给定一个整数数组arr []和一个整数K ，任务是计算GCD等于K的所有三胞胎。
例子：
Input: arr[] = {1, 4, 8, 14, 20}, K = 2
Output: 3
Explanation:
Triplets (4, 14, 20), (8, 14, 20) and (4, 8, 14) have GCD equal to 2.
Input: arr[] = {1, 2, 3, 4, 5}, K = 7
Output: 0
方法：
请按照以下步骤解决问题：
- 维护一个数组cnt [i] ，该数组表示GCD = i时数组中三元组的数目。
- 现在，要找到cnt [i] ，首先要计算存储在另一个数组mul [i]中的i的所有倍数。
- 现在，从mul [i]中选择任意三个值。选择三个值的方式为mul [i] C 3 ，将此值存储在cnt [i]中。
- 但是，它也包括具有GCD三胞胎我的倍数。所以我们再次迭代所有i的倍数，并从CNT减去CNT [j]的[I]其中j为i的倍数。
- 最终返回cnt [K] 。
下面是上述方法的实现：
C++
// C++ program to count the
// number of triplets in the
// array with GCD equal to K
#include
using namespace std;
const int MAXN = 1e6 + 1;
// frequency array
int freq[MAXN] = { 0 };
// mul[i] stores the count
// of multiples of i
int mul[MAXN] = { 0 };
// cnt[i] stores the count
// of triplets with gcd = i
int cnt[MAXN] = { 0 };
// Return nC3
int nC3(int n)
{
if (n < 3)
return 0;
return (n * (n - 1) * (n - 2)) / 6;
}
// Function to count and return
// the number of triplets in the
// array with GCD equal to K
void count_triplet(vector arr,
int N, int K)
{
for (int i = 0; i < N; i++) {
// Store frequency of
// array elements
freq[arr[i]]++;
}
for (int i = 1; i <= 1000000; i++) {
for (int j = i; j <= 1000000;
j += i) {
// Store the multiples of
// i present in the array
mul[i] += freq[j];
}
// Count triplets with gcd
// equal to any multiple of i
cnt[i] = nC3(mul[i]);
}
// Remove all triplets which have gcd
// equal to a multiple of i
for (int i = 1000000; i >= 1; i--) {
for (int j = 2 * i; j <= 1000000;
j += i) {
cnt[i] -= cnt[j];
}
}
cout << "Number of triplets "
<< "with GCD " << K;
cout << " are " << cnt[K];
}
// Driver Program
int main()
{
vector arr = { 1, 7, 12, 6,
15, 9 };
int N = 6, K = 3;
count_triplet(arr, N, K);
return 0;
} Java
// Java program to count the
// number of triplets in the
// array with GCD equal to K
class GFG{
static int MAXN = 1000001;
// Frequency array
static int freq[] = new int[MAXN];
// mul[i] stores the count
// of multiples of i
static int mul[] = new int[MAXN];
// cnt[i] stores the count
// of triplets with gcd = i
static int cnt[] = new int[MAXN];
// Return nC3
static int nC3(int n)
{
if (n < 3)
return 0;
return (n * (n - 1) * (n - 2)) / 6;
}
// Function to count and return
// the number of triplets in the
// array with GCD equal to K
static void count_triplet(int[] arr,
int N, int K)
{
int i, j;
for(i = 0; i < N; i++)
{
// Store frequency of
// array elements
freq[arr[i]]++;
}
for(i = 1; i <= 1000000; i++)
{
for(j = i; j <= 1000000; j += i)
{
// Store the multiples of
// i present in the array
mul[i] += freq[j];
}
// Count triplets with gcd
// equal to any multiple of i
cnt[i] = nC3(mul[i]);
}
// Remove all triplets which have gcd
// equal to a multiple of i
for(i = 1000000; i >= 1; i--)
{
for(j = 2 * i; j <= 1000000; j += i)
{
cnt[i] -= cnt[j];
}
}
System.out.print("Number of triplets " +
"with GCD " + K);
System.out.print(" are " + cnt[K]);
}
// Driver code
public static void main (String []args)
{
int []arr = { 1, 7, 12, 6, 15, 9 };
int N = 6, K = 3;
count_triplet(arr, N, K);
}
}
// This code is contributed by chitranayalPython3
# Python3 program to count the number of
# triplets in the array with GCD equal to K
MAXN = int(1e6 + 1)
# Frequency array
freq = [0] * MAXN
# mul[i] stores the count
# of multiples of i
mul = [0] * MAXN
# cnt[i] stores the count
# of triplets with gcd = i
cnt = [0] * MAXN
# Return nC3
def nC3(n):
if(n < 3):
return 0
return (n * (n - 1) * (n - 2)) / 6
# Function to count and return
# the number of triplets in the
# array with GCD equal to K
def count_triplet(arr, N, K):
for i in range(N):
# Store frequency of
# array elements
freq[arr[i]] += 1
for i in range(1, 1000000 + 1):
for j in range(i, 1000000 + 1, i):
# Store the multiples of
# i present in the array
mul[i] += freq[j]
# Count triplets with gcd
# equal to any multiple of i
cnt[i] = nC3(mul[i])
# Remove all triplets which have gcd
# equal to a multiple of i
for i in range(1000000, 0, -1):
for j in range(2 * i, 1000000 + 1, i):
cnt[i] -= cnt[j]
print("Number of triplets with GCD"
" {0} are {1}".format(K, int(cnt[K])))
# Driver Code
if __name__ == '__main__':
arr = [ 1, 7, 12, 6, 15, 9 ]
N = 6
K = 3
count_triplet(arr, N, K)
# This code is contributed by Shivam SinghC#
// C# program to count the
// number of triplets in the
// array with GCD equal to K
using System;
class GFG{
static int MAXN = 1000001;
// Frequency array
static int []freq = new int[MAXN];
// mul[i] stores the count
// of multiples of i
static int []mul = new int[MAXN];
// cnt[i] stores the count
// of triplets with gcd = i
static int []cnt = new int[MAXN];
// Return nC3
static int nC3(int n)
{
if (n < 3)
return 0;
return (n * (n - 1) * (n - 2)) / 6;
}
// Function to count and return
// the number of triplets in the
// array with GCD equal to K
static void count_triplet(int[] arr,
int N, int K)
{
int i, j;
for(i = 0; i < N; i++)
{
// Store frequency of
// array elements
freq[arr[i]]++;
}
for(i = 1; i <= 1000000; i++)
{
for(j = i; j <= 1000000; j += i)
{
// Store the multiples of
// i present in the array
mul[i] += freq[j];
}
// Count triplets with gcd
// equal to any multiple of i
cnt[i] = nC3(mul[i]);
}
// Remove all triplets which have gcd
// equal to a multiple of i
for(i = 1000000; i >= 1; i--)
{
for(j = 2 * i; j <= 1000000; j += i)
{
cnt[i] -= cnt[j];
}
}
Console.Write("Number of triplets " +
"with GCD " + K);
Console.Write(" are " + cnt[K]);
}
// Driver code
public static void Main (string []args)
{
int []arr = { 1, 7, 12, 6, 15, 9 };
int N = 6, K = 3;
count_triplet(arr, N, K);
}
}
// This code is contributed by Ritik Bansal输出：
Number of triplets with GCD 3 are 4
时间复杂度： O(N * log N)
辅助空间： O(N)