建造给定等级N的纸牌屋所需的纸牌数量

给定数字N ，任务是找到制作N个等级的纸牌屋所需的纸牌数。

例子：

Input: N = 3 
Output: 15 
From the above image, it is clear that for the House of Cards for 3 levels 15 cards are needed
Input: N = 2 
Output: 7 
 

方法：

1. 如果我们仔细观察，将形成如下所示的系列，其中第i个项目表示构成i个等级的金字塔所需的三角形卡片的数量：

2, 7, 15, 26, 40, 57, 77, 100, 126, 155………and so on.

2. 上面的序列是一种差异序列的方法，其中AP中的差异为5、8、11、14……。等等。
3. 因此，序列的第n个术语将是：

nth term = 2 + {5 + 8 + 11 +14 +.....(n-1) terms}
         = 2 + (n-1)*(2*5+(n-1-1)*3)/2
         = 2 + (n-1)*(10+(n-2)*3)/2
         = 2 + (n-1)*(10+3n-6)/2
         = 2 + (n-1)*(3n+4)/2
         = n*(3*n+1)/2;

2. 因此，建造N个等级的纸牌屋所需的纸牌数为：

$n*(3*n+1)/2$

下面是上述方法的实现：

CPP

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function to find number of cards needed
int noOfCards(int n)
{
    return n * (3 * n + 1) / 2;
}
 
// Driver Code
int main()
{
    int n = 3;
    cout << noOfCards(n) << ", ";
    return 0;
}

Java

// Java implementation of the above approach
import java.lang.*;
 
class GFG
{
    // Function to find number of cards needed
    public static int noOfCards(int n)
    {
        return n * (3 * n + 1) / 2;
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int n = 3;
        System.out.print(noOfCards(n));
    }
}
 
// This code is contributed by shubhamsingh10

Python3

# Python3 implementation of the above approach
 
# Function to find number of cards needed
def noOfCards(n):
    return n * (3 * n + 1) // 2
 
# Driver Code
n = 3
print(noOfCards(n))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of the above approach
using System;
 
class GFG
{
    // Function to find number of cards needed
    public static int noOfCards(int n)
    {
        return n * (3 * n + 1) / 2;
    }
      
    // Driver Code
    public static void Main(String []args)
    {
        int n = 3;
        Console.Write(noOfCards(n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the above approach
 
// Function to find number of cards needed
function noOfCards(n)
{
    return parseInt(n * (3 * n + 1) / 2);
}
 
// Driver Code
var n = 3;
document.write(noOfCards(n));
 

输出：

15

时间复杂度： O(1)