给定N个整数的数组arr [] ，任务是找到对数(arr [i]，arr [j]) ，以使arr [i] * arr [j]是一个完美的平方。
例子：
Input: arr[] = { 1, 2, 4, 8, 5, 6}
Output: 2
Explanation:
The pairs such that product of element is perfect square are (1, 4) and (8, 2).
Input: arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }
Output: 4
Explanation:
The pairs such that product of element is perfect square are (1, 4), (1, 9), (2, 8) and (4, 9).
天真的方法：
从1到n运行两个循环，并计算所有对(i，j) ，其中arr [i] * arr [j]是一个理想平方。该方法的时间复杂度将为O(N 2 ) 。
高效方法：
arr []中的每个整数都可以用以下形式表示：
arr[i] = k*x ..............(1)
where k is not divisible by any perfect square other than 1,
and x = perfect square,
脚步：
- 以等式(1)的形式表示每个元素。
- 然后，对于每对(ARR [I]，编曲[j]的)在ARR []可表示为：
arr[i] = ki*x; arr[j] = kj*y; where x and y are perfect square对于对(arr [i]，arr [j]) ，当且仅当k i = k j时， arr [i]和arr [j]的乘积可以是理想平方。
- 使用Eratosthenes筛子为数组arr []中的每个元素预先计算k的值。
- 将每个元素的k频率存储在map的arr []中。
- 因此，对的总数由频率大于1的元素形成的对的数目给定。
- 由n个元素构成的对的总数由下式给出：
Number of Pairs = (f*(f-1))/2 where f is the frequency of an element.
下面是上述方法的实现：
C++
// C++ program to calculate the number of
// pairs with product is perfect square
#include
using namespace std;
// Prime[] array to calculate Prime Number
int prime[100001] = { 0 };
// Array k[] to store the value of k for
// each element in arr[]
int k[100001] = { 0 };
// For value of k, Sieve function is
// implemented
void Sieve()
{
// Initialize k[i] to i
for (int i = 1; i < 100001; i++)
k[i] = i;
// Prime Sieve
for (int i = 2; i < 100001; i++) {
// If i is prime then remove all
// factors of prime from it
if (prime[i] == 0)
for (int j = i; j < 100001; j += i) {
// Update that j is not
// prime
prime[j] = 1;
// Remove all square divisors
// i.e. if k[j] is divisible
// by i*i then divide it by i*i
while (k[j] % (i * i) == 0)
k[j] /= (i * i);
}
}
}
// Function that return total count
// of pairs with pefect square product
int countPairs(int arr[], int n)
{
// Map used to store the frequency of k
unordered_map freq;
// Store the frequency of k
for (int i = 0; i < n; i++) {
freq[k[arr[i]]]++;
}
int sum = 0;
// The total number of pairs is the
// summation of (fi * (fi - 1))/2
for (auto i : freq) {
sum += ((i.second - 1) * i.second) / 2;
}
return sum;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 4, 8, 5, 6 };
// Size of arr[]
int n = sizeof(arr) / sizeof(int);
// To pre-compute the value of k
Sieve();
// Function that return total count
// of pairs with perfect square product
cout << countPairs(arr, n) << endl;
return 0;
} Java
// Java program to calculate the number of
// pairs with product is perfect square
import java.util.*;
class GFG{
// Prime[] array to calculate Prime Number
static int []prime = new int[100001];
// Array k[] to store the value of k for
// each element in arr[]
static int []k = new int[100001];
// For value of k, Sieve function is
// implemented
static void Sieve()
{
// Initialize k[i] to i
for (int i = 1; i < 100001; i++)
k[i] = i;
// Prime Sieve
for (int i = 2; i < 100001; i++) {
// If i is prime then remove all
// factors of prime from it
if (prime[i] == 0)
for (int j = i; j < 100001; j += i) {
// Update that j is not
// prime
prime[j] = 1;
// Remove all square divisors
// i.e. if k[j] is divisible
// by i*i then divide it by i*i
while (k[j] % (i * i) == 0)
k[j] /= (i * i);
}
}
}
// Function that return total count
// of pairs with pefect square product
static int countPairs(int arr[], int n)
{
// Map used to store the frequency of k
HashMap freq = new HashMap();
// Store the frequency of k
for (int i = 0; i < n; i++) {
