Divisibility | Practice problem Hackerearth solution

 Problem:-

You are provided an array $A$ of size $N$ that contains non-negative integers. Your task is to determine whether the number that is formed by selecting the last digit of all the N numbers is divisible by $10$.

Note: View the sample explanation section for more clarification.

Input format

• First line: A single integer $N$ denoting the size of array $A$
• Second line: $N$ space-separated integers.

Output format

If the number is divisible by $10$, then print $Yes$. Otherwise, print $No$.

Constraints
$$\begin{gathered} 1\le N\le {10}^5 \\ 0\le A[i]\le {10}^5 \end{gathered}$$

Sample Input

5
85 25 65 21 84

Sample Output

No

Time Limit: 1

Memory Limit: 256

Source Limit:

Explanation

Last digit of $85$ is $5$.
Last digit of $25$ is $5$.
Last digit of $65$ is $5$.
Last digit of $21$ is $1$.
Last digit of $84$ is $4$.
Therefore the number formed is $55514$ which is not divisible by $10$.

Code:-

  Here I am going to give you two solution first one is on the basis of C language and second one is on the basis of c++ language which you can submit in c++14 and c++17 also

Solution 1 ( C language):-

include <stdio.h>

int main(){

    int N = 0;

    scanf("%d", &N);

    

    long data[N];

    for(auto i=0; i<N; i++)

     scanf("%ld", &data[i]);

    char ans[3];

    

    if (data[N-1] % 10 == 0)

        printf("Yes");

    else

        printf("No");

return 0;

}

Solution 2 ( C++ language):-

 This solution is based on the c++ language and you can submit ib c++14 and c++17 also.
In this solution first three lines of the main function is only for the decreasing the time of execution of the program..

ios_base::sync_with_stdio(false);

cin.tie(NULL);

cout.tie(NULL);

This is your choice that you want to use this or not but in some cases the code may take more time in execution and that time we need it .

#include<bits/stdc++.h>

using namespace std;

int main()

{

ios_base::sync_with_stdio(false);

cin.tie(NULL);

    cout.tie(NULL);

int n,i;

cin>>n;

int a[n] , b[n];

for(i=0;i<n;i++)

{

cin>>a[i];

}

for(i=0;i<n;i++)

{

b[i] = a[i]%10;

}

if(b[n-1] == 0)

cout<<"Yes\n";

else

cout<<"No\n";

}

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