给定N个整数的数组arr [] ，其中| arr [i]≤1000 |对于所有有效的我。任务是计算来自数组的整数对，它们的平均值也存在于同一数组中，即(arr [i]，arr [j])为有效对(arr [i] + arr [j]) / 2也必须存在于数组中。
例子：
Input: arr[] = {2, 1, 3}
Output: 1
Only valid pair is (1, 3) as (1 + 3) / 2 = 2 is also present in the array.
Input: arr[] = {4, 2, 5, 1, 3, 5}
Output: 7
方法：制作一个频率阵列，存储每个阵列元素的频率。请记住，如果数组也包含负数，那么我们必须将频率数组的大小取为原始大小的两倍。更新频率阵列后，有两种情况：
- 如果freq [i]> 0，则所需对的总数将为count =(freq [i] *(freq [i] – 1))/ 2 。
- 对于每对(freq [i]，freq [j]) ，其中freq [i]> 0 ， freq [j]> 0和freq [(i + j)/ 2]> 0，则所需对的总数将为be count =(freq [i] * freq [j]) 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
const int N = 1000;
// Function to return the count
// of valid pairs
int countPairs(int arr[], int n)
{
// Frequency array
// Twice the original size to hold
// negative elements as well
int size = (2 * N) + 1;
int freq[size] = { 0 };
// Update the frequency of each
// of the array element
for (int i = 0; i < n; i++) {
int x = arr[i];
// If say x = -1000 then we will place
// the frequency of -1000 at
// (-1000 + 1000 = 0) a[0] index
freq[x + N]++;
}
// To store the count of valid pairs
int ans = 0;
// Remember we will check only for (even, even)
// or (odd, odd) pairs of indexes as the average
// of two consecutive elements is
// a floating point number
for (int i = 0; i < size; i++) {
if (freq[i] > 0) {
ans += ((freq[i]) * (freq[i] - 1)) / 2;
for (int j = i + 2; j < 2001; j += 2) {
if (freq[j] > 0 && (freq[(i + j) / 2] > 0)) {
ans += (freq[i] * freq[j]);
}
}
}
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 4, 2, 5, 1, 3, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countPairs(arr, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
static int N = 1000;
// Function to return the count
// of valid pairs
static int countPairs(int arr[], int n)
{
// Frequency array
// Twice the original size to hold
// negative elements as well
int size = (2 * N) + 1;
int freq[] = new int[size];
// Update the frequency of each
// of the array element
for (int i = 0; i < n; i++)
{
int x = arr[i];
// If say x = -1000 then we will place
// the frequency of -1000 at
// (-1000 + 1000 = 0) a[0] index
freq[x + N]++;
}
// To store the count of valid pairs
int ans = 0;
// Remember we will check only for (even, even)
// or (odd, odd) pairs of indexes as the average
// of two consecutive elements is
// a floating point number
for (int i = 0; i < size; i++)
{
if (freq[i] > 0)
{
ans += ((freq[i]) * (freq[i] - 1)) / 2;
for (int j = i + 2; j < 2001; j += 2)
{
if (freq[j] > 0 && (freq[(i + j) / 2] > 0))
{
ans += (freq[i] * freq[j]);
}
}
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {4, 2, 5, 1, 3, 5};
int n = arr.length;
System.out.println(countPairs(arr, n));
}
}
// This code has been contributed by 29AjayKumarPython3
# Python 3 implementation of the approach
N = 1000
# Function to return the count
# of valid pairs
def countPairs(arr, n):
# Frequency array
# Twice the original size to hold
# negative elements as well
size = (2 * N) + 1
freq = [0 for i in range(size)]
# Update the frequency of each
# of the array element
for i in range(n):
x = arr[i]
# If say x = -1000 then we will place
# the frequency of -1000 at
# (-1000 + 1000 = 0) a[0] index
freq[x + N] += 1
# To store the count of valid pairs
ans = 0
# Remember we will check only for (even, even)
# or (odd, odd) pairs of indexes as the average
# of two consecutive elements is
# a floating point number
for i in range(size):
if (freq[i] > 0):
ans += int(((freq[i]) * (freq[i] - 1)) / 2)
for j in range(i + 2, 2001, 2):
if (freq[j] > 0 and
(freq[int((i + j) / 2)] > 0)):
ans += (freq[i] * freq[j])
return ans
# Driver code
if __name__ == '__main__':
arr = [4, 2, 5, 1, 3, 5]
n = len(arr)
print(countPairs(arr, n))
# This code is contributed by
# Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
static int N = 1000;
// Function to return the count
// of valid pairs
static int countPairs(int []arr, int n)
{
// Frequency array
// Twice the original size to hold
// negative elements as well
int size = (2 * N) + 1;
int []freq = new int[size];
// Update the frequency of each
// of the array element
for (int i = 0; i < n; i++)
{
int x = arr[i];
// If say x = -1000 then we will place
// the frequency of -1000 at
// (-1000 + 1000 = 0) a[0] index
freq[x + N]++;
}
// To store the count of valid pairs
int ans = 0;
// Remember we will check only for (even, even)
// or (odd, odd) pairs of indexes as the average
// of two consecutive elements is
// a floating point number
for (int i = 0; i < size; i++)
{
if (freq[i] > 0)
{
ans += ((freq[i]) * (freq[i] - 1)) / 2;
for (int j = i + 2; j < 2001; j += 2)
{
if (freq[j] > 0 && (freq[(i + j) / 2] > 0))
{
ans += (freq[i] * freq[j]);
}
}
}
}
return ans;
}
// Driver code
public static void Main()
{
int []arr = {4, 2, 5, 1, 3, 5};
int n = arr.Length;
Console.WriteLine(countPairs(arr, n));
}
}
// This code is contributed by AnkitRai01输出：
7