按位“或”为奇数的对的数量
给定一个大小为 n 的数组 A[]，任务是找出存在多少对(I，j)，使得 A[i]或 A[j]为奇数。 例 :
Input : N = 4
A[] = { 5, 6, 2, 8 }
Output : 3
Explanation :
Since pair of A[] = ( 5, 6 ), ( 5, 2 ), ( 5, 8 ),
( 6, 2 ), ( 6, 8 ), ( 2, 8 )
5 OR 6 = 7, 5 OR 2 = 7, 5 OR 8 = 13
6 OR 2 = 6, 6 OR 8 = 14, 2 OR 8 = 10
Total pair A( i, j ) = 6 and Odd = 3
Input : N = 7
A[] = {8, 6, 2, 7, 3, 4, 9}
Output :15
一个简单的解决方案是检查每一对并找到按位或，然后用按位或计数所有这样的对为奇数。 以下是上述办法的实施情况:
C++
// C++ program to count pairs with odd OR
#include <iostream>
using namespace std;
// Function to count pairs with odd OR
int findOddPair(int A[], int N)
{
int oddPair = 0;
for (int i = 0; i < N; i++) {
for (int j = i + 1; j < N; j++) {
// find OR operation
// check odd or odd
if ((A[i] | A[j]) % 2 != 0)
oddPair++;
}
}
// return count of odd pair
return oddPair;
}
// Driver Code
int main()
{
int A[] = { 5, 6, 2, 8 };
int N = sizeof(A) / sizeof(A[0]);
cout << findOddPair(A, N) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count pairs
// with odd OR
class GFG
{
// Function to count pairs with odd OR
static int findOddPair(int A[], int N)
{
int oddPair = 0;
for (int i = 0; i < N; i++)
{
for (int j = i + 1; j < N; j++)
{
// find OR operation
// check odd or odd
if ((A[i] | A[j]) % 2 != 0)
oddPair++;
}
}
// return count of odd pair
return oddPair;
}
// Driver Code
public static void main(String []args)
{
int A[] = { 5, 6, 2, 8 };
int N = A.length;
System.out.println(findOddPair(A, N));
}
}
// This code is contributed by ANKITRAI1
Python 3
# Python3 program to count pairs with odd OR
# Function to count pairs with odd OR
def findOddPair(A, N):
oddPair = 0
for i in range(0, N):
for j in range(i+1, N):
# find OR operation
# check odd or odd
if ((A[i] | A[j]) % 2 != 0):
oddPair+=1
# return count of odd pair
return oddPair
# Driver Code
def main():
A = [ 5, 6, 2, 8 ]
N = len(A)
print(findOddPair(A, N))
if __name__ == '__main__':
main()
# This code is contributed by PrinciRaj1992
C
// C# program to count pairs
// with odd OR
using System;
public class GFG{
// Function to count pairs with odd OR
static int findOddPair(int[] A, int N)
{
int oddPair = 0;
for (int i = 0; i < N; i++)
{
for (int j = i + 1; j < N; j++)
{
// find OR operation
// check odd or odd
if ((A[i] | A[j]) % 2 != 0)
oddPair++;
}
}
// return count of odd pair
return oddPair;
}
// Driver Code
static public void Main (){
int []A = { 5, 6, 2, 8 };
int N = A.Length;
Console.WriteLine(findOddPair(A, N));
}
}
//This code is contributed by ajit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
//PHP program to count pairs with odd OR
// Function to count pairs with odd OR
function findOddPair($A, $N)
{
$oddPair = 0;
for ($i = 0; $i < $N; $i++) {
for ($j = $i + 1; $j < $N; $j++) {
// find OR operation
// check odd or odd
if (($A[$i] | $A[$j]) % 2 != 0)
$oddPair++;
}
}
// return count of odd pair
return $oddPair;
}
// Driver Code
$A = array (5, 6, 2, 8 );
$N = sizeof($A) / sizeof($A[0]);
echo findOddPair($A, $N),"\n";
#This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript program to count pairs with odd OR
// Function to count pairs with odd OR
function findOddPair(A, N)
{
let oddPair = 0;
for (let i = 0; i < N; i++)
{
for (let j = i + 1; j < N; j++)
{
// find OR operation
// check odd or odd
if ((A[i] | A[j]) % 2 != 0)
oddPair++;
}
}
// return count of odd pair
return oddPair;
}
// Driver Code
let A = [ 5, 6, 2, 8 ];
let N = A.length;
document.write(findOddPair(A, N));
// This code is contributed by souravmahato348.
</script>
Output:
3
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