[L,R]范围内的数字计数，其中至少包含一个除以K的数字 _R]范围内的数字计数

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 + = 1return count# Driver code l = 20 r = 35 k = 45print (findCount(l, r, k))# This code is contributed by Mohit Kumar

//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

【[L,R]范围内的数字计数，其中至少包含一个除以K的数字】输出如下：