Check whether a very large number of the given form is a multiple of 3.

Consider a very long K-digit number N with digits d0, d1, …, dK-1 (in decimal notation; d0 is the most significant and dK-1 the least significant digit). This number is so large that it can’t be given or written down explicitly; instead, only its starting digits are given and a way to construct the remainder of the number.

Specifically, you are given d0 and d1; for each i ≥2, di is the sum of all preceding (more significant) digits, modulo 10, more formally – Determine if N is a multiple of 3.

Constraints:
2 ≤K ≤1012
1 ≤d0 ≤9
0 ≤d1 ≤9

Examples:

Input : K = 13, d0 = 8, d1 = 1

Output : YES

Explanation: The whole number N is 8198624862486, which is divisible by 3,
so the answer is YES.

Input :  K = 5, d0 = 3, d1 = 4

Output : NO

Explanation: The whole number N is 34748, which is not divisible by 3,
so the answer is NO.

Method 1 (Brute Force)
We can apply the brute force method to calculate the whole number N by using the condition given for constructing the number iteratively (sum of preceding numbers modulo 10) and check whether the number is divisible by 3 or not. But since the number of digits (K) can be as large as 1012, we can’t store it as an integer since it will be very larger than the maximum range of ‘long long int’. Hence below is an efficient method to determine if N is a multiple of 3.

Method 2 (Efficient)
The key idea behind the solution is the fact that the digits start to repeat after some time in a cycle of length 4. Firstly, we will find the sum of all the digits and then determine if it is divisible by 3 or NOT.

We know d0 and d1.
d2 = ( d0 + d1 ) % 10

d3 = ( d2 + d1 + d0 ) % 10 = (( d0 + d1) % 10 + d0 + d1) % 10 = 2 * ( d0 + d1 ) % 10

Similarly,
d4 = ( d3 + d2 + d1 + d0 ) % 10 = 4 * ( d0 + d1 ) % 10

d5 = ( d4 + d3 + d2 + d1 + d0 ) % 10 = 8 * ( d0 + d1 ) % 10

d6 = ( d5 + … + d1 + d0 ) % 10 = 16 * (d0 + d3) % 10 = 6 * ( d0 + d1 ) % 10

d7 = ( d6 + … + d1 + d0 ) % 10 = 12 * ( d0 + d1 ) % 10 = 2 * ( d0 + d1 ) % 10

If we keep on finding on di, we will see that that the resultant is just looping around the same values (2, 4, 8, 6).
Here the cycle length is 4 and d2 is not present in the cycle. Hence after d2 the cycle starts forming in length of 4 starting from any value in (2, 4, 8, 6) but in the same order giving a sum of S = 2 + 4 + 8 + 6 = 20 for consecutive four digits. Thus, the total sum of digits for the whole number is = d0 + d1 + d2 + S*(k – 3)/4 + x, where first three terms will be covered by d0, d1, d2
and after that groups of 4 will be covered by S. But since (k – 3) may be not a multiple of 4, some remaining digits will be left which is covered by x which can be calculated by running a loop as those number of terms will be less than 4.
For e.g.

When K = 13,
sum of digits = d0 + d1 + d2 + S * (13 – 3) / 4 + x = d0 + d1 + d2 + S * 2 + x,

where first S will have d3, d4, d5, d6 and second S will have d7, d8, d9, d10 and
x = d11 + d12

• d11 = 2 * ( d0 + d1 ) % 10
• d12 = 4 * ( d0 + d1 ) % 10

Below is the implementation of above idea :

C++

 // CPP Program to determine if // number N of given form is  // divisible by 3 or not  #include using namespace std;    // function returns true if number N is   // divisible by 3 otherwise false,  // dig0 - most significant digit  // dig1 - 2nd most significant digit // K - number of digits  bool multipleOfThree(int K, int dig0, int dig1) {     // sum of digits     long long int sum = 0;        // store the sum of first two digits     // modulo 10 in a temporary variable     int temp = (dig0 + dig1) % 10;        sum = dig0 + dig1;        // if the number N is a two digit number     if (K == 2) {         if (sum % 3 == 0)             return true;         else             return false;     }        // add temp to sum to get the sum     // of first three digits which are     // not a part of cycle     sum += temp;        // get the number of groups in cycle     long long int numberofGroups = (K - 3) / 4;        // get the remaining number of digits     int remNumberofDigits = (K - 3) % 4;        // add sum of 20 for each group (2, 4, 8, 6)     sum += (numberofGroups * 20);        // find the remaining sum of remaining digits     for (int i = 0; i < remNumberofDigits; i++) {         temp = (2 * temp) % 10;         sum += temp;     }        // check if it is divisible by 3 or not     if (sum % 3 == 0)         return true;     else         return false; }    // Driver Code int main() {     int K = 5, dig0 = 3, dig1 = 4;     if (multipleOfThree(K, dig0, dig1))         cout << "YES" << endl;     else         cout << "NO" << endl;     return 0; }

