最大化数组中索引 K 处的元素，其总和为 M，相邻元素之间的差值至多为 1

给定一个正整数N ，任务是构造一个长度为N的数组，并找到索引K 处的最大值，使得所有数组元素的总和至多为M并且任意两个连续数组元素之间的绝对差为大多数1 .

例子：

Input: N = 3, M = 7, K = 1
Output: 3
Explanation: 
According to the given constraints, the array with values {2, 3, 2}maximizes the value at index 1. Therefore, the required output is 3.

Input: N = 3, M = 8, K = 1
Output: 3

方法：想法是在索引K处实现最大值，并减少其他元素的总和以满足数组总和的标准，最多为M 。请按照以下步骤操作：

• 设索引K处的值为X 。所以K – 1处的元素是X – 1 ， K – 2处的元素是X – 2等等。
• 在索引0 处，值为X – K。类似地，在索引K + 1 处，值是X – 1 ，依此类推，直到索引N – 1处的X – (N – K – 1) 。
• 因此，要在索引K处获得最大值，数组结构将是X – K, X – (K – 1), …., X – 2, X – 1, X, X – 1, X – 2, …。 ., X – (N – K – 1) .
• 所以等式排列后，就变成了K * X – (1 + 2 + …. + K) + X + (N – K – 1) * X – (1 + 2 + …. + (N – K – 1) ) ≤ M 。

按照以下步骤求解上述方程：

• 计算(1 + 2 + …. + K)使用K * (K + 1) / 2和(1 + 2 + ….. + (N – K – 1))使用(N – K – 1) * (N – K) / 2并分别存储在S1和S2 中。
• 这将方程简化为X * (K + 1 + N – K – 1) ≤ M + S1 + S2 。
• 现在，可以通过计算(M + S1 + S2) / N来获得X的最大值。

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
// possible value at index K
void maxValueAtIndexK(int N, int K, int M)
{
    // Stores the sum of elements
    // in the left and right of index K
    int S1 = 0, S2 = 0;
 
    S1 = K * (K + 1) / 2;
    S2 = (N - K - 1) * (N - K) / 2;
 
    // Stores the maximum
    // possible value at index K
    int X = (M + S1 + S2) / N;
 
    // Print the answer
    cout << X;
}
 
// Driver Code
int main()
{
    // Given N, K & M
    int N = 3, K = 1, M = 7;
    maxValueAtIndexK(N, K, M);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to calculate the maximum
// possible value at index K
static void maxValueAtIndexK(int N, int K,
                             int M)
{
     
    // Stores the sum of elements
    // in the left and right of index K
    int S1 = 0, S2 = 0;
 
    S1 = K * (K + 1) / 2;
    S2 = (N - K - 1) * (N - K) / 2;
 
    // Stores the maximum
    // possible value at index K
    int X = (M + S1 + S2) / N;
 
    // Print the answer
    System.out.println(X);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given N, K & M
    int N = 3, K = 1, M = 7;
     
    maxValueAtIndexK(N, K, M);
}
}
 
// This code is contributed by Dharanendra L V

Python3

# Python program for the above approach
 
# Function to calculate the maximum
# possible value at index K
def maxValueAtIndexK(N, K, M):
 
    # Stores the sum of elements
    # in the left and right of index K
    S1 = 0; S2 = 0;
    S1 = K * (K + 1) // 2;
    S2 = (N - K - 1) * (N - K) // 2;
 
    # Stores the maximum
    # possible value at index K
    X = (M + S1 + S2) // N;
 
    # Prthe answer
    print(X);
 
# Driver Code
if __name__ == '__main__':
 
    # Given N, K & M
    N = 3; K = 1; M = 7;
    maxValueAtIndexK(N, K, M);
 
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to calculate the maximum
// possible value at index K
static void maxValueAtIndexK(int N, int K,
                             int M)
{
     
    // Stores the sum of elements
    // in the left and right of index K
    int S1 = 0, S2 = 0;
 
    S1 = K * (K + 1) / 2;
    S2 = (N - K - 1) * (N - K) / 2;
 
    // Stores the maximum
    // possible value at index K
    int X = (M + S1 + S2) / N;
 
    // Print the answer
    Console.WriteLine(X);
}
 
// Driver Code
static public void Main()
{
     
    // Given N, K & M
    int N = 3, K = 1, M = 7;
     
    maxValueAtIndexK(N, K, M);
}
}
 
// This code is contributed by Dharanendra L V

Javascript

// JavaScript program for
// the above approach
  
// Function to calculate the maximum
// possible value at index K
function maxValueAtIndexK(N, K, M)
{
    // Stores the sum of elements
    // in the left and right of index K
    let S1 = 0, S2 = 0;
 
    S1 = K * (K + 1) / 2;
    S2 = (N - K - 1) * (N - K) / 2;
 
    // Stores the maximum
    // possible value at index K
    let X = (M + S1 + S2) / N;
 
    // Print the answer
     document.write(X);
}
 
// Driver Code
  
    // Given N, K & M
    let N = 3, K = 1, M = 7;
    maxValueAtIndexK(N, K, M);
  

输出：

3

时间复杂度： O(1)
辅助空间： O(1)

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