Dada una array arr[] de longitud N, cuente el número de pares (i, j) tales que arr[i] * arr[j] = arr[i] + arr[j] y 0 <= i < j <= n _ También se da que los elementos de la array pueden ser cualquier número entero positivo, incluido el cero.
Ejemplos:
Input : arr[] = {2, 0, 3, 2, 0}
Output : 2
Input : arr[] = {1, 2, 3, 4}
Output : 0
Solución simple:
podemos generar todos los pares posibles de la array y contar aquellos pares que satisfacen la condición dada.
A continuación se muestra la implementación del enfoque anterior:
CPP
// C++ program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
long countPairs(int arr[], int n)
{
long count = 0;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Increment count if condition satisfy
if (arr[i] * arr[j] == arr[i] + arr[j])
count++;
}
}
// Return count of pairs
return count;
}
// Driver code
int main()
{
int arr[] = { 2, 0, 3, 2, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
// Get and print count of pairs
cout << countPairs(arr, n);
return 0;
}
Java
// Java program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
class GFG {
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
static long countPairs(int arr[], int n)
{
long count = 0;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Increment count if condition satisfy
if (arr[i] * arr[j] == arr[i] + arr[j])
count++;
}
}
// Return count of pairs
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 0, 3, 2, 0 };
int n = arr.length;
// Get and print count of pairs
System.out.println(countPairs(arr, n));
}
}
Python3
# Python3 program to count pairs (i, j) # such that arr[i] * arr[j] = arr[i] + arr[j] # Function to return the count of pairs(i, j) # such that arr[i] * arr[j] = arr[i] + arr[j] def countPairs(arr, n) : count = 0; for i in range(n - 1) : for j in range(i + 1, n) : # Increment count if condition satisfy if (arr[i] * arr[j] == arr[i] + arr[j]) : count += 1; # Return count of pairs return count; # Driver code if __name__ == "__main__" : arr = [ 2, 0, 3, 2, 0 ]; n = len(arr); # Get and print count of pairs print(countPairs(arr, n)); # This code is contributed by AnkitRai01
C#
// C# program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
using System;
class GFG {
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
static long countPairs(int[] arr, int n)
{
long count = 0;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Increment count if condition satisfy
if (arr[i] * arr[j] == arr[i] + arr[j])
count++;
}
}
// Return count of pairs
return count;
}
// Driver code
public static void Main(string[] args)
{
int[] arr = { 2, 0, 3, 2, 0 };
int n = arr.Length;
// Get and print count of pairs
Console.WriteLine(countPairs(arr, n));
}
}
Javascript
<script>
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
function countPairs(arr, n)
{
let count = 0;
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
// Increment count if condition satisfy
if (arr[i] * arr[j] == arr[i] + arr[j])
count++;
}
}
// Return count of pairs
return count;
}
let arr = [ 2, 0, 3, 2, 0 ];
let n = arr.length;
document.write(countPairs(arr, n));
// This code is contributed by khatriharsh281</script>
CPP
// C++ program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
long countPairs(int arr[], int n)
{
int countZero = 0;
int countTwo = 0;
// Count number of 0's and 2's in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 0)
countZero++;
else if (arr[i] == 2)
countTwo++;
}
// Total pairs due to occurrence of 0's
long pair0 = (countZero * (countZero - 1)) / 2;
// Total pairs due to occurrence of 2's
long pair2 = (countTwo * (countTwo - 1)) / 2;
// Return count of all pairs
return pair0 + pair2;
}
// Driver code
int main()
{
int arr[] = { 2, 0, 3, 2, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
// Get and print count of pairs
cout << countPairs(arr, n);
return 0;
}
Java
// Java program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
class GFG {
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
static long countPairs(int arr[], int n)
{
int countZero = 0;
int countTwo = 0;
// Count number of 0's and 2's in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 0)
countZero++;
else if (arr[i] == 2)
countTwo++;
}
// Total pairs due to occurrence of 0's
long pair0 = (countZero * (countZero - 1)) / 2;
// Total pairs due to occurrence of 2's
long pair2 = (countTwo * (countTwo - 1)) / 2;
// Return count of all pairs
return pair0 + pair2;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 0, 3, 2, 0 };
int n = arr.length;
// Get and print count of pairs
System.out.println(countPairs(arr, n));
}
}
Python3
# Python3 program to count pairs (i, j) # such that arr[i] * arr[j] = arr[i] + arr[j] # Function to return the count of pairs(i, j) # such that arr[i] * arr[j] = arr[i] + arr[j] def countPairs(arr, n): countZero = 0; countTwo = 0; # Count number of 0's and 2's in the array for i in range(n) : if (arr[i] == 0) : countZero += 1; elif (arr[i] == 2) : countTwo += 1; # Total pairs due to occurrence of 0's pair0 = (countZero * (countZero - 1)) // 2; # Total pairs due to occurrence of 2's pair2 = (countTwo * (countTwo - 1)) // 2; # Return count of all pairs return pair0 + pair2; # Driver code if __name__ == "__main__" : arr = [ 2, 0, 3, 2, 0 ]; n = len(arr); # Get and print count of pairs print(countPairs(arr, n)); # This code is contributed by AnkitRai01
C#
// C# program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
using System;
class GFG {
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
static long countPairs(int[] arr, int n)
{
int countZero = 0;
int countTwo = 0;
// Count number of 0's and 2's in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 0)
countZero++;
else if (arr[i] == 2)
countTwo++;
}
// Total pairs due to occurrence of 0's
long pair0 = (countZero * (countZero - 1)) / 2;
// Total pairs due to occurrence of 2's
long pair2 = (countTwo * (countTwo - 1)) / 2;
// Return count of all pairs
return pair0 + pair2;
}
// Driver code
public static void Main(string[] args)
{
int[] arr = { 2, 0, 3, 2, 0 };
int n = arr.Length;
// Get and print count of pairs
Console.WriteLine(countPairs(arr, n));
}
}
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