c语言位运算符的使用_C语言程序使用位运算符检查给定数字是否为回文

c语言位运算符的使用

Problem statement: Write a C program to check whether a number (binary representation) is palindrome or not using bitwise operators. Maximum input is 255..

问题陈述：编写一个C程序以使用按位运算符检查数字(二进制表示形式)是否为回文。 最大输入为255 。

Solution: We can use bitwise operator here to solve the problem.

解决方案：我们可以在这里使用按位运算符来解决问题。

Pre-requisite: Input number n

前提条件 ：输入数字n

Input Example:

输入示例：

    Input number: 24
Binary representation: 00011000
Thus it's a palindrome
Input number 153:
Binary representation: 10011001
Thus it's a palindrome
Input number: 25
Binary representation: 00011001
Thus it's not a palindrome

Algorithm:

算法：

1)  Take the input.
2)  Create an array of length 8 to store 8 bit binary representation 
of input number
3)  Use bitwise operators to convert into binary from
a)  Initialize i to 7 (8-1)
b)  Find ith bit & store in the array
array[i]=n & 1
c)  Right shift n & decrement i
n=n>>1
d)  IF n=0
Break
ELSE
Repeat step b-d
4)  Check the array where the binary representation is stored, 
for palindrome
a)  Set j= 0 & k=7 (8-1, 8 is the MAX SIZE)
b)  IF (array [j]!= array [k])
It's not a palindrome
c)  IF j < k
Increment j& decrement k
Repeat b, c
ELSE
It's a palindrome

Example with explanation:

带有说明的示例：

Input no: 24
Converting to binary representation using bitwise operator
Initially,
N=24
Array[8]={0};
0	0	0	0	0	0	0	0 (LSB)
i=7
-----------------------------------------------------------------
first iteration,
array[7]=n&1 = 0 (refer to bitwise operators and 
their working for understanding the outcome)
0	0	0	0	0	0	0	0 (current bit)
n=n>>1=12
i=6
-----------------------------------------------------------------
second iteration,
array [6]=n&1 = 0 
0	0	0	0	0	0	0 (current bit)	0 
n=n>>1=6
i=5
-----------------------------------------------------------------
third iteration,
array [5]=n&1 = 0 
0	0	0	0	0	0 (current bit)	0 	0 
n=n>>1=3
i=4
-----------------------------------------------------------------
fourth iteration,
array [4]=n&1 = 1 
0	0	0	0	1(current bit)	0 	0 	0 
n=n>>1=1
i=3
-----------------------------------------------------------------
fifth iteration,
array [3]=n&1 = 1 
0	0	0	1 (current bit)	1	0 	0 	0 
n=n>>1=0
i=2
-----------------------------------------------------------------
sixth iteration,
n=0
so no more processing
thus the array finally becomes which is 
the binary representation
0	0	0	1 	1	0 	0 	0  (LSB)
Checking palindrome
Initially,
j=0 & k=7 (8-1)
-----------------------------------------------------------------
First iteration
Array[j] == array [k] (both 0)
j=j+1=1
k=k-1=6
j<k
-----------------------------------------------------------------
Second iteration
Array[j] == array [k] (both 0)
j=j+1=2
k=k-1=5
j<k
-----------------------------------------------------------------
Third iteration
Array[j] == array [k] (both 0)
j=j+1=3
k=k-1=4
j<k
-----------------------------------------------------------------
Fourth iteration
Array[j] == array [k] (both 1)
j=j+1=4
k=k-1=3
j>k
thus, stops processing & prints it's a palindrome

C Implementation

C实施

#include <stdio.h>
#define SIZE 8
int main()
{
unsigned int n;
printf("enter the no ( max range 255)\n");
scanf("%d",&n);
int c[SIZE]={0};
int i=SIZE-1;
printf("binary representation is: ");
while(n!=0){
c[i--]=n&1;
n=n>>1;
}
for(int j=0;j<SIZE;j++)
printf("%d",c[j]);
printf("\n");
for(int j=0,k=SIZE-1;j<k;j++,k--){
if(c[j]!=c[k]){
printf("Not palindrome\n");
return 0;
}
}
printf("it's palindrome\n");
return 0;
}

Output

输出量

First run:
enter the no ( max range 255)
153
binary representation is: 10011001
it's palindrome
Second run:
enter the no ( max range 255)
24
binary representation is: 00011000
it's palindrome
Third run:
enter the no ( max range 255)
-8
binary representation is: 11111000
Not palindrome