在 N 个男孩中分配 C 个糖果，使得收到的最大和最小糖果之间的差异为 K

给定两个整数N和C ，代表男孩和糖果的数量，以及一个整数K ，任务是计算任何男孩收到的最大和最小糖果数量，使它们之间的差为K 。

例子：

Input: N = 4, C = 12, K = 3
Output:
Maximum = 5
Minimum = 2
Explanation:
Distribute the {2, 2, 3, 5} candies among N (= 4) boys.
Therefore, the difference between the maximum and minimum is 5 – 2 = 3 (= K).

Input: N = 2, C = 8, K = 2
Output:
Maximum = 5
Minimum = 3.

方法：根据以下观察可以解决给定的问题：

1. 如果 K >= C ：所有C糖果都可以分配给^第一个男孩。因此，糖果的最大数量为C ，最小数量为0 。
2. 否则，将糖果分发给保持最大和最小计数≤ K 的男孩。

Illustration: Refer to the table below to observe the distribution pattern of candies.

Boy A	Boy B	Boy C	Difference
K	0	0	K
K	1	1	K-1
K+1	1	1	K
K+1	2	2	K-1
K+2	2	2	K

 Initially, distribute K candies to the 1^st boy.
Now, distribute remaining C – K candies to each boy line-wise starting form 2nd boy to Nth boy and then, again from 1^st to Nth and so on.                                                                                              

请按照以下步骤解决问题：

1. 初始化两个变量，比如最大值和最小值，以存储一个男孩可以拥有的最大和最小糖果数量。
2. 如果 N = 1：设置最大值 = C和最小值 = C
3. 如果 K >= C：设置最大值= C和最小值= 0
4. 否则，如果K < C ，则设置最大值 = K和最小值 = 0 。现在，请按照以下步骤操作：
   1. 初始化一个变量，比如continue_candy，并将其设置为C – K。
   2. 将 ( remain_candy/N ) 添加到最大值和最小值。
   3. 如果（ remain_candy % N == N-1 ），增加最小值。
5. 打印最小值和最大值。

下面是上述方法的实现：

C++

// C++ program for
// the above approach
#include 
using namespace std;
  
// Function to calculate the
// maximum and minimum number
// of candies a boy can possess
int max_min(int N, int C, int K)
{
    int maximum, minimum;
  
    if (N == 1) {
  
        // All candies will be
        // given to one boy
        maximum = minimum = C;
    }
  
    else if (K >= C) {
  
        // All the candies will
        // be given to 1 boy
        maximum = C;
        minimum = 0;
    }
  
    else {
  
        // Give K candies to 1st
        // boy initially
        maximum = K;
        minimum = 0;
  
        // Count remaining candies
        int remain_candy = C - K;
  
        maximum += remain_candy / N;
        minimum = remain_candy / N;
  
        // If the last candy of remaining candies
        // is given to the last boy, i.e Nth boy
        if (remain_candy % N == N - 1) {
  
            // Increase minimum count
            minimum++;
        }
    }
  
    cout << "Maximum = " << maximum << endl;
    cout << "Minimum = " << minimum;
  
    return 0;
}
  
// Driver Code
int main()
{
    int N = 4;
    int C = 12;
    int K = 3;
  
    max_min(N, C, K);
  
    return 0;
}

Java

// Java program for the above approach
class GFG{
      
// Function to calculate the
// maximum and minimum number
// of candies a boy can possess
static void max_min(int N, int C, int K)
{
    int maximum, minimum;
  
    if (N == 1)
    {
          
        // All candies will be
        // given to one boy
        maximum = minimum = C;
    }
  
    else if (K >= C) 
    {
          
        // All the candies will
        // be given to 1 boy
        maximum = C;
        minimum = 0;
    }
  
    else 
    {
          
        // Give K candies to 1st
        // boy initially
        maximum = K;
        minimum = 0;
  
        // Count remaining candies
        int remain_candy = C - K;
  
        maximum += remain_candy / N;
        minimum = remain_candy / N;
  
