第N个非平方数

给定n ，找到自然数(1、2、3、4、5、6…)中不是完美平方的第n个数

例子：

Input : 3
Output : 5
First three non-square numbers are 2, 3
and 5

Input : 5
Output : 7

Input : 16
Output : 20

查看问题陈述，我们可以得出一种简单明了的暴力方法。我们可以从n = 1开始，然后开始检查它们是否是一个完美的正方形。因此，我们可以得出第n个非平方数。
但是，上述方法非常慢，因为它每次都搜索小于目标的每个数字。
我们可以看到正在考虑的序列是2、3、5、6、7、8、10、11、12、13、14、15、17…。
通过检查，我们可以得出此序列中第n个数字的恒定时间公式。
$ans(n) = n + \left \lfloor \frac{1}{2} + \sqrt{n}\right \rfloor$ 。
公式的正确性可以通过数学归纳原理来证明。

上面公式的实现如下。

C++

// CPP program to find n-th non-square number.
#include 
 
using namespace std;
 
// function to find the nth Non-Square Number
int findNthNonSquare(int n)
{
    // conversion from int to long double is
    // necessary in order to preserve decimal
    // places after square root.
    long double x = (long double)n;
 
    // calculating the result
    long double ans = x + floor(0.5 + sqrt(x));
 
    return (int)ans;
}
 
// Driver code
int main()
{
    // initializing the term number
    int n = 16;
 
    // Print the result
    cout << "The " << n << "th Non-Square number is ";
    cout << findNthNonSquare(n);
 
    return 0;
}

Java

// Java program to find
// n-th non-square number.
import java.io.*;
import java.util.*;
import java.lang.*;
 
class GFG
{
     
// function to find the
// nth Non-Square Number
static int findNthNonSquare(int n)
{
    // conversion from int to
    // long double is necessary
    // in order to preserve decimal
    // places after square root.
    double x = (double)n;
 
    // calculating the result
    double ans = x + Math.floor(0.5 +
                     Math.sqrt(x));
 
    return (int)ans;
}
 
// Driver code
public static void main(String[] args)
{
    // initializing
    // the term number
    int n = 16;
 
    // Print the result
    System.out.print("The " + n +
                     "th Non-Square number is ");
    System.out.print(findNthNonSquare(n));
}
}

Python3

# Python3 program to find n-th
# non-square number.
import math
 
# function to find the nth
# Non-Square Number
def findNthNonSquare(n):
 
    # conversion from int to long
    # double is necessary in order
    # to preserve decimal places
    # after square root.
    x = n;
 
    # calculating the result
    ans = x + math.floor(0.5 + math.sqrt(x));
 
    return int(ans);
 
# Driver code
 
# initializing the term number
n = 16;
 
# Print the result
print("The", n, "th Non-Square number is",
                     findNthNonSquare(n));
 
# This code is contributed by mits

C#

// C# program to find
// n-th non-square number.
using System;
 
class GFG
{
     
// function to find the
// nth Non-Square Number
static int findNthNonSquare(int n)
{
    // conversion from int
    // to long double is
    // necessary in order
    // to preserve decimal
    // places after square
    // root.
    double x = (double)n;
 
    // calculating the result
    double ans = x + Math.Floor(0.5 +
                     Math.Sqrt(x));
 
    return (int)ans;
}
 
// Driver code
public static void Main()
{
    // initializing
    // the term number
    int n = 16;
 
    // Print the result
    Console.Write("The " + n +
                  "th Non-Square " +
                      "number is ");
    Console.Write(findNthNonSquare(n));
}
}
 
// This code is contributed
// by anuj_67.

Javascript

 
// Javascript program to find
// n-th non-square number.
 
// Function to find the
// nth Non-Square Number
function findNthNonSquare(n)
{
     
    // Conversion from var to
    // var var is necessary
    // in order to preserve decimal
    // places after square root.
    var x =  n;
 
    // Calculating the result
    var ans = x + Math.floor(0.5 + Math.sqrt(x));
 
    return parseInt(ans);
}
 
// Driver code
 
// Initializing
// the term number
var n = 16;
 
// Print the result
document.write("The " + n + "th Non-Square number is ");
document.write(findNthNonSquare(n));
 
// This code is contributed by todaysgaurav
 

输出：

The 16th Non-Square number is 20

时间复杂度$\mathcal{O}(1)$ 。
空间复杂度$\mathcal{O}(1)$

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