给定数字n，请找出其小数表示形式中至少一位与数字n匹配的除数的数量。
例子：
Input : n = 10
Output: 2
Explanation: numbers are 1 and 10, 1 and 10
both have a nimbus of 1 digit common with n = 10.
Input : n = 15
Output: 3
Explanation: the numbers are 1, 5, 15, all of them
have a minimum of 1 digit common.
天真的方法是从1迭代到n，并检查所有除数，然后使用散列确定是否有任何数字与n匹配。
时间复杂度：O(n)
辅助空间：O(1)
一种有效的方法是从1迭代到sqrt(n)并检查所有除数i和n / i，如果两者都不同，我们检查i和n / i是否存在匹配项，如果是，我们只需加1答案。我们使用散列来存储是否存在数字。
下面是上述方法的实现
C++
// C++ program to count divisors of n that have at least
// one digit common with n
#include
using namespace std;
// function to return true if any digit of m is
// present in hash[].
bool isDigitPresent(int m, bool hash[])
{
// check till last digit
while (m) {
// if number is also present in original number
// then return true
if (hash[m % 10])
return true;
m = m / 10;
}
// if no number matches then return 1
return false;
}
// Count the no of divisors that have at least 1 digits
// same
int countDivisibles(int n)
{
// Store digits present in n in a hash[]
bool hash[10] = { 0 };
int m = n;
while (m) {
// marks that the number is present
hash[m % 10] = true;
// last digit removed
m = m / 10;
}
// loop to traverse from 1 to sqrt(n) to
// count divisors
int ans = 0;
for (int i = 1; i <= sqrt(n); i++) {
// if i is the factor
if (n % i == 0) {
// call the function to check if any
// digits match or not
if (isDigitPresent(i, hash))
ans++;
if (n / i != i) {
// if n/i != i then a different number,
// then check it also
if (isDigitPresent(n/i, hash))
ans++;
}
}
}
// return the answer
return ans;
}
// driver program to test the above function
int main()
{
int n = 15;
cout << countDivisibles(n);
return 0;
} Java
// Java program to count divisors of n that
// have at least one digit common with n
class GFG {
// function to return true if any digit
// of m is present in hash[].
static boolean isDigitPresent(int m,
boolean hash[])
{
// check till last digit
while (m > 0) {
// if number is also present
// in original number then
// return true
if (hash[m % 10])
return true;
m = m / 10;
}
// if no number matches then
// return 1
return false;
}
// Count the no of divisors that
// have at least 1 digits same
static int countDivisibles(int n)
{
// Store digits present in n
// in a hash[]
boolean hash[] = new boolean[10];
int m = n;
while (m > 0) {
// marks that the number
// is present
hash[m % 10] = true;
// last digit removed
m = m / 10;
}
// loop to traverse from 1 to
// sqrt(n) to count divisors
int ans = 0;
for (int i = 1; i <= Math.sqrt(n);
i++)
{
// if i is the factor
if (n % i == 0) {
// call the function to
// check if any digits
// match or not
if (isDigitPresent(i, hash))
ans++;
if (n / i != i) {
// if n/i != i then a
// different number,
// then check it also
if (isDigitPresent(n/i, hash))
ans++;
}
}
}
// return the answer
return ans;
}
//driver code
public static void main (String[] args)
{
int n = 15;
System.out.print(countDivisibles(n));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python3 program to count divisors
# of n that have at least
# one digit common with n
import math
# Function to return true if any
# digit of m is present in hash[].
def isDigitPresent(m, Hash):
# check till last digit
while (m):
# if number is also present in original
# number then return true
if (Hash[m % 10]):
return True
m = m // 10
# if no number matches then return 1
return False
# Count the no of divisors that
# have at least 1 digits same
def countDivisibles(n):
# Store digits present in n in a hash[]
Hash = [False for i in range(10)]
m = n
while (m):
# marks that the number is present
Hash[m % 10] = True
# last digit removed
m = m // 10
# loop to traverse from 1 to sqrt(n) to
# count divisors
ans = 0
for i in range(1, int(math.sqrt(n)) + 1):
# if i is the factor
if (n % i == 0):
# call the function to check if any
# digits match or not
if (isDigitPresent(i, Hash)):
ans += 1
if (n // i != i):
# if n/i != i then a different number,
# then check it also
if (isDigitPresent(n // i, Hash)):
ans += 1
# return the answer
return ans
# Driver Code
n = 15
print(countDivisibles(n))
# This code is contributed by Anant Agarwal.C#
// Program to count divisors of
// n that have at least one digit
// common with n
using System;
class GFG {
// function to return true if
// any digit of m is present
// in hash[].
static bool isDigitPresent(int m, bool[] hash)
{
// check till last digit
while (m > 0) {
// if number is also present in the
// original number then return true
if (hash[m % 10])
return true;
m = m / 10;
}
// if no number matches then return 1
return false;
}
// Count the no of divisors that have
// at least 1 digits same
static int countDivisibles(int n)
{
// Store digits present in n in a hash[]
bool[] hash = new bool[10];
int m = n;
while (m > 0) {
// marks that the number is present
hash[m % 10] = true;
// last digit removed
m = m / 10;
}
// loop to traverse from 1 to sqrt(n) to
// count divisors
int ans = 0;
for (int i = 1; i <= Math.Sqrt(n); i++) {
// if i is the factor
if (n % i == 0) {
// call the function to check if any
// digits match or not
if (isDigitPresent(i, hash))
ans++;
if (n / i != i) {
// if n/i != i then a different number,
// then check it also
if (isDigitPresent(n / i, hash))
ans++;
}
}
}
// return the answer
return ans;
}
// Driver code
public static void Main()
{
int n = 15;
Console.Write(countDivisibles(n));
}
}
// This code is contributed by Anant Agarwal.PHP
输出：
3
时间复杂度： O(sqrt n)
辅助空间： O(1)