原文:
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 arr: 元素 1152 的所有数字都不能被 9、8、7、6、4、3 整除。 对于 arr: 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，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 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
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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