Dados dos números enteros N y M , la tarea es contar todos los posibles pares de números enteros (i, j) (1 ≤ i ≤ N, 1 ≤ j ≤ M) tales que i + j sea par.
Ejemplos:
Entrada: N = 6, M = 4
Salida: 12
Explicación: Los pares (1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3 ), (4, 2), (4, 4), (5, 1), (5, 3), (6, 2), (6, 4) cumplen la condición requerida. Por lo tanto, la cuenta es 12.Entrada: N = 2 y M = 8
Salida: 8
Enfoque ingenuo: el enfoque más simple para resolver este problema es atravesar el rango [1, M] para cada número en ese rango, atravesar el rango [1, N] y generar todos los pares posibles. Para cada par posible, comprueba si su suma es un número par o no. Si se encuentra que es cierto, incremente el conteo. Finalmente, imprima el conteo obtenido.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to count pairs with even sum
int countEvenPairs(int N, int M)
{
// Stores the count of pairs
// with even sum
int count = 0;
// Traverse the range 1 to N
for(int i = 1; i <= N; i++)
{
// Traverse the range 1 to M
for(int j = 1; j <= M; j++)
{
// Check if the sum of the
// pair (i, j) is even or not
if ((i + j) % 2 == 0)
{
// Update count
count++;
}
}
}
// Return the count
return count;
}
// Driver Code
int main()
{
int N = 4;
int M = 6;
cout << countEvenPairs(N, M) << endl;
return 0;
}
// This code is contributed by akhilsaini
Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to count pairs with even sum
public static int countEvenPairs(
int N, int M)
{
// Stores the count of pairs
// with even sum
int count = 0;
// Traverse the range 1 to N
for (int i = 1; i <= N; i++) {
// Traverse the range 1 to M
for (int j = 1; j <= M; j++) {
// Check if the sum of the
// pair (i, j) is even or not
if ((i + j) % 2 == 0) {
// Update count
count++;
}
}
}
// Return the count
return count;
}
// Driver Code
public static void main(String[] args)
{
int N = 4;
int M = 6;
System.out.print(
countEvenPairs(N, M));
}
}
Python3
# Python3 program for the above approach # Function to count pairs with even sum def countEvenPairs(N, M): # Stores the count of pairs # with even sum count = 0 # Traverse the range 1 to N for i in range(1, N + 1): # Traverse the range 1 to M for j in range(1, M + 1): # Check if the sum of the # pair (i, j) is even or not if ((i + j) % 2 == 0): # Update count count += 1 # Return the count return count # Driver Code if __name__ == '__main__': N = 4 M = 6 print(countEvenPairs(N, M)) # This code is contributed by akhilsaini
C#
// C# program for the above approach
using System;
class GFG{
// Function to count pairs with even sum
public static int countEvenPairs(int N, int M)
{
// Stores the count of pairs
// with even sum
int count = 0;
// Traverse the range 1 to N
for(int i = 1; i <= N; i++)
{
// Traverse the range 1 to M
for(int j = 1; j <= M; j++)
{
// Check if the sum of the
// pair (i, j) is even or not
if ((i + j) % 2 == 0)
{
// Update count
count++;
}
}
}
// Return the count
return count;
}
// Driver Code
public static void Main()
{
int N = 4;
int M = 6;
Console.WriteLine(
countEvenPairs(N, M));
}
}
// This code is contributed by akhilsaini
Javascript
<script>
// Javascript program for the above approach
// Function to count pairs with even sum
function countEvenPairs(N, M)
{
// Stores the count of pairs
// with even sum
var count = 0;
// Traverse the range 1 to N
for(var i = 1; i <= N; i++)
{
// Traverse the range 1 to M
for(var j = 1; j <= M; j++)
{
// Check if the sum of the
// pair (i, j) is even or not
if ((i + j) % 2 == 0)
{
// Update count
count++;
}
}
}
// Return the count
return count;
}
// Driver code
var N = 4;
var M = 6;
document.write(countEvenPairs(N, M));
// This code is contributed by Kirti
</script>
Producción:
12
Complejidad de tiempo: O(N*M)
Complejidad de espacio: O(1)
Enfoque eficiente: el enfoque anterior se puede optimizar en función de las siguientes observaciones:
- número par + número par = número par
- número impar + número impar = número par
Siga los pasos a continuación para resolver el problema:
- Inicialice dos variables, digamos nEven y nOdd , para almacenar el recuento de enteros pares e impares hasta N .
