Create a Number by Doubling Each Digit

The program must accept an integer N as the input. The task is to create a new number by doubling each digit of N and return the new number. If doubling a digit results in a number greater than 9, then only the unit’s place of the result must be considered.

Boundary Condition(s):
1 <= N <= 10^9

Input Format:
The first line contains the integer N.

Output Format:
The first line contains the new number formed by doubling each digit of N.

Example Input/Output 1:
Input:
248

Output:
416

Explanation:
By doubling each digit, we get the sequences: 4, 16, 16. Using only the unit’s place from each result, we get the new number 416.

Example Input/Output 2:
Input:
359

Output:
6108

Explanation:
By doubling each digit, we get the sequences: 6, 10, 18. Using only the unit’s place from each result, we get the new number 6108.

n = input().strip()
result = ''.join(str((int(ch) * 2) % 10) for ch in n)
print(result)

#include<stdio.h>
#include<string.h>

int main() {
    char n[12];
    scanf("%s", n);
    int len = strlen(n);
    for(int i = 0; i < len; i++) {
        printf("%d", (2 * (n[i] - '0')) % 10);
    }
    return 0;
}

#include <iostream>
#include <string>
using namespace std;

int main() {
    string n;
    cin >> n;
    for(char ch : n) {
        cout << (2 * (ch - '0')) % 10;
    }
    return 0;
}

import java.util.Scanner;

public class DoubleEachDigit {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String n = sc.next();
        for(char ch : n.toCharArray()) {
            System.out.print((2 * (ch - '0')) % 10);
        }
    }
}