总和为完美平方的子数组

给定一个大小为N的数组arr [] ，任务是打印所有和为完美平方的子数组的开始和结束索引。

例子：

Input: arr[] = {65, 79, 81}
Output: (0, 1) (0, 2) (2, 2)
Explanation: 
Subarray sum whose start and end index is (0, 1) = 65 + 79 = 144 = 12² 
Subarray sum whose start and end index is (0, 2} = 65 + 79 + 81 = 225 = 15² 
Subarray sum whose start and end index is {2, 2} = 81 = 9²

Input: arr[] = {1, 2, 3, 4, 5}
Output: (0, 0) (1, 3) (3, 3) (3, 4) 

方法：可以使用“前缀和数组”技术解决该问题。这个想法是使用前缀求和数组来找到所有子数组的和。对于每个子数组，请检查子数组的总和是否为理想平方。如果发现对任何子数组都是正确的，则打印该子数组的开始和结束索引。请按照以下步骤解决问题。

1. 初始化一个变量，例如currSubSum，以存储当前的子数组总和。
2. 遍历数组以生成给定数组的所有可能的子数组。
3. 计算每个子数组的总和，对于每个子数组的总和，检查它是否为理想平方。
4. 如果发现对任何子数组都是正确的，则打印子数组的开始和结束索引。

下面是上述方法的实现：

C++14

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to print the start and end
// indices of all subarrays whose sum
// is a perfect square
void PrintIndexes(int arr[], int N)
{
 
    for (int i = 0; i < N; i++) {
 
        // Stores the current
        // subarray sum
        int currSubSum = 0;
 
        for (int j = i; j < N; j++) {
 
            // Update current subarray sum
            currSubSum += arr[j];
 
            // Stores the square root
            // of currSubSum
            int sq = sqrt(currSubSum);
 
            // Check if currSubSum is
            // a perfect square or not
            if (sq * sq == currSubSum) {
                cout << "(" << i << ", "
                     << j << ") ";
            }
        }
    }
}
 
// Driver Code
int main()
{
 
    int arr[] = { 65, 79, 81 };
    int N = sizeof(arr) / sizeof(arr[0]);
    PrintIndexes(arr, N);
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
     
// Function to print the start and end
// indices of all subarrays whose sum
// is a perfect square
static void PrintIndexes(int arr[], int N)
{
    for(int i = 0; i < N; i++)
    {
         
        // Stores the current
        // subarray sum
        int currSubSum = 0;
   
        for(int j = i; j < N; j++)
        {
             
            // Update current subarray sum
            currSubSum += arr[j];
   
            // Stores the square root
            // of currSubSum
            int sq = (int)Math.sqrt(currSubSum);
   
            // Check if currSubSum is
            // a perfect square or not
            if (sq * sq == currSubSum)
            {
                System.out.print("(" + i + "," +
                                 j + ")" + " ");
            }
        }
    }
}
 
// Driver code
public static void main (String[] args)
throws java.lang.Exception
{
    int arr[] = { 65, 79, 81 };
     
    PrintIndexes(arr, arr.length);
}
}
 
// This code is contributed by bikram2001jha

Python3

# Python3 program to implement
# the above approach
import math
 
# Function to print the start and end
# indices of all subarrays whose sum
# is a perfect square
def PrintIndexes(arr, N):
  
    for i in range(N):
  
        # Stores the current
        # subarray sum
        currSubSum = 0
  
        for j in range(i, N, 1):
             
            # Update current subarray sum
            currSubSum += arr[j]
  
            # Stores the square root
            # of currSubSum
            sq = int(math.sqrt(currSubSum))
  
            # Check if currSubSum is
            # a perfect square or not
            if (sq * sq == currSubSum):
                print("(", i, ",",
                           j, ")", end = " ")
                            
# Driver Code
arr = [ 65, 79, 81 ]
N = len(arr)
 
PrintIndexes(arr, N)
 
# This code is contributed by sanjoy_62

C#

// C# program to implement
// the above approach
using System;
class GFG{
     
// Function to print the start
// and end indices of all
// subarrays whose sum
// is a perfect square
static void PrintIndexes(int []arr,
                         int N)
{
  for(int i = 0; i < N; i++)
  {
    // Stores the current
    // subarray sum
    int currSubSum = 0;
 
    for(int j = i; j < N; j++)
    {
      // Update current subarray sum
      currSubSum += arr[j];
 
      // Stores the square root
      // of currSubSum
      int sq = (int)Math.Sqrt(currSubSum);
 
      // Check if currSubSum is
      // a perfect square or not
      if (sq * sq == currSubSum)
      {
        Console.Write("(" + i + "," +
                      j + ")" + " ");
      }
    }
  }
}
 
// Driver code
public static void Main(String[] args)
{
  int []arr = {65, 79, 81};
  PrintIndexes(arr, arr.Length);
}
}
 
// This code is contributed by shikhasingrajput

输出：

(0, 1) (0, 2) (2, 2)

时间复杂度： O(N ² )
辅助空间： O(1)