以任意给定数字结尾的范围[L，R]中的数字计数

给定的范围内[L，R]，则任务是从最后的数字或者是2，3或9的范围内找到号码的计数。

例子：

Input: L = 1, R = 3
Output: 2
2 and 3 are the only valid numbers.

Input: L = 11, R = 33
Output: 8

方法：初始化计数器的计数= 0，并为范围中的每个元素运行一个循环，如果当前元素以任何给定的数字结尾，则增加计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count
// of the required numbers
int countNums(int l, int r)
{
    int cnt = 0;
  
    for (int i = l; i <= r; i++)
    {
  
        // Last digit of the current number
        int lastDigit = (i % 10);
  
        // If the last digit is equal to
        // any of the given digits
        if ((lastDigit % 10) == 2 || (lastDigit % 10) == 3
            || (lastDigit % 10) == 9) 
        {
            cnt++;
        }
    }
  
    return cnt;
}
  
// Driver code
int main() 
{
    int l = 11, r = 33;
    cout << countNums(l, r) ;
}
      
// This code is contributed by AnkitRai01

Java

// Java implementation of the approach
class GFG {
  
    // Function to return the count
    // of the required numbers
    static int countNums(int l, int r)
    {
        int cnt = 0;
  
        for (int i = l; i <= r; i++) {
  
            // Last digit of the current number
            int lastDigit = (i % 10);
  
            // If the last digit is equal to
            // any of the given digits
            if ((lastDigit % 10) == 2 || (lastDigit % 10) == 3
                || (lastDigit % 10) == 9) {
                cnt++;
            }
        }
  
        return cnt;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int l = 11, r = 33;
        System.out.print(countNums(l, r));
    }
}

Python3

# Python3 implementation of the approach
  
# Function to return the count
# of the required numbers
def countNums(l, r) :
    cnt = 0;
  
    for i in range(l, r + 1) :
  
        # Last digit of the current number
        lastDigit = (i % 10);
  
        # If the last digit is equal to
        # any of the given digits
        if ((lastDigit % 10) == 2 or (lastDigit % 10) == 3
            or (lastDigit % 10) == 9) :
          
            cnt += 1;
  
    return cnt;
  
# Driver code
if __name__ == "__main__" : 
      
    l = 11; r = 33;
      
    print(countNums(l, r)) ;
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach 
using System;
  
class GFG 
{ 
  
    // Function to return the count 
    // of the required numbers 
    static int countNums(int l, int r) 
    { 
        int cnt = 0; 
  
        for (int i = l; i <= r; i++) 
        { 
  
            // Last digit of the current number 
            int lastDigit = (i % 10); 
  
            // If the last digit is equal to 
            // any of the given digits 
            if ((lastDigit % 10) == 2 || 
                (lastDigit % 10) == 3 || 
                (lastDigit % 10) == 9)
            { 
                cnt++; 
            } 
        } 
        return cnt; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int l = 11, r = 33; 
        Console.Write(countNums(l, r)); 
    } 
}
  
// This code is contributed by AnkitRai01

输出：

8

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