正方形为N位的最小数字

给定数字N ，任务是找到平方数为N位数的最小数字。

例子：

Input: N = 2
Output: 4
Explanation:
3² = 9, which has 1 digit.
4² = 16, which has 2 digits.
Hence, 4 is the smallest number whose square has N digits.

Input: N = 3
Output: 10
Explanation:
10² = 100, which has 3 digits.

天真的方法：解决问题的最简单方法是从每个数字的平方开始计算，并计算其平方中的位数。打印第一个数字，该数字的平方是N位数字。
时间复杂度： O(√(10 ^N ))

有效方法：要解决此问题，我们需要进行以下观察：

The smallest number whose square has 1 digit = 1
The smallest number whose square has 2 digits = 4
The smallest number whose square has 3 digits = 10
The smallest number whose square has 4 digits = 32
The smallest number whose square has 5 digits = 100

Hence, these numbers form a series 1, 4, 10, 32, 100, 317, …….

现在，我们需要找到该系列的^第N^个项的公式。

该系列的术语可以用以下形式表示：

If N = 1, Smallest number possible is 1.
$T(1) = \lceil 10^{\frac{1-1}{2}}\rceil = 1$

If N = 2, Smallest number possible is 41.
$T(2) = \lceil 10^{\frac{2-1}{2}}\rceil = 4$

If N = 3, Smallest number possible is 10.
$T(3) = \lceil 10^{\frac{3 -1}{2}}\rceil = 10$

因此，我们可以得出结论：该系列的^第N^个可以表示为

$T(N) = \lceil 10^{\frac{N-1}{2}}\rceil$

因此，为了解决该问题，我们只需要为给定的整数N计算ceil(10 ^{(N – 1)/ 2} ) 。

下面是上述方法的实现：

C++

// C++ Program to find the smallest
// number whose square has N digits
  
#include 
using namespace std;
  
// Function to return smallest number
// whose square has N digits
int smallestNum(int N)
{
  
    // Calculate N-th term of the series
    float x = pow(10.0, (N - 1) / 2.0);
    return ceil(x);
}
  
// Driver Code
int main()
{
    int N = 4;
    cout << smallestNum(N);
  
    return 0;
}

Java

// Java program for above approach 
class GFG{ 
  
// Function to return smallest number
// whose square has N digits
static int smallestNum(int N)
{
   
    // Calculate N-th term of the series
    float x = (float)(Math.pow(10, (N - 1) / 2.0));
    return (int)(Math.ceil(x));
}
  
// Driver code 
public static void main(String[] args) 
{ 
    int N = 4;
    System.out.print(smallestNum(N)); 
} 
} 
  
// This code is contributed by spp

Python3

# Python3 Program to find the smallest
# number whose square has N digits
import math;
  
# Function to return smallest number
# whose square has N digits
def smallestNum(N):
  
    # Calculate N-th term of the series
    x = pow(10.0, (N - 1) / 2.0);
    return math.ceil(x);
  
# Driver Code
N = 4;
print(smallestNum(N));
  
# This code is contributed by Code_mech

C#

// C# program for above approach 
using System;
class GFG{ 
  
// Function to return smallest number
// whose square has N digits
static int smallestNum(int N)
{
  
    // Calculate N-th term of the series
    float x = (float)(Math.Pow(10, (N - 1) / 2.0));
    return (int)(Math.Ceiling(x));
}
  
// Driver code 
public static void Main() 
{ 
    int N = 4;
    Console.Write(smallestNum(N)); 
} 
} 
  
// This code is contributed by Code_Mech

输出：

32

时间复杂度： O(log(N))
辅助空间： O(1)