中心正方形编号
给定一个数字 n ，任务是找到第 n 个中心方块数。
中心正方形数字是一个中心图形数字，它给出了一个正方形中的点的数量，一个点在中心，所有其他点围绕着连续正方形层中的中心点。N 第个中心方块数可以用公式n2+(n-1)2计算。
示例:
Input : n = 2
Output : 5
Input : n = 9
Output : 145
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Finding n-th Centered Square Number If we take a closer look, we can notice that the n-th Centered Square Number can be seen as the sum of two consecutive square numbers (1 dot, 4 dots, 9 dots, 16 dots, etc).
我们可以用下面的公式找到第 n 个中心平方数。
第 n 个中心方块数= n 2 + (n-1) 2
下面是实现:
C++
``` // C++ program to find nth // Centered square number.
include
using namespace std;
// Function to calculate Centered // square number function int centered_square_num(int n) { // Formula to calculate nth // Centered square number return n * n + ((n - 1) * (n - 1)); }
// Driver Code int main() { int n = 7; cout << n << "th Centered square number: "; cout << centered_square_num(n); return 0; } ```
Java 语言(一种计算机语言，尤用于创建网站)
``` // Java program to find nth Centered square // number import java.io.*;
class GFG {
// Function to calculate Centered // square number function static int centered_square_num(int n) { // Formula to calculate nth // Centered square number return n * n + ((n - 1) * (n - 1)); }
// Driver Code public static void main (String[] args) { int n = 7; System.out.print( n + "th Centered" + " square number: " + centered_square_num(n)); } }
// This code is contributed by anuj_67. ```
Python 3
```
Python program to find nth
Centered square number.
Function to calculate Centered
square number function
def centered_square_num(n):
# Formula to calculate nth # Centered square number return n * n + ((n - 1) * (n - 1))
Driver Code
n = 7 print("%sth Centered square number: " %n, centered_square_num(n))
```
C
``` // C# program to find nth // Centered square number. using System;
public class GFG {
// Function to calculate Centered // square number function static int centered_square_num(int n) { // Formula to calculate nth // Centered square number return n * n + ((n - 1) * (n - 1)); }
// Driver Code
static public void Main (){ int n = 7; Console.WriteLine( n + "th Centered" + " square number: " + centered_square_num(n)); } }
// This code is contributed by anuj_67. ```
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
``` <?php // PHP program to find nth // Centered square number
// Function to calculate Centered // square number function function centered_square_num( $n) { // Formula to calculate nth // Centered square number return $n * $n + (($n - 1) * ($n - 1)); }
// Driver Code $n = 7; echo $n , "th Centered square number: "; echo centered_square_num($n);
// This code is contributed by anuj_67. ?> ```
Output :
``` 7th Centered square number: 85
```
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Check if N is centred-square-number or not:
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前几个居中的方形数字是:
1，5，13，25，41，61，85，113，145，181、………………
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由于第 n 个中心平方数由
H(n) = n * n + ((n - 1) * (n - 1))给出 * 该公式表明第 N 个中心平方数二次依赖于 N，因此，试求 N = H(n)方程的正整数根。
``` H(n) = nth centered-square-number number N = Given Number
Solve for n: H(n) = N n * n + ((n - 1) * (n - 1)) = N
Applying Shridharacharya Formula The positive root of equation (i) n = (9 + sqrt(36*N+45))/18;
```
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获取 n 后，检查是否为整数。如果 n–floor(n)为 0，则 n 为整数。
下面是上述方法的实现:
卡片打印处理机（Card Print Processor 的缩写）
```
include
using namespace std;
bool centeredSquare_number(int N) { float n = (9 + sqrt(36*N+45))/18; return (n - (int) n) == 0; }
int main() { int i = 13; cout<<centeredSquare_number(i); return 0; } ```
Java 语言(一种计算机语言，尤用于创建网站)
``` // Java Code implementation of the above approach class GFG {
static int centeredSquare_number(int N) { float n = (9 + (float)Math.sqrt(36*N+45))/18; if (n - (int) n == 0) return 1; else return 0; }
// Driver code public static void main (String[] args) { int i = 13; System.out.println(centeredSquare_number(i)); }
}
// This code is contributed by Yash_R ```
Python 3
```
Python3 implementation of the above approach
from math import sqrt
def centeredSquare_number(N) :
n = (9 + sqrt(36 * N + 45))/18; if (n - int(n)) == 0 : return 1 else : return 0
Driver Code
if name == "main" :
i = 13; print(centeredSquare_number(i));
This code is contributed by Yash_R
```
C
``` // C# Code implementation of the above approach using System;
class GFG {
static int centeredSquare_number(int N) { float n = (9 + (float)Math.Sqrt(36 * N + 45))/18; if (n - (int) n == 0) return 1; else return 0; }
// Driver code public static void Main (String[] args) { int i = 13; Console.WriteLine(centeredSquare_number(i)); }
}
// This code is contributed by Yash_R ```
Output:
``` 0
```
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参考:T2】https://en.wikipedia.org/wiki/Centered_square_number
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