对数组中的对进行计数，其和可被K整除

给定一个数组A []和正整数K ，任务是计算总和可被K整除的数组中的对总数。
注意：此问题是此版本的通用版本
例子：

Input : A[] = {2, 2, 1, 7, 5, 3}, K = 4
Output : 5
Explanation : 
There are five pairs possible whose sum
is divisible by '4' i.e., (2, 2), 
(1, 7), (7, 5), (1, 3) and (5, 3)

Input : A[] = {5, 9, 36, 74, 52, 31, 42}, K = 3
Output : 7

天真的方法：最简单的方法是遍历数组的每一对，但使用两个嵌套的for循环并计算总和可被“ K”整除的那些对。该方法的时间复杂度为O(N ² )。
高效的方法：一种有效的方法是使用哈希技术。我们将根据元素(值mod K)将其分为多个存储桶。当数字除以K时，余数可以是0、1、2，最高为(k-1)。因此，取一个数组，说大小为K的freq [] (用零初始化)并增加freq [A [i]％K]的值，以便我们可以计算出除以k的余数j的值的数量。

C++

// C++ Program to count pairs
// whose sum divisible by 'K'
#include 
using namespace std;
 
// Program to count pairs whose sum divisible
// by 'K'
int countKdivPairs(int A[], int n, int K)
{
    // Create a frequency array to count
    // occurrences of all remainders when
    // divided by K
    int freq[K] = { 0 };
 
    // Count occurrences of all remainders
    for (int i = 0; i < n; i++)
        ++freq[A[i] % K];
 
    // If both pairs are divisible by 'K'
    int sum = freq[0] * (freq[0] - 1) / 2;
 
    // count for all i and (k-i)
    // freq pairs
    for (int i = 1; i <= K / 2 && i != (K - i); i++)
        sum += freq[i] * freq[K - i];
    // If K is even
    if (K % 2 == 0)
        sum += (freq[K / 2] * (freq[K / 2] - 1) / 2);
    return sum;
}
 
// Driver code
int main()
{
 
    int A[] = { 2, 2, 1, 7, 5, 3 };
    int n = sizeof(A) / sizeof(A[0]);
    int K = 4;
    cout << countKdivPairs(A, n, K);
 
    return 0;
}

Java

// Java program to count pairs
// whose sum divisible by 'K'
import java.util.*;
 
class Count {
    public static int countKdivPairs(int A[], int n, int K)
    {
        // Create a frequency array to count
        // occurrences of all remainders when
        // divided by K
        int freq[] = new int[K];
 
        // Count occurrences of all remainders
        for (int i = 0; i < n; i++)
            ++freq[A[i] % K];
 
        // If both pairs are divisible by 'K'
        int sum = freq[0] * (freq[0] - 1) / 2;
 
        // count for all i and (k-i)
        // freq pairs
        for (int i = 1; i <= K / 2 && i != (K - i); i++)
            sum += freq[i] * freq[K - i];
        // If K is even
        if (K % 2 == 0)
            sum += (freq[K / 2] * (freq[K / 2] - 1) / 2);
        return sum;
    }
    public static void main(String[] args)
    {
        int A[] = { 2, 2, 1, 7, 5, 3 };
        int n = 6;
        int K = 4;
        System.out.print(countKdivPairs(A, n, K));
    }
}

Python3

# Python3 code to count pairs whose
# sum is divisible by 'K'
 
# Function to count pairs whose
# sum is divisible by 'K'
def countKdivPairs(A, n, K):
     
    # Create a frequency array to count
    # occurrences of all remainders when
    # divided by K
    freq = [0] * K
     
    # Count occurrences of all remainders
    for i in range(n):
        freq[A[i] % K]+= 1
         
    # If both pairs are divisible by 'K'
    sum = freq[0] * (freq[0] - 1) / 2;
     
    # count for all i and (k-i)
    # freq pairs
    i = 1
    while(i <= K//2 and i != (K - i) ):
        sum += freq[i] * freq[K-i]
        i+= 1
 
    # If K is even
    if( K % 2 == 0 ):
        sum += (freq[K//2] * (freq[K//2]-1)/2);
     
    return int(sum)
 
# Driver code
A = [2, 2, 1, 7, 5, 3]
n = len(A)
K = 4
print(countKdivPairs(A, n, K))

C#

// C# program to count pairs
// whose sum divisible by 'K'
using System;
 
class Count
{
    public static int countKdivPairs(int []A, int n, int K)
    {
        // Create a frequency array to count
        // occurrences of all remainders when
        // divided by K
        int []freq = new int[K];
 
        // Count occurrences of all remainders
        for (int i = 0; i < n; i++)
            ++freq[A[i] % K];
 
        // If both pairs are divisible by 'K'
        int sum = freq[0] * (freq[0] - 1) / 2;
 
        // count for all i and (k-i)
        // freq pairs
        for (int i = 1; i <= K / 2 && i != (K - i); i++)
            sum += freq[i] * freq[K - i];
             
        // If K is even
        if (K % 2 == 0)
            sum += (freq[K / 2] * (freq[K / 2] - 1) / 2);
        return sum;
    }
     
    // Driver code
    static public void Main ()
    {
        int []A = { 2, 2, 1, 7, 5, 3 };
        int n = 6;
        int K = 4;
        Console.WriteLine(countKdivPairs(A, n, K));
    }
}
 
// This code is contributed by akt_mit.

输出 ：

5

时间复杂度： O(N)
辅助空间： O(K)