        // If the last candy of remaining candies
        // is given to the last boy, i.e Nth boy
        if (remain_candy % N == N - 1) 
        {
              
            // Increase minimum count
            minimum++;
        }
    }
    System.out.println("Maximum = " + maximum);
    System.out.println("Minimum = " + minimum);
}
  
// Driver code
public static void main(String[] args)
{
    int N = 4;
    int C = 12;
    int K = 3;
  
    max_min(N, C, K);
}
}
  
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
  
# Function to calculate the
# maximum and minimum number
# of candies a boy can possess
def max_min(N, C, K):
      
    maximum = 0
    minimum = 0
      
    if (N == 1):
          
        # All candies will be
        # given to one boy
        maximum = minimum = C
  
    elif (K >= C):
          
        # All the candies will
        # be given to 1 boy
        maximum = C
        minimum = 0
  
    else:
          
        # Give K candies to 1st
        # boy initially
        maximum = K
        minimum = 0
  
        # Count remaining candies
        remain_candy = C - K
  
        maximum += remain_candy // N
        minimum = remain_candy // N
  
    # If the last candy of remaining candies
    # is given to the last boy, i.e Nth boy
    if (remain_candy % N == N - 1):
          
        # Increase minimum count
        minimum += 1
  
    print("Maximum = {}".format(maximum))
    print("Minimum = {}".format(minimum))
  
# Driver code
N = 4
C = 12
K = 3
  
max_min(N, C, K)
  
# This code is contributed by abhinavjain194

C#

// C# program for the above approach
using System;
  
class GFG{
      
// Function to calculate the
// maximum and minimum number
// of candies a boy can possess
static void max_min(int N, int C, int K)
{
    int maximum, minimum;
  
    if (N == 1) 
    {
          
        // All candies will be
        // given to one boy
        maximum = minimum = C;
    }
  
    else if (K >= C) 
    {
          
        // All the candies will
        // be given to 1 boy
        maximum = C;
        minimum = 0;
    }
  
    else 
    {
          
        // Give K candies to 1st
        // boy initially
        maximum = K;
        minimum = 0;
  
        // Count remaining candies
        int remain_candy = C - K;
  
        maximum += remain_candy / N;
        minimum = remain_candy / N;
  
        // If the last candy of remaining candies
        // is given to the last boy, i.e Nth boy
        if (remain_candy % N == N - 1)
        {
              
            // Increase minimum count
            minimum++;
        }
    }
    Console.WriteLine("Maximum = " + maximum);
    Console.WriteLine("Minimum = " + minimum);
}
  
// Driver code
static void Main()
{
    int N = 4;
    int C = 12;
    int K = 3;
  
    max_min(N, C, K);
}
}
  
// This code is contibuted by abhinavjain194

Javascript

  
// JavaScript program to implement
// the above approach
  
// Function to calculate the
// maximum and minimum number
// of candies a boy can possess
function max_min(N, C, K)
{
    let maximum, minimum;
   
    if (N == 1)
    {
           
        // All candies will be
        // given to one boy
        maximum = minimum = C;
    }
   
    else if (K >= C)
    {
           
        // All the candies will
        // be given to 1 boy
        maximum = C;
        minimum = 0;
    }
   
    else
    {
           
        // Give K candies to 1st
        // boy initially
        maximum = K;
        minimum = 0;
   
        // Count remaining candies
        let remain_candy = C - K;
   
        maximum += Math.floor(remain_candy / N);
        minimum = Math.floor(remain_candy / N);
   
        // If the last candy of remaining candies
        // is given to the last boy, i.e Nth boy
        if (remain_candy % N == N - 1)
        {
               
            // Increase minimum count
            minimum++;
        }
    }
    document.write("Maximum = " + maximum + "<br/>");
    document.write("Minimum = " + minimum);
}
  
  
// Driver code
  
    let N = 4;
    let C = 12;
    let K = 3;
   
    max_min(N, C, K);
             

输出：

Maximum = 5
Minimum = 2

时间复杂度： O(1)
辅助空间： O(1)