在3人中分配N项的方式计数，其中1人获得最大奖励

给定整数N ，任务是找到在3个人之间分配N的总数，使得：

• 恰好一个人在这三个人中获得的物品数最多。
• 每个人至少获得1件物品。

例子：

Input: N = 5 
Output: 3 
Explanation: 
3 way distribute the item among 3 people are {1, 1, 3}, {1, 3, 1} and {3, 1, 1}. 
Distributions like {1, 2, 2} or {2, 1, 2} are not valid as two persons are getting the maximum.
Input: N = 10 
Output: 33 
Explanation: 
For the Input N = 10 there are 33 ways of distribution. 
 

方法：
为了解决上述问题，我们必须观察到，如果N <4 ，那么这种分布是不可能的。
对于所有N≥4的值，请按照以下步骤解决问题：

• (N – 1)*(N – 2)/ 2给出了在3个人中分配N项的总方法数。
• 初始化变量s = 0 ，该变量存储无法进行分配的方式的数量。
• 迭代两个嵌套循环，其中i介于[2，N – 3]和j之间，直至i ：
  • 对于每次迭代，请检查N = 2 * i + j是否为2，即最多可以接收2个元素
  • 如果是这样，则将s加1 。如果N可被3整除，则将s更新3 * s + 1 。否则，更新为3 * s 。
• 最后，返回ans –作为在三个人之间分配N个项目的总数。

下面是上述方法的实现：

C++

// C++ program to find the number
// of ways to distribute N item
// among three people such
// that one person always gets
// the maximum value
 
#include 
using namespace std;
 
// Function to find the number
// of ways to distribute N
// items among 3 people
int countWays(int N)
{
    // No distribution
    // possible
    if (N < 4)
        return 0;
 
    // Total number of ways to
    // distribute N items
    // among 3 people
    int ans = ((N - 1) * (N
                          - 2))
              / 2;
 
    // Store the number of
    // distributions which
    // are not possible
    int s = 0;
 
    for (int i = 2; i <= N - 3;
         i++) {
        for (int j = 1; j < i;
             j++) {
 
            // Count possiblities
            // of two persons
            // receiving the
            // maximum
            if (N == 2 * i + j)
                s++;
        }
    }
 
    // If N is divisible by 3
    if (N % 3 == 0)
        s = 3 * s + 1;
 
    else
        s = 3 * s;
 
    // Return the final
    // count of ways
    // to distribute
    return ans - s;
}
 
// Driver Code
int main()
{
    int N = 10;
    cout << countWays(N);
    return 0;
}

Java

// Java program to find the number
// of ways to distribute N item
// among three people such
// that one person always gets
// the maximum value
class GFG{
 
// Function to find the number
// of ways to distribute N
// items among 3 people
static int countWays(int N)
{
    // No distribution
    // possible
    if (N < 4)
        return 0;
 
    // Total number of ways to
    // distribute N items
    // among 3 people
    int ans = ((N - 1) * (N - 2)) / 2;
 
    // Store the number of
    // distributions which
    // are not possible
    int s = 0;
 
    for (int i = 2; i <= N - 3; i++)
    {
        for (int j = 1; j < i; j++)
        {
 
            // Count possiblities
            // of two persons
            // receiving the
            // maximum
            if (N == 2 * i + j)
                s++;
        }
    }
 
    // If N is divisible by 3
    if (N % 3 == 0)
        s = 3 * s + 1;
    else
        s = 3 * s;
 
    // Return the final
    // count of ways
    // to distribute
    return ans - s;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 10;
    System.out.println(countWays(N));
}
}
 
// This code is contributed by rock_cool

Python3

# Python3 program to find the number
# of ways to distribute N item
# among three people such
# that one person always gets
# the maximum value
 
# Function to find the number
# of ways to distribute N
# items among 3 people
def countWays(N):
     
    # No distribution
    # possible
    if (N < 4):
        return 0
 
    # Total number of ways to
    # distribute N items
    # among 3 people
    ans = ((N - 1) * (N - 2)) // 2
 
    # Store the number of
    # distributions which
    # are not possible
    s = 0
 
    for i in range( 2, N - 2, 1):
        for j in range( 1, i, 1):
             
            # Count possiblities
            # of two persons
            # receiving the
            # maximum
            if (N == 2 * i + j):
                s += 1
 
    # If N is divisible by 3
    if (N % 3 == 0):
        s = 3 * s + 1
 
    else:
        s = 3 * s
 
    # Return the final
    # count of ways
    # to distribute
    return ans - s
 
# Driver Code
N = 10
 
print (countWays(N))
     
# This code is contributed by sanjoy_62

C#

// C# program to find the number
// of ways to distribute N item
// among three people such
// that one person always gets
// the maximum value
using System;
class GFG{
 
// Function to find the number
// of ways to distribute N
// items among 3 people
static int countWays(int N)
{
    // No distribution
    // possible
    if (N < 4)
        return 0;
 
    // Total number of ways to
    // distribute N items
    // among 3 people
    int ans = ((N - 1) * (N - 2)) / 2;
 
    // Store the number of
    // distributions which
    // are not possible
    int s = 0;
 
    for (int i = 2; i <= N - 3; i++)
    {
        for (int j = 1; j < i; j++)
        {
 
            // Count possiblities
            // of two persons
            // receiving the
            // maximum
            if (N == 2 * i + j)
                s++;
        }
    }
 
    // If N is divisible by 3
    if (N % 3 == 0)
        s = 3 * s + 1;
    else
        s = 3 * s;
 
    // Return the final
    // count of ways
    // to distribute
    return ans - s;
}
 
// Driver Code
public static void Main()
{
    int N = 10;
    Console.Write(countWays(N));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
    // Javascript program to find the number
    // of ways to distribute N item
    // among three people such
    // that one person always gets
    // the maximum value
     
    // Function to find the number
    // of ways to distribute N
    // items among 3 people
    function countWays(N)
    {
        // No distribution
        // possible
        if (N < 4)
            return 0;
 
        // Total number of ways to
        // distribute N items
        // among 3 people
        let ans = ((N - 1) * (N - 2)) / 2;
 
        // Store the number of
        // distributions which
        // are not possible
        let s = 0;
 
        for (let i = 2; i <= N - 3; i++) {
            for (let j = 1; j < i; j++) {
 
                // Count possiblities
                // of two persons
                // receiving the
                // maximum
                if (N == 2 * i + j)
                    s++;
            }
        }
 
        // If N is divisible by 3
        if (N % 3 == 0)
            s = 3 * s + 1;
 
        else
            s = 3 * s;
 
        // Return the final
        // count of ways
        // to distribute
        return ans - s;
    }
 
    let N = 10;
    document.write(countWays(N));
     

输出：

33

时间复杂度： O(N ² )
辅助空间： O(1)