计算总和等于N的非负三元组

给定整数N。任务是找到非负整数的不同有序三元组(a，b，c)的数量，以使a + b + c = N。
例子：

Input : N = 2 
Output : 6 
Triplets are : (0, 0, 2), (1, 0, 1), (0, 1, 1), (2, 0, 0), (0, 2, 0), (1, 1, 0)
Input : N = 50 
Output : 1326 
 

方法 ：
首先，很容易看出，对于每个非负整数N ，方程( a + b = N)可以由(a + 1)个(N + 1)对不同的有序对来满足。现在我们可以分配0到N之间的c值，然后可以找到a + b的有序对。它将形成一系列N + 1个自然数，其总和将得出三元组的数量。
下面是上述方法的实现：

C++

// CPP program to find triplets count
#include 
using namespace std;
 
// Function to find triplets count
int triplets(int N)
{
    // Sum of first n+1 natural numbers
    return ((N + 1) * (N + 2)) / 2;
}
 
// Driver code
int main()
{
    int N = 50;
     
    // Function call
    cout << triplets(N);
     
    return 0;
}

Java

// Java program to find triplets count
class GFG
{
     
// Function to find triplets count
static int triplets(int N)
{
    // Sum of first n+1 natural numbers
    return ((N + 1) * (N + 2)) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 50;
 
    System.out.println(triplets(N));
}
}
 
// This code is contributed
// by PrinciRaj1992

Python3

# Python3 program to find triplets count
 
# Function to find triplets count
def triplets(N):
 
    # Sum of first n+1 natural numbers
    return ((N + 1) * (N + 2)) // 2;
 
# Driver code
N = 50;
     
# Function call
print(triplets(N))
 
# This code is contributed by nidhi

C#

// C# program to find triplets count
using System;
 
class GFG
{
     
// Function to find triplets count
static int triplets(int N)
{
    // Sum of first n+1 natural numbers
    return ((N + 1) * (N + 2)) / 2;
}
 
// Driver code
public static void Main()
{
    int N = 50;
 
    Console.WriteLine(triplets(N));
}
}
 
// This code is contributed
// by anuj_67..

Javascript

 
    // Javascript program to find triplets count   
     
    // Function to find triplets count
    function triplets(N)
    {
        // Sum of first n+1 natural numbers
        return ((N + 1) * (N + 2)) / 2;
    }
     
    let N = 50;
    document.write(triplets(N));
     

输出：

1326

时间复杂度： O(1)