给定分别由N和M个整数组成的两个数组A []和B [] ，任务是对对(A [i]，B [j])进行计数，以使它们的不同素因数的乘积为偶数。
例子：
Input: A[] = {1, 2, 3}, B[] = {4, 5, 6}, N = 3, M = 3
Output: 2
Explanation:
- Replacing all array elements with the count of their distinct prime factors modifies the arrays to A[] = {0, 1, 1} and B[] = {1, 1, 2}.
- Therefore, total pairs with even product are {{2, 6}, {3, 6}}.
Input: A[] = {1, 7}, B[] = {5, 6}, N = 2, M = 2
Output: 1
Explanation:
- Replacing all array elements with the count of their distinct prime factors modifies the arrays to A[] = {0, 1} and B[] = {1, 2}.
- Therefore, total pairs with even product is {7, 6}.
简单方法：最简单的方法是将生成的所有可能的对(A [I]，B [j])从两个阵列以及对于每一对，计算的不同的素因子华氏度数组元素的计数和检查，如果他们的产品是连或不。如果发现为真，则增加此类对的计数。
时间复杂度： O(N 5/2 )
辅助空间： O(1)
高效方法：可以通过以下方法优化上述方法 预先计算所有数字的不同素数的计数 取决于 两个数组中最大的元素，并使用两个数字的乘积的以下属性：
- Odd * Odd = Odd
- Even * Odd = Even
- Odd * Even = Even
- Even * Even = Even
请按照以下步骤解决问题：
- 首先，计算最大为MAX的所有数字的不同素数，并将其存储在vector
中，例如countDistinct。 - 初始化两个变量，例如evenCount和奇数计数，以将元素的计数存储为B []中数组元素的不同素数的偶数和奇数计数。
- 遍历数组B []。如果countDistinct [B [i]] = 0 ，则跳过此步骤。如果是奇数，增量oddCount 1。否则，增量evenCount一个。
- 将变量EvenPairs初始化为0 。
- 通过evenCount遍历数组A []和增量evenPairs如果countDistinct [A [I]]为奇数。
- 否则，将evenPairs增加evenCount +奇数。
- 打印evenPairs的值。
下面是上述方法的实现：
C++
// C++ implementation of
// the above appraoch
#include
using namespace std;
#define MAX 1000000
// Function to calculate count of
// distinct prime factors of a number
void countOfPrimefactors(vector& CountDistinct)
{
bool prime[MAX + 1];
for (int i = 0; i <= MAX; i++) {
CountDistinct[i] = 0;
prime[i] = true;
}
// Seive of Eratosthenes
for (long long int i = 2; i <= MAX; i++) {
if (prime[i] == true) {
CountDistinct[i] = 1;
for (long long int j = i * 2; j <= MAX;
j += i) {
CountDistinct[j]++;
prime[j] = false;
}
}
}
}
// Function to count pairs with even
// product of distinct prime factors
int CountEvenPair(int A[], int B[], int N, int M)
{
// Stores count of
// distinct prime factors
vector countDistinct(MAX + 1);
countOfPrimefactors(countDistinct);
// Stores the count of numbers
// with even prime factors in B[]
int evenCount = 0;
// Stores the count of numbers
// with odd prime factors in B[]
int oddCount = 0;
// Even Product Pairs
int evenPairs = 0;
// Traverse the array B[]
for (int i = 0; i < M; i++) {
// Since, product has to be
// positive i.e > 0
if (countDistinct[B[i]] == 0)
continue;
// If count of prime factors is odd
if (countDistinct[B[i]] & 1) {
// Increment oddCount by 1
oddCount++;
}
else {
// Increment evenCount by 1
evenCount++;
}
}
for (int i = 0; i < N; i++) {
// Since, product has to be
// positive i.e > 0
if (countDistinct[A[i]] == 0)
continue;
// If count of prime factors is odd
if (countDistinct[A[i]] & 1) {
// odd * even = even
evenPairs += (evenCount);
}
// If count of prime factors is even
else {
// even * odd = even
// even * even = even
evenPairs += evenCount + oddCount;
}
}
return evenPairs;
}
// Driver Code
int main()
{
int A[] = { 1, 2, 3 };
int B[] = { 4, 5, 6 };
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
cout << CountEvenPair(A, B, N, M);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
static int MAX = 1000000;
// Function to calculate count of
// distinct prime factors of a number
static void countOfPrimefactors(int[] CountDistinct)
{
boolean[] prime = new boolean[MAX + 1];
for (int i = 0; i <= MAX; i++) {
CountDistinct[i] = 0;
prime[i] = true;
}
// Seive of Eratosthenes
for (int i = 2; i <= MAX; i++) {
if (prime[i] == true) {
CountDistinct[i] = 1;
for (int j = i * 2; j <= MAX;
j += i) {
CountDistinct[j]++;
prime[j] = false;
}
}
}
}
// Function to count pairs with even
// product of distinct prime factors
static int CountEvenPair(int A[], int B[], int N, int M)
{
// Stores count of
// distinct prime factors
int[] countDistinct = new int[(MAX + 1)];
countOfPrimefactors(countDistinct);
// Stores the count of numbers
// with even prime factors in B[]
int evenCount = 0;
// Stores the count of numbers
// with odd prime factors in B[]
int oddCount = 0;
// Even Product Pairs
int evenPairs = 0;
// Traverse the array B[]
for (int i = 0; i < M; i++) {
// Since, product has to be
// positive i.e > 0
if (countDistinct[B[i]] == 0)
continue;
// If count of prime factors is odd
if ((countDistinct[B[i]] & 1) != 0) {
// Increment oddCount by 1
