Street numbers#6278

Street numbers (Талдоо)

Two houses are on the same side of the street if their numbers are both odd or both even.

If the sum of the numbers $n + m$ n + m is even, then $n$ and $m$ have the same parity.
If the sum of the numbers $n + m$ n + m is odd, then $n$ and $m$ have different parities.

The condition for checking whether $n$ and $m$ have the same parity can be expressed as:

i f ((n i s e v e n an d m i s e v e n) ∣∣ (n i s o dd an d m i s o dd))

if~((n~is~even~and~m~is~even)~||~(n~is~odd~and~m~is~odd))

This can be simplified to:

i f ((n + m) % 2 == 0)

if~((n + m)~\%~2 == 0)

Algorithm implementation

Read the input data.

scanf("%d %d",&n,&m);

If the sum of the numbers $n + m$ n + m is even, then $n$ and $m$ have the same parity. In this case print $1$ . Otherwise print $0$ .

if ((n + m) % 2 == 0) puts("1");
else puts("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 m = con.nextInt();

    if ((n + m) % 2 == 0) 
      System.out.println("1");
    else 
      System.out.println("0");

    con.close();
  }
}

Java

Python implementation

Read the input data.

n, m = map(int, input().split())

If the sum of the numbers $n + m$ n + m is even, then $n$ and $m$ have the same parity. In this case print $1$ . Otherwise print $0$ .

if (n + m) % 2 == 0:
  print("1")
else:
  print("0")