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

1N1050A[i]105

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[N1] % 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[n1] == 0)
cout<<“Yesn”;
else
cout<<“Non”;
}

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