给定数N ,任务是找到第N个兆数。
A Megagon number is a class of figurate numbers. It has a 1000000-sided polygon called Megagon. The N-th Megagon number count’s the 1000000 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few Megagonol numbers are 1, 1000000, 2999997, 5999992, 9999985, 14999976, …
例子:
Input: N = 2
Output: 1000000
Explanation:
The second Megagonol number is 1000000.
Input: N = 3
Output: 2999997
方法:第N个百万像素数由以下公式给出:
- 侧多边形的第N个项=
- 因此1000000个多边形的第N个项是
下面是上述方法的实现:
C++
// C++ implementation for the
// above approach
#include
using namespace std;
// Function to find the
// nth Megagon Number
int MegagonNum(int n)
{
return (999998 * n * n - 999996 * n) / 2;
}
// Driver Code
int main()
{
int n = 3;
cout << MegagonNum(n);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to find the
// nth Megagon Number
static int MegagonNum(int n)
{
return (999998 * n * n - 999996 * n) / 2;
}
// Driver code
public static void main(String[] args)
{
int n = 3;
System.out.print(MegagonNum(n));
}
}
// This code is contributed by shubham
Python3
# Python3 implementation for the
# above approach
# Function to find the
# nth Megagon Number
def MegagonNum(n):
return (999998 * n * n - 999996 * n) // 2;
# Driver Code
n = 3;
print(MegagonNum(n));
# This code is contributed by Code_Mech
C#
// C# program for the above approach
using System;
class GFG{
// Function to find the
// nth Megagon Number
static int MegagonNum(int n)
{
return (999998 * n * n - 999996 * n) / 2;
}
// Driver code
public static void Main(String[] args)
{
int n = 3;
Console.Write(MegagonNum(n));
}
}
// This code is contributed by sapnasingh4991
Javascript
输出:
2999997
参考: https : //en.wikipedia.org/wiki/Megagon