisorSum(int n){
       int sum = 0;
       for(int i = 1; i <= (int)Math.sqrt(n); i++) {
           if(n%i==0){
               sum = sum + i + n/i;
           }
       }
       if((int)Math.sqrt(n)*(int)Math.sqrt(n)==n){
           sum -= Math.sqrt(n);
       }
        return sum;
    }
}

    • nuhman + 2 comments

      I din't understand why you are doing the

      if((int)Math.sqrt(n)*(int)Math.sqrt(n)==n){
           sum -= Math.sqrt(n);
       }

      part. Can you explain?

      • impex_ua + 0 comments

        If n is square number, you have to minus sqrt n. Just math.

      • toraju + 1 comment

        You can optimize that one step to the following... to avoid correction for doubble addition, when input number is an exact square.

        public int divisorSum(int n){
       int sum = 0;int i;
       for( i = 1; i <(int)Math.sqrt(n); i++) {
           if(n%i==0){
               sum = sum + i + n/i;
           }
       }
       if(i*i == n){
           sum = sum+i;
       }
        return sum;
    }

        • Ankurjain04 + 0 comments

          sum = sum+i; should be sum = sum-i;