计算三元组 (a, b, c) 使得 a + b, b + c 和 a + c 都可以被 K 整除2套

给定两个正整数N和K ，任务是计算三元组(a, b, c) 的数量，使得0 < a, b, c < N and (a + b) , (b + c) and (c) + a)都是K 的倍数。

例子：

Input: N = 2, K = 2
Output: 2
Explanation: All possible triplets that satisfy the given property are (1, 1, 1) and (2, 2, 2).
Therefore, the total count is 2.

Input: N = 3, K = 2
Output: 9

朴素的方法：请参阅上一篇文章，了解解决此问题的最简单方法。
时间复杂度： O(N ³ )
辅助空间： O(1)

高效的方法：上述方法也可以基于以下观察进行优化：

• 给定条件由同余公式表示：

=> a + b ≡ b + c ≡ c + a ≡ 0(mod K)
=> a+b ≡ b+c (mod K)
=> a ≡ c(mod K)

• 不使用同余公式也可以观察到上述关系：
  • 因为(a + b)是K的倍数，而(c + b)是K的倍数。因此， (a + b) − (c + b) = a – c也是K的倍数，即a ≡ b ≡ c (mod K) 。
  • 因此，表达式可以进一步评估为：

=> a + b ≡ 0 (mod K)
=> a + a ≡ 0 (mod K) (since a is congruent to b)
=> 2a ≡ 0 (mod K)

根据以上观察，可以计算出以下两种情况的结果：

• 如果K 是奇数，那么a ≡ b ≡ c ≡ 0(mod K)因为所有三个都是全等的，并且三元组的总数可以计算为(N / K) ³ 。
• 如果K 是偶数，则K可被2和a ≡ 0 (mod K) 、 b ≡ 0 (mod K)和c ≡ 0 (mod K)整除。因此，三胞胎的总数可以计算为(N / K) ³ ((N + (K/2)) / K) ³ 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// Function to count the number of
// triplets from the range [1, N - 1]
// having sum of all pairs divisible by K
int countTriplets(int N, int K)
{
    // If K is even
    if (K % 2 == 0) {
        long long int x = N / K;
        long long int y = (N + (K / 2)) / K;
 
        return x * x * x + y * y * y;
    }
 
    // Otherwise
    else {
        long long int x = N / K;
        return x * x * x;
    }
}
 
// Driver Code
int main()
{
    int N = 2, K = 2;
    cout << countTriplets(N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to count the number of
// triplets from the range [1, N - 1]
// having sum of all pairs divisible by K
static int countTriplets(int N, int K)
{
     
    // If K is even
    if (K % 2 == 0)
    {
        int x = N / K;
        int y = (N + (K / 2)) / K;
 
        return x * x * x + y * y * y;
    }
 
    // Otherwise
    else
    {
        int x = N / K;
        return x * x * x;
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 2, K = 2;
     
    System.out.print(countTriplets(N, K));
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
 
# Function to count the number of
# triplets from the range [1, N - 1]
# having sum of all pairs divisible by K
def countTriplets(N, K):
 
    # If K is even
    if (K % 2 == 0):
        x = N // K
        y = (N + (K // 2)) // K
 
        return x * x * x + y * y * y
 
    # Otherwise
    else:
        x = N // K
        return x * x * x
 
# Driver Code
if __name__ == "__main__":
 
    N = 2
    K = 2
     
    print(countTriplets(N, K))
 
# This code is contributed by ukasp

Javascript

 
// Javascript program for the above approach 
 
// Function to count the number of
// triplets from the range [1, N - 1]
// having sum of all pairs divisible by K
function countTriplets(N, K)
{
     
    // If K is even
    if (K % 2 == 0)
    {
        var x = parseInt(N / K);
        var y = parseInt((N + (K / 2)) / K);
 
        return x * x * x + y * y * y;
    }
 
    // Otherwise
    else
    {
        var x = parseInt(N / K);
        return x * x * x;
    }
}
 
// Driver Code
var N = 2, K = 2;
 
document.write(countTriplets(N, K));
 
// This code is contributed by Ankita saini
 

C#

// C# program for the above approach
 
using System;
 
class GFG{
 
// Function to count the number of
// triplets from the range [1, N - 1]
// having sum of all pairs divisible by K
static int countTriplets(int N, int K)
{
     
    // If K is even
    if (K % 2 == 0)
    {
        int x = N / K;
        int y = (N + (K / 2)) / K;
 
        return x * x * x + y * y * y;
    }
 
    // Otherwise
    else
    {
        int x = N / K;
        return x * x * x;
    }
}
 
// Driver Code
    static void Main() {
       int N = 2, K = 2;
     
       Console.Write(countTriplets(N, K));
    }
}
 
// This code is contributed by SoumikMondal

输出：

2

时间复杂度： O(1)
辅助空间： O(1)

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