Editorial

Algorithm Analysis

Let $n$ be the input number. Let's identify its hundreds digit $b$ and tens digit $c$ :

$b = n /100%10;$
$c = n /10%10;$

The answer will be the number $res = 10 * b + c$ .

The task can also be solved by extracting the digits of the number $n$ using formatted input.

Algorithm Implementation

Read the input number $n$ . Extract the hundreds digit $b$ and the tens digit $c$ .

scanf("%d", &n);
b = n / 100 % 10;
c = n / 10 % 10;

Calculate and print the result.

res = 10 * b + c;
printf("%d\n", res);

Algorithm Implementation – Formatted Input

Extract the digits of thousands, hundreds, tens, and units.

scanf("%1d%1d%1d%1d",&a,&b,&c,&d);

Print the hundreds and tens digits.

printf("%d%d\n",b,c);

Implementation Using a Character Array

#include <stdio.h>

char *s;

int main(void)
{
    s = new char[5];
    gets(s);
    s[3] = 0;
    puts(s+1);
    delete[] s;
    return 0;
}

Java Implementation

import java.util.*;

public class Main
{
    public static void main(String[] args)
    {
        Scanner con = new Scanner(System.in);
        int n = con.nextInt();
        
        int b = n / 100 % 10;
        int c = n / 10 % 10;
        
        int res = 10 * b + c;
        System.out.print(res);
        con.close();
    }
}

Java Implementation – String

import java.util.*;

public class Main
{
    public static void main(String[] args)
    {
        Scanner con = new Scanner(System.in);
        String s = con.nextLine();
        System.out.printf("%c%c\n",s.charAt(1),s.charAt(2));
    }
}

Python Implementation

n = int(input())

b = n // 100 % 10
c = n // 10 % 10

res = 10 * b + c
print(res)