Java

 // Java Program to determine if // number N of given form is  // divisible by 3 or not  import java.io.*;    public class GFG {    // function returns true if number N is  // divisible by 3 otherwise false,  // dig0 - most significant digit  // dig1 - 2nd most significant digit // K - number of digits  static boolean multipleOfThree(int K,                                 int dig0,                                 int dig1) {            // sum of digits     long sum = 0;        // store the sum of first two digits     // modulo 10 in a temporary variable     int temp = (dig0 + dig1) % 10;        sum = dig0 + dig1;        // if the number N is a two digit number     if (K == 2) {         if (sum % 3 == 0)             return true;         else             return false;     }        // add temp to sum to get the sum     // of first three digits which are     // not a part of cycle     sum += temp;        // get the number of groups in cycle     long numberofGroups = (K - 3) / 4;        // get the remaining number of digits     int remNumberofDigits = (K - 3) % 4;        // add sum of 20 for each group (2, 4, 8, 6)     sum += (numberofGroups * 20);        // find the remaining sum of      // remaining digits     for (int i = 0; i < remNumberofDigits; i++)     {         temp = (2 * temp) % 10;         sum += temp;     }        // check if it is divisible by 3 or not     if (sum % 3 == 0)         return true;     else         return false; }        // Driver Code     static public void main (String[] args)     {             int K = 5, dig0 = 3, dig1 = 4;             if (multipleOfThree(K, dig0, dig1))                 System.out.println("Yes");             else                 System.out.println("No");                } }    // This code is contributed by vt_m.

Python 3

 # Python 3 Program to determine if # number N of given form is  # divisible by 3 or not     # function returns true if number N  # is divisible by 3 otherwise false,  # dig0 - most significant digit  # dig1 - 2nd most significant digit # K - number of digits  def multipleOfThree(K, dig0, dig1):            # sum of digits     sum = 0        # store the sum of first two digits     # modulo 10 in a temporary variable     temp = (dig0 + dig1) % 10        sum = dig0 + dig1        # if the number N is a      # two digit number     if (K == 2) :         if (sum % 3 == 0):             return True         else:             return False               # add temp to sum to get the sum     # of first three digits which are     # not a part of cycle     sum += temp        # get the number of groups in cycle     numberofGroups = (K - 3) // 4        # get the remaining number of digits     remNumberofDigits = (K - 3) % 4        # add sum of 20 for each     # group (2, 4, 8, 6)     sum += (numberofGroups * 20)        # find the remaining sum of      # remaining digits     for i in range(remNumberofDigits) :         temp = (2 * temp) % 10         sum += temp        # check if it is divisible      # by 3 or not     if (sum % 3 == 0):         return True     else:         return False    # Driver Code if __name__ == "__main__":     K = 5     dig0 = 3     dig1 = 4     if (multipleOfThree(K, dig0, dig1)):         print("Yes")     else:         print("No")    # This code is contributed by ChitraNayal

C#

 // C# Program to determine if // number N of given form is  // divisible by 3 or not  using System;    class GFG {    // function returns true if number N is  // divisible by 3 otherwise false,  // dig0 - most significant digit  // dig1 - 2nd most significant digit // K - number of digits  static bool multipleOfThree(int K,                              int dig0,                              int dig1) {     // sum of digits     long sum = 0;        // store the sum of first two digits     // modulo 10 in a temporary variable     int temp = (dig0 + dig1) % 10;        sum = dig0 + dig1;        // if the number N is     // a two digit number     if (K == 2) {         if (sum % 3 == 0)             return true;         else             return false;     }        // add temp to sum to get the sum     // of first three digits which are     // not a part of cycle     sum += temp;        // get the number of groups in cycle     long numberofGroups = (K - 3) / 4;        // get the remaining number of digits     int remNumberofDigits = (K - 3) % 4;        // add sum of 20 for each group (2, 4, 8, 6)     sum += (numberofGroups * 20);        // find the remaining sum of      // remaining digits     for (int i = 0; i < remNumberofDigits; i++)     {         temp = (2 * temp) % 10;         sum += temp;     }        // check if it is divisible by 3 or not     if (sum % 3 == 0)         return true;     else         return false; }        // Driver Code     static public void Main(String[] args)     {         int K = 5, dig0 = 3, dig1 = 4;         if (multipleOfThree(K, dig0, dig1))             Console.WriteLine("Yes");         else             Console.WriteLine("No");     } }    // This code is contributed by vt_m.

PHP



Output:

No

Time Complexity: O(1)

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