- Inicialice dos variables, digamos mEven y mOdd , para almacenar el recuento de enteros pares e impares hasta M .
- Finalmente, cuente el número requerido de pares usando la fórmula:
cuenta = nPar * mPar + nOdd * mOdd
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count even pairs
int countEvenPairs(int N, int M)
{
// Stores count of pairs having even sum
int count = 0;
// Stores count of even numbers up to N
int nEven = floor(N / 2);
// Stores count of odd numbers up to N
int nOdd = ceil(N / 2);
// Stores count of even numbers up to M
int mEven = floor(M / 2);
// Stores count of odd numbers up to M
int mOdd = ceil(M / 2);
count = nEven * mEven + nOdd * mOdd;
// Return the count
return count;
}
// Driver Code
int main()
{
int N = 4;
int M = 6;
cout << countEvenPairs(N, M);
return 0;
}
// This code is contributed by Dharanendra L V
Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to count even pairs
public static int countEvenPairs(
int N, int M)
{
// Stores count of pairs having even sum
int count = 0;
// Stores count of even numbers up to N
int nEven = (int)Math.floor((double)N / 2);
// Stores count of odd numbers up to N
int nOdd = (int)Math.ceil((double)N / 2);
// Stores count of even numbers up to M
int mEven = (int)Math.floor((double)M / 2);
// Stores count of odd numbers up to M
int mOdd = (int)Math.ceil((double)M / 2);
count = nEven * mEven + nOdd * mOdd;
// Return the count
return count;
}
// Driver Code
public static void main(String[] args)
{
int N = 4;
int M = 6;
System.out.print(countEvenPairs(N, M));
}
}
Python3
# Python3 program for the above approach import math # Function to count even pairs def countEvenPairs(N, M): # Stores count of pairs having even sum count = 0; # Stores count of even numbers up to N nEven = int(math.floor(N / 2)); # Stores count of odd numbers up to N nOdd = int(math.ceil(N / 2)); # Stores count of even numbers up to M mEven = int(math.floor(M / 2)); # Stores count of odd numbers up to M mOdd = int(math.ceil(M / 2)); count = nEven * mEven + nOdd * mOdd; # Return the count return count; # Driver Code if __name__ == '__main__': N = 4; M = 6; print(countEvenPairs(N, M)); # This code is contributed by 29AjayKumar
C#
// C# program for the above approach
using System;
class GFG
{
// Function to count even pairs
public static int countEvenPairs(int N, int M)
{
// Stores count of pairs having even sum
int count = 0;
// Stores count of even numbers up to N
int nEven = (int)Math.Floor((double)N / 2);
// Stores count of odd numbers up to N
int nOdd = (int)Math.Ceiling((double)N / 2);
// Stores count of even numbers up to M
int mEven = (int)Math.Floor((double)M / 2);
// Stores count of odd numbers up to M
int mOdd = (int)Math.Ceiling((double)M / 2);
count = nEven * mEven + nOdd * mOdd;
// Return the count
return count;
}
// Driver Code
public static void Main(String[] args)
{
int N = 4;
int M = 6;
Console.Write(countEvenPairs(N, M));
}
}
// This code is contributed by shikhasingrajput
Javascript
<script>
// Javascript program for the above approach
// Function to count even pairs
function countEvenPairs(N, M)
{
// Stores count of pairs having even sum
let count = 0;
// Stores count of even numbers up to N
nEven = parseInt(Math.floor(N / 2));
// Stores count of odd numbers up to N
nOdd = parseInt(Math.ceil(N / 2));
// Stores count of even numbers up to M
mEven = parseInt(Math.floor(M / 2));
// Stores count of odd numbers up to M
mOdd = parseInt(Math.ceil(M / 2));
count = nEven * mEven + nOdd * mOdd;
// Return the count
return count;
}
// Driver Code
let N = 4;
let M = 6;
document.write(countEvenPairs(N, M));
// This code is contributed by mohan1240760
</script>
Producción:
12
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)