oddCount++;
}
else {
// Increment evenCount by 1
evenCount++;
}
}
for (int i = 0; i < N; i++) {
// Since, product has to be
// positive i.e > 0
if (countDistinct[A[i]] == 0)
continue;
// If count of prime factors is odd
if ((countDistinct[A[i]] & 1) != 0) {
// odd * even = even
evenPairs += (evenCount);
}
// If count of prime factors is even
else {
// even * odd = even
// even * even = even
evenPairs += evenCount + oddCount;
}
}
return evenPairs;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 1, 2, 3 };
int B[] = { 4, 5, 6 };
int N = A.length;
int M = B.length;
System.out.println(CountEvenPair(A, B, N, M));
}
}
// This code is contributed by sanjoy_62.Python3
# Python 3 implementation of
# the above appraoch
MAX = 1000000
# Function to calculate count of
# distinct prime factors of a number
def countOfPrimefactors(CountDistinct):
global MAX
prime = [0 for i in range(MAX + 1)]
for i in range(MAX+1):
CountDistinct[i] = 0
prime[i] = True
# Seive of Eratosthenes
for i in range(2,MAX+1,1):
if (prime[i] == True):
CountDistinct[i] = 1
for j in range(i * 2,MAX+1,i):
CountDistinct[j] += 1
prime[j] = False
# Function to count pairs with even
# product of distinct prime factors
def CountEvenPair(A, B, N, M):
global MAX
# Stores count of
# distinct prime factors
countDistinct = [0 for i in range(MAX + 1)]
countOfPrimefactors(countDistinct)
# Stores the count of numbers
# with even prime factors in B[]
evenCount = 0
# Stores the count of numbers
# with odd prime factors in B[]
oddCount = 0
# Even Product Pairs
evenPairs = 0
# Traverse the array B[]
for i in range(M):
# Since, product has to be
# positive i.e > 0
if (countDistinct[B[i]] == 0):
continue
# If count of prime factors is odd
if (countDistinct[B[i]] & 1):
# Increment oddCount by 1
oddCount += 1
else:
# Increment evenCount by 1
evenCount += 1
for i in range(N):
# Since, product has to be
# positive i.e > 0
if (countDistinct[A[i]] == 0):
continue
# If count of prime factors is odd
if (countDistinct[A[i]] & 1):
# odd * even = even
evenPairs += (evenCount)
# If count of prime factors is even
else:
# even * odd = even
# even * even = even
evenPairs += evenCount + oddCount
return evenPairs
# Driver Code
if __name__ == '__main__':
A = [1, 2, 3]
B = [4, 5, 6]
N = len(A)
M = len(B)
print(CountEvenPair(A, B, N, M))
# This code is contributed by SURENDRA_GANGWAR.C#
// C# program for the above approach
using System;
public class GFG
{
static int MAX = 1000000;
// Function to calculate count of
// distinct prime factors of a number
static void countOfPrimefactors(int[] CountDistinct)
{
bool[] prime = new bool[MAX + 1];
for (int i = 0; i <= MAX; i++) {
CountDistinct[i] = 0;
prime[i] = true;
}
// Seive of Eratosthenes
for (int i = 2; i <= MAX; i++) {
if (prime[i] == true) {
CountDistinct[i] = 1;
for (int j = i * 2; j <= MAX;
j += i) {
CountDistinct[j]++;
prime[j] = false;
}
}
}
}
// Function to count pairs with even
// product of distinct prime factors
static int CountEvenPair(int []A, int []B, int N, int M)
{
// Stores count of
// distinct prime factors
int[] countDistinct = new int[(MAX + 1)];
countOfPrimefactors(countDistinct);
// Stores the count of numbers
// with even prime factors in B[]
int evenCount = 0;
// Stores the count of numbers
// with odd prime factors in B[]
int oddCount = 0;
// Even Product Pairs
int evenPairs = 0;
// Traverse the array B[]
for (int i = 0; i < M; i++) {
// Since, product has to be
// positive i.e > 0
if (countDistinct[B[i]] == 0)
continue;
// If count of prime factors is odd
if ((countDistinct[B[i]] & 1) != 0) {
// Increment oddCount by 1
oddCount++;
}
else {
// Increment evenCount by 1
evenCount++;
}
}
for (int i = 0; i < N; i++) {
// Since, product has to be
// positive i.e > 0
if (countDistinct[A[i]] == 0)
continue;
// If count of prime factors is odd
if ((countDistinct[A[i]] & 1) != 0) {
// odd * even = even
evenPairs += (evenCount);
}
// If count of prime factors is even
else {
// even * odd = even
// even * even = even
evenPairs += evenCount + oddCount;
}
}
return evenPairs;
}
// Driver Code
public static void Main(string[] args)
{
int []A = { 1, 2, 3 };
int []B = { 4, 5, 6 };
int N = A.Length;
int M = B.Length;
Console.WriteLine(CountEvenPair(A, B, N, M));
}
}
// This code is contributed by AnkThon输出：
2时间复杂度： O(N * log(N))
辅助空间： O(N)