CONCEPTS USED:
Basic Mathematics
DIFFICULTY LEVEL:
Easy
PROBLEM STATEMENT(SIMPLIFIED):
Given
Nseats, where English paper students are evenly placed(0,2,4,...)and Hindi paper students are oddly placed(1,3,5,...). Print the roll number of English paper students.
See original problem statement here
For Example :
N = 5
Arr[] = [1, 2, 3, 4, 5]
Output: 1 3 5
Explanation: Students with roll number 1 3 5 are placed on even indexes i.e. 0 2 4.SOLVING APPROACH:
Keep traversing the array and print the elements that are placed on even indices as the English paper students are sitting on roll numbers like
0,2,4etc.If the current index is a multiple of
2, then print it else move ahead by referring Master algorithms online.
SOLUTIONS:
#include <stdio.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int arr[n];
for(int i=0;i<n;i++)
{
scanf("%d",&arr[i]);
if(i%2==0)
{
printf("%d ",arr[i]);
}
}
printf("\n");
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;cin>>t;
while(t--)
{
int n;cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
if(i%2==0)
{
cout<<arr[i]<<" ";
}
}
cout<<"\n";
}
return 0;
}
import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[]) throws IOException {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t!=0)
{
int n = sc.nextInt();
int arr[] = new int[n];
for(int i=0;i<n;i++)
{
arr[i] = sc.nextInt();
}
for(int i=0;i<n;i++)
{
if(i%2==0)
{
System.out.println(arr[i]);
}
}
System.out.println("");
t--;
}
}
}