由无限连接给定数组形成的 Array 的前 M 个元素的总和

给定一个由N个整数和一个正整数M组成的数组arr[] ，任务是找到由给定数组arr[]的无限级联形成的数组的前M个元素的总和。

例子：

Input: arr[] = {1, 2, 3}, M = 5
Output: 9
Explanation:
The array formed by the infinite concatenation of the given array arr[] is of the form {1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, … }.
The sum of the first M(= 5) elements of the array is 1 + 2 + 3 + 1 + 2 = 9.

Input: arr[] = {1}, M = 7
Output: 7

方法：给定的问题可以通过使用模运算符 (%)来解决 并将给定的数组视为循环数组，并相应地找到前M个元素的总和。请按照以下步骤解决此问题：

• 初始化一个变量，比如sum为0以存储新数组的前M个元素的总和。
• 使用变量i在[0, M – 1]范围内迭代，并将sum的值增加arr[i%N] 。
• 完成上述步骤后，打印sum的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the sum of first
// M numbers formed by the infinite
// concatenation of the array A[]
int sumOfFirstM(int A[], int N, int M)
{
    // Stores the resultant sum
    int sum = 0;
 
    // Iterate over the range [0, M - 1]
    for (int i = 0; i < M; i++) {
 
        // Add the value A[i%N] to sum
        sum = sum + A[i % N];
    }
 
    // Return the resultant sum
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3 };
    int M = 5;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sumOfFirstM(arr, N, M);
 
    return 0;
}

Java

// Java program for the above approach
 
import java.io.*;
import java.lang.*;
 
class GFG {
 
  // Function to find the sum of first
  // M numbers formed by the infinite
  // concatenation of the array A[]
  public static int sumOfFirstM(int A[], int N, int M)
  {
     
    // Stores the resultant sum
    int sum = 0;
 
    // Iterate over the range [0, M - 1]
    for (int i = 0; i < M; i++) {
 
      // Add the value A[i%N] to sum
      sum = sum + A[i % N];
    }
 
    // Return the resultant sum
    return sum;
  }
 
  // Driver Code
  public static void main(String[] args) {
    int arr[] = { 1, 2, 3 };
    int M = 5;
    int N = arr.length;
    System.out.println(sumOfFirstM(arr, N, M));
 
  }
}
 
// This code is contributed by gfgking.

Python3

# Python3 program for the above approach
 
# Function to find the sum of first
# M numbers formed by the infinite
# concatenation of the array A[]
def sumOfFirstM(A, N, M):
     
    # Stores the resultant sum
    sum = 0
 
    # Iterate over the range [0, M - 1]
    for i in range(M):
         
        # Add the value A[i%N] to sum
        sum = sum + A[i % N]
 
    # Return the resultant sum
    return sum
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 2, 3 ]
    M = 5
    N = len(arr)
     
    print(sumOfFirstM(arr, N, M))
     
# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;
class GFG {
 
    // Function to find the sum of first
    // M numbers formed by the infinite
    // concatenation of the array A[]
    static int sumOfFirstM(int[] A, int N, int M)
    {
 
        // Stores the resultant sum
        int sum = 0;
 
        // Iterate over the range [0, M - 1]
        for (int i = 0; i < M; i++) {
 
            // Add the value A[i%N] to sum
            sum = sum + A[i % N];
        }
 
        // Return the resultant sum
        return sum;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 2, 3 };
        int M = 5;
        int N = arr.Length;
        Console.WriteLine(sumOfFirstM(arr, N, M));
    }
}
 
// This code is contributed by subhammahato348.

Javascript

  
        // JavaScript program for the above approach
 
 
        // Function to find the sum of first
        // M numbers formed by the infinite
        // concatenation of the array A[]
        function sumOfFirstM(A, N, M) {
            // Stores the resultant sum
            let sum = 0;
            // Iterate over the range [0, M - 1]
            for (let i = 0; i < M; i++) {
 
                // Add the value A[i%N] to sum
                sum = sum + A[i % N];
            }
 
            // Return the resultant sum
            return sum;
        }
 
        // Driver Code
 
        let arr = [1, 2, 3];
        let M = 5;
        let N = arr.length;
        document.write(sumOfFirstM(arr, N, M));
 
 
 
  // This code is contributed by Potta Lokesh
   

输出：

9

时间复杂度： O(M)
辅助空间： O(1)