Multiple of 3 CodeChef Solution

Hello Programmers In this post, you will know how to solve the Multiple of 3 CodeChef Solution. Which is a part of CodeChef DSA Learning Series.

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One more thing to add, don’t directly look for the solutions, first try to solve the problems of Codechef by yourself. If you find any difficulty after trying several times, then you can look for solutions.

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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 Format

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 Format

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

Sample Input

3
5 3 4
13 8 1
760399384224 5 1

Sample 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.

Multiple of 3 CodeChef Solution in Java

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);
        }
    }
}

Multiple of 3 CodeChef Solution in CPP

#include <bits/stdc++.h>
#define mid(l,u) ((l+u)/2)
#define MOD 1000000007
#define INF 10000000000000
#define int long long
using namespace std;
bool mulofthree(int n, int a, int b){
    int sum = a+b;
    if(sum==1 || sum%10==0 || sum%10==5) return false;
    if(n==2) sum = a+b;
    else{
        sum+=sum%10;
        int rem = (n-3)>=0 ? (n-3)%4 : 0;
        int q = (n-3)>=0 ? (n-3)/4 : 0;
        sum+=q*20;
        while(rem--){
            sum+=sum%10;
        }
    }
    if(sum % 3==0) return true;
    else return false;
}
signed main()
{
    int t;
    cin>>t;
    while(t--){
        int n, a, b;
        cin>>n>>a>>b;
        if(mulofthree(n,a,b)) cout<<"YES\n";
        else cout<<"NO\n";
    }
    return 0;
}

Multiple of 3 CodeChef Solution in Python

Comming soon

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Disclaimer: The above Problem (Multiple of 3 CodeChef) is generated by CodeChef but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purpose.

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