三元组的数量，以使每个值小于N并且每对和为K的倍数

给定两个整数N和K。找出三元组(a，b，c)的数量，使0≤a ，b，c≤N并且(a + b) ， (b + c)和(c + a)是K的倍数。

例子：

Input: N = 3, K = 2 
Output: 9 
Triplets possible are: 
{(1, 1, 1), (1, 1, 3), (1, 3, 1) 
(1, 3, 3), (2, 2, 2), (3, 1, 1) 
(3, 1, 1), (3, 1, 3), (3, 3, 3)}

Input: N = 5, K = 3 
Output: 1 
Only possible triplet is (3, 3, 3) 
 

方法：假设(a + b) ， (b + c)和(c + a)是K的倍数。因此，我们可以说(a + b)％K = 0 ， (b + c)％K = 0和(c + a)％K = 0 。
如果a属于K的x模类，则使用第一个条件， b应该在第(K – x)模类中。
从第二个条件可以看出， c属于K的x模类。现在，由于a和c属于同一模类，因此它们必须满足第三种关系，即(a + c)％K = 0 。仅当x = 0或x = K / 2时才有可能。
当K为奇数整数时， x = K / 2无效。

因此，为了解决该问题，从0数元素到N的^第在模类和(K / 2)^个模类K的0的数目。

• 如果K为奇数，则结果为cnt [0] ³
• 如果K为偶数，则结果为cnt [0] ³ + cnt [K / 2] ³ 。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to return the number of triplets
int NoofTriplets(int N, int K)
{
    int cnt[K];
 
    // Initializing the count array
    memset(cnt, 0, sizeof(cnt));
 
    // Storing the frequency of each modulo class
    for (int i = 1; i <= N; i += 1) {
        cnt[i % K] += 1;
    }
 
    // If K is odd
    if (K & 1)
        return cnt[0] * cnt[0] * cnt[0];
 
    // If K is even
    else {
        return (cnt[0] * cnt[0] * cnt[0]
                + cnt[K / 2] * cnt[K / 2] * cnt[K / 2]);
    }
}
 
// Driver Code
int main()
{
    int N = 3, K = 2;
 
    // Function Call
    cout << NoofTriplets(N, K);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;
 
class GFG
{
 
    // Function to return the number of triplets
    static int NoofTriplets(int N, int K)
    {
        int[] cnt = new int[K];
 
        // Initializing the count array
        Arrays.fill(cnt, 0, cnt.length, 0);
 
        // Storing the frequency of each modulo class
        for (int i = 1; i <= N; i += 1)
        {
            cnt[i % K] += 1;
        }
 
        // If K is odd
        if ((K & 1) != 0)
        {
            return cnt[0] * cnt[0] * cnt[0];
        }
        // If K is even
        else
        {
            return (cnt[0] * cnt[0] * cnt[0]
                    + cnt[K / 2] * cnt[K / 2] * cnt[K / 2]);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int N = 3, K = 2;
 
        // Function Call
        System.out.println(NoofTriplets(N, K));
    }
}
 
// This code is contributed by Princi Singh

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to return the number of triplets
    static int NoofTriplets(int N, int K)
    {
        int[] cnt = new int[K];
 
        // Initializing the count array
        Array.Fill(cnt, 0, cnt.Length, 0);
 
        // Storing the frequency of each modulo class
        for (int i = 1; i <= N; i += 1)
        {
            cnt[i % K] += 1;
        }
 
        // If K is odd
        if ((K & 1) != 0)
        {
            return cnt[0] * cnt[0] * cnt[0];
        }
        // If K is even
        else
        {
            return (cnt[0] * cnt[0] * cnt[0]
                    + cnt[K / 2] * cnt[K / 2] * cnt[K / 2]);
        }
    }
 
    // Driver Code
    static public void Main ()
    {
            int N = 3, K = 2;
 
        // Function Call
        Console.Write(NoofTriplets(N, K));
    }
}
 
// This code is contributed by ajit

Python3

# Python3 implementation of the above approach
 
# Function to return the number of triplets
def NoofTriplets(N, K) :
     
    # Initializing the count array
    cnt = [0]*K;
 
    # Storing the frequency of each modulo class
    for i in range(1, N + 1) :
        cnt[i % K] += 1;
 
    # If K is odd
    if (K & 1) :
        rslt = cnt[0] * cnt[0] * cnt[0];
        return rslt
 
    # If K is even
    else :
        rslt = (cnt[0] * cnt[0] * cnt[0] +
                cnt[K // 2] * cnt[K // 2] * cnt[K // 2]);
        return rslt
 
# Driver Code
if __name__ == "__main__" :
 
    N = 3; K = 2;
 
    # Function Call
    print(NoofTriplets(N, K));
 
# This code is contributed by AnkitRai01

Javascript

 
// Javascript implementation of the above approach
 
// Function to return the number of triplets
function NoofTriplets(N, K)
{
    let cnt = Array(K);
 
    for(let i = 0; i < K; i++) 
        cnt[i] = 0;
 
    // Storing the frequency of
    // each modulo class
    for(let i = 1; i <= N; i += 1)
    {
        cnt[i % K] += 1;
    }
 
    // If K is odd
    if (K & 1)
        return cnt[0] * cnt[0] * cnt[0];
 
    // If K is even
    else
    {
        return (cnt[0] * cnt[0] * cnt[0] +
                 cnt[K / 2] * cnt[K / 2] *
                 cnt[K / 2]);
    }
}
 
// Driver Code
let N = 3;
let K = 2;
 
// Function Call
document.write(NoofTriplets(N, K));
 
// This code is contributed by mohit kumar 29
 

输出：

9

时间复杂度： O(N)