计算给定范围内包含给定数字作为后缀的数字

给定三个整数A、L和R ，任务是计算L到R范围内的数字，其中包含A作为其后缀。

例子：

Input: A = 2, L = 2, R = 20
Output: 2
Explanation: 
Only two possible numbers from the given range that satisfies the conditions are 2 and 12.

Input: A = 25, L = 25, R = 273
Output: 3
Explanation: 
The three numbers from the given range that satisfies the conditions are 25, 125 and 225.

朴素方法：解决问题的最简单方法是遍历L到R范围内的数字，并检查数字是否以A结尾。对于所有发现为真的数字，将这些数字的计数加 1。最后，打印最终计数。

时间复杂度： O(B)
辅助空间： O(log ₂ R)

高效的方法：要优化上述方法，请按照以下步骤操作：

• 计算并存储A的位数并将其存储在变量中，例如count。
• 将 10 乘以A的位数，即 10 次^计数。
• 检查每个循环 10^计数是否在[L, R]范围内。如果发现为真，则将计数增加 1。
• 打印count的最终值。

下面是上述方法的实现：

C++

// C++ Program of the
// above approach
 
#include 
 
using namespace std;
 
// Function to count the number
// ends with given number in range
void countNumEnds(int A, int L, int R)
{
    int temp, count = 0, digits;
    int cycle;
 
    // Find number of digits in A
    digits = log10(A) + 1;
 
    // Find the power of 10
    temp = pow(10, digits);
    cycle = temp;
 
    while (temp <= R) {
 
        if (temp >= L)
            count++;
 
        // Incrementing the A
        temp += cycle;
    }
 
    cout << count;
}
 
// Driver Code
int main()
{
    int A = 2, L = 2, R = 20;
 
    // Function Call
    countNumEnds(A, L, R);
}

Java

// Java Program of the
// above approach
class GFG{
 
// Function to count the number
// ends with given number in range
static void countNumEnds(int A,
                         int L, int R)
{
  int temp, count = 0, digits;
  int cycle;
 
  // Find number of digits in A
  digits = (int) (Math.log10(A) + 1);
 
  // Find the power of 10
  temp = (int) Math.pow(10, digits);
  cycle = temp;
 
  while (temp <= R)
  {
    if (temp >= L)
      count++;
 
    // Incrementing the A
    temp += cycle;
  }
  System.out.print(count);
}
 
// Driver Code
public static void main(String[] args)
{
  int A = 2, L = 2, R = 20;
 
  // Function Call
  countNumEnds(A, L, R);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program of the
# above approach
from math import log10
 
# Function to count the number
# ends with given number in range
def countNumEnds(A, L, R):
 
    count = 0
 
    # Find number of digits in A
    digits = int(log10(A) + 1)
 
    # Find the power of 10
    temp = int(pow(10, digits))
    cycle = temp
 
    while(temp <= R):
        if(temp >= L):
            count += 1
 
        # Incrementing the A
        temp += cycle
 
    print(count)
 
# Driver Code
A = 2
L = 2
R = 20
 
# Function call
countNumEnds(A, L, R)
 
# This code is contributed by Shivam Singh

C#

// C# program of the
// above approach
using System;
 
class GFG{
 
// Function to count the number
// ends with given number in range
static void countNumEnds(int A, int L,
                         int R)
{
    int temp, count = 0, digits;
    int cycle;
     
    // Find number of digits in A
    digits = (int)(Math.Log10(A) + 1);
     
    // Find the power of 10
    temp = (int)Math.Pow(10, digits);
    cycle = temp;
     
    while (temp <= R)
    {
        if (temp >= L)
        count++;
     
        // Incrementing the A
        temp += cycle;
    }
    Console.Write(count);
}
 
// Driver Code
public static void Main(String[] args)
{
    int A = 2, L = 2, R = 20;
     
    // Function call
    countNumEnds(A, L, R);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// JavaScript program for
// the above approach
  
// Function to count the number
// ends with given number in range
function countNumEnds(A, L, R)
{
  let temp = 0, count = 0, digits = 0;
  let cycle = 0;
   
  // Find number of digits in A
  digits = Math.round(Math.log10(A)) + 1;
   
  // Find the power of 10
  temp = Math.round(Math.pow(10, digits));
  cycle = temp;
   
  while (temp <= R)
  {
    if (temp >= L)
      count++;
   
    // Incrementing the A
    temp += cycle;
  }
  document.write(count);
}
 
// Driver code
 
  let A = 2, L = 2, R = 20;
   
  // Function Call
  countNumEnds(A, L, R);
   
  // This code is contributed by splevel62.

输出：

2

时间复杂度： O(N)，其中 N 是范围。
辅助空间： O(1)