位数之和为N的平方的最小数字

给定整数N ，任务是找到位数之和为N ²的最小数字。
例子：

Input: N = 4 
Output: 79 
2⁴ = 16 
sum of digits of 79 = 76
Input: N = 6 
Output: 9999 
2¹⁰ = 1024 which has 4 digits 
 

方法：想法是找到位数为N的平方的最小数字的通用术语。

// First Few terms 
First Term = 1 // N = 1
Second Term = 4 // N = 2
Third Term = 9 // N = 3
Fourth Term = 79 // N = 4
.
.
Nth Term:

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

(n^2 \% 9 + 1) * 10 ^ {\lfloor n^2/9 \rfloor} - 1 

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function to return smallest
// number whose sum of digits is n^2
int smallestNum(int n)
{
  cout<

---

Java

// Java implementation of the above approach
import java.util.*;
 
class GFG{
 
// Function to return smallest
// number whose sum of digits is n^2
static int smallestNum(int n)
{
    return (int)((n * n % 9 + 1) *
         Math.pow(10, n * n / 9) - 1);
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 4;
     
    System.out.print(smallestNum(n));
}
}
 
// This code is contributed by Rajput-Ji

Python 3

# Python implementation of the above approach
 
# Function to return smallest
# number whose sum of digits is n^2
def smallestNum(n):
 
    return ((n * n % 9 + 1)  *
          pow(10, int(n * n / 9)) - 1)
 
# Driver Code
 
# Given N
N = 4
 
print(smallestNum(N))
 
# This code is contributed by Vishal Maurya.

C#

// C# implementation of the above approach
using System;
 
class GFG{
 
// Function to return smallest
// number whose sum of digits is n^2
static int smallestNum(int n)
{
    return (int)((n * n % 9 + 1) *
         Math.Pow(10, n * n / 9) - 1);
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 4;
     
    Console.Write(smallestNum(n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// Javascript implementation of the above approach
 
    // Function to return smallest
    // number whose sum of digits is n^2
    function smallestNum( n) {
        return parseInt (n * n % 9 + 1) * Math.pow(10, parseInt(n*n / 9)) -1;
    }
 
    // Driver Code
      
        let n = 4;
 
        document.write(smallestNum(n));
     
 
// This code contributed by aashish1995

输出：

79