通过使用最大为N的数字来检查Pascal的三角形是否可以用于一个完整的图层

给定数字N ，任务是确定是否可以通过使用总数N整数(如果可能)来制作具有完整图层的Pascal三角形，如果可能，请打印是，否则请打印否。

注意： Pascal三角形是二项式系数的三角形阵列。以下是帕斯卡三角形的前6行。

1  
1 1 
1 2 1 
1 3 3 1 
1 4 6 4 1 
1 5 10 10 5 1 

从最顶层开始的Pascal三角形中，有1个整数，在从下到上的每一层中，该层的大小都增加了1。

例子：

Input: N = 10
Output: Yes
Explanation: 
You can use 1, 2, 3 and 4 integers to make first, second, third, and fourth layer of pascal’s triangle respectively and also N = 10 satisfy by using (1 + 2 + 3 + 4) integers on each layer = 10.
Input: N = 5 
Output: No
Explanation: 
You can use 1 and 2 integers to make first and second layer respectively and after that you have only 2 integers left and you can’t make 3rd layer complete as that layer required 3 integers.

方法：在这里，我们从第一层开始在每一层上都使用整数1、2、3，…，因此，只有在可以用1 + 2 +的总和表示N的情况下，才能使Pascal三角形完整。

1. 前X个整数的总和由下式给出

$1 + 2 + 3 + 4 + .... + X = \frac{X*(X + 1)}{2}$

1. 当且仅当我们仅能通过使用N个整数来制作帕斯卡的三角形$N = \frac{X*(X + 1)}{2}$其中X必须是一个正整数。因此，我们必须检查x是否存在正整数值。
2. 为了从第二步确定X的值，我们可以推导公式为：

$X = \frac{\sqrt{8*N + 1} - 1}{2}$

1. 如果X的值等于N的给定值，则可以使Pascal Triangle成为Pascal Triangle。否则，我们将无法制作Pascal Triangle。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if Pascaltriangle
// can be made by N integers
void checkPascaltriangle(int N)
{
    // Find X
    double x = (sqrt(8 * N + 1) - 1) / 2;
 
    // If x is integer
    if (ceil(x) - x == 0)
        cout << "Yes";
 
    else
        cout << "No";
}
 
// Driver Code
int main()
{
    // Given number N
    int N = 10;
 
    // Function Call
    checkPascaltriangle(N);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to check if Pascaltriangle
// can be made by N integers
static void checkPascaltriangle(int N)
{
     
    // Find X
    double x = (Math.sqrt(8 * N + 1) - 1) / 2;
 
    // If x is integer
    if (Math.ceil(x) - x == 0)
        System.out.print("Yes");
    else
        System.out.print("No");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given number N
    int N = 10;
 
    // Function call
    checkPascaltriangle(N);
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 program for the above approach
import math
 
# Function to check if Pascaltriangle
# can be made by N integers
def checkPascaltriangle(N):
     
    # Find X
    x = (math.sqrt(8 * N + 1) - 1) / 2
 
    # If x is integer
    if (math.ceil(x) - x == 0):
        print("Yes")
    else:
        print("No")
 
# Driver Code
 
# Given number N
N = 10
 
# Function call
checkPascaltriangle(N)
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if Pascaltriangle
// can be made by N integers
static void checkPascaltriangle(int N)
{
     
    // Find X
    double x = (Math.Sqrt(8 * N + 1) - 1) / 2;
 
    // If x is integer
    if (Math.Ceiling(x) - x == 0)
        Console.Write("Yes");
    else
        Console.Write("No");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given number N
    int N = 10;
 
    // Function call
    checkPascaltriangle(N);
}
}
 
// This code is contributed by amal kumar choubey

Javascript

 
      // JavaScript program for the above approach
 
      // Function to check if Pascaltriangle
      // can be made by N integers
      function checkPascaltriangle(N) {
        // Find X
        var x = (Math.sqrt(8 * N + 1) - 1) / 2;
 
        // If x is integer
        if (Math.ceil(x) - x == 0)
        document.write("Yes");
        else
        document.write("No");
      }
 
      // Driver Code
 
      // Given number N
      var N = 10;
 
      // Function Call
      checkPascaltriangle(N);
       
    

输出：

Yes

时间复杂度： O(sqrt(N))
辅助空间： O(1)