计算范围内的数字数量，它与q的乘积不相等
给定一系列数字[l, r]和一个整数q。 任务是对给定范围内的所有此类数字进行计数，以使该数字的任何数字与乘积中具有给定数字q的任何数字都不匹配。
示例：
Input : l = 10, r = 12, q = 2
Output : 1
10*2 = 20 which has 0 as same digit
12*2 = 24 which as 2 as same digit
11*2 = 22 no same digit
Input : l = 5, r = 15, q = 2
Output : 9
来源：高盛面试集 46
的想法是从l到r循环运行，以生成范围内的所有数字，并将每个此类数字n转换为q，即使用to_string()方法对字符串进行n * q，然后使用基本字符串哈希检查string2中是否存在string2中的任何字符。
下面是上述方法的实现：
C++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if all of the digits
// in a number and it's product with q
// are unequal or not
bool checkIfUnequal(int n, int q)
{
// convert first number into string
string s1 = to_string(n);
int a[26] = { 0 };
// Insert elements from 1st number
// to hash
for (int i = 0; i < s1.size(); i++)
a[s1[i] - '0']++;
// Calculate corresponding product
int prod = n * q;
// Convert the product to string
string s2 = to_string(prod);
// Using the hash check if any digit of
// product matches with the digits of
// input number
for (int i = 0; i < s2.size(); i++)
{
// If yes, return false
if (a[s2[i] - '0'])
return false;
}
// Return true
return true;
}
// Function to count numbers in the range [l, r]
// such that all of the digits of the number and
// it's product with q are unequal
int countInRange(int l, int r, int q)
{
int count = 0;
for (int i = l; i <= r; i++) {
// check for every number between l and r
if (checkIfUnequal(i, q))
count++;
}
return count;
}
// Driver Code
int main()
{
int l = 10, r = 12, q = 2;
// Function Call
cout << countInRange(l, r, q);
return 0;
}
Java
// Java program for above approach
class GfG {
// Function to check if all of the digits
// in a number and it's product with q
// are unequal or not
static boolean checkIfUnequal(int n, int q)
{
// convert first number into string
String s1 = Integer.toString(n);
int a[] = new int[26];
// Insert elements from 1st number
// to hash
for (int i = 0; i < s1.length(); i++)
a[s1.charAt(i) - '0']++;
// Calculate corresponding product
int prod = n * q;
// Convert the product to string
String s2 = Integer.toString(prod);
// Using the hash check if any digit of
// product matches with the digits of
// input number
for (int i = 0; i < s2.length(); i++)
{
// If yes, return false
if (a[s2.charAt(i) - '0'] > 0)
return false;
}
// else, return true
return true;
}
// Function to count numbers in the range [l, r]
// such that all of the digits of the number and
// it's product with q are unequal
static int countInRange(int l, int r, int q)
{
int count = 0;
for (int i = l; i <= r; i++) {
// check for every number between l and r
if (checkIfUnequal(i, q))
count++;
}
return count;
}
// Driver Code
public static void main(String[] args)
{
int l = 10, r = 12, q = 2;
// Function Call
System.out.println(countInRange(l, r, q));
}
}
Python3
# Python 3 program for above approach
# Function to check if all of the digits
# in a number and it's product with q
# are unequal or not
def checkIfUnequal(n, q):
# convert first number into string
s1 = str(n)
a = [0 for i in range(26)]
# Insert elements from 1st number
# to hash
for i in range(0, len(s1), 1):
a[ord(s1[i]) - ord('0')] += 1
# Calculate corresponding product
prod = n * q
# Convert the product to string
s2 = str(prod)
# Using the hash check if any digit of
# product matches with the digits of
# input number
for i in range(0, len(s2), 1):
# If yes, return false
if (a[ord(s2[i]) - ord('0')]):
return False
# Return true
return True
# Function to count numbers in the range [l, r]
# such that all of the digits of the number and
# it's product with q are unequal
def countInRange(l, r, q):
count = 0
for i in range(l, r + 1, 1):
# check for every number between l and r
if (checkIfUnequal(i, q)):
count += 1
return count
# Driver Code
if __name__ == '__main__':
l = 10
r = 12
q = 2
# Function call
print(countInRange(l, r, q))
# This code is contributed by
# Sahil_Shelangia
C
// C# program for above approach
using System;
class GfG {
// Function to check if all of the digits
// in a number and it's product with q
// are unequal or not
static bool checkIfUnequal(int n, int q)
{
// convert first number into string
string s1 = n.ToString();
int[] a = new int[26];
// Insert elements from 1st number
// to hash
for (int i = 0; i < s1.Length; i++)
a[s1[i] - '0']++;
// Calculate corresponding product
int prod = n * q;
// Convert the product to string
string s2 = prod.ToString();
// Using the hash check if any digit of
// product matches with the digits of
// input number
for (int i = 0; i < s2.Length; i++)
{
// If yes, return false
if (a[s2[i] - '0'])
return false;
}
// Else, return true
return true;
}
// Function to count numbers in the range [l, r]
// such that all of the digits of the number and
// it's product with q are unequal
static int countInRange(int l, int r, int q)
{
int count = 0;
for (int i = l; i <= r; i++)
{
// check for every number between l and r
if (checkIfUnequal(i, q))
count++;
}
return count;
}
// Driver Code
public static void Main()
{
int l = 10, r = 12, q = 2;
// Function call
Console.WriteLine(countInRange(l, r, q));
}
}
// This code is contributed bt Archana_kumari
PHP
// PHP program for above code
// Function to check if all of the digits
// in a number and it's product with q
// are unequal or not
function checkIfUnequal($n, $q)
{
// convert first number into string
$s1 = strval($n);
$a = array_fill(0, 26, NULL);
// Insert elements from 1st number
// to hash
for ($i = 0; $i < strlen($s1); $i++)
$a[ord($s1[$i]) - ord('0')]++;
// Calculate corresponding product
$prod = $n * $q;
// Convert the product to string
$s2 = strval($prod);
// Using the hash check if any digit of
// product matches with the digits of
// input number
for ($i = 0; $i < strlen($s2); $i++)
{
// If yes, return false
if ($a[ord($s2[$i]) - ord('0')])
return false;
}
// Else, return true
return true;
}
// Function to count numbers in the range
// [l, r] such that all of the digits of the
// number and it's product with q are unequal
function countInRange($l, $r, $q)
{
$count = 0;
for ($i = $l; $i <= $r; $i++)
{
// check for every number between l and r
if (checkIfUnequal($i, $q))
$count++;
}
return $count;
}
// Driver Code
$l = 10;
$r = 12;
$q = 2;
// Function call
echo countInRange($l, $r, $q);
// This code is contributed by ita_c
?>
输出：
1
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