打印每个数组元素的数字，不分割该元素的任何数字
原文:https://www . geeksforgeeks . org/print-digits-for-每个数组元素不除该元素的任何数字/
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】中找到所有不除 arr[i] 中任何数字的数字。
示例:
输入: arr[] = {4162，1152，99842} 输出: 4162->5 7 8 9 1152->3 4 6 7 8 9 99842->5 6 7 解释: 为 arr[0] ( = 4162): 无位数 对于 arr1: 元素 1152 的所有数字都不能被 9、8、7、6、4、3 整除。 对于 arr2: 元素 99842 的所有数字都不能被 7、6、5 整除。
输入:arr[]= { 2021 } T3】输出:T5】2021->3 4 5 6 7 8 9
方法: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 迭代范围【2，9】，如果元素arr【I】中不存在任何可被 i 整除的数字，则打印数字 i 。
- 否则，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 <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
java 描述语言
<script>
// javascript program for the above approach
// Function to find digits for each array
// element that doesn't divide any digit
// of the that element
function indivisibleDigits(arr , N) {
// Traverse the array arr
for (i = 0; i < N; i++) {
document.write(arr[i] + ": ");
// Iterate over the range [2, 9]
for (j = 2; j < 10; j++) {
var temp = arr[i];
// Stores if there exists any digit
// in arr[i] which is divisible by j
var 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 = parseInt(temp/10);
}
// If the digit j doesn't
// divide any digit of arr[i]
if (flag) {
document.write(j + " ");
}
}
document.write("<br/>");
}
}
// Driver Code
var arr = [ 4162, 1152, 99842 ];
var N = arr.length;
indivisibleDigits(arr, N);
// This code contributed by aashish1995
</script>
Output:
4162: 5 7 8 9
1152: 3 4 6 7 8 9
99842: 5 6 7
时间复杂度:O(10 * N * log10N) T8】辅助空间: O(1)
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