用接下来的K个元素的总和替换圆形数组中的每个元素

给定一个由N个整数和一个整数K组成的圆形数组arr [] ，任务是在执行以下操作后打印该数组：

• 如果K为非负数，则用接下来的K个元素的总和替换A [i] 。
• 如果K为负，则将其替换为先前K个元素的总和。

A cyclic array means the next element of the last element of the array is the first element and the previous element of the first element is the last element.

例子：

Input: arr[] = {4, 2, -5, 11}, K = 3
Output: 8 10 17 1
Explanation: 
Step 1: For index 0, replace arr[0] = arr[1] + arr[2] + arr[3] = 2 + (-5) + 11 = 8 
Step 2: For index 1, replace arr[1] = arr[2] + arr[3] + arr[0] = (-5) + 11 + 4 = 10 
Step 3: For index 2, replace arr[2] = arr[3] + arr[0] + arr[1] = 11 + 4 + 2 = 17 
Step 4: For index 3, replace arr[3] = arr[0] + arr[1] + arr[2] = 4 + 2 + -5 = 1 
Therefore, the resultant array is {8, 10, 17, 1}

Input: arr[] = {2, 5, 7, 9}, K = -2
Output: 16 11 7 12
Explanation: 
Step 1: For index 0, replace arr[0] = arr[3] + arr[2] = 9 + 7 = 16 
Step 2: For index 1, replace arr[1] = arr[0] + arr[3] = 2 + 9 = 11 
Step 3: For index 2, replace arr[2] = arr[1] + arr[0] = 5 + 2 = 7 
Step 4: For index 3, replace arr[3] = arr[2] + arr[1] = 7 + 5 = 12 
Therefore, the resultant array is {16, 11, 7, 12}

天真的方法：解决问题的最简单方法是遍历数组，并根据每个数组元素的K值遍历下一个或上一个K元素，并打印它们的和。请按照以下步骤解决问题：

• 如果K的值为负，则反转数组并找到接下来的K个元素的总和。
• 如果K的值为非负数，则只需为每个元素找到接下来的K个元素的总和。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
#define pb push_back
 
// Function to find the sum of next
// or previous k array elements
vector sumOfKelementsUtil(
    vector& a, int x)
{
    // Size of the array
    int n = a.size();
 
    int count, k, temp;
 
    // Stores the result
    vector ans;
 
    // Iterate the given array
    for (int i = 0; i < n; i++) {
 
        count = 0;
        k = i + 1;
        temp = 0;
 
        // Traverse the k elements
        while (count < x) {
 
            temp += a[k % n];
            count++;
            k++;
        }
 
        // Push the K elements sum
        ans.pb(temp);
    }
 
    // Return the resultant vector
    return ans;
}
 
// Function that prints the sum of next
// K element for each index in the array
void sumOfKelements(vector& arr,
                    int K)
{
    // Size of the array
    int N = (int)arr.size();
 
    // Stores the result
    vector ans;
 
    // If key is negative,
    // reverse the array
    if (K < 0) {
        K = K * (-1);
 
        // Reverse the array
        reverse(arr.begin(),
                arr.end());
 
        // Find the resultant vector
        ans = sumOfKelementsUtil(arr, K);
 
        // Reverse the array again to
        // get the original sequence
        reverse(ans.begin(), ans.end());
    }
 
    // If K is positive
    else {
        ans = sumOfKelementsUtil(arr, K);
    }
 
    // Print the answer
    for (int i = 0; i < N; i++) {
        cout << ans[i] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    vector arr = { 4, 2, -5, 11 };
    int K = 3;
 
    // Function Calll
    sumOfKelements(arr, K);
 
    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{
     
//reverse array
static Vector reverse(Vector a)
{
  int i, n = a.size(), t;
  for (i = 0; i < n / 2; i++)
  {
    t = a.elementAt(i);
    a.set(i, a.elementAt(n - i - 1));
    a.set(n - i - 1, t);
  }
  return a;
}
 
// Function to find the sum of next
// or previous k array elements
static Vector sumOfKelementsUtil(
       Vectora, int x)
{
  // Size of the array
  int n = a.size();
 
  int count, k, temp;
 
  // Stores the result
  Vector ans = new Vector<>();
 
  // Iterate the given array
  for (int i = 0; i < n; i++)
  {
    count = 0;
    k = i + 1;
    temp = 0;
 
    // Traverse the k elements
    while (count < x)
    {
      temp += a.elementAt(k % n);
      count++;
      k++;
    }
 
