找到位置 i 来拆分 Array，使得直到 i-1 的前缀总和，直到 i+1 的后缀总和在 GP 中，共同比率为 K
给定一个数组arr[]和一个正整数K 。任务是找到arr[]中元素的位置 say i ，使得前缀总和直到i-1 ， i和后缀总和直到i+1在几何级数中，具有共同比率K 。
例子：
Input: arr[] = { 5, 1, 4, 20, 6, 15, 9, 10 }, K = 2
Output: 4
Explanation: The following operations are performed to get required GP.
Sum of element from position 1 to 3 is 5 + 1 + 4 = 10 and from 5 to 8 is 6 + 15 + 9 + 10 = 40.
And element at position 4 is 20.
Therefore10, 20, 40 is a Geometric Progression series with common ratio K.
Input: arr[] ={ -3, 5, 0, 2, 1, 25, 25, 100 }, K = 5
Output: 6
方法：给定的问题可以通过使用线性搜索和基本前缀和来解决。请按照以下步骤解决给定的问题。
- 如果数组的大小小于 3，则不可能有序列，因此只需返回 -1。
- 初始化一个变量，比如arrSum以存储arr[]的所有元素的总和。
- 计算数组arr[]的总和并将其存储在arrSum中。
- 如果arrSum % R != 0 ，则返回0 。其中R = K * K + 1 + K + 1 。
- 初始化一个变量mid = K * (Sum / R)以存储 GP 系列的中间元素，其公比为K。
- 取一个变量说temp来存储临时结果。
- 使用变量i将arr[]从索引1迭代到 ( arr[] 的大小) – 2 。
- 温度 = 温度 + arr[i-1]
- 如果arr[i] = 中
- 如果temp = mid/k ，则返回(i+1)作为答案。
- 否则返回0 。
- 如果循环终止并且arr[]中没有元素等于 mid 然后简单地返回0 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if there is
// an element forming G.P. series
// having common ratio k
int checkArray(int arr[], int N, int k)
{
// If size of array is less than
// three then return -1
if (N < 3)
return -1;
// Initialize the variables
int i, Sum = 0, temp = 0;
// Calculate total sum of array
for (i = 0; i < N; i++)
Sum += arr[i];
int R = (k * k + k + 1);
if (Sum % R != 0)
return 0;
// Calculate Middle element of G.P. series
int Mid = k * (Sum / R);
// Iterate over the range
for (i = 1; i < N - 1; i++) {
// Store the first element of G.P.
// series in the variable temp
temp += arr[i - 1];
if (arr[i] == Mid) {
// Return position of middle element
// of the G.P. series if the first
// element is in G.P. of common ratio k
if (temp == Mid / k)
return i + 1;
// Else return 0
else
return 0;
}
}
// if middle element is not found in arr[]
return 0;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 5, 1, 4, 20, 6, 15, 9, 10 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 2;
cout << checkArray(arr, N, K) << endl;
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to check if there is
// an element forming G.P. series
// having common ratio k
static int checkArray(int arr[], int N, int k)
{
// If size of array is less than
// three then return -1
if (N < 3)
return -1;
// Initialize the variables
int i, Sum = 0, temp = 0;
// Calculate total sum of array
for (i = 0; i < N; i++)
Sum += arr[i];
int R = (k * k + k + 1);
if (Sum % R != 0)
return 0;
// Calculate Middle element of G.P. series
int Mid = k * (Sum / R);
// Iterate over the range
for (i = 1; i < N - 1; i++) {
// Store the first element of G.P.
// series in the variable temp
temp += arr[i - 1];
if (arr[i] == Mid) {
// Return position of middle element
// of the G.P. series if the first
// element is in G.P. of common ratio k
if (temp == Mid / k)
return i + 1;
// Else return 0
else
return 0;
}
}
// if middle element is not found in arr[]
return 0;
}
// Driver Code
public static void main (String[] args) {
// Given array
int arr[] = { 5, 1, 4, 20, 6, 15, 9, 10 };
int N = arr.length;
int K = 2;
System.out.println(checkArray(arr, N, K));
}
}
// This code is contributed by Dharanendra L V.Python3
# python program for the above approach
# Function to check if there is
# an element forming G.P. series
# having common ratio k
def checkArray(arr, N, k):
# If size of array is less than
# three then return -1
if (N < 3):
return -1
# Initialize the variables
Sum = 0
temp = 0
# Calculate total sum of array
for i in range(0, N):
Sum += arr[i]
R = (k * k + k + 1)
if (Sum % R != 0):
return 0
# Calculate Middle element of G.P. series
Mid = k * (Sum // R)
# Iterate over the range
for i in range(1, N-1):
# Store the first element of G.P.
# series in the variable temp
temp += arr[i - 1]
if (arr[i] == Mid):
# Return position of middle element
# of the G.P. series if the first
# element is in G.P. of common ratio k
if (temp == Mid // k):
return i + 1
# Else return 0
else:
return 0
# if middle element is not found in arr[]
return 0
# Driver Code
if __name__ == "__main__":
# Given array
arr = [5, 1, 4, 20, 6, 15, 9, 10]
N = len(arr)
K = 2
print(checkArray(arr, N, K))
# This code is contributed by rakeshsahniC#
// C# program for the above approach
using System;
class GFG {
// Function to check if there is
// an element forming G.P. series
// having common ratio k
static int checkArray(int[] arr, int N, int k)
{
// If size of array is less than
// three then return -1
if (N < 3)
return -1;
// Initialize the variables
int i, Sum = 0, temp = 0;
// Calculate total sum of array
for (i = 0; i < N; i++)
Sum += arr[i];
int R = (k * k + k + 1);
if (Sum % R != 0)
return 0;
// Calculate Middle element of G.P. series
int Mid = k * (Sum / R);
// Iterate over the range
for (i = 1; i < N - 1; i++) {
// Store the first element of G.P.
// series in the variable temp
temp += arr[i - 1];
if (arr[i] == Mid) {
// Return position of middle element
// of the G.P. series if the first
// element is in G.P. of common ratio k
if (temp == Mid / k)
return i + 1;
// Else return 0
else
return 0;
}
}
// if middle element is not found in arr[]
return 0;
}
// Driver Code
public static void Main(string[] args)
{
// Given array
int[] arr = { 5, 1, 4, 20, 6, 15, 9, 10 };
int N = arr.Length;
int K = 2;
Console.WriteLine(checkArray(arr, N, K));
}
}
// This code is contributed by ukasp.Javascript
输出
4时间复杂度：O(N)
辅助空间：O(1)