打印带有数字根号D的包含K位数字的数字

给定数字根“ D”和位数“ K”。任务是打印一个数字，该数字包含K个数字，其数字根等于D。如果不存在此数字，则打印“ -1”。

例子：

Input: D = 4, K = 4
Output: 4000
No. of digits is 4.
Sum of digits is also 4.

Input:  D = 0, K = 1
Output: 0

方法：解决此问题的一个关键观察结果是，将任意数量的0附加到一个数字不会改变其数字根。因此， D后面跟(K-1)0是一个简单的解决方案。

Special case when D is 0 and K is not 1 does not have a solution since the only number with digital root 0 is 0 itself. 

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to find a number
void printNumberWithDR(int k, int d)
{
 
    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        cout << "-1";
 
    else {
        cout << d;
        k--;
 
        // Print k-1 zeroes
        while (k--)
            cout << "0";
    }
}
 
// Driver code
int main()
{
    int k = 4, d = 4;
 
    printNumberWithDR(k, d);
 
    return 0;
}

Java

// Java implementation of the above approach
 
import java.io.*;
 
class GFG {
 
 
// Function to find a number
static void printNumberWithDR(int k, int d)
{
 
    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        System.out.print( "-1");
 
    else {
        System.out.print(d);
        k--;
 
        // Print k-1 zeroes
        while (k-->0)
            System.out.print( "0");
    }
}
 
// Driver code
 
    public static void main (String[] args) {
            int k = 4, d = 4;
 
    printNumberWithDR(k, d);
    }
}
 
 
//This code is contributed by inder_verma..

Python3

# Python3 implementation of
# the above approach
 
# Function to find a number
def printNumberWithDR( k, d ) :
 
    # If d is 0, k has to be 1
    if d == 0 and k != 1 :
        print(-1, end = "")
 
    else :
        print(d, end = "")
        k -= 1
 
        # Print k-1 zeroes
        while k :
             
            print(0,end = "")
            k -= 1
             
 
# Driver code    
if __name__ == "__main__" :
 
    k, d = 4, 4
 
    # Function call
    printNumberWithDR( k, d )
             
# This code is contributed by
# ANKITRAI1

C#

// C# implementation of the above approach
using System;
 
class GFG {
     
// Function to find a number
static void printNumberWithDR(int k, int d)
{
 
    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        Console.Write( "-1");
 
    else {
         
        Console.Write(d);
        k--;
 
        // Print k-1 zeroes
        while (k-->0)
            Console.Write( "0");
    }
}
 
// Driver code
static public void Main ()
{
    int k = 4, d = 4;
 
    printNumberWithDR(k, d);
}
}
 
// This code is contributed by ajit.

Javascript

 
// Javascript implementation of the above approach
 
// Function to find a number
function printNumberWithDR(k, d)
{
     
    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        document.write("-1");
 
    else
    {
        document.write(d);
        k--;
         
        // Print k-1 zeroes
        while (k-->0)
            document.write("0");
    }
}
  
// Driver Code
var k = 4, d = 4;
 
printNumberWithDR(k, d);
 
// This code is contributed by Ankita saini
                     

输出：

4000

时间复杂度： O(K)