Create Number from Multiplication of Digits

The program must accept an integer N as the input. The task is to form a new number by multiplying each of the digits of N. If any of the digits is 0, omit that digit from multiplication.

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 multiplying the digits of N.

Example Input/Output 1:
Input:
12345

Output:
120

Explanation:
The multiplication of digits in 12345 is 1 * 2 * 3 * 4 * 5 = 120.

Example Input/Output 2:
Input:
1052

Output:
10

Explanation:
The multiplication of non-zero digits in 1052 is 1 * 5 * 2 = 10.

n = input().strip()
result = 1
for ch in n:
    if ch != '0':
        result *= int(ch)
print(result)

#include<stdio.h>

int main() {
    char n[12];
    scanf("%s", n);
    int result = 1;
    for(int i = 0; n[i] != ''; i++) {
        if(n[i] != '0') {
            result *= (n[i] - '0');
        }
    }
    printf("%d", result);
    return 0;
}

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

int main() {
    string n;
    cin >> n;
    int result = 1;
    for(char ch : n) {
        if(ch != '0') {
            result *= (ch - '0');
        }
    }
    cout << result;
    return 0;
}

import java.util.Scanner;

public class MultiplyDigits {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String n = sc.next();
        int result = 1;
        for(char ch : n.toCharArray()) {
            if(ch != '0') {
                result *= (ch - '0');
            }
        }
        System.out.println(result);
    }
}