Multiple of 3 CodeChef Solution – DSA LEARNING SERIES

Today we will be Solving Multiple of 3 CodeChef Problem which is a part of CodeChef DSA Learning Series.

Task

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 we can’t give it to you on the input explicitly; instead, you are only given its starting digits 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, the following formula must hold:

Determine if N is a multiple of 3.

Input

The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.

The first and only line of each test case contains three space-separated integers K, d0 and d1.

Output

For each test case, print a single line containing the string “YES” (without quotes) if the number N is a multiple of 3 or “NO” (without quotes) otherwise.

Constraints

• 1 ≤ T ≤ 1000
• 2 ≤ K ≤ 1012
• 1 ≤ d0 ≤ 9
• 0 ≤ d1 ≤ 9

Example

Input

3
5 3 4
13 8 1
760399384224 5 1

Output

NO
YES
YES

Explanation

Example case 1: The whole number N is 34748, which is not divisible by 3, so the answer is NO.

Example case 2: The whole number N is 8198624862486, which is divisible by 3, so the answer is YES.

Solutions – Multiple of 3 CodeChef Solution

import java.util.Scanner;

class MULTHREE {
    public static void solve(long k,int d0,int d1){
        StringBuffer sb=new StringBuffer(""+d0+d1);
        long sum=d1+d0;

        if(k>=3){
            if(sum%5!=0){
                sum=sum+(sum%10);
                for(long i=(k-3)%4;i>0;i--){
                    sum=sum+(sum%10);
                }
                sum=sum+((k-3)/4)*20;
            }else {
                System.out.println("NO");
                return;
            }
        }

        if(sum%3==0){
            System.out.println("YES");
        }else {
            System.out.println("NO");
        }


    }
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int T=sc.nextInt();
        while (T-->0){
           int d0,d1;
           long k;
           k=sc.nextLong();
           d0=sc.nextInt();
           d1=sc.nextInt();
           solve(k,d0,d1);
        }
    }
}

C++

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

int main() {



	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

	int t ;

	cin >> t;

	while(t--){

		ll k;int d0,d1;
		cin >> k >> d0 >> d1;
		int s = d0+d1;

		int c = (2*s)%10 + (4*s)%10 +(6*s)%10 +(8*s)%10;
		ll cycles = (k-3)/4;
		ll ans = 0;
		if(k==2){
			ans = s;
		} 
		else{
			ans = s+(s%10) + (c *1LL *cycles);
		
		int left = (k-3)-(cycles*4);
		int p=2;
		for(int i=1;i<=left;i++){
			ans += (p*s)%10;
			p=(p*2)%10;
		}
	}

	
		if(ans%3==0) cout << "YES\n";

		else      cout << "NO\n";
	}
	return 0;
}

Disclaimer: The above Problem (Multiple of 3) is generated by CodeChef but the Solution is provided by CodingBroz.

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