寻求平价的计划

奇偶校验：数字的奇偶校验是指它是否包含奇数或偶数个1位。如果该数字包含奇数个1位，则具有“奇校验”，如果包含偶数个1位，则具有“偶校验”。
以下解决方案的主要思想是–循环，当n不为0时，在循环中取消设置位之一并反转奇偶校验。

Algorithm: getParity(n)
1. Initialize parity = 0
2. Loop while n != 0      
      a. Invert parity 
             parity = !parity
      b. Unset rightmost set bit
             n = n & (n-1)
3. return parity

Example:
 Initialize: n = 13 (1101)   parity = 0

n = 13 & 12  = 12 (1100)   parity = 1
n = 12 & 11 = 8  (1000)   parity = 0
n = 8 & 7 = 0  (0000)    parity = 1

程序：

C++

// C++ program to find parity
// of an integer
# include
# define bool int
using namespace std;
 
// Function to get parity of number n. It returns 1
// if n has odd parity, and returns 0 if n has even
// parity
bool getParity(unsigned int n)
{
    bool parity = 0;
    while (n)
    {
        parity = !parity;
        n     = n & (n - 1);
    }    
    return parity;
}
 
/* Driver program to test getParity() */
int main()
{
    unsigned int n = 7;
    cout<<"Parity of no "<

---

C

// C program to find parity
// of an integer
# include 
# define  bool int
 
/* Function to get parity of number n. It returns 1
   if n has odd parity, and returns 0 if n has even
   parity */
bool getParity(unsigned int n)
{
    bool parity = 0;
    while (n)
    {
        parity = !parity;
        n      = n & (n - 1);
    }       
    return parity;
}
 
/* Driver program to test getParity() */
int main()
{
    unsigned int n = 7;
    printf("Parity of no %d = %s",  n,
             (getParity(n)? "odd": "even"));
     
    getchar();
    return 0;
}

Java

// Java program to find parity
// of an integer
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger;
 
class GFG
 {
    /* Function to get parity of number n.
    It returns 1 if n has odd parity, and
    returns 0 if n has even parity */
    static boolean getParity(int n)
    {
        boolean parity = false;
        while(n != 0)
        {
            parity = !parity;
            n = n & (n-1);
        }
        return parity;
         
    }
     
    /* Driver program to test getParity() */
    public static void main (String[] args)
    {
        int n = 12;
        System.out.println("Parity of no " + n + " = " +
                         (getParity(n)? "odd": "even"));
    }
}
/* This code is contributed by Amit khandelwal*/

Python3

# Python3 code to get parity.
 
# Function to get parity of number n.
# It returns 1 if n has odd parity,
# and returns 0 if n has even parity
def getParity( n ):
    parity = 0
    while n:
        parity = ~parity
        n = n & (n - 1)
    return parity
 
# Driver program to test getParity()
n = 7
print ("Parity of no ", n," = ",
     ( "odd" if getParity(n) else "even"))
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to find parity of an integer
using System;
 
class GFG {
     
    /* Function to get parity of number n.
    It returns 1 if n has odd parity, and
    returns 0 if n has even parity */
    static bool getParity(int n)
    {
        bool parity = false;
        while(n != 0)
        {
            parity = !parity;
            n = n & (n-1);
        }
        return parity;
         
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 7;
        Console.Write("Parity of no " + n
                 + " = " + (getParity(n)?
                          "odd": "even"));
    }
}
 
// This code is contributed by nitin mittal.

Javascript

 
// Javascript program to find parity
// of an integer
 
// Function to get parity of number n.
// It returns 1 if n has odd parity, and
// returns 0 if n has even parity
function getParity(n)
{
    var parity = false;
    while(n != 0)
    {
        parity = !parity;
        n = n & (n - 1);
    }
    return parity;
}
     
// Driver code
var n = 7;
document.write("Parity of no " + n + " = " +
              (getParity(n) ? "odd": "even"));
  
// This code is contributed by Kirti
 

输出：

Parity of no 7 = odd

通过使用查找表可以优化上述解决方案。有关详细信息，请参阅Bit Twiddle Hacks [1st reference]。
时间复杂度：以上算法所花费的时间与设置的位数成正比。最糟糕的情况是O(Log n)。
用途：奇偶校验用于错误检测和加密。
使用XOR和表查找计算数字的奇偶校验
参考：
http://graphics.stanford.edu/~seander/bithacks.html#ParityNaive –最后检查日期为2009年5月30日。