Dados dos enteros positivos X e Y , la tarea es contar pares con una suma par posible eligiendo cualquier entero del rango 1 a X y otro entero del rango 1 a Y .
Ejemplos:
Entrada: X = 2, Y = 3
Salida: 3
Explicación: Todos estos pares posibles son {1, 1}, {1, 3}, {2, 2}.
Enfoque: siga los pasos a continuación para resolver el problema:
- Inicialice una variable, digamos cntXEvenNums , para almacenar el recuento de números pares en el rango [1, X] obtenido al dividir X entre 2 .
- Inicialice una variable, digamos cntXOddNums , para almacenar el recuento de números impares en el rango [1, X] obtenido al dividir (X + 1) entre 2 .
- Inicialice una variable, digamos cntYEvenNums , para almacenar el recuento de números pares entre 1 y Y dividiendo Y por 2 .
- Inicialice una variable, digamos cntYOddNums , para almacenar el recuento de números impares entre 1 y Y dividiendo (Y + 1) por 2 .
- Inicialice una variable, digamos cntPairs , para almacenar el recuento de pares de sumas pares multiplicando cntXEvenNums con cntYEvenNums y multiplicando cntXOddNums con cntYOddNums y encuentre la suma de ambos.
- Finalmente, imprime el valor de cntPairs obtenido.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count even
// sum pairs in the given range
long long cntEvenSumPairs(long long X, long long Y)
{
// Stores the count of even
// numbers between 1 to X
long long cntXEvenNums = X / 2;
// Stores the count of odd
// numbers between 1 to X
long long cntXOddNums = (X + 1) / 2;
// Stores the count of even
// numbers between 1 to Y
long long cntYEvenNums = Y / 2;
// Stores the count of odd
// numbers between 1 to Y
long long cntYOddNums = (Y + 1) / 2;
// Stores the count of
// pairs having even sum
long long cntPairs
= (cntXEvenNums * 1LL * cntYEvenNums)
+ (cntXOddNums * 1LL * cntYOddNums);
// Returns the count of pairs
// having even sum
return cntPairs;
}
// Driver Code
int main()
{
long long X = 2;
long long Y = 3;
cout << cntEvenSumPairs(X, Y);
return 0;
}
Java
// Java program to implement
// the above approach
import java.io.*;
class GFG {
// Function to count maximum even
// sum pairs in the given range
static long cntEvenSumPairs(long X, long Y)
{
// Stores the count of even
// numbers between 1 to X
long cntXEvenNums = X / 2;
// Stores the count of odd
// numbers between 1 to X
long cntXOddNums = (X + 1) / 2;
// Stores the count of even
// numbers between 1 to Y
long cntYEvenNums = Y / 2;
// Stores the count of odd
// numbers between 1 to Y
long cntYOddNums = (Y + 1) / 2;
// Stores the count of
// pairs having even sum
long cntPairs = (cntXEvenNums * cntYEvenNums)
+ (cntXOddNums * cntYOddNums);
// Returns the count of pairs
// having even sum
return cntPairs;
}
// Driver Code
public static void main(String[] args)
{
long X = 2;
long Y = 3;
System.out.println(cntEvenSumPairs(X, Y));
}
}
Python
# Python program to implement # the above approach # Function to count even # sum pairs in the given range def cntEvenSumPairs(X, Y): # Stores the count of even # numbers between 1 to X cntXEvenNums = X / 2 # Stores the count of odd # numbers between 1 to X cntXOddNums = (X + 1) / 2 # Stores the count of even # numbers between 1 to Y cntYEvenNums = Y / 2 # Stores the count of odd # numbers between 1 to Y cntYOddNums = (Y + 1) / 2 # Stores the count of # pairs having even sum cntPairs = ((cntXEvenNums * cntYEvenNums) + (cntXOddNums * cntYOddNums)) # Returns the count of pairs # having even sum return cntPairs # Driver code X = 2 Y = 3 print(cntEvenSumPairs(X, Y)) # This code is contributed by hemanth gadarla
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to count maximum even
// sum pairs in the given range
static long cntEvenSumPairs(long X, long Y)
{
// Stores the count of even
// numbers between 1 to X
long cntXEvenNums = X / 2;
// Stores the count of odd
// numbers between 1 to X
long cntXOddNums = (X + 1) / 2;
// Stores the count of even
// numbers between 1 to Y
long cntYEvenNums = Y / 2;
// Stores the count of odd
// numbers between 1 to Y
long cntYOddNums = (Y + 1) / 2;
// Stores the count of
// pairs having even sum
long cntPairs = (cntXEvenNums * cntYEvenNums) +
(cntXOddNums * cntYOddNums);
// Returns the count of pairs
// having even sum
return cntPairs;
}
// Driver Code
public static void Main(string[] args)
{
long X = 2;
long Y = 3;
Console.WriteLine(cntEvenSumPairs(X, Y));
}
}
// This code is contributed by chitranayal
Javascript
<script>
// Javascript program to implement
// the above approach
// Function to count maximum even
// sum pairs in the given range
function cntEvenSumPairs(X , Y) {
// Stores the count of even
// numbers between 1 to X
var cntXEvenNums = parseInt(X / 2);
// Stores the count of odd
// numbers between 1 to X
var cntXOddNums = parseInt((X + 1) / 2);
// Stores the count of even
// numbers between 1 to Y
var cntYEvenNums = parseInt(Y / 2);
// Stores the count of odd
// numbers between 1 to Y
var cntYOddNums =parseInt( (Y + 1) / 2);
// Stores the count of
// pairs having even sum
var cntPairs = (cntXEvenNums * cntYEvenNums) +
(cntXOddNums * cntYOddNums);
// Returns the count of pairs
// having even sum
return cntPairs;
}
// Driver Code
var X = 2;
var Y = 3;
document.write(cntEvenSumPairs(X, Y));
// This code contributed by Rajput-Ji
</script>
Producción:
3
Complejidad de Tiempo : O(1)
Espacio Auxiliar : O(1)
Publicación traducida automáticamente
Artículo escrito por dharanendralv23 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA