给定两个由N个整数组成的数组A []和B [] ，任务是对具有偶数和奇数乘积的对(A [i]，B [j])进行计数，并打印两个计数中的最大值。
例子：
Input: A[] = {1, 2, 3}, B[] = {4, 5, 6}
Output: 7
Explanation:
Pairs with odd product count are: (1, 5) (3, 5). Therefore, number of pairs with odd product = 2.
Pairs with even product count are: (1, 4) (1, 6) (2, 4) (2, 5) (2, 6) (3, 4) (3, 6). Therefore, number of pairs with even product = 7
Therefore, the maximum of the two counts is 7. Hence, the output 7.
Input: A[] = {1, 5, 9}, B[] = {1, 7, 3}
Output: 9
Explanation:
Pairs with odd product count are: (1, 1) (1, 7) (1, 3) (5, 1) (5, 7) (5, 3) (9, 1) (9, 7) (9, 3). Therefore, number of pairs with odd product = 9
No pair with even product count exists.
Therefore, the output 9.
天真的方法：最简单的方法是生成所有可能的对，并检查它们的乘积是偶数还是奇数，并相应地增加其各自的计数。检查所有对之后，打印这两个计数中的最大值。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效的解决方案：可以通过观察仅乘以一对奇数整数得出奇数来优化上述方法。否则，将生成一个偶数整数。请按照以下步骤解决问题：
- 初始化变量oddcount ， evencount ， 和oddCountB至0存储奇数产物对，甚至产品对的计数，和奇数号的在阵列B上分别[]计数。
- 遍历数组B [] ，如果元素为奇数，则递增oddCountB 。现在， (N –奇数计数B)的值给出了数组B []中偶数个元素的数量。
- 遍历数组A []并执行以下操作：
- 如果当前元素为偶数，则将N增加偶数计数，因为将B []中的任何数字相乘将得出偶数乘积对。
- 否则，递增evencount通过–因为在B中的偶数(N oddCountB)[]当与A []奇数相乘，得到(N – oddCountB)对和由oddCountB增量oddcount，因为在B []乘以奇数在A []中带有奇数的数字将给出奇数乘积对的计数。
- 完成上述步骤后，打印出奇数和偶数计数的最大值作为结果。
下面是上述方法的实现：
C++
// C++ program for above approach
#include
using namespace std;
// Function to return the maximum
// count of odd count pair or
// even count pair
int maxcountpair(int A[], int B[], int N)
{
// Stores odd count, even count
// pairs and odd numbers in B[]
int oddcount = 0, evencount = 0;
int oddCountB = 0;
// Traverse the array B[]
for (int i = 0; i < N; i++) {
// If B[i] is an odd number
if (B[i] & 1)
// Increment oddCountB by 1
oddCountB += 1;
}
// Traverse the array A[]
for (int i = 0; i < N; i++) {
// If A[i] is an odd number
if (A[i] & 1) {
oddcount += oddCountB;
evencount += (N - oddCountB);
}
// If A[i] is even
else
evencount += N;
}
// Return maximum count of odd
// and even pairs
return max(oddcount, evencount);
}
// Driver Code
int main()
{
int A[] = { 1, 2, 3 };
int B[] = { 4, 5, 6 };
int N = sizeof(A) / sizeof(A[0]);
// Function Call
cout << maxcountpair(A, B, N);
return 0;
} C
// C program for above approach
#include
// Function to return the
// maximum of two numbers
int max(int num1, int num2)
{
return (num1 > num2) ? num1 : num2;
}
// Function to return the maximum count of
// odd count pair or even count pair
int maxcountpair(int A[], int B[], int N)
{
// Stores odd count, even count
// pairs and odd numbers in B[]
int oddcount = 0, evencount = 0;
int oddCountB = 0;
// Traverse the array B[]
for(int i = 0; i < N; i++)
{
// If B[i] is an odd number
if (B[i] & 1)
// Increment oddCountB by 1
oddCountB += 1;
}
// Traverse the array A[]
for(int i = 0; i < N; i++)
{
// If A[i] is an odd number
if (A[i] & 1)
{
oddcount += oddCountB;
evencount += (N - oddCountB);
}
// If A[i] is even
else
evencount += N;
}
// Return maximum count of odd
// and even pairs
return max(oddcount, evencount);
}
// Driver Code
int main()
{
int A[] = { 1, 2, 3 };
int B[] = { 4, 5, 6 };
int N = sizeof(A) / sizeof(A[0]);
printf("%d", maxcountpair(A, B, N));
return 0;
}
// This code is contributed by sourav singh Java
// Java program for above approach
import java.io.*;
class GFG{
// Function to return the maximum count of
// odd count pair or even count pair
static int maxcountpair(int A[], int B[],
int N)
{
// Stores odd count, even count
// pairs and odd numbers in B[]
int oddcount = 0, evencount = 0;
int oddCountB = 0;
// Traverse the array B[]
for(int i = 0; i < N; i++)
{
// If B[i] is an odd number
if (B[i] % 2 == 1)
// Increment oddCountB by 1
oddCountB += 1;
}
// Traverse the array A[]
for(int i = 0; i < N; i++)
{
// If A[i] is an odd number
if (A[i] % 2 == 1)
{
oddcount += oddCountB;
evencount += (N - oddCountB);
}
// If A[i] is even
else
evencount += N;
}
// Return maximum count of odd
// and even pairs
return (oddcount > evencount) ? oddcount
: evencount;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 1, 2, 3 };
int B[] = { 4, 5, 6 };
int N = A.length;
System.out.print(maxcountpair(A, B, N));
}
}
// This code is contributed by sourav singhPython3
# Python3 program for above approach
# Function to return the maximum count of
# odd count pair or even count pair
def maxcountpair(A, B, N):
# Function to return the maximum count of
# odd count pair or even count pair
oddcount = 0
evencount = 0
oddCountB = 0
# Traverse the array, B[]
for i in range(0, N):
# If B[i] is an odd number
if B[i] % 2 == 1:
oddCountB += 1
# Traverse the array, A[]
for i in range(0, N):
# If A[i] is an odd number
if A[i] % 2 == 1:
oddcount += oddCountB
evencount += (N - oddCountB)
else:
evencount += N
# Return maximum count of odd
# and even pairs
return max(oddcount, evencount)
# Driver Code
A = [ 1, 2, 3 ]
B = [ 4, 5, 6 ]
N = len(A)
print(maxcountpair(A, B, N))
# This code is contributed by sourav singhC#
// C# program for above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to return the maximum count of
// odd count pair or even count pair
static int maxcountpair(int[] A, int[] B,
int N)
{
// Stores odd count, even count
// pairs and odd numbers in B[]
int oddcount = 0, evencount = 0;
int oddCountB = 0;
// Traverse the array B[]
for(int i = 0; i < N; i++)
{
// If B[i] is an odd number
if (B[i] % 2 == 1)
// Increment oddCountB by 1
oddCountB += 1;
}
// Traverse the array A[]
for(int i = 0; i < N; i++)
{
// If A[i] is an odd number
if (A[i] % 2 == 1)
{
oddcount += oddCountB;
evencount += (N - oddCountB);
}
// If A[i] is even
else
evencount += N;
}
// Return maximum count of odd
// and even pairs
return (oddcount > evencount) ? oddcount
: evencount;
}
// Driver Code
public static void Main(String[] args)
{
int[] A = { 1, 2, 3 };
int[] B = { 4, 5, 6 };
int N = A.Length;
// Function call
Console.Write(maxcountpair(A, B, N));
}
}
// This code is contributed by sourav singhJavascript
输出：
7时间复杂度： O(N)
辅助空间： O(1)