给定一个由N个正整数组成的数组arr [] ，每个数组元素arr [i]的任务是从[ 0，9]中查找所有数字，这些数字不除掉arr [i]中存在的任何数字。
例子：
Input: arr[] = {4162, 1152, 99842}
Output:
4162 -> 5 7 8 9
1152 -> 3 4 6 7 8 9
99842 -> 5 6 7
Explanation:
For arr[0] ( = 4162): None of the digits of the element 4162 are divisible by 5, 7, 8, 9.
For arr[1]( = 1152): None of the digits of the element 1152 are divisible by 9, 8, 7, 6, 4, 3.
For arr[2]( = 99842): None of the digits of the element 99842 are divisible by 7, 6, 5.
Input: arr[] = {2021}
Output:
2021 -> 3 4 5 6 7 8 9
方法：请按照以下步骤解决问题：
- 遍历给定数组arr []并执行以下步骤：
- 使用变量i遍历[ 2，9 ]的范围，如果元素arr [i]中不存在可被i整除的数字，则打印数字i 。
- 否则，请继续进行下一次迭代。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find digits for each array
// element that doesn't divide any digit
// of the that element
void indivisibleDigits(int arr[], int N)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
int num = 0;
cout << arr[i] << ": ";
// Iterate over the range [2, 9]
for (int j = 2; j < 10; j++) {
int temp = arr[i];
// Stores if there exists any digit
// in arr[i] which is divisible by j
bool flag = true;
while (temp > 0) {
// If any digit of the number
// is divisible by j
if ((temp % 10) != 0
&& (temp % 10) % j == 0) {
flag = false;
break;
}
temp /= 10;
}
// If the digit j doesn't
// divide any digit of arr[i]
if (flag) {
cout << j << ' ';
}
}
cout << endl;
}
}
// Driver Code
int main()
{
int arr[] = { 4162, 1152, 99842 };
int N = sizeof(arr) / sizeof(arr[0]);
indivisibleDigits(arr, N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to find digits for each array
// element that doesn't divide any digit
// of the that element
static void indivisibleDigits(int[] arr, int N)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++)
{
System.out.print(arr[i] + ": ");
// Iterate over the range [2, 9]
for (int j = 2; j < 10; j++)
{
int temp = arr[i];
// Stores if there exists any digit
// in arr[i] which is divisible by j
boolean flag = true;
while (temp > 0) {
// If any digit of the number
// is divisible by j
if ((temp % 10) != 0
&& (temp % 10) % j == 0) {
flag = false;
break;
}
temp /= 10;
}
// If the digit j doesn't
// divide any digit of arr[i]
if (flag) {
System.out.print(j + " ");
}
}
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 4162, 1152, 99842 };
int N = arr.length;
indivisibleDigits(arr, N);
}
}
// This code is contributed by sanjoy_62.Python3
# Python3 program for the above approach
# Function to find digits for each array
# element that doesn't divide any digit
# of the that element
def indivisibleDigits(arr, N) :
# Traverse the array arr[]
for i in range(N):
num = 0
print(arr[i], end = ' ')
# Iterate over the range [2, 9]
for j in range(2, 10):
temp = arr[i]
# Stores if there exists any digit
# in arr[i] which is divisible by j
flag = True
while (temp > 0) :
# If any digit of the number
# is divisible by j
if ((temp % 10) != 0
and (temp % 10) % j == 0) :
flag = False
break
temp //= 10
# If the digit j doesn't
# divide any digit of arr[i]
if (flag) :
print(j, end = ' ')
print()
# Driver Code
arr = [ 4162, 1152, 99842 ]
N = len(arr)
indivisibleDigits(arr, N)
# This code is contributed by susmitakundugoaldanga.C#
// C# program for the above approach
using System;
class GFG
{
// Function to find digits for each array
// element that doesn't divide any digit
// of the that element
static void indivisibleDigits(int[] arr, int N)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++)
{
Console.Write(arr[i] + ": ");
// Iterate over the range [2, 9]
for (int j = 2; j < 10; j++)
{
int temp = arr[i];
// Stores if there exists any digit
// in arr[i] which is divisible by j
bool flag = true;
while (temp > 0) {
// If any digit of the number
// is divisible by j
if ((temp % 10) != 0
&& (temp % 10) % j == 0) {
flag = false;
break;
}
temp /= 10;
}
// If the digit j doesn't
// divide any digit of arr[i]
if (flag) {
Console.Write(j + " ");
}
}
Console.WriteLine();
}
}
// Driver Code
public static void Main()
{
int[] arr = { 4162, 1152, 99842 };
int N = arr.Length;
indivisibleDigits(arr, N);
}
}
// This code is contributed by rishavmahato348.输出：
4162: 5 7 8 9
1152: 3 4 6 7 8 9
99842: 5 6 7时间复杂度： O(10 * N * log 10 N)
辅助空间： O(1)