给定范围内最后 K 位数字相等的整数个数

给定一个从L到R的范围和一个整数K ，任务是计算给定范围内的整数个数，使它们的最后K个数字相等。

例子：

Input: L = 49, R = 101, K=2
Output: 6
Explanation: There are 6 possible integers t.e., 55, 66, 77, 88, 99 and 100 such that their last K(i.e., 2) digits are equal.

Input: L = 10, R = 20, K=2
Output: 1

有效方法：可以观察到，在1到X范围内，最后K位等于整数z （即 i % 10 ^K = z）的整数i的计数为(X – z)/10 ^K + 1 .使用此观察，可以使用以下步骤解决上述问题：

1. 假设intCount(X, K)表示从1到X的最后K位数相等的整数的计数。
2. 要计算intCount(X, K) ，请遍历所有具有K位数字的z的可能性（即{00…0, 11…1, 22…2, 33…3, 44…4, …., 99…9 } ）在公式 (X – z)/10 ^K +1 中并保持它们的总和是所需的值。
3. 因此， L到R范围内的整数计数可以通过intCount(R, K) – intCount(L-1, K)获得。

下面是上述方法的实现：

C++

// C++ Program of the above approach
 
#include 
using namespace std;
 
// Function to return the count of
// integers from 1 to X having the
// last K digits as equal
int intCount(int X, int K)
{
    // Stores the total count of integers
    int ans = 0;
 
    // Loop to iterate over all
    // possible values of z
    for (int z = 0; z < pow(10, K);
         z += (pow(10, K) - 1) / 9) {
 
        // Terminate the loop when z > X
        if (z > X)
            break;
 
        // Add count of integers with
        // last K digits equal to z
        ans += ((X - z) / pow(10, K) + 1);
    }
 
    // Return count
    return ans;
}
 
// Function to return the count of
// integers from L to R having the
// last K digits as equal
int intCountInRange(int L, int R, int K)
{
    return (intCount(R, K)
            - intCount(L - 1, K));
}
 
// Driver Code
int main()
{
    int L = 49;
    int R = 101;
    int K = 2;
 
    // Function Call
    cout << intCountInRange(L, R, K);
 
    return 0;
}

Java

// Java Program of the above approach
import java.util.*;
 
class GFG{
 
// Function to return the count of
// integers from 1 to X having the
// last K digits as equal
static int intCount(int X, int K)
{
    // Stores the total count of integers
    int ans = 0;
 
    // Loop to iterate over all
    // possible values of z
    for (int z = 0; z < Math.pow(10, K);
         z += (Math.pow(10, K) - 1) / 9) {
 
        // Terminate the loop when z > X
        if (z > X)
            break;
 
        // Add count of integers with
        // last K digits equal to z
        ans += ((X - z) / Math.pow(10, K) + 1);
    }
 
    // Return count
    return ans;
}
 
// Function to return the count of
// integers from L to R having the
// last K digits as equal
static int intCountInRange(int L, int R, int K)
{
    return (intCount(R, K)
            - intCount(L - 1, K));
}
 
// Driver Code
public static void main(String[] args)
{
    int L = 49;
    int R = 101;
    int K = 2;
 
    // Function Call
    System.out.print(intCountInRange(L, R, K));
 
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function to return the count of
# integers from 1 to X having the
# last K digits as equal
def intCount(X, K):
     
    # Stores the total count of integers
    ans = 0
 
    # Loop to iterate over all
    # possible values of z
    for z in range(0, int(pow(10, K)),
                     int((pow(10, K) - 1) / 9)):
 
        # Terminate the loop when z > X
        if (z > X):
            break
 
        # Add count of integers with
        # last K digits equal to z
        ans += int((X - z) / int(pow(10, K)) + 1)
     
    # Return count
    return ans
 
# Function to return the count of
# integers from L to R having the
# last K digits as equal
def intCountInRange(L, R, K):
     
    return(intCount(R, K) - intCount(L - 1, K))
 
# Driver Code
L = 49
R = 101
K = 2
 
# Function Call
print(intCountInRange(L, R, K))
 
# This code is contributed by sanjoy_62

C#

// C# Program of the above approach
using System;
 
class GFG{
 
// Function to return the count of
// integers from 1 to X having the
// last K digits as equal
static int intCount(int X, int K)
{
    // Stores the total count of integers
    int ans = 0;
 
    // Loop to iterate over all
    // possible values of z
    for (int z = 0; z < Math.Pow(10, K);
         z += ((int)Math.Pow(10, K) - 1) / 9) {
 
        // Terminate the loop when z > X
        if (z > X)
            break;
 
        // Add count of integers with
        // last K digits equal to z
        ans += ((X - z) / (int)Math.Pow(10, K) + 1);
    }
 
    // Return count
    return ans;
}
 
// Function to return the count of
// integers from L to R having the
// last K digits as equal
static int intCountInRange(int L, int R, int K)
{
    return (intCount(R, K)
            - intCount(L - 1, K));
}
 
// Driver Code
public static void Main(String[] args)
{
    int L = 49;
    int R = 101;
    int K = 2;
 
    // Function Call
    Console.Write(intCountInRange(L, R, K));
 
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

 
// JavaScript program for the above approach
 
// Function to return the count of
// integers from 1 to X having the
// last K digits as equal
function intCount(X, K)
{
     
    // Stores the total count of integers
    let ans = 0;
 
    // Loop to iterate over all
    // possible values of z
    for(let z = 0; z < Math.pow(10, K);
            z += Math.floor((Math.pow(10, K) - 1) / 9))
    {
         
        // Terminate the loop when z > X
        if (z > X)
            break;
 
        // Add count of integers with
        // last K digits equal to z
        ans += Math.floor(((X - z) /
               Math.pow(10, K) + 1));
    }
 
    // Return count
    return ans;
}
 
// Function to return the count of
// integers from L to R having the
// last K digits as equal
function intCountInRange(L, R, K)
{
    return(intCount(R, K) -
           intCount(L - 1, K));
}
 
// Driver Code
let L = 49;
let R = 101;
let K = 2;
 
// Function Call
document.write(intCountInRange(L, R, K));
 
// This code is contributed by Potta Lokesh
 

输出

6

时间复杂度： O(log K)
空间复杂度： O(1)