最大化每个Array元素的模量总和

给定一个由N个正整数组成的数组A [] ，任务是找到以下各项的最大可能值：

F(M) = M % A[0] + M % A[1] + …. + M % A[N -1] where M can be any integer value

例子：

Input: arr[] = {3, 4, 6} 
Output: 10 
Explanation: 
The maximum sum occurs for M = 11. 
(11 % 3) + (11 % 4) + (11 % 6) = 2 + 3 + 5 = 10
Input: arr[] = {2, 5, 3} 
Output:7 
Explanation: 
The maximum sum occurs for M = 29. 
(29 % 2) + (29 % 5) + (29 % 3) = 1 + 4 + 2 = 7. 
 

方法：
请按照以下步骤解决问题：

1. 计算所有阵列元素的LCM。
2. 如果M等于阵列的LCM，则F(M)= 0即F(M)的最小可能值。这是因为，对于每个^第i^个索引， M％a [i]始终为0。
3. 对于M =数组元素的LCM – 1， F(M)被最大化。这是因为，对于每个^第i^个索引， M％a [i]等于a [i] – 1 ，这是可能的最大值。
4. 因此， F(M)的最大可能值可以是数组元素的总和– N。

下面是上述方法的实现：

C++

// C++ program to find the
// maximum sum of modulus
// with every array element
#include 
using namespace std;
 
// Function to return the
// maximum sum of modulus
// with every array element
int maxModulosum(int a[], int n)
{
    int sum = 0;
 
    // Sum of array elements
    for (int i = 0; i < n; i++) {
        sum += a[i];
    }
 
    // Return the answer
    return sum - n;
}
 
// Driver Program
int main()
{
    int a[] = { 3, 4, 6 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << maxModulosum(a, n);
 
    return 0;
}

Java

// Java program to find the maximum
// sum of modulus with every array
// element
import java.io.*;
 
class GFG{
 
// Function to return the maximum
// sum of modulus with every array
// element
static int maxModulosum(int a[], int n)
{
    int sum = 0;
     
    // Sum of array elements
    for(int i = 0; i < n; i++)
    {
       sum += a[i];
    }
     
    // Return the answer
    return sum - n;
}
     
// Driver Code
public static void main (String[] args)
{
    int a[] = new int[]{ 3, 4, 6 };
    int n = a.length;
     
    System.out.println(maxModulosum(a, n));
}
}
 
// This code is contributed by Shubham Prakash

Python3

# Python3 program to find the
# maximum sum of modulus
# with every array element
 
# Function to return the
# maximum sum of modulus
# with every array element
def maxModulosum(a, n):
 
    sum1 = 0;
 
    # Sum of array elements
    for i in range(0, n):
        sum1 += a[i];
     
    # Return the answer
    return sum1 - n;
 
# Driver Code
a = [ 3, 4, 6 ];
n = len(a);
print(maxModulosum(a, n));
 
# This code is contributed by Code_Mech

C#

// C# program to find the maximum
// sum of modulus with every array
// element
using System;
class GFG{
 
// Function to return the maximum
// sum of modulus with every array
// element
static int maxModulosum(int []a, int n)
{
    int sum = 0;
     
    // Sum of array elements
    for(int i = 0; i < n; i++)
    {
        sum += a[i];
    }
     
    // Return the answer
    return sum - n;
}
     
// Driver Code
public static void Main(String[] args)
{
    int []a = new int[]{ 3, 4, 6 };
    int n = a.Length;
     
    Console.Write(maxModulosum(a, n));
}
}
 
// This code is contributed
// by shivanisinghss2110

Javascript

 
    // Javascript program to find the
    // maximum sum of modulus
    // with every array element
     
    // Function to return the
    // maximum sum of modulus
    // with every array element
    function maxModulosum(a, n)
    {
        let sum = 0;
 
        // Sum of array elements
        for (let i = 0; i < n; i++) {
            sum += a[i];
        }
 
        // Return the answer
        return sum - n;
    }
      
    let a = [ 3, 4, 6 ];
    let n = a.length;
    document.write(maxModulosum(a, n));
 

输出：

10

时间复杂度： O(N)
辅助空间： O(1)