a + b + c = n 的非负积分解的数量

给定一个数 n，找出多种方法可以将 3 个非负整数相加，使它们的和为 n。
例子 ：

Input : n = 1
Output : 3
There are four ways to get sum 1.
(1, 0, 0), (0, 1, 0) and (0, 0, 1)

Input : n = 2
Output : 6
There are six ways to get sum 2.
(2, 0, 0), (0, 2, 0), (0, 0, 2), (1, 1, 0)
(1, 0, 1) and (0, 1, 1)

Input : n = 3
Output : 10
There are ten ways to get sum 10.
(3, 0, 0), (0, 3, 0), (0, 0, 3), (1, 2, 0)
(1, 0, 2), (0, 1, 2), (2, 1, 0), (2, 0, 1),
(0, 2, 1) and (1, 1, 1)

方法 1 [蛮力：O(n ³ ) ]
一个简单的解决方案是使用三个循环来考虑所有三元组。对于每个三元组，检查其总和是否等于 n。如果总和为 n，则增加解的计数。
下面是实现。

C++

// A naive C++ solution to count solutions of
// a + b + c = n
#include
using namespace std;
 
// Returns count of solutions of a + b + c = n
int countIntegralSolutions(int n)
{
    // Initialize result
    int result = 0;
 
    // Consider all triplets and increment
    // result whenever sum of a triplet is n.
    for (int i=0; i<=n; i++)
      for (int j=0; j<=n-i; j++)
          for (int k=0; k<=(n-i-j); k++)
             if (i + j + k == n)
              result++;
 
    return result;
}
 
// Driver code
int main()
{
    int n = 3;
    cout <<  countIntegralSolutions(n);
    return 0;
}

Java

// A naive Java solution to count
// solutions of a + b + c = n
import java.io.*;
 
class GFG
{
    // Returns count of solutions of a + b + c = n
    static int countIntegralSolutions(int n)
    {
        // Initialize result
        int result = 0;
     
        // Consider all triplets and increment
        // result whenever sum of a triplet is n.
        for (int i = 0; i <= n; i++)
        for (int j = 0; j <= n - i; j++)
            for (int k = 0; k <= (n - i - j); k++)
                if (i + j + k == n)
                result++;
     
        return result;
    }
 
    // Driver code
    public static void main (String[] args)
    {
        int n = 3;
        System.out.println( countIntegralSolutions(n));
     
    }
}
 
// This article is contributed by vt_m

Python3

# Python3 code to count
# solutions of a + b + c = n
 
# Returns count of solutions
# of a + b + c = n
def countIntegralSolutions (n):
 
    # Initialize result
    result = 0
     
    # Consider all triplets and
    # increment result whenever
    # sum of a triplet is n.
    for i in range(n + 1):
        for j in range(n + 1):
            for k in range(n + 1):
                if i + j + k == n:
                    result += 1
     
    return result
     
# Driver code
n = 3
print(countIntegralSolutions(n))
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// A naive C# solution to count
// solutions of a + b + c = n
using System;
 
class GFG {
     
    // Returns count of solutions
    // of a + b + c = n
    static int countIntegralSolutions(int n)
    {
         
        // Initialize result
        int result = 0;
     
        // Consider all triplets and increment
        // result whenever sum of a triplet is n.
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= n - i; j++)
                for (int k = 0; k <= (n - i - j); k++)
                    if (i + j + k == n)
                    result++;
     
        return result;
    }
 
    // Driver code
    public static void Main ()
    {
        int n = 3;
        Console.Write(countIntegralSolutions(n));
     
    }
}
 
// This article is contributed by Smitha.

Javascript

 
// Javascript program solution to count
// solutions of a + b + c = n
 
  // Returns count of solutions of a + b + c = n
    function countIntegralSolutions(n)
    {
        // Initialize result
        let result = 0;
       
        // Consider all triplets and increment
        // result whenever sum of a triplet is n.
        for (let i = 0; i <= n; i++)
        for (let j = 0; j <= n - i; j++)
            for (let k = 0; k <= (n - i - j); k++)
                if (i + j + k == n)
                result++;
       
        return result;
    }
 
// Driver Code
 
    let n = 3;
    document.write(countIntegralSolutions(n));
 

输出：

10

方法 2 [直接公式：O(1)]
如果我们仔细观察这个模式，我们会发现解的数量是 ((n+1) * (n+2)) / 2。 这个问题等价于分配 n + 1 个相同的球(对于 0 到n) 在三个框中，解为ⁿ⁺² C ₂ 。一般来说，如果有 m 个变量(或框)和 n 个可能的值，则公式变为^n+m-1 C _m-1 。

C++

// A naive C++ solution to count solutions of
// a + b + c = n
#include
using namespace std;
 
// Returns count of solutions of a + b + c = n
int countIntegralSolutions(int n)
{
    return ((n+1)*(n+2))/2;
}
 
// Driver code
int main()
{
    int n = 3;
    cout <<  countIntegralSolutions(n);
    return 0;
}

Java

// Java solution to count
// solutions of a + b + c = n
import java.io.*;
 
class GFG
{
    // Returns count of solutions
    // of a + b + c = n
    static int countIntegralSolutions(int n)
    {
    return ((n + 1) * (n + 2)) / 2;
         
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 3;
        System.out.println ( countIntegralSolutions(n));
         
    }
}
// This article is contributed by vt_m

蟒蛇3

# Python3 solution to count
# solutions of a + b + c = n
 
# Returns count of solutions
# of a + b + c = n
def countIntegralSolutions (n):
    return int(((n + 1) * (n + 2)) / 2)
     
# Driver code
n = 3
print(countIntegralSolutions(n))
 
# This code is contributed by "Sharad_Bhardwaj".

C＃

// C# solution to count
// solutions of a + b + c = n
using System;
 
class GFG {
     
    // Returns count of solutions
    // of a + b + c = n
    static int countIntegralSolutions(int n)
    {
        return ((n + 1) * (n + 2)) / 2;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int n = 3;
        Console.Write ( countIntegralSolutions(n));
         
    }
}
 
// This code is contributed by parashar.

Javascript

 
// A naive JavaScript solution
// to count solutions of
// a + b + c = n
 
// Returns count of solutions of a + b + c = n
function countIntegralSolutions(n)
{
    return Math.floor(((n+1)*(n+2))/2);
}
 
// Driver code
 
    let n = 3;
    document.write(countIntegralSolutions(n));
 
 
// This code is contributed by Surbhi Tyagi.
 

输出 ：

10