    // Push the K elements sum
    ans.add(temp);
  }
 
  // Return the resultant vector
  return ans;
}
 
// Function that prints the sum of next
// K element for each index in the array
static void sumOfKelements(Vector arr,
                           int K)
{
  // Size of the array
  int N = (int)arr.size();
 
  // Stores the result
  Vector ans = new Vector<>();
 
  // If key is negative,
  // reverse the array
  if (K < 0)
  {
    K = K * (-1);
 
    // Reverse the array
    arr =  reverse(arr);
 
    // Find the resultant vector
    ans = sumOfKelementsUtil(arr, K);
 
    // Reverse the array again to
    // get the original sequence
    ans = reverse(ans);
  }
 
  // If K is positive
  else
  {
    ans = sumOfKelementsUtil(arr, K);
  }
 
  // Print the answer
  for (int i = 0; i < N; i++)
  {
    System.out.print(ans.elementAt(i) + " ");
  }
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    Vector arr = new Vector<>();
    arr.add(4);
    arr.add(2);
    arr.add(-5);
    arr.add(11);
    int K = 3;
 
    // Function Calll
    sumOfKelements(arr, K);
 
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program for
# the above approach
 
# Function to find the sum of next
# or previous k array elements
def sumOfKelementsUtil(a, x):
 
    # Size of the array
    n = len(a)
     
    # Stores the result
    ans = []
 
    # Iterate the given array
    for i in range(n):
        count = 0
        k = i + 1
        temp = 0
 
        # Traverse the k elements
        while (count < x):
            temp += a[k % n]
            count += 1
            k += 1
 
        # Push the K elements sum
        ans.append(temp)
    
    # Return the resultant vector
    return ans
 
# Function that prints the
# sum of next K element for
# each index in the array
def sumOfKelements(arr, K):
 
    # Size of the array
    N = len(arr)
 
    #Stores the result
    ans = []
 
    # If key is negative,
    # reverse the array
    if (K < 0):
        K = K * (-1)
 
        # Reverse the array
        arr =  arr.reverse()
         
        # Find the resultant vector
        ans = sumOfKelementsUtil(arr, K)
 
        #Reverse the array again to
        # get the original sequence
        ans = ans.reverse()
 
    # If K is positive
    else:
        ans = sumOfKelementsUtil(arr, K)
   
    # Print the answer
    for i in range(N):
        print (ans[i], end = " ")
    
# Driver Code
if __name__ == "__main__":
   
    # Given array arr[]
    arr = [4, 2, -5, 11]
    K = 3
 
    # Function Calll
    sumOfKelements(arr, K)
 
# This code is contributed by Chitranayal

C#

// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
     
// Reverse array
static List reverse(List a)
{
  int i, n = a.Count, t;
  for (i = 0; i < n / 2; i++)
  {
    t = a[i];
    a[i] = a[n - i - 1];
    a[n - i - 1] = t;
  }
  return a;
}
 
// Function to find the sum of next
// or previous k array elements
static List sumOfKelementsUtil(
       Lista, int x)
{
  // Size of the array
  int n = a.Count;
 
  int count, k, temp;
 
  // Stores the result
  List ans = new List();
 
  // Iterate the given array
  for (int i = 0; i < n; i++)
  {
    count = 0;
    k = i + 1;
    temp = 0;
 
    // Traverse the k elements
    while (count < x)
    {
      temp += a[k % n];
      count++;
      k++;
    }
 
    // Push the K elements sum
    ans.Add(temp);
  }
 
  // Return the resultant vector
  return ans;
}
 
// Function that prints the sum of next
// K element for each index in the array
static void sumOfKelements(List arr,
                           int K)
{
  // Size of the array
  int N = (int)arr.Count;
 
  // Stores the result
  List ans = new List();
 
  // If key is negative,
  // reverse the array
  if (K < 0)
  {
    K = K * (-1);
 
    // Reverse the array
    arr =  reverse(arr);
 
    // Find the resultant vector
    ans = sumOfKelementsUtil(arr, K);
 
    // Reverse the array again to
    // get the original sequence
    ans = reverse(ans);
  }
 
  // If K is positive
  else
  {
    ans = sumOfKelementsUtil(arr, K);
  }
 
  // Print the answer
  for (int i = 0; i < N; i++)
  {
    Console.Write(ans[i] + " ");
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array []arr
  List arr = new List();
  arr.Add(4);
  arr.Add(2);
  arr.Add(-5);
  arr.Add(11);
  int K = 3;
 
