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  1. /**
  2.  * Continuos Sub-Sequence divisible by a number
  3.  * @author Prateek
  4.  */
  5.  class SumSubSequence {
  6.  
  7. 	public static void main(String[] args) {
  8. 		//int[] arr={1, 1, 9 , 7, 12 , 5 , 8 , 2, 7, 2, 10, 2, 3}; 
  9. 		//int divisor=13;
  10. 		int[] arr={1 , 1 , 3 , 2 , 4, 1 ,4 , 5};
  11. 		int divisor=5;
  12. 		solve(arr,divisor);
  13. 	}
  14.  
  15. 	/**
  16. 	 * Sub-Routine to count numnber of sub-sequnces
  17. 	 * @param d: the given divisor
  18. 	 */
  19. 	public static int solve(int[] arr, int d)
  20. 	{
  21. 		int Answer = 0;
  22. 		int[] hash = new int[d];
  23.  
  24. 		int sum = 0;
  25. 		int val;
  26. 		int num;
  27.  
  28. 		for (int i = 0; i < arr.length; i++) {
  29. 			num = arr[i];
  30.  
  31. 			if(num % d == 0) // counte numbes which are divisible by divisor
  32. 				Answer ++ ;
  33.  
  34. 			sum +=  num; 
  35. 			val = sum % d;
  36.  
  37. 			if(val<0) //handle negative numbers
  38. 				val = val * (-1);
  39.  
  40. 			hash[val] = hash[val] + 1;
  41. 		}
  42.  
  43. 		int size=hash.length ;
  44. 		for (int i = 0; i < size; i++) 	{
  45. 			int count = hash[i];
  46.  
  47. 			if(count > 1)
  48. 				Answer = Answer + count * (count -1)/2 ;
  49. 		}
  50. 		System.out.println(Answer+hash[0]);
  51. 		return Answer+hash[0];
  52. 	}	
  53. }
  54.  
Success #stdin #stdout 0.06s 380224KB
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https://ideone.com/A1pODr
Continuos Sub-Sequence divisible by a number
language:
Java (12.0.1)
created:
10 years ago
visibility:
public

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