最大化阵列的模和

给定一个由N个正整数组成的数组arr [] ，任务是找到一个非负整数M的最大值∑(M mod arr [i]) ，其中arr [i]是任何数组元素。

例子：

Input: arr[] = {3, 4, 6} 
Output: 10 
Explanation: For M = 11, (11 mod 3) + (11 mod 4) + (11 mod 6) =10

Input: arr[]={7, 46, 11, 20, 11} 
Output: 90

方法：请按照以下步骤解决问题：

• 由于A MOD B是余当A除以B，那么表达式Σ的最大值(M MOD改编[I])为：

(M mod Arr[0]) + (M mod Arr[1]) +. . . + (M mod Arr[N-1]) = (Arr[0] − 1) + (Arr[1] − 1) + · · · + (Arr[N-1]− 1)

• 考虑到K = Arr [0]×Arr [1]×···×Arr [n – 1] ，则(K mod Arr [i])= 0对于范围[0，N – 1]中的每个i
• 因此，((K-1)mod Arr [i])= Arr [i]-1。因此，对于M = K-1 ，可以获得最佳结果。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to caluclate maximum modulo
// sum possible from a given array
int MaximumModuloSum(int Arr[], int N)
{
    // Stores the sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        sum += Arr[i] - 1;
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 4, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << MaximumModuloSum(arr, N);
 
    return 0;
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    public static int MaximumModuloSum(int Arr[], int N)
    {
       
        // Stores the sum
        int sum = 0;
 
        // Traverse the array
        for (int i = 0; i < N; i++) {
            sum += Arr[i] - 1;
        }
        return sum;
    }
    public static void main(String[] args)
    {
        int arr[] = { 3, 4, 6 };
        int N = 3;
        System.out.println(MaximumModuloSum(arr, N));
    }
}
 
// This code is contributed by aditya7409.

Python3

# Python 3 Program to implement
# the above approach
 
# Function to caluclate maximum modulo
# sum possible from a given array
def MaximumModuloSum( Arr, N):
 
    # Stores the sum
    sum = 0;
 
    # Traverse the array
    for i in range( N ):
        sum += Arr[i] - 1;
     
    return sum;
 
# Driver Code
if __name__ == "__main__":
   
    arr = [ 3, 4, 6 ];
    N = len(arr)
    print(MaximumModuloSum(arr, N))
 
    # This code is contributed by chitranayal.

C#

// C# program for the above approach
using System;
class GFG
{
 
  public static int MaximumModuloSum(int[] Arr, int N)
  {
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
      sum += Arr[i] - 1;
    }
    return sum;
  }
 
  // Driver code
  static void Main()
  {
    int[] arr = { 3, 4, 6 };
    int N = 3;
    Console.WriteLine(MaximumModuloSum(arr, N));
  }
}
 
// This code is contributed by susmitakundugoaldanga.

输出：

10

时间复杂度： O(N)
辅助空间： O(1)