查找大于给定数字的下一个完美平方

给定数字N，任务是找到下一个大于N的理想平方。
例子：

Input: N = 6
Output: 9
9 is a greater number than 6 and
is also a perfect square

Input: N = 9
Output: 16

方法：

1. 找出给定N的平方根。
2. 使用C++中的下限函数计算其下限值。
3. 然后添加1。
4. 打印该数字的正方形。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
#include
using namespace std;
 
// Function to find the next perfect square
int nextPerfectSquare(int N)
{
    int nextN = floor(sqrt(N)) + 1;
 
    return nextN * nextN;
}
 
// Driver Code
int main()
{
    int n = 35;
 
    cout << nextPerfectSquare(n);
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
     
// Function to find the
// next perfect square
static int nextPerfectSquare(int N)
{
    int nextN = (int)Math.floor(Math.sqrt(N)) + 1;
 
    return nextN * nextN;
}
 
// Driver Code
public static void main(String args[])
{
    int n = 35;
 
    System.out.println (nextPerfectSquare(n));
}
}
 
// This code is contributed by Subhadeep

Python3

# Python3 implementation of above approach
 
import math
#Function to find the next perfect square
 
def nextPerfectSquare(N):
 
    nextN = math.floor(math.sqrt(N)) + 1
 
    return nextN * nextN
 
if __name__=='__main__':
    N = 35
    print(nextPerfectSquare(N))
 
# this code is contributed by Surendra_Gangwar

C#

// C# implementation of above approach
using System;
 
class GFG
{
     
// Function to find the
// next perfect square
static int nextPerfectSquare(int N)
{
    int nextN = (int)Math.Floor(Math.Sqrt(N)) + 1;
 
    return nextN * nextN;
}
 
// Driver Code
public static void Main()
{
    int n = 35;
 
    Console.WriteLine(nextPerfectSquare(n));
}
}
 
// This code is contributed
// by Shashank

Javascript

// Javascript implementation of above approach
 
// Function to find the next perfect square
function nextPerfectSquare(N)
{
    let nextN = Math.floor(Math.sqrt(N)) + 1;
 
    return nextN * nextN;
}
 
// Driver Code
let n = 35;
 
document.write(nextPerfectSquare(n));
 
// This code is contributed by souravmahato348.

输出：

36

时间复杂度： O(1)

辅助空间： O(1)