if(freq.containsKey(k[arr[i]])) {
freq.put(k[arr[i]], freq.get(k[arr[i]])+1);
}
else
freq.put(k[arr[i]], 1);
}
int sum = 0;
// The total number of pairs is the
// summation of (fi * (fi - 1))/2
for (Map.Entry i : freq.entrySet()){
sum += ((i.getValue() - 1) * i.getValue()) / 2;
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 4, 8, 5, 6 };
// Size of arr[]
int n = arr.length;
// To pre-compute the value of k
Sieve();
// Function that return total count
// of pairs with perfect square product
System.out.print(countPairs(arr, n) +"\n");
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to calculate the number
# of pairs with product is perfect square
# prime[] array to calculate Prime Number
prime = [0] * 100001
# Array to store the value of k
# for each element in arr[]
k = [0] * 100001
# For value of k, Sieve implemented
def Sieve():
# Initialize k[i] to i
for i in range(1, 100001):
k[i] = i
# Prime sieve
for i in range(2, 100001):
# If i is prime then remove all
# factors of prime from it
if (prime[i] == 0):
for j in range(i, 100001, i):
# Update that j is not prime
prime[j] = 1
# Remove all square divisors
# i.e if k[j] is divisible by
# i*i then divide it by i*i
while (k[j] % (i * i) == 0):
k[j] /= (i * i)
# Function that return total count of
# pairs with perfect square product
def countPairs (arr, n):
# Store the frequency of k
freq = dict()
for i in range(n):
if k[arr[i]] in freq.keys():
freq[k[arr[i]]] += 1
else:
freq[k[arr[i]]] = 1
Sum = 0
# The total number of pairs is the
# summation of (fi * (fi - 1))/2
for i in freq:
Sum += (freq[i] * (freq[i] - 1)) / 2
return Sum
# Driver code
arr = [ 1, 2, 4, 8, 5, 6 ]
# Length of arr
n = len(arr)
# To pre-compute the value of k
Sieve()
# Function that return total count
# of pairs with perfect square product
print(int(countPairs(arr, n)))
# This code is contributed by himanshu77C#
// C# program to calculate the number of
// pairs with product is perfect square
using System;
using System.Collections.Generic;
class GFG{
// Prime[] array to calculate Prime Number
static int []prime = new int[100001];
// Array k[] to store the value of k for
// each element in []arr
static int []k = new int[100001];
// For value of k, Sieve function is
// implemented
static void Sieve()
{
// Initialize k[i] to i
for (int i = 1; i < 100001; i++)
k[i] = i;
// Prime Sieve
for (int i = 2; i < 100001; i++) {
// If i is prime then remove all
// factors of prime from it
if (prime[i] == 0)
for (int j = i; j < 100001; j += i) {
// Update that j is not
// prime
prime[j] = 1;
// Remove all square divisors
// i.e. if k[j] is divisible
// by i*i then divide it by i*i
while (k[j] % (i * i) == 0)
k[j] /= (i * i);
}
}
}
// Function that return total count
// of pairs with pefect square product
static int countPairs(int []arr, int n)
{
// Map used to store the frequency of k
Dictionary freq = new Dictionary();
// Store the frequency of k
for (int i = 0; i < n; i++) {
if(freq.ContainsKey(k[arr[i]])) {
freq[k[arr[i]]] = freq[k[arr[i]]]+1;
}
else
freq.Add(k[arr[i]], 1);
}
int sum = 0;
// The total number of pairs is the
// summation of (fi * (fi - 1))/2
foreach (KeyValuePair i in freq){
sum += ((i.Value - 1) * i.Value) / 2;
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 2, 4, 8, 5, 6 };
// Size of []arr
int n = arr.Length;
// To pre-compute the value of k
Sieve();
// Function that return total count
// of pairs with perfect square product
Console.Write(countPairs(arr, n) +"\n");
}
}
// This code is contributed by PrinciRaj1992 输出：
2
时间复杂度： O(N * log(log N))