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

川长思鸟来 2023-03-05 04:29 13阅读 0赞

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][bitwise operator]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

> 翻译自: [https://www.includehelp.com/c-programs/check-number-is-palindrome-or-not-using-bitwise-operator.aspx][https_www.includehelp.com_c-programs_check-number-is-palindrome-or-not-using-bitwise-operator.aspx]

c语言位运算符的使用

[bitwise operator]: https://www.includehelp.com/c/bitwise-operators-and-their-working-with-examples-in-c.aspx
[https_www.includehelp.com_c-programs_check-number-is-palindrome-or-not-using-bitwise-operator.aspx]: https://www.includehelp.com/c-programs/check-number-is-palindrome-or-not-using-bitwise-operator.aspx