要添加到N或从N减去的最小数，以使其成为一个完美平方

给定数字N ，找到需要添加到N或从N减去的最小数，以使其成为一个完美的平方。如果要加数字，则用+号打印；否则，如果要减数字，则用-号打印。

例子：

Input: N = 14
Output: 2
Nearest perfect square before 14 = 9
Nearest perfect square after 14 = 16
Therefore 2 needs to be added to 14 to get the closest perfect square

Input: N = 18
Output: -2
Nearest perfect square before 18 = 16
Nearest perfect square after 18 = 25
Therefore 2 needs to be subtracted from 18 to get the closest perfect square

方法：

1. 获取号码。
2. 找到数字的平方根，然后将结果转换为整数。
3. 将double值转换为整数后，它将包含N之前的完美平方的根，即floor(square root(N)) 。
4. 然后找到该数字的平方，它将是N之前的完美平方。
5. 在N之后找到理想平方的根，即ceil(square root(N)) 。
6. 然后找到该数字的平方，它将是N之后的完美平方。
7. 检查底值的平方是否最接近N或ceil值。
8. 如果底值的平方最接近N，则用-符号打印差异。否则，打印出ceil值的平方和带有+号的N之间的差。

下面是上述方法的实现：

C++

// C++ implementation of the approach
  
#include 
using namespace std;
  
// Function to return the Least number
int nearest(int n)
{
  
    // Get the perfect square
    // before and after N
    int prevSquare = sqrt(n);
    int nextSquare = prevSquare + 1;
    prevSquare = prevSquare * prevSquare;
    nextSquare = nextSquare * nextSquare;
  
    // Check which is nearest to N
    int ans
        = (n - prevSquare) < (nextSquare - n)
              ? (prevSquare - n)
              : (nextSquare - n);
  
    // return the result
    return ans;
}
  
// Driver code
int main()
{
    int n = 14;
    cout << nearest(n) << endl;
  
    n = 16;
    cout << nearest(n) << endl;
  
    n = 18;
    cout << nearest(n) << endl;
  
    return 0;
}

Java

// Java implementation of the approach
class GFG {
          
    // Function to return the Least number
    static int nearest(int n)
    {
      
        // Get the perfect square
        // before and after N
        int prevSquare = (int)Math.sqrt(n);
        int nextSquare = prevSquare + 1;
        prevSquare = prevSquare * prevSquare;
        nextSquare = nextSquare * nextSquare;
      
        // Check which is nearest to N
        int ans = (n - prevSquare) < (nextSquare - n)? (prevSquare - n): (nextSquare - n);
      
        // return the result
        return ans;
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int n = 14;
        System.out.println(nearest(n));
      
        n = 16;
        System.out.println(nearest(n)) ;
      
        n = 18;
        System.out.println(nearest(n)) ;
      
    }
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach 
from math import sqrt
  
# Function to return the Least number 
def nearest(n) : 
  
    # Get the perfect square 
    # before and after N 
    prevSquare = int(sqrt(n)); 
    nextSquare = prevSquare + 1; 
    prevSquare = prevSquare * prevSquare; 
    nextSquare = nextSquare * nextSquare; 
  
    # Check which is nearest to N 
    ans    = (prevSquare - n) if (n - prevSquare) < (nextSquare - n) else (nextSquare - n); 
  
    # return the result 
    return ans; 
  
# Driver code 
if __name__ == "__main__" : 
  
    n = 14; 
    print(nearest(n)) ; 
  
    n = 16; 
    print(nearest(n)); 
  
    n = 18; 
    print(nearest(n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
  
class GFG {
          
    // Function to return the Least number
    static int nearest(int n)
    {
      
        // Get the perfect square
        // before and after N
        int prevSquare = (int)Math.Sqrt(n);
        int nextSquare = prevSquare + 1;
        prevSquare = prevSquare * prevSquare;
        nextSquare = nextSquare * nextSquare;
      
        // Check which is nearest to N
        int ans = (n - prevSquare) < (nextSquare - n)? (prevSquare - n): (nextSquare - n);
      
        // return the result
        return ans;
    }
      
    // Driver code
    public static void Main (string[] args)
    {
        int n = 14;
        Console.WriteLine(nearest(n));
      
        n = 16;
        Console.WriteLine(nearest(n)) ;
      
        n = 18;
        Console.WriteLine(nearest(n)) ;
      
    }
}
  
// This code is contributed by AnkitRai01

输出：

2
0
-2