高效地找到数字的奇偶性

给定整数N。任务是编写程序以查找给定数字的奇偶校验。
注意：数字奇偶校验用于定义数字中的置位总数(二进制表示形式为1位)是偶数还是奇数。如果一个数字的二进制表示形式中的置位总数为偶数，则该数字被称为具有偶数奇偶校验，否则，它将具有奇数奇偶校验。

例子：

Input : N = 13 
Output : Odd Parity
Explanation:
Binary representation of 13 is (1101)

Input : N = 9 (1001)
Output : Even Parity

通过执行以下操作，可以有效地计算由32位表示的数字的奇偶校验。
假设给定数字为x，然后执行以下操作：

• y = x ^(x >> 1)
• y = y ^(y >> 2)
• y = y ^(y >> 4)
• y = y ^(y >> 8)
• y = y ^(y >> 16)

现在，y中最右边的位将表示x的奇偶校验。如果最右边的位是1，则x将具有奇数奇偶校验，如果它是0，则x将具有偶数奇偶校验。
因此，为了提取y的最后一位，请对y与1进行按位与运算。

为什么这样做?

Consider that we want to find the parity of n = 150 = 1001 0110 (in binary).

1. Let’s divide this number into two parts and xor them and assign it to n: n = n ^ (n >> 4) = 1001 ^ 0110 = 1111.

Dissimilar bits result in a 1 bit in the result while similar bits result in a 0. We have basically considered all 8 bits to arrive at this intermediate result. So, effectively we have nullified even parities within the number.

Now repeat step 1 again until you end up with a single bit.
n = n ^ (n >> 2) = 11 ^ 11 = 00
n = n ^ (n >> 1) = 0 ^ 0 = 0
Final result = n & 1 = 0

Another example:

n = 1000 0101
n = n ^ (n >> 4) = 1000 ^ 0101 = 1101
n = n ^ (n >> 2) = 11 ^ 01 = 10
n = n ^ (n >> 1) = 1 ^ 0 = 1
Final result = n & 1 = 1

if(y&1==1)
    odd Parity
else
    even Parity

下面是上述方法的实现：

C++

// Program to find the parity of a given number
#include 
 
using namespace std;
 
// Function to find the parity
bool findParity(int x)
{
    int y = x ^ (x >> 1);
    y = y ^ (y >> 2);
    y = y ^ (y >> 4);
    y = y ^ (y >> 8);
    y = y ^ (y >> 16);
 
    // Rightmost bit of y holds the parity value
    // if (y&1) is 1 then parity is odd else even
    if (y & 1)
        return 1;
    return 0;
}
 
// Driver code
int main()
{
    (findParity(9)==0)?cout<<"Even Parity\n":
                            cout<<"Odd Parity\n";
     
    (findParity(13)==0)?cout<<"Even Parity\n":
                            cout<<"Odd Parity\n";
     
    return 0;
}

Java

// Program to find the
// parity of a given number
import java.io.*;
 
class GFG
{
 
// Function to find the parity
static boolean findParity(int x)
{
    int y = x ^ (x >> 1);
        y = y ^ (y >> 2);
        y = y ^ (y >> 4);
        y = y ^ (y >> 8);
        y = y ^ (y >> 16);
 
    // Rightmost bit of y holds
    // the parity value
    // if (y&1) is 1 then parity
    // is odd else even
    if ((y & 1) > 0)
        return true;
    return false;
}
 
// Driver code
public static void main (String[] args)
{
    if((findParity(9) == false))
        System.out.println("Even Parity");
    else
        System.out.println("Odd Parity");
     
    if(findParity(13) == false)
        System.out.println("Even Parity");
    else
        System.out.println("Odd Parity");
}
}
 
// This Code is Contributed by chandan_jnu.

Python3

# Program to find the
# parity of a given number
 
# Function to find the parity
def findParity(x):
    y = x ^ (x >> 1);
    y = y ^ (y >> 2);
    y = y ^ (y >> 4);
    y = y ^ (y >> 8);
    y = y ^ (y >> 16);
 
    # Rightmost bit of y holds
    # the parity value if (y&1)
    # is 1 then parity is odd
    # else even
    if (y & 1):
        return 1;
    return 0;
 
# Driver code
if(findParity(9) == 0):
    print("Even Parity");
else:
    print("Odd Parity\n");
 
if(findParity(13) == 0):
    print("Even Parity");
else:
    print("Odd Parity");
     
# This code is contributed by mits

C#

// Program to find the
// parity of a given number
using System;
 
class GFG
{
 
// Function to find the parity
static bool findParity(int x)
{
    int y = x ^ (x >> 1);
        y = y ^ (y >> 2);
        y = y ^ (y >> 4);
        y = y ^ (y >> 8);
        y = y ^ (y >> 16);
 
    // Rightmost bit of y holds
    // the parity value
    // if (y&1) is 1 then parity
    // is odd else even
    if ((y & 1) > 0)
        return true;
    return false;
}
 
// Driver code
public static void Main ()
{
    if((findParity(9) == false))
        Console.WriteLine("Even Parity");
    else
        Console.WriteLine("Odd Parity");
     
    if(findParity(13) == false)
        Console.WriteLine("Even Parity");
    else
        Console.WriteLine("Odd Parity");
}
}
 
// This Code is Contributed
// by chandan_jnu

PHP

> 1);
    $y = $y ^ ($y >> 2);
    $y = $y ^ ($y >> 4);
    $y = $y ^ ($y >> 8);
    $y = $y ^ ($y >> 16);
 
    // Rightmost bit of y holds
    // the parity value if (y&1)
    // is 1 then parity is odd
    // else even
    if ($y & 1)
        return 1;
    return 0;
}
 
// Driver code
(findParity(9) == 0) ?
 print("Even Parity\n"):
 print("Odd Parity\n");
 
(findParity(13) == 0) ?
print("Even Parity\n"):
print("Odd Parity\n");
     
// This Code is Contributed by mits
?>

输出：

Even Parity
Odd Parity

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。