计算所有数字，直到N，最后一位是M

给定两个正数M和N ，任务是从范围[1，N]中查找所有以M为最后一位的所有数字的计数。

例子：

Input: M = 5, N = 15 
Output: 2 
Explanation: 
Only 2 numbers(5 and 15) from the range [1, 15] ends with the digit ‘5’.
Input: M = 1, N = 100 
Output: 10

天真的方法：最简单的方法是在1到N的范围内进行迭代，并检查最后一位是否等于M。如果发现为真，则增加计数。最后，打印获得的计数。
时间复杂度： O(N)
辅助空间： O(1)
高效方法：为了优化上述方法，该思想基于以下事实：以每个数字结尾的数字的计数将一直相同，直到10的最大倍数小于N (例如x )为止。因此，其计数为(N / 10)。现在，简化任务以计算以M结尾的在x和N之间的数字的计数。
步骤如下：

• 初始化一个变量以存储总数，例如total_count 。
• 将(N / 10)添加到总数中。
• 使用以下公式计算x以存储最大10的倍数，该最大倍数小于N ：

 
 

x = (N / 10) * 10

 

• 现在，计算以M结尾的介于x和N之间的数字的数量。
• 将此计数添加到total_count中。打印最终值 获得的total_count 。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
  
#include 
using namespace std;
  
// Function to count the numbers
// ending with M
int getCount(int N, int M)
{
  
    // Stores the count of
    // numbers required
    int total_count = 0;
  
    // Calculate count upto
    // nearest power of 10
    total_count += (N / 10);
  
    // Computing the value of x
    int x = (N / 10) * 10;
  
    // Adding the count of numbers
    // ending at M from x to N
    if ((N - x) >= M) 
    {
        total_count = total_count + 1;
    }
    return total_count;
}
  
// Driver Code
int main()
{
    int N = 100, M = 1;
  
    // Function Call
    cout << getCount(N, M);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
  
class GFG{
  
// Function to count the numbers
// ending with M
static int getCount(int N, int M)
{
  
    // Stores the count of
    // numbers required
    int total_count = 0;
  
    // Calculate count upto
    // nearest power of 10
    total_count += (N / 10);
  
    // Computing the value of x
    int x = (N / 10) * 10;
  
    // Adding the count of numbers
    // ending at M from x to N
    if ((N - x) >= M) 
    {
        total_count = total_count + 1;
    }
    return total_count;
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 100, M = 1;
  
    // Function call
    System.out.print(getCount(N, M));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 Program to implement
# the above approach
  
# Function to count the numbers
# ending with M 
def getCount(N, M):
  
    # Stores the count of
    # numbers required
    total_count = 0
  
    # Calculate count upto
    # nearest power of 10
    total_count += N // 10
  
    # Computing the value of x
    x = (N // 10) * 10
  
    # Adding the count of numbers
    # ending at M from x to N
    if((N - x) >= M):
        total_count = total_count + 1
  
    return total_count
  
# Driver Code
N = 100
M = 1
  
# Function call
print(getCount(N, M))
  
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
class GFG{
  
    // Function to count the 
    // numbers ending with M
    static int getCount(int N, int M)
    {
  
        // Stores the count of
        // numbers required
        int total_count = 0;
  
        // Calculate count upto
        // nearest power of 10
        total_count += (N / 10);
  
        // Computing the value of x
        int x = (N / 10) * 10;
  
        // Adding the count of numbers
        // ending at M from x to N
        if ((N - x) >= M) 
        {
            total_count = total_count + 1;
        }
        return total_count;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int N = 100, M = 1;
  
        // Function call
        Console.Write(getCount(N, M));
    }
}
  
// This code is contributed by shikhasingrajput

输出：

10

时间复杂度： O(1)
辅助空间： O(1)