Find if the given number is present in the infinite sequence or not

Given three integers A, B and C. In an infinite sequence, A is the first number, C is the common difference (Si – Si – 1 = C). The task is to check if the number B will appear in the sequence or not.

Examples:

Input: A = 1, B = 7, C = 3
Output: Yes
The sequence will be 1, 4, 7, 10, …

Input: A = 1, B = -4, C = 5
Output: No

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: There are two cases:

1. When C = 0, print Yes if A = B else No as the sequence will consist only the number A
2. When C > 0, for any non-negative integer k the equation B = A + k * C must be satisfied i.e. (B – A) / C must be a non-negative integer.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    // Function that returns true if // the sequence will contain B bool doesContainB(int a, int b, int c) {     if (a == b)         return true;        if ((b - a) * c > 0 && (b - a) % c == 0)         return true;        return false; }    // Driver code int main() {     int a = 1, b = 7, c = 3;        if (doesContainB(a, b, c))         cout << "Yes";     else         cout << "No";        return 0; }

Java

 // Java implementation of the approach class GFG  {        // Function that returns true if      // the sequence will contain B      static boolean doesContainB(int a, int b, int c)      {         if (a == b)          {             return true;         }            if ((b - a) * c > 0 && (b - a) % c == 0)          {             return true;         }            return false;     }        // Driver code      public static void main(String[] args)      {         int a = 1, b = 7, c = 3;            if (doesContainB(a, b, c))          {             System.out.println("Yes");         }          else          {             System.out.println("No");         }     } }    // This code contributed by Rajput-Ji

Python3

 # Python3 implementation of the approach    # Function that returns true if # the sequence will contain B def doesContainB(a, b, c):     if (a == b):         return True        if ((b - a) * c > 0 and (b - a) % c == 0):         return True        return False    # Driver code if __name__ == '__main__':     a, b, c = 1, 7, 3        if (doesContainB(a, b, c)):         print("Yes")     else:         print("No")    # This code is contributed by 29AjayKumar

C#

 // C# implementation of the approach using System;    class GFG  {        // Function that returns true if      // the sequence will contain B      static bool doesContainB(int a, int b, int c)      {         if (a == b)          {             return true;         }            if ((b - a) * c > 0 && (b - a) % c == 0)          {             return true;         }            return false;     }        // Driver code      public static void Main()      {         int a = 1, b = 7, c = 3;            if (doesContainB(a, b, c))          {             Console.WriteLine("Yes");         }          else         {             Console.WriteLine("No");         }     } }    /* This code contributed by PrinciRaj1992 */

PHP

 0 &&          (\$b - \$a) % \$c == 0)         return true;        return false; }    // Driver code \$a = 1; \$b = 7; \$c = 3;    if (doesContainB(\$a, \$b, \$c))     echo "Yes"; else     echo "No";    // This code is contributed // by Akanksha Rai ?>

Output:

Yes

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