从给定数组中找到原始数组，其中第 I 个元素是前 I 个元素的平均值
原文:https://www . geesforgeks . org/find-原始数组-从给定数组-其中-with-element-是第一个 i-elements 的平均值/
给定一个长度为 N、的数组 arr[] ，任务是找到原始数组，使得给定数组中的每个和元素( arr[i] )都是原始数组中第一个 i 元素的平均值。
示例:
输入:arr = { 4 3 3 3 } T3】输出:4 2 3 3 T6】解释: (4) / 1 = 1，(4+2) / 2 = 3，(4+2+3) / 3 = 3，(4+2+3+3) / 4 = 3
输入:arr = { 2 6 8 10 } T3】输出:2 10 12 16 T6】解释: (2) / 1 = 2，(2+10) / 2 = 6，(2+10+12) / 3 = 8，(2+10+12+16) / 4 = 10
方法:给定的问题可以使用数学方法来解决。请遵循以下步骤:
- 初始化一个变量和到数组的第一个元素 arr
- 从第二个索引到最后迭代数组 arr ，每次迭代:
- 将当前元素arr【I】乘以当前索引+ 1 (i + 1) ，并从中减去 sum 的值
- 将得到的当前元素添加到变量和中
- 修改后返回结果数组，因为它将是原始数组
C++
// C++ implementation for the above approach
#include <iostream>
using namespace std;
// Function to find the original
// array from the modified array
void findOriginal(int arr[], int N)
{
// Initialize the variable sum
// with the first element of array
int sum = arr[0];
for (int i = 1; i < N; i++) {
// Calculate original element
// from average of first i elements
arr[i] = (i + 1) * arr[i] - sum;
// Add current element to sum
sum += arr[i];
}
// Print the array
for (int i = 0; i < N; i++) {
cout << arr[i] << " ";
}
}
// Driver function
int main()
{
int arr[] = { 2, 6, 8, 10 };
int N = sizeof(arr) / sizeof(arr[0]);
// Call the function
findOriginal(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation for the above approach
class GFG
{
// Function to find the original
// array from the modified array
static void findOriginal(int arr[], int N)
{
// Initialize the variable sum
// with the first element of array
int sum = arr[0];
for (int i = 1; i < N; i++) {
// Calculate original element
// from average of first i elements
arr[i] = (i + 1) * arr[i] - sum;
// Add current element to sum
sum += arr[i];
}
// Print the array
for (int i = 0; i < N; i++) {
System.out.print(arr[i] + " ");
}
}
// Driver function
public static void main(String [] args)
{
int [] arr = new int [] { 2, 6, 8, 10 };
int N = arr.length;
// Call the function
findOriginal(arr, N);
}
}
// This code is contributed by ihritik
Python 3
# Python implementation for the above approach
# Function to find the original
# array from the modified array
def findOriginal(arr, N):
# Initialize the variable sum
# with the first element of array
sum = arr[0]
for i in range(1, N):
# Calculate original element
# from average of first i elements
arr[i] = (i + 1) * arr[i] - sum
# Add current element to sum
sum = sum + arr[i]
# Print the array
for i in range (0, N):
print(arr[i], end=" ")
# Driver function
arr= [ 2, 6, 8, 10 ]
N = len(arr)
# Call the function
findOriginal(arr, N)
# This code is contributed by ihritik
C
// C# program for above approach
using System;
class GFG {
// Function to find the original
// array from the modified array
static void findOriginal(int[] arr, int N)
{
// Initialize the variable sum
// with the first element of array
int sum = arr[0];
for (int i = 1; i < N; i++) {
// Calculate original element
// from average of first i elements
arr[i] = (i + 1) * arr[i] - sum;
// Add current element to sum
sum += arr[i];
}
// Print the array
for (int i = 0; i < N; i++)
Console.Write(arr[i] + " ");
}
// Driver Code
public static void Main(String[] args)
{
int N = 4;
int[] arr = { 2, 6, 8, 10 };
findOriginal(arr, N);
}
}
// This code is contributed by dwivediyash
java 描述语言
<script>
// JavaScript implementation for the above approach
// Function to find the original
// array from the modified array
function findOriginal(arr, N)
{
// Initialize the variable sum
// with the first element of array
var sum = arr[0];
for (var i = 1; i < N; i++) {
// Calculate original element
// from average of first i elements
arr[i] = (i + 1) * arr[i] - sum;
// Add current element to sum
sum += arr[i];
}
// Print the array
for (var i = 0; i < N; i++) {
document.write(arr[i] + " ");
}
}
// Driver code
var arr = [ 2, 6, 8, 10 ];
var N = arr.length;
// Call the function
findOriginal(arr, N);
// This code is contributed by AnkThon
</script>
Output
2 10 12 16
时间复杂度:O(N) T3】辅助空间: O(1)
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