Today we will be Solving Multiple of 3 CodeChef Problem which is a part of CodeChef DSA Learning Series.
Multiple of 3 CodeChef Solution
Problem
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
Disclaimer: The above Problem (Multiple of 3 CodeChef ) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.
