Repdigit数字

Repdigit Number是一个数字N ，在以B为基数的表示中，所有数字均相等。
某些Repdigit号码是：

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55…. 
 

检查N是否为重复数字

给定数字N ，任务是检查N是否为基数B中的Repdigit数字。如果N是基数B中的代表数字，则打印“是”，否则打印“否” 。
例子：

Input: N = 2000, B = 7 
Output: Yes 
Explanation: 
2000 in base 7 is 5555 which has all digits equal.
Input: N = 112, B = 10 
Output: No 
 

方法：

• 一对一地找到基数B中N的所有数字。
• 将每个数字与其前一个数字进行比较。
• 如果任何数字不等于前一个数字，则返回false。
• 否则返回true。

下面是上述方法的实现：

C++

// C++ implementation to check
// if a number is Repdigit
 
#include 
using namespace std;
 
// Function to check if a number
// is a Repdigit number
bool isRepdigit(int num, int b)
{
    // To store previous digit (Assigning
    // initial value which is less than any
    // digit)
    int prev = -1;
 
    // Traverse all digits from right to
    // left and check if any digit is
    // smaller than previous.
    while (num) {
        int digit = num % b;
        num /= b;
        if (prev != -1 && digit != prev)
            return false;
        prev = digit;
    }
 
    return true;
}
 
// Driver code
int main()
{
    int num = 2000, base = 7;
    isRepdigit(num, base) ? cout << "Yes"
                          : cout << "No";
    return 0;
}

Java

// Java implementation to check
// if a number is Repdigit
class GFG{
 
// Function to check if a number
// is a Repdigit number
static boolean isRepdigit(int num, int b)
{
    // To store previous digit (Assigning
    // initial value which is less than any
    // digit)
    int prev = -1;
 
    // Traverse all digits from right to
    // left and check if any digit is
    // smaller than previous.
    while (num != 0)
    {
        int digit = num % b;
        num /= b;
        if (prev != -1 && digit != prev)
            return false;
        prev = digit;
    }
    return true;
}
 
// Driver code
public static void main(String args[])
{
    int num = 2000, base1 = 7;
    if(isRepdigit(num, base1))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Code_Mech

Python3

# Python3 implementation to check
# if a number is Repdigit
 
# Function to check if a number
# is a Repdigit number
def isRepdigit(num, b) :
     
    # To store previous digit (Assigning
    # initial value which is less than any
    # digit)
    prev = -1
 
    # Traverse all digits from right to
    # left and check if any digit is
    # smaller than previous.
    while (num) :
        digit = num % b
        num //= b
        if (prev != -1 and digit != prev) :
            return False
        prev = digit
    return True
 
# Driver code
num = 2000
base = 7
if(isRepdigit(num, base)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Vishal Maurya.

C#

// C# implementation to check
// if a number is Repdigit
using System;
class GFG{
 
// Function to check if a number
// is a Repdigit number
static bool isRepdigit(int num, int b)
{
    // To store previous digit (Assigning
    // initial value which is less than any
    // digit)
    int prev = -1;
 
    // Traverse all digits from right to
    // left and check if any digit is
    // smaller than previous.
    while (num != 0)
    {
        int digit = num % b;
        num /= b;
        if (prev != -1 && digit != prev)
            return false;
        prev = digit;
    }
    return true;
}
 
// Driver code
public static void Main()
{
    int num = 2000, base1 = 7;
    if(isRepdigit(num, base1))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// Javascript implementation to check
// if a number is Repdigit
 
 
    // Function to check if a number
    // is a Repdigit number
    function isRepdigit( num ,b) {
        // To store previous digit (Assigning
        // initial value which is less than any
        // digit)
        let prev = -1;
 
        // Traverse all digits from right to
        // left and check if any digit is
        // smaller than previous.
        while (num != 0) {
            let digit = num % b;
            num = parseInt(num/b);
            if (prev != -1 && digit != prev)
                return false;
            prev = digit;
        }
        return true;
    }
 
    // Driver code
      
        let num = 2000, base1 = 7;
        if (isRepdigit(num, base1)) {
            document.write("Yes");
        } else {
            document.write("No");
        }
 
// This code contributed by gauravrajput1
 

输出：

Yes

时间复杂度： O(n)
参考：http://www.numbersaplenty.com/set/repdigit/