满足给定条件的数组中的对
给定一个数组 arr[] ，任务是计算数组中所有有效的对。如果func(arr[I])+func(arr[j])= func(XOR(arr[I]，arr[j]) ) 其中 func(x) 返回 x 中设置位的个数，则一对 (arr[i]，arr[j]) 被认为是有效的。
示例:
输入: arr[] = {2，3，4，5，6} 输出: 3 (2，4)、(2，5)和(3，4)是唯一有效的对。
输入: arr[] = {12，13，34，25，6 } T3】输出: 4
方法:迭代每一个可能的对，检查该对是否满足给定的条件。如果条件满足，则更新计数=计数+ 1 。最后打印计数。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number
// of set bits in n
int setBits(int n)
{
int count = 0;
while (n) {
n = n & (n - 1);
count++;
}
return count;
}
// Function to return the count of required pairs
int countPairs(int a[], int n)
{
int count = 0;
for (int i = 0; i < n - 1; i++) {
// Set bits for first element of the pair
int setbits_x = setBits(a[i]);
for (int j = i + 1; j < n; j++) {
// Set bits for second element of the pair
int setbits_y = setBits(a[j]);
// Set bits of the resultant number which is
// the XOR of both the elements of the pair
int setbits_xor_xy = setBits(a[i] ^ a[j]);
// If the condition is satisfied
if (setbits_x + setbits_y == setbits_xor_xy)
// Increment the count
count++;
}
}
// Return the total count
return count;
}
// Driver code
int main()
{
int a[] = { 2, 3, 4, 5, 6 };
int n = sizeof(a) / sizeof(a[0]);
cout << countPairs(a, n);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function to return the number
// of set bits in n
static int setBits(int n)
{
int count = 0;
while (n > 0)
{
n = n & (n - 1);
count++;
}
return count;
}
// Function to return the count of
// required pairs
static int countPairs(int a[], int n)
{
int count = 0;
for (int i = 0; i < n - 1; i++)
{
// Set bits for first element
// of the pair
int setbits_x = setBits(a[i]);
for (int j = i + 1; j < n; j++)
{
// Set bits for second element
// of the pair
int setbits_y = setBits(a[j]);
// Set bits of the resultant number which is
// the XOR of both the elements of the pair
int setbits_xor_xy = setBits(a[i] ^ a[j]);
// If the condition is satisfied
if (setbits_x + setbits_y == setbits_xor_xy)
// Increment the count
count++;
}
}
// Return the total count
return count;
}
// Driver code
public static void main (String[] args)
{
int []a = { 2, 3, 4, 5, 6 };
int n = a.length;
System.out.println(countPairs(a, n));
}
}
// This code is contributed by ajit.
Python 3
# Python 3 implementation of the approach
# Function to return the number
# of set bits in n
def setBits(n):
count = 0
while (n):
n = n & (n - 1)
count += 1
return count
# Function to return the count
# of required pairs
def countPairs(a, n):
count = 0
for i in range(0, n - 1, 1):
# Set bits for first element
# of the pair
setbits_x = setBits(a[i])
for j in range(i + 1, n, 1):
# Set bits for second element
# of the pair
setbits_y = setBits(a[j])
# Set bits of the resultant number
# which is the XOR of both the
# elements of the pair
setbits_xor_xy = setBits(a[i] ^ a[j]);
# If the condition is satisfied
if (setbits_x +
setbits_y == setbits_xor_xy):
# Increment the count
count += 1
# Return the total count
return count
# Driver code
if __name__ == '__main__':
a = [2, 3, 4, 5, 6]
n = len(a)
print(countPairs(a, n))
# This code is contributed by
# Sanjit_Prasad
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the number
// of set bits in n
static int setBits(int n)
{
int count = 0;
while (n > 0)
{
n = n & (n - 1);
count++;
}
return count;
}
// Function to return the count of
// required pairs
static int countPairs(int []a, int n)
{
int count = 0;
for (int i = 0; i < n - 1; i++)
{
// Set bits for first element
// of the pair
int setbits_x = setBits(a[i]);
for (int j = i + 1; j < n; j++)
{
// Set bits for second element
// of the pair
int setbits_y = setBits(a[j]);
// Set bits of the resultant number which is
// the XOR of both the elements of the pair
int setbits_xor_xy = setBits(a[i] ^ a[j]);
// If the condition is satisfied
if (setbits_x + setbits_y == setbits_xor_xy)
// Increment the count
count++;
}
}
// Return the total count
return count;
}
// Driver code
public static void Main()
{
int []a = { 2, 3, 4, 5, 6 };
int n = a.Length;
Console.Write(countPairs(a, n));
}
}
// This code is contributed
// by Akanksha Rai
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the number
// of set bits in n
function setBits($n)
{
$count = 0;
while ($n)
{
$n = $n & ($n - 1);
$count++;
}
return $count;
}
// Function to return the count of
// required pairs
function countPairs(&$a, $n)
{
$count = 0;
for ($i = 0; $i < $n - 1; $i++)
{
// Set bits for first element
// of the pair
$setbits_x = setBits($a[$i]);
for ($j = $i + 1; $j < $n; $j++)
{
// Set bits for second element of the pair
$setbits_y = setBits($a[$j]);
// Set bits of the resultant number which is
// the XOR of both the elements of the pair
$setbits_xor_xy = setBits($a[$i] ^ $a[$j]);
// If the condition is satisfied
if ($setbits_x +
$setbits_y == $setbits_xor_xy)
// Increment the count
$count++;
}
}
// Return the total count
return $count;
}
// Driver code
$a = array(2, 3, 4, 5, 6 );
$n = sizeof($a) / sizeof($a[0]);
echo countPairs($a, $n);
// This code is contributed by ita_c
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the number
// of set bits in n
function setBits(n)
{
let count = 0;
while (n > 0)
{
n = n & (n - 1);
count++;
}
return count;
}
// Function to return the count of
// required pairs
function countPairs(a, n)
{
let count = 0;
for(let i = 0; i < n - 1; i++)
{
// Set bits for first element
// of the pair
let setbits_x = setBits(a[i]);
for(let j = i + 1; j < n; j++)
{
// Set bits for second element
// of the pair
let setbits_y = setBits(a[j]);
// Set bits of the resultant number which is
// the XOR of both the elements of the pair
let setbits_xor_xy = setBits(a[i] ^ a[j]);
// If the condition is satisfied
if (setbits_x + setbits_y == setbits_xor_xy)
// Increment the count
count++;
}
}
// Return the total count
return count;
}
// Driver code
let a = [ 2, 3, 4, 5, 6 ];
let n = a.length;
document.write(countPairs(a, n));
// This code is contributed by unknown2108
</script>
Output:
3
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