找到最小的完美平方数A，使N + A也是一个完美平方数

给定正数N。任务是找出最小的理想平方数A，使N + A也是理想平方数或返回-1。
例子：

Input: N = 3
Output: 1
Explanation: 
As 1 + 3 = 4 = 22

Input: N=1
Output: -1

天真的方法：
从{1,2,3,4,5…}遍历M，并检查(N + M * M)是否为理想的平方数。
高效方法：

• 关于观察，我们有一个等式：

• N + (X * X) = (M * M) where N is given and M and X are unknown.
• We can rearrange it and get: 
  • N = (M * M) – (X * X)
  • N = (M + X) * (M – X)

• 现在我们可以看到，要获得N，我们需要找到N的因数.N的因数可以在O(N)时间中获得。但是通过这种方法可以将其优化为O(N ^ 1/2)。
• 令N的因子为a，b =(N / a)。因此，从上面的等式a =(M – X)和b =(M + X)，求解后，我们可以获得X =(b – a)/ 2的值。

下面是上述方法的实现：

C++

// C++ code to find out the smallest
// perfect square X which when added to N
// yields another perfect square number.
#include
using namespace std;
 
long SmallestPerfectSquare(long N){
     
    // X is the smallest perfect
    // square number
    long X = (long)1e9;
    long ans;
     
    // Loop from 1 to square root of N
    for(int i = 1; i < sqrt(N); i++)
    {
     
        // Condition to check whether i
        // is factor of N or not
        if (N % i == 0)
        {
            long a = i;
            long b = N / i;
     
            // Condition to check whether
            // factors satisfies the
            // equation or not
            if((b - a != 0) && ((b - a) % 2 == 0))
            {
                         
                // Stores minimum value
                X = min(X, (b - a) / 2);
            }
        }
    }
     
    // Return if X * X if X is not equal
    // to 1e9 else return -1
    if (X != 1e9)
        ans = X * X;
    else
        ans = -1;
             
    return ans;
}
     
// Driver code
int main()
{
    long N = 3;
    cout << SmallestPerfectSquare(N);
    return 0;
}
 
// This code is contributed by AnkitRai01

Java

// Java code to find out the smallest
// perfect square X which when added to N
// yields another perfect square number.
 
public class GFG {
     
 
    static long SmallestPerfectSquare(long N)
    {
     
        // X is the smallest perfect
        // square number
        long X = (long)1e9;
        long ans;
     
        // Loop from 1 to square root of N
        for(int i = 1; i < Math.sqrt(N); i++){
     
            // Condition to check whether i
            // is factor of N or not
            if (N % i == 0){
                long a = i ;
                long b = N / i;
     
                // Condition to check whether
                // factors satisfies the
                // equation or not
                if ((b - a != 0) && ((b - a) % 2 == 0)){
                         
                    // Stores minimum value
                    X = Math.min(X, (b - a) / 2) ;
                }
            }
        }
     
        // Return if X * X if X is not equal
        // to 1e9 else return -1
        if (X != 1e9)
            ans = X * X;
        else
            ans = -1;
             
        return ans;
    }
     
    // Driver code
    public static void main (String[] args){
        long N = 3;
         
        System.out.println(SmallestPerfectSquare(N)) ;
     
        }
}
// This code is contributed by AnkitRai01

Python3

# Python3 code to find out the smallest
# perfect square X which when added to N
# yields another perfect square number.
import math
def SmallestPerfectSquare(N):
 
    # X is the smallest perfect
    # square number
    X = 1e9
 
    # Loop from 1 to square root of N
    for i in range(1, int(math.sqrt(N)) + 1):
 
        # Condition to check whether i
        # is factor of N or not
        if N % i == 0:
            a = i
            b = N // i 
 
            # Condition to check whether 
            # factors satisfies the
            # equation or not
            if b - a != 0 and (b - a) % 2 == 0:
                    
                # Stores minimum value
                 X = min(X, (b - a) // 2)
 
    # Return if X * X if X is not equal
    # to 1e9 else return -1
    return(X * X if X != 1e9 else -1)
 
# Driver code
if __name__ == "__main__" : 
   
    N = 3
   
    print(SmallestPerfectSquare(N))

C#

// C# code to find out the smallest
// perfect square X which when added to N
// yields another perfect square number.
using System;
 
class GFG {
     
    static long SmallestPerfectSquare(long N)
    {
        // X is the smallest perfect
        // square number
        long X = (long)1e9;
        long ans;
     
        // Loop from 1 to square root of N
        for(int i = 1; i < Math.Sqrt(N); i++){
     
            // Condition to check whether i
            // is factor of N or not
            if (N % i == 0)
            {
                long a = i;
                long b = N / i;
     
                // Condition to check whether
                // factors satisfies the
                // equation or not
                if ((b - a != 0) && ((b - a) % 2 == 0))
                {
                    // Stores minimum value
                    X = Math.Min(X, (b - a) / 2);
                }
            }
        }
     
        // Return if X*X if X is not equal
        // to 1e9 else return -1
        if (X != 1e9)
            ans = X * X;
        else
            ans = -1;
             
        return ans;
    }
     
    // Driver code
    public static void Main (string[] args)
    {
        long N = 3;
        Console.WriteLine(SmallestPerfectSquare(N));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// JavaScript code to find out the smallest
// perfect square X which when added to N
// yields another perfect square number. 
 
function SmallestPerfectSquare(N){
     
    // X is the smallest perfect
    // square number
    let X = 1e9;
    let ans;
     
    // Loop from 1 to square root of N
    for(let i = 1; i < Math.sqrt(N); i++)
    {
     
        // Condition to check whether i
        // is factor of N or not
        if (N % i == 0)
        {
            let a = i;
            let b = N / i;
     
            // Condition to check whether
            // factors satisfies the
            // equation or not
            if((b - a != 0) && ((b - a) % 2 == 0))
            {
                         
                // Stores minimum value
                X = Math.min(X, (b - a) / 2);
            }
        }
    }
     
    // Return if X * X if X is not equal
    // to 1e9 else return -1
    if (X != 1e9)
        ans = X * X;
    else
        ans = -1;
             
    return ans;
}
     
// Driver code
 
    let N = 3;
    document.write(SmallestPerfectSquare(N)); 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

1

时间复杂度： sqrt(N)

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