本文概述
- 建议:在继续解决方案之前, 请先在"实践"上解决它。
- C ++
- Python3
- 的PHP
1 -->
parity of the set is odd0 -->
parity of the set is even
例子:
Input : 254Output : Odd ParityExplanation : Binary of 254 is 11111110. There are 7 ones. Thus, parity is odd.Input : 1742346774Output : Even
推荐:请在"实践首先, 在继续解决方案之前。方法1 :(幼稚的方法)
我们已经讨论过这种方法
这里
.
方法2:(有效)
前提条件:
查表
,
异或魔术
如果我们将数字S分为两部分1和S2这样S = S1小号2。如果我们知道S的奇偶性1和S2, 我们可以使用以下事实来计算S的奇偶校验:
- 如果S1和S2具有相同的奇偶校验, 即它们都具有偶数个位数或奇数个位数, 它们的并集S将具有偶数个位数。
- 因此, S的奇偶性是S的奇偶性的异或1和S2
步骤如下:
1. Create a look-up table for 8-bit numbers ( 0 to 255 )Parity of 0 is 0.Parity of 1 is 1....Parity of 255 is 0.2. Break the number into 8-bit chunkswhile performing XOR operations.3. Check for the result in the table forthe 8-bit number.
由于32位或64位数字包含恒定数量的字节, 因此上述步骤需要O(1)时间。
范例:
1. Take 32-bit number : 17423467742. Calculate Binary of the number : 011001111101101000011010000101103. Split the 32-bit binary representation into 16-bit chunks :0110011111011010 | 0001101000010110 4. Compute X-OR :0110011111011010^ 0001101000010110___________________= 01111101110011005. Split the 16-bit binary representation into 8-bit chunks : 01111101 | 110011006. Again, Compute X-OR :01111101^ 11001100___________________= 1011000110110001 is 177 in decimal. Check for its parity in look-up table :Even number of 1 = Even parity.Thus, Parity of 1742346774 is even.
下面是实现适用于32位和64位数字。
C ++
// CPP program to illustrate Compute the parity of a
// number using XOR
#include <
bits/stdc++.h>
// Generating the look-up table while pre-processing
#define P2(n) n, n ^ 1, n ^ 1, n
#define P4(n) P2(n), P2(n ^ 1), P2(n ^ 1), P2(n)
#define P6(n) P4(n), P4(n ^ 1), P4(n ^ 1), P4(n)
#define LOOK_UP P6(0), P6(1), P6(1), P6(0)// LOOK_UP is the macro expansion to generate the table
unsigned int table[256] = { LOOK_UP };
// Function to find the parity
int Parity( int num)
{
// Number is considered to be of 32 bits
int max = 16;
// Dividing the number into 8-bit
// chunks while performing X-OR
while (max >
= 8) {
num = num ^ (num >
>
max);
max = max / 2;
}// Masking the number with 0xff (11111111)
// to produce valid 8-bit result
return table[num &
0xff];
}// Driver code
int main()
{
unsigned int num = 1742346774;
// Result is 1 for odd parity, 0 for even parity
bool result = Parity(num);
// Printing the desired result
result ? std::cout <
<
"Odd Parity" :
std::cout <
<
"Even Parity" ;
return 0;
}
Python3
# Python3 program to illustrate Compute the
# parity of a number using XOR # Generating the look-up table while
# pre-processing
def P2(n, table):
table.extend([n, n ^ 1 , n ^ 1 , n])
def P4(n, table):
return (P2(n, table), P2(n ^ 1 , table), P2(n ^ 1 , table), P2(n, table))
def P6(n, table):
return (P4(n, table), P4(n ^ 1 , table), P4(n ^ 1 , table), P4(n, table))
def LOOK_UP(table):
return (P6( 0 , table), P6( 1 , table), P6( 1 , table), P6( 0 , table)) # LOOK_UP is the macro expansion to
# generate the table
table = [ 0 ] * 256
LOOK_UP(table)# Function to find the parity
def Parity(num) :# Number is considered to be
# of 32 bits
max = 16# Dividing the number o 8-bit
# chunks while performing X-OR
while ( max >
= 8 ):
num = num ^ (num >
>
max )
max = max / / 2# Masking the number with 0xff (11111111)
# to produce valid 8-bit result
return table[num &
0xff ] # Driver code
if __name__ = = "__main__" :
num = 1742346774# Result is 1 for odd parity, # 0 for even parity
result = Parity(num)
print ( "Odd Parity" ) if result else print ( "Even Parity" )# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)
的PHP
<
?php
// PHP program to illustrate
// Compute the parity of a
// number using XOR/* Generating the look-up
table while pre-processing
#define P2(n) n, n ^ 1, n ^ 1, n
#define P4(n) P2(n), P2(n ^ 1), P2(n ^ 1), P2(n)
#define P6(n) P4(n), P4(n ^ 1), P4(n ^ 1), P4(n)
#define LOOK_UP P6(0), P6(1), P6(1), P6(0)LOOK_UP is the macro expansion
to generate the table
$table = array(LOOK_UP );
*/// Function to find
// the parity
function Parity( $num )
{
global $table ;
// Number is considered
// to be of 32 bits
$max = 16;
// Dividing the number
// into 8-bit chunks
// while performing X-OR
while ( $max >
= 8)
{
$num = $num ^ ( $num >
>
$max );
$max = (int) $max / 2;
}// Masking the number with
// 0xff (11111111) to produce
// valid 8-bit result
return $table [ $num &
0xff];
}// Driver code
$num = 1742346774;
// Result is 1 for odd
// parity, 0 for even parity
$result = Parity( $num );
// Printing the desired result
if ( $result == true)
echo "Odd Parity" ;
else
echo "Even Parity" ;
// This code is contributed by ajit
?>
输出如下:
Even Parity
时间复杂度:O(1)。请注意, 32位或64位数字具有固定的字节数(如果是32位, 则为4, 如果是64位, 则为8)。
【算法设计(使用XOR和表查找计算数字的奇偶校验)】如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。
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