[L,R]范围内的数字计数，其中至少包含一个除以K的数字

本文概述

• C ++
• Java
• Python3
• C#

给定三个正整数L, R和K。任务是找出在[L, R]范围内至少包含一个能除数K的数字的所有数。

例子：

输入：L = 5, R = 11, K = 10
输出：3 5、10和11就是这样的数字。
输入：L = 32, R = 38, K = 13
输出：0

方法:初始化count = 0，对于[L, R]范围内的每个元素，检查它是否包含至少一个除k的数字，如果是，则增加计数。

下面是上述方法的实现：

C ++

//C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
//Function that returns true if num
//contains at least one digit
//that divides k
bool digitDividesK( int num, int k)
{
     while (num) {
  
         //Get the last digit
         int d = num % 10;
  
         //If the digit is non-zero
         //and it divides k
         if (d != 0 and k % d == 0)
             return true ;
  
         //Remove the last digit
         num = num /10;
     }
  
     //There is no digit in num
     //that divides k
     return false ;
}
  
//Function to return the required
//count of elements from the given
//range which contain at least one
//digit that divides k
int findCount( int l, int r, int k)
{
  
     //To store the result
     int count = 0;
  
     //For every number from the range
     for ( int i = l; i <= r; i++) {
  
         //If any digit of the current
         //number divides k
         if (digitDividesK(i, k))
             count++;
     }
     return count;
}
  
//Driver code
int main()
{
     int l = 20, r = 35;
     int k = 45;
  
     cout <<findCount(l, r, k);
  
     return 0;
}

Java

//Java implementation of the approach 
  
class GFG
{
     //Function that returns true if num 
     //contains at least one digit 
     //that divides k 
     static boolean digitDividesK( int num, int k) 
     { 
         while (num != 0 ) 
         { 
      
             //Get the last digit 
             int d = num % 10 ; 
      
             //If the digit is non-zero 
             //and it divides k 
             if (d != 0 && k % d == 0 ) 
                 return true ; 
      
             //Remove the last digit 
             num = num /10 ; 
         } 
      
         //There is no digit in num 
         //that divides k 
         return false ; 
     } 
      
     //Function to return the required 
     //count of elements from the given 
     //range which contain at least one 
     //digit that divides k 
     static int findCount( int l, int r, int k) 
     { 
      
         //To store the result 
         int count = 0 ; 
      
         //For every number from the range 
         for ( int i = l; i <= r; i++) 
         { 
      
             //If any digit of the current 
             //number divides k 
             if (digitDividesK(i, k)) 
                 count++; 
         } 
         return count; 
     } 
      
     //Driver code 
     public static void main(String []args)
     { 
         int l = 20 , r = 35 ; 
         int k = 45 ; 
      
         System.out.println(findCount(l, r, k)); 
     } 
}
  
//This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
  
# Function that returns true if num
# contains at least one digit
# that divides k
def digitDividesK(num, k):
     while (num):
  
         # Get the last digit
         d = num % 10
  
         # If the digit is non-zero
         # and it divides k
         if (d ! = 0 and k % d = = 0 ):
             return True
  
         # Remove the last digit
         num = num //10
  
     # There is no digit in num
     # that divides k
     return False
  
# Function to return the required
# count of elements from the given
# range which contain at least one
# digit that divides k
def findCount(l, r, k):
  
     # To store the result
     count = 0
  
     # For every number from the range
     for i in range (l, r + 1 ):
  
         # If any digit of the current
         # number divides k
         if (digitDividesK(i, k)):
             count + = 1
  
     return count
  
# Driver code
l = 20
r = 35
k = 45
  
print (findCount(l, r, k))
  
# This code is contributed by Mohit Kumar

C#

//C# implementation of the approach 
using System;
  
class GFG
{
     //Function that returns true if num 
     //contains at least one digit 
     //that divides k 
     static bool digitDividesK( int num, int k) 
     { 
         while (num != 0) 
         { 
      
             //Get the last digit 
             int d = num % 10; 
      
             //If the digit is non-zero 
             //and it divides k 
             if (d != 0 && k % d == 0) 
                 return true ; 
      
             //Remove the last digit 
             num = num /10; 
         } 
      
         //There is no digit in num 
         //that divides k 
         return false ; 
     } 
      
     //Function to return the required 
     //count of elements from the given 
     //range which contain at least one 
     //digit that divides k 
     static int findCount( int l, int r, int k) 
     { 
      
         //To store the result 
         int count = 0; 
      
         //For every number from the range 
         for ( int i = l; i <= r; i++) 
         { 
      
             //If any digit of the current 
             //number divides k 
             if (digitDividesK(i, k)) 
                 count++; 
         } 
         return count; 
     } 
      
     //Driver code 
     public static void Main()
     { 
         int l = 20, r = 35; 
         int k = 45; 
      
         Console.WriteLine(findCount(l, r, k)); 
     } 
}
  
//This code is contributed by AnkitRai01

输出如下：

10

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