查找一个N位数，使其不能被其任何位数整除

给定整数N，任务是找到一个N位数，以使其不能被其任何位数整除。

注意：每个N值可以有多个答案。
例子：

Input: N = 4
Output: 6789
Explanation:
As the number 6789 is not divisible by any of its digits that is 6, 7, 8 and 9 and it is also a four digit number, Hence it can be the desired number.
Input: N = 2
Output: 57
Explanation:
As the number 57 is not divisible by any of its digits that is 5 and 7 and it is also a 2-digit number, Hence it can be the desired number.

方法：问题中的主要观察结果是2和3是不相除的数字，数字“ 23、233、2333，…”也不能被2或3整除。因此，对于任何N-位数，最高有效位数为2，其余位数为3以获取所需的数字。

算法：

• 检查N的值是否等于1，则不可能有这样的数字，因此返回-1。
• 否则，初始化变量num ，将数字存储为2。
• 运行一个从1到N的循环，然后对于每次迭代将该数字乘以10并加3。

  num = (num * 10) + 3 
  

下面是上述方法的实现：

C++

// C++ implementation to find a
// N-digit number such that the number
// it is not divisible by its digits
  
#include 
using namespace std;
  
typedef long long int ll;
  
// Function to find the number
// such that it is not divisible
// by its digits
void solve(ll n)
{
    // Base Cases
    if (n == 1)
    {
        cout << -1;
    }
    else {
          
        // First Digit of the
        // number will be 2
        int num = 2;
          
        // Next digits of the numbers
        for (ll i = 0; i < n - 1; i++) {
            num = (num * 10) + 3;
        }
        cout << num;
    }
}
  
// Driver Code
int main()
{
    ll n = 4;
      
    // Function Call
    solve(n);
}

Java

// Java implementation to find a
// N-digit number such that the number
// it is not divisible by its digits
class GFG {
  
    long ll;
      
    // Function to find the number
    // such that it is not divisible
    // by its digits
    static void solve(long n)
    {
        // Base Cases
        if (n == 1)
        {
            System.out.println(-1);
        }
        else {
              
            // First Digit of the
            // number will be 2
            int num = 2;
              
            // Next digits of the numbers
            for (long i = 0; i < n - 1; i++) {
                num = (num * 10) + 3;
            }
            System.out.println(num);
        }
    }
      
    // Driver Code
    public static void main (String[] args)
    {
        long n = 4;
          
            // Function Call
            solve(n);
    }
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation to find a 
# N-digit number such that the number 
# it is not divisible by its digits 
  
# Function to find the number 
# such that it is not divisible 
# by its digits 
def solve(n) : 
  
    # Base Cases 
    if (n == 1) :
  
        print(-1); 
      
    else :
          
        # First Digit of the 
        # number will be 2 
        num = 2; 
          
        # Next digits of the numbers 
        for i in range(n - 1) : 
            num = (num * 10) + 3; 
           
        print(num); 
  
# Driver Code 
if __name__ == "__main__" : 
  
    n = 4; 
      
    # Function Call 
    solve(n); 
      
# This code is contributed by AnkitRai01

C#

// C# implementation to find a
// N-digit number such that the number
// it is not divisible by its digits
using System;
  
class GFG {
   
    long ll;
       
    // Function to find the number
    // such that it is not divisible
    // by its digits
    static void solve(long n)
    {
        // Base Cases
        if (n == 1)
        {
            Console.WriteLine(-1);
        }
        else {
               
            // First Digit of the
            // number will be 2
            int num = 2;
               
            // Next digits of the numbers
            for (long i = 0; i < n - 1; i++) {
                num = (num * 10) + 3;
            }
            Console.WriteLine(num);
        }
    }
       
    // Driver Code
    public static void Main(String[] args)
    {
        long n = 4;
           
            // Function Call
            solve(n);
    }
}
  
// This code is contributed by sapnasingh4991

输出：

2333

性能分析：

• 时间复杂度： O(N)。
• 辅助空间： O(1)。