检查编号是否为回文数，是否在基数B中

给定整数N ，任务是检查是否

$N_B$

(碱基B中的N )是否为回文。

例子：

Input: N = 5, B = 2 
Output: Yes 
Explanation: 
(5)₁₀ = (101)₂ which is palindrome. Therefore, the required output is Yes.
Input: N = 4, B = 2 
Output: No 
 

处理方法：可以通过检查是否反转的十进制值来解决该问题。

$N_B$

是否等于N。请按照以下步骤解决问题。

1. 初始化变量rev = 0以存储N的倒数。
2. 提取的数字

$N_B$

1. 按N％B 。
2. 对于的每个数字

$N_B$

1. 更新rev = rev * B + N％B
2. 最后，检查N是否等于rev

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to check if N in
// base B is palindrome or not
int checkPalindromeB(int N, int B)
{
    // Stores the reverse of N
    int rev = 0;
 
    // Stores the value of N
    int N1 = N;
 
    // Extract all the digits of N
    while (N1) {
        // Generate its reverse
        rev = rev * B + N1 % B;
        N1 = N1 / B;
    }
 
    return N == rev;
}
 
// Driver Code
int main()
{
    int N = 5, B = 2;
    if (checkPalindromeB(N, B)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to check if N in
// base B is palindrome or not
static boolean checkPalindromeB(int N,
                                int B)
{
     
    // Stores the reverse of N
    int rev = 0;
 
    // Stores the value of N
    int N1 = N;
 
    // Extract all the digits of N
    while (N1 > 0)
    {
         
        // Generate its reverse
        rev = rev * B + N1 % B;
        N1 = N1 / B;
    }
    return N == rev;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 5, B = 2;
     
    if (checkPalindromeB(N, B))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Dewanti

Python3

# Python3 program to implement
# the above approach
 
# Function to check if N in
# base B is palindrome or not
def checkPalindromeB(N, B):
 
    # Stores the reverse of N
    rev = 0;
 
    # Stores the value of N
    N1 = N;
 
    # Extract all the digits of N
    while (N1 > 0):
 
        # Generate its reverse
        rev = rev * B + N1 % B;
        N1 = N1 // B;
     
    return N == rev;
 
# Driver code
if __name__ == '__main__':
    N = 5; B = 2;
 
    if (checkPalindromeB(N, B)):
        print("Yes");
    else:
        print("No");
 
# This code is contributed by Princi Singh

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to check if N in
// base B is palindrome or not
static bool checkPalindromeB(int N,
                             int B)
{
     
    // Stores the reverse of N
    int rev = 0;
 
    // Stores the value of N
    int N1 = N;
 
    // Extract all the digits of N
    while (N1 > 0)
    {
         
        // Generate its reverse
        rev = rev * B + N1 % B;
        N1 = N1 / B;
    }
    return N == rev;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 5, B = 2;
     
    if (checkPalindromeB(N, B))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to check if N in
// base B is palindrome or not
function checkPalindromeB(N, B)
{
    // Stores the reverse of N
    var rev = 0;
 
    // Stores the value of N
    var N1 = N;
 
    // Extract all the digits of N
    while (N1) {
        // Generate its reverse
        rev = rev * B + N1 % B;
        N1 = parseInt(N1 / B);
    }
 
    return N == rev;
}
 
// Driver Code
var N = 5, B = 2;
if (checkPalindromeB(N, B)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
 

输出：

Yes

时间复杂度： O(log _B N)
辅助空间： O(1)