序号为K且连续元素之间的绝对差的最小和为最小

给定两个整数N和K ，任务是找到一个长度为N的整数序列，以使序列中所有元素的总和为K，并且所有连续元素之间的绝对差之和最小。打印此最小化的总和。

例子：

Input: N = 3, K = 56
Output: 1
The sequence is {19, 19, 18} and the sum of absolute
differences of all the consecutive elements is
|19 – 19| + |19 – 18| = 0 + 1 = 1 which is the minimum possible.

Input: N = 12, K = 48
Output: 0
The sequence is {4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}.

方法：可能有两种情况：

1. 当K％N = 0时，答案将为0，因为K可以平均分为N个部分，即序列中的每个元素都将相等。
2. 当K％N！= 0时，答案将为1，因为可以按以下方式排列序列：
   • (K –(K％N))可被N整除，因此可以在所有N个部分中平均分配。
   • 剩下的(K％N)值可以以最小化连续绝对差的方式进行划分，即在第一个或最后一个(K％N)元素上加1 ，并且序列的类型为{x，x， x，x，y，y，y，y}得出的最小和为1 。

下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimized sum
int minimum_sum(int n, int k)
{
  
    // If k is divisible by n
    // then the answer will be 0
    if (k % n == 0)
        return 0;
  
    // Else the answer will be 1
    return 1;
}
  
// Driver code
int main()
{
    int n = 3, k = 56;
  
    cout << minimum_sum(n, k);
  
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
// Function to return the minimized sum
static int minimum_sum(int n, int k)
{
  
    // If k is divisible by n
    // then the answer will be 0
    if (k % n == 0)
        return 0;
  
    // Else the answer will be 1
    return 1;
}
  
// Driver code
public static void main (String[] args) 
{
  
    int n = 3, k = 56;
    System.out.println (minimum_sum(n, k));
}
}
  
// This code is contributed By @ajit_23

Python3

# Python3 implementation of the approach
  
# Function to return the minimized sum
def minimum_sum(n, k):
  
    # If k is divisible by n
    # then the answer will be 0
    if (k % n == 0):
        return 0;
  
    # Else the answer will be 1
    return 1
  
# Driver code
  
n = 3
k = 56
  
print(minimum_sum(n, k))
  
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;
  
class GFG
{
          
// Function to return the minimized sum
static int minimum_sum(int n, int k)
{
  
    // If k is divisible by n
    // then the answer will be 0
    if (k % n == 0)
        return 0;
  
    // Else the answer will be 1
    return 1;
}
  
// Driver code
static public void Main ()
{
    int n = 3, k = 56;
    Console.Write(minimum_sum(n, k));
}
}
  
// This code is contributed By Tushil

输出：

1