计算用给定数组的偶积构造数组的方法，以使相同索引元素的绝对差最大为1

给定大小为N的数组A [] ，任务是计算构造大小为N的数组B []的方式数，以使相同索引元素的绝对差必须小于或等于1 ，即abs(A [i] – B [i]) ≤1 ，并且数组B []的元素乘积必须为偶数。

例子：

Input: A[] = { 2, 3 } 
Output: 7 
Explanation: 
Possible values of the array B[] are { { 1, 2 }, { 1, 4 }, { 2, 2 }, { 2, 4 }, { 3, 2 }, { 3, 4 } } 
Therefore, the required output is 7.

Input: A[] = { 90, 52, 56, 71, 44, 8, 13, 30, 57, 84 } 
Output: 58921 
 

方法：想法是首先计算构造数组B []的方法数，以使abs(A [i] – B [i])<= 1 ，然后删除那些元素乘积不为偶数的数组数字。请按照以下步骤解决问题：

• 使得abs(A [i] – B [i])<= 1的B [i]可能值为{A [i]，A [i] + 1，A [i] – 1} 。因此，构造abs(A [i] – B [i])小于或等于1的数组B []的总数为3 ^N。
• 遍历数组并将偶数计数存储在数组A []中，例如X。
• 如果A [i]是偶数，则(A [i] – 1)和(A [i] +1)必须是奇数。因此，乘积不是偶数的构造数组B []的总数为2 ^X。
• 最后，打印( 3 ^N – 2 ^X )的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find count the ways to construct
// an array, B[] such that abs(A[i] - B[i]) <=1
// and product of elements of B[] is even
void cntWaysConsArray(int A[], int N)
{
 
    // Stores count of arrays B[] such
    // that abs(A[i] - B[i]) <=1
    int total = 1;
 
    // Stores count of arrays B[] whose
    // product of elements is not even
    int oddArray = 1;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Update total
        total = total * 3;
 
        // If A[i] is an even number
        if (A[i] % 2 == 0) {
 
            // Update oddArray
            oddArray *= 2;
        }
    }
 
    // Print 3^N - 2^X
    cout << total - oddArray << "\n";
}
 
// Driver Code
int main()
{
    int A[] = { 2, 4 };
    int N = sizeof(A) / sizeof(A[0]);
 
    cntWaysConsArray(A, N);
 
    return 0;
}

Java

// Java Program to implement the
// above approach
import java.util.*;
class GFG
{
  
// Function to find count the ways to construct
// an array, B[] such that abs(A[i] - B[i]) <=1
// and product of elements of B[] is even
static void cntWaysConsArray(int A[], int N)
{
 
    // Stores count of arrays B[] such
    // that abs(A[i] - B[i]) <=1
    int total = 1;
 
    // Stores count of arrays B[] whose
    // product of elements is not even
    int oddArray = 1;
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
 
        // Update total
        total = total * 3;
 
        // If A[i] is an even number
        if (A[i] % 2 == 0)
        {
 
            // Update oddArray
            oddArray *= 2;
        }
    }
 
    // Print 3^N - 2^X
    System.out.println( total - oddArray);
}
  
// Driver Code
public static void main(String[] args)
{
    int A[] = { 2, 4 };
    int N = A.length;
    cntWaysConsArray(A, N);
}
}
 
// This code is contributed by code_hunt.

Python3

# Python3 program to implement
# the above approach
 
# Function to find count the ways to construct
# an array, B[] such that abs(A[i] - B[i]) <=1
# and product of elements of B[] is even
def cntWaysConsArray(A, N) :
 
    # Stores count of arrays B[] such
    # that abs(A[i] - B[i]) <=1
    total = 1;
 
    # Stores count of arrays B[] whose
    # product of elements is not even
    oddArray = 1;
 
    # Traverse the array
    for i in range(N) :
 
        # Update total
        total = total * 3;
 
        # If A[i] is an even number
        if (A[i] % 2 == 0) :
 
            # Update oddArray
            oddArray *= 2;
   
    # Print 3^N - 2^X
    print(total - oddArray);
 
# Driver Code
if __name__ == "__main__" :
    A = [ 2, 4 ];
    N = len(A);
    cntWaysConsArray(A, N);
     
    # This code is contributed by AnkThon

C#

// C# program to implement the
// above approach
using System;
 
class GFG{
  
// Function to find count the ways to construct
// an array, []B such that abs(A[i] - B[i]) <=1
// and product of elements of []B is even
static void cntWaysConsArray(int []A, int N)
{
     
    // Stores count of arrays []B such
    // that abs(A[i] - B[i]) <=1
    int total = 1;
 
    // Stores count of arrays []B whose
    // product of elements is not even
    int oddArray = 1;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Update total
        total = total * 3;
 
        // If A[i] is an even number
        if (A[i] % 2 == 0)
        {
             
            // Update oddArray
            oddArray *= 2;
        }
    }
 
    // Print 3^N - 2^X
    Console.WriteLine(total - oddArray);
}
  
// Driver Code
public static void Main(String[] args)
{
    int []