  // Function Calll
  sumOfKelements(arr, K);
}
}
 
// This code is contributed by Rajput-Ji

输出

8 10 17 1 

时间复杂度： O(N * K)，其中N是给定数组的长度，K是给定整数。
辅助空间： O(N)

高效的方法：为了优化上述方法，其思想是使用前缀和。请按照以下步骤解决问题：

1. 如果K为负，则反转给定数组并将K乘以-1 。
2. 计算给定数组的前缀和并将其存储在pre []中。
3. 创建数组ans []来存储每个元素的答案。
4. 对于每个索引i ，如果i + K小于N ，则更新ans [i] = pre [i + k] – pre [i]
5. 否则，将所有从索引i到N – 1的元素的总和相加，找到剩余的K –(N – 1 – i)个元素，因为在上述步骤中已经添加了(N – 1 – i)个元素。
6. 将所有元素的总和相加((K – N – 1 – i)/ N)次，并将给定数组的剩余元素之和从0到((K –(N – 1 – i))％N)– 1 。
7. 在计算完数组中每个元素的答案之后，请检查数组是否先前已反转。如果是，请反转ans []数组。
8. 完成上述步骤后，打印存储在数组ans []中的所有元素。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include
using namespace std;
 
// Function to print the
// required resultant array
void sumOfKElements(int arr[], int n,
                    int k)
{
     
    // Reverse the array
    bool rev = false;
 
    if (k < 0)
    {
        rev = true;
        k *= -1;
        int l = 0, r = n - 1;
 
        // Traverse the range
        while (l < r)
        {
            int tmp = arr[l];
            arr[l] = arr[r];
            arr[r] = tmp;
            l++;
            r--;
        }
    }
 
    // Store prefix sum
    int dp[n] = {0};
    dp[0] = arr[0];
 
    // Find the prefix sum
    for(int i = 1; i < n; i++)
    {
        dp[i] += dp[i - 1] + arr[i];
    }
 
    // Store the answer
    int ans[n] = {0};
 
    // Calculate the answers
    for(int i = 0; i < n; i++)
    {
        if (i + k < n)
            ans[i] = dp[i + k] - dp[i];
        else
        {
             
            // Count of remaining elements
            int x = k - (n - 1 - i);
 
            // Add the sum of all elements
            // y times
            int y = x / n;
 
            // Add the remaining elements
            int rem = x % n;
 
            // Update ans[i]
            ans[i] = dp[n - 1] - dp[i] +
                 y * dp[n - 1] + (rem - 1 >= 0 ?
                   dp[rem - 1] : 0);
        }
    }
 
    // If array is reversed print
    // ans[] in reverse
    if (rev)
    {
        for(int i = n - 1; i >= 0; i--)
        {
            cout << ans[i] << " ";
        }
    }
    else
    {
        for(int i = 0; i < n; i++)
        {
            cout << ans[i] << " ";
        }
    }
}
 
// Driver Code
int main()
{
     
    // Given array arr[]
    int arr[] = { 4, 2, -5, 11 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given K
    int K = 3;
 
    // Function
    sumOfKElements(arr, N, K);
}
 
// This code is contributed by SURENDRA_GANGWAR

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to print the
    // required resultant array
    static void sumOfKElements(
        int arr[], int n, int k)
    {
 
        // Reverse the array
        boolean rev = false;
 
        if (k < 0) {
 
            rev = true;
            k *= -1;
            int l = 0, r = n - 1;
 
            // Traverse the range
            while (l < r) {
 
                int tmp = arr[l];
                arr[l] = arr[r];
                arr[r] = tmp;
                l++;
                r--;
            }
        }
 
        // Store prefix sum
        int dp[] = new int[n];
        dp[0] = arr[0];
 
        // Find the prefix sum
        for (int i = 1; i < n; i++) {
 
            dp[i] += dp[i - 1] + arr[i];
        }
 
        // Store the answer
        int ans[] = new int[n];
 
        // Calculate the answers
        for (int i = 0; i < n; i++) {
 
            if (i + k < n)
                ans[i] = dp[i + k] - dp[i];
            else {
 
                // Count of remaining elements
                int x = k - (n - 1 - i);
 
                // Add the sum of all elements
                // y times
                int y = x / n;
 
                // Add the remaining elements
                int rem = x % n;
 
