在中心连接 n 边多边形顶点形成的循环数
给定一个 N 边正多边形，我们已经连接了多边形中心的所有顶点，从而将多边形分成 N 个相等的部分。我们的任务是计算多边形中的循环总数。 注:一个周期是在同一点上开始和结束的闭环。 示例:**
输入: N = 3 输出: 7 解释:
当一个 3 边多边形在中心被顶点连接时，我们可以得到 7 个循环，如图所示。 输入: N = 5 输出: 21 解释:
当一个 5 边的多边形在中心被顶点连接时，我们得到 21 个可能的循环，如图所示。
方法:对于上面提到的问题，我们应该计算分割后给定多边形中可能的闭环总数。该方法基于数学模型。由于多边形的划分，将会有已经创建的 N 个循环。 N 块中的一个将与其余(N–1)块形成一个循环。其余(N–1)块将与其他(N–2)块形成循环。所以我们的总循环数可以用下面给出的公式计算出来:
总周期= N+1 (N–1)+(N–1)(N–2) 总周期= 2 * N–1)+(N–1)*(N–2)
以下是上述方法的实现:
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
Java 语言(一种计算机语言，尤用于创建网站)
// 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));
}
}
Python 3
# 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
C
// C# program for the above approach
using System;
class GFG {
// Function to calculate number of cycles
static 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
static void Main()
{
// Given N
int N = 5;
// Function Call
Console.WriteLine(findCycles(N));
}
}
// This code is contributed by divyesh072019
java 描述语言
<script>
// Javascript program for the above approach
// Function to calculate number of cycles
function findCycles(N)
{
let res = 0;
let finalResult = 0;
let val = 2 * N - 1;
// BigInteger is used here
// if N=10^9 then multiply
// will result into value
// greater than 10^18
let s = val;
// BigInteger multiply function
res = (N - 1) * (N - 2);
finalResult = res + s;
// Return the final result
return finalResult;
}
// Given N
let N = 5;
// Function Call
document.write(findCycles(N));
</script>
Output:
21
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