Given an **N-sided regular polygon**, we have connected all the vertices at the center of the polygon, thus dividing the polygon into N equal parts. Our task is to the Count of the total number of cycles in the polygon.**Note:** A Cycle is a closed-loop starting and ending on same point.**Examples:**

Input:N = 3Output:7Explanation:

When a 3 sided polygon is connected by vertices at the center then we get 7 cycles possible for it as shown in the image.

Input:N = 5Output:21Explanation:

When a 5 sided polygon is connected by vertices at the center then we get 21 cycles possible for it as shown in the image.

**Approach:** To the problem mentioned above we are supposed to count the total number of closed loops possible in the given polygon after division. The approach is based upon **Mathematical Pattern**. There will be **N cycles** already created due to the division of polygon. One out of **N blocks** will form a cycle with rest **(N – 1)** blocks. The remaining **(N – 1)** blocks will form cycle with other **(N – 2)** blocks. So the total cycles we have can be found out using the formula given below:

Total Cycles = N + 1 * (N – 1) + (N – 1) * (N – 2)

Total Cycles = 2 * N – 1) + (N – 1) * (N – 2)

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Function to calculate number of cycles` `int` `findCycles(` `int` `N)` `{` ` ` `int` `res = 0;` ` ` `int` `finalResult = 0;` ` ` `int` `val = 2 * N - 1;` ` ` ` ` `// BigInteger is used here` ` ` `// if N=10^9 then multiply` ` ` `// will result into value` ` ` `// greater than 10^18` ` ` `int` `s = val;` ` ` ` ` `// BigInteger multiply function` ` ` `res = (N - 1) * (N - 2);` ` ` `finalResult = res + s;` ` ` ` ` `// Return the final result` ` ` `return` `finalResult;` `}` `// Driver code` `int` `main()` `{` ` ` `// Given N` ` ` `int` `N = 5;` ` ` ` ` `// Function Call` ` ` `cout << findCycles(N) << endl; ` ` ` `return` `0;` `}` `// This code is contributed by divyeshrabadiya07` |

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## Java

`// Java program for the above approach` `import` `java.util.*;` `import` `java.math.*;` `class` `GFG {` ` ` `// Function to calculate number of cycles` ` ` `static` `BigInteger findCycles(` `int` `N)` ` ` `{` ` ` `BigInteger res, finalResult;` ` ` `long` `val = ` `2` `* N - ` `1` `;` ` ` `String st = String.valueOf(val);` ` ` `// BigInteger is used here` ` ` `// if N=10^9 then multiply` ` ` `// will result into value` ` ` `// greater than 10^18` ` ` `BigInteger str = ` `new` `BigInteger(st);` ` ` `String n1 = String.valueOf((N - ` `1` `));` ` ` `String n2 = String.valueOf((N - ` `2` `));` ` ` `BigInteger a = ` `new` `BigInteger(n1);` ` ` `BigInteger b = ` `new` `BigInteger(n2);` ` ` `// BigInteger multiply function` ` ` `res = a.multiply(b);` ` ` `finalResult = res.add(str);` ` ` `// Return the final result` ` ` `return` `finalResult;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` ` ` `main(String args[]) ` `throws` `Exception` ` ` `{` ` ` `// Given N` ` ` `int` `N = ` `5` `;` ` ` `// Function Call` ` ` `System.out.println(findCycles(N));` ` ` `}` `}` |

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## Python3

`# Python3 program for the above approach` ` ` `# Function to calculate number of cycles` `def` `findCycles(N):` ` ` `res ` `=` `0` ` ` `finalResult ` `=` `0` ` ` `val ` `=` `2` `*` `N ` `-` `1` `;` ` ` `# BigInteger is used here` ` ` `# if N=10^9 then multiply` ` ` `# will result into value` ` ` `# greater than 10^18` ` ` `s ` `=` `val` ` ` `# BigInteger multiply function` ` ` `res ` `=` `(N ` `-` `1` `) ` `*` `(N ` `-` `2` `)` ` ` `finalResult ` `=` `res ` `+` `s;` ` ` `# Return the final result` ` ` `return` `finalResult;` `# Driver Code` `if` `__name__` `=` `=` `'__main__'` `:` ` ` ` ` `# Given N` ` ` `N ` `=` `5` `;` ` ` `# Function Call` ` ` `print` `(findCycles(N));` ` ` `# This code is contributed by pratham76` |

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**Output:**

21

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