给定整数数组arr [] ，任务是找到子数组的总数，以使偶数位置的元素之和与奇数位置的元素之和相等。
例子：
Input: arr[] = {1, 2, 3, 4, 1}
Output: 1
Explanation:
{3, 4, 1} is the only subarray in which sum of elements at even position {3, 1} = sum of element at odd position {4}
Input: arr[] = {2, 4, 6, 4, 2}
Output: 2
Explanation:
There are two subarrays {2, 4, 6, 4} and {4, 6, 4, 2}.
方法：想法是生成所有可能的子数组。对于形成的每个子数组，找到偶数索引处的元素之和，然后减去奇数索引处的元素。如果总和为0，则计数此子数组，否则检查下一个子数组。
下面是上述方法的实现：
C++
// C program for the above approach
#include
using namespace std;
// Function to count subarrays in
// which sum of elements at even
// and odd positions are equal
void countSubarrays(int arr[], int n)
{
// Initialize variables
int count = 0;
// Iterate over the array
for(int i = 0; i < n; i++)
{
int sum = 0;
for(int j = i; j < n; j++)
{
// Check if position is
// even then add to sum
// then add it to sum
if ((j - i) % 2 == 0)
sum += arr[j];
// Else subtract it to sum
else
sum -= arr[j];
// Increment the count
// if the sum equals 0
if (sum == 0)
count++;
}
}
// Print the count of subarrays
cout << " " << count ;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 2, 4, 6, 4, 2 };
// Size of the array
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
countSubarrays(arr, n);
return 0;
}
// This code is contributed by shivanisinghss2110 C
// C program for the above approach
#include
// Function to count subarrays in
// which sum of elements at even
// and odd positions are equal
void countSubarrays(int arr[], int n)
{
// Initialize variables
int count = 0;
// Iterate over the array
for(int i = 0; i < n; i++)
{
int sum = 0;
for(int j = i; j < n; j++)
{
// Check if position is
// even then add to sum
// then add it to sum
if ((j - i) % 2 == 0)
sum += arr[j];
// Else subtract it to sum
else
sum -= arr[j];
// Increment the count
// if the sum equals 0
if (sum == 0)
count++;
}
}
// Print the count of subarrays
printf("%d", count);
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 2, 4, 6, 4, 2 };
// Size of the array
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
countSubarrays(arr, n);
return 0;
}
// This code is contributed by piyush3010 Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function to count subarrays in
// which sum of elements at even
// and odd positions are equal
static void countSubarrays(int arr[],
int n)
{
// Initialize variables
int count = 0;
// Iterate over the array
for (int i = 0; i < n; i++) {
int sum = 0;
for (int j = i; j < n; j++) {
// Check if position is
// even then add to sum
// then add it to sum
if ((j - i) % 2 == 0)
sum += arr[j];
// else subtract it to sum
else
sum -= arr[j];
// Increment the count
// if the sum equals 0
if (sum == 0)
count++;
}
}
// Print the count of subarrays
System.out.println(count);
}
// Driver Code
public static void
main(String[] args)
{
// Given array arr[]
int arr[] = { 2, 4, 6, 4, 2 };
// Size of the array
int n = arr.length;
// Function call
countSubarrays(arr, n);
}
}Python3
# Python3 program for the above approach
# Function to count subarrays in
# which sum of elements at even
# and odd positions are equal
def countSubarrays(arr, n):
# Initialize variables
count = 0
# Iterate over the array
for i in range(n):
sum = 0
for j in range(i, n):
# Check if position is
# even then add to sum
# hen add it to sum
if ((j - i) % 2 == 0):
sum += arr[j]
# else subtract it to sum
else:
sum -= arr[j]
# Increment the count
# if the sum equals 0
if (sum == 0):
count += 1
# Print the count of subarrays
print(count)
# Driver Code
if __name__ == '__main__':
# Given array arr[]
arr = [ 2, 4, 6, 4, 2 ]
# Size of the array
n = len(arr)
# Function call
countSubarrays(arr, n)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
class GFG{
// Function to count subarrays in
// which sum of elements at even
// and odd positions are equal
static void countSubarrays(int []arr, int n)
{
// Initialize variables
int count = 0;
// Iterate over the array
for(int i = 0; i < n; i++)
{
int sum = 0;
for(int j = i; j < n; j++)
{
// Check if position is
// even then add to sum
// then add it to sum
if ((j - i) % 2 == 0)
sum += arr[j];
// else subtract it to sum
else
sum -= arr[j];
// Increment the count
// if the sum equals 0
if (sum == 0)
count++;
}
}
// Print the count of subarrays
Console.WriteLine(count);
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 2, 4, 6, 4, 2 };
// Size of the array
int n = arr.Length;
// Function call
countSubarrays(arr, n);
}
}
// This code is contributed by 29AjayKumar输出：
2
时间复杂度： O(N 2 )
辅助空间： O(1)