                // Update ans[i]
                ans[i] = dp[n - 1]
                         - dp[i]
                         + y * dp[n - 1]
                         + (rem - 1 >= 0 ? dp[rem - 1] : 0);
            }
        }
 
        // If array is reversed print
        // ans[] in reverse
        if (rev) {
            for (int i = n - 1; i >= 0; i--) {
                System.out.print(ans[i] + " ");
            }
        }
        else {
            for (int i = 0; i < n; i++) {
                System.out.print(ans[i] + " ");
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array arr[]
        int arr[] = { 4, 2, -5, 11 };
 
        int N = arr.length;
 
        // Given K
        int K = 3;
 
        // Function
        sumOfKElements(arr, N, K);
    }
}

Python3

# Python3 program for
# the above approach
 
# Function to prthe
# required resultant array
def sumOfKElements(arr, n, k):
   
    # Reverse the array
    rev = False;
 
    if (k < 0):
 
        rev = True;
        k *= -1;
        l = 0;
        r = n - 1;
 
        # Traverse the range
        while (l < r):
            tmp = arr[l];
            arr[l] = arr[r];
            arr[r] = tmp;
            l += 1;
            r -= 1;
 
    # Store prefix sum
    dp = [0] * n;
    dp[0] = arr[0];
 
    # Find the prefix sum
    for i in range(1, n):
        dp[i] += dp[i - 1] + arr[i];
 
    # Store the answer
    ans = [0] * n;
 
    # Calculate the answers
    for i in range(n):
        if (i + k < n):
            ans[i] = dp[i + k] - dp[i];
        else:
 
            # Count of remaining
            # elements
            x = k - (n - 1 - i);
 
            # Add the sum of
            # all elements y times
            y = x // n;
 
            # Add the remaining
            # elements
            rem = x % n;
 
            # Update ans[i]
            ans[i] = (dp[n - 1] - dp[i] +
                      y * dp[n - 1] +
                      (dp[rem - 1]
                      if rem - 1 >= 0
                      else 0));
 
    # If array is reversed
    # print ans in reverse
    if (rev):
        for i in range(n - 1, -1, -1):
            print(ans[i], end = " ");
 
    else:
        for i in range(n):
            print(ans[i], end = " ");
 
# Driver Code
if __name__ == '__main__':
   
    # Given array arr
    arr = [4, 2, -5, 11];
 
    N = len(arr);
 
    # Given K
    K = 3;
 
    # Function
    sumOfKElements(arr, N, K);
 
# This code is contributed by 29AjayKumar

C＃

// C# program for
// the above approach
using System;
class GFG {
 
// Function to print the
// required resultant array
static void sumOfKElements(int []arr,
                           int n, int k)
{
  // Reverse the array
  bool rev = false;
 
  if (k < 0)
  {
    rev = true;
    k *= -1;
    int l = 0, r = n - 1;
 
    // Traverse the range
    while (l < r)
    {
      int tmp = arr[l];
      arr[l] = arr[r];
      arr[r] = tmp;
      l++;
      r--;
    }
  }
 
  // Store prefix sum
  int []dp = new int[n];
  dp[0] = arr[0];
 
  // Find the prefix sum
  for (int i = 1; i < n; i++)
  {
    dp[i] += dp[i - 1] + arr[i];
  }
 
  // Store the answer
  int []ans = new int[n];
 
  // Calculate the answers
  for (int i = 0; i < n; i++)
  {
    if (i + k < n)
      ans[i] = dp[i + k] - dp[i];
    else
    {
      // Count of remaining elements
      int x = k - (n - 1 - i);
 
      // Add the sum of all elements
      // y times
      int y = x / n;
 
      // Add the remaining elements
      int rem = x % n;
 
      // Update ans[i]
      ans[i] = dp[n - 1] - dp[i] +
               y * dp[n - 1] +
               (rem - 1 >= 0 ?
                dp[rem - 1] : 0);
    }
  }
 
  // If array is reversed print
  // ans[] in reverse
  if (rev)
  {
    for (int i = n - 1; i >= 0; i--)
    {
      Console.Write(ans[i] + " ");
    }
  }
  else
  {
    for (int i = 0; i < n; i++)
    {
      Console.Write(ans[i] + " ");
    }
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array []arr
  int []arr = {4, 2, -5, 11};
 
  int N = arr.Length;
 
  // Given K
  int K = 3;
 
  // Function
  sumOfKElements(arr, N, K);
}
}
 
// This code is contributed by Princi Singh

输出

8 10 17 1 

时间复杂度： O(N)
辅助空间： O(N)