6278. City numbers

As you may be aware, houses with even street numbers are typically situated on one side of the street, while those with odd numbers are positioned on the opposite side. Determine whether the houses numbered n and m are situated on the same side of the street.

Input. Two integers n and m (1 ≤ n, m ≤ 100).

Output. Print 1 if the houses with numbers n and m are located on the same side of the street, and 0 otherwise.

Sample input	Sample output
1 2	0

SOLUTION

conditional statement

Algorithm analysis

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 is even, then n and m have the same parity.

· If the sum of the numbers 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:

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

This can be simplified to:

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

Algorithm implementation

Read the input data.

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

If the sum of the numbers 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();

}

}

Python implementation

Read the input data.

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

If the sum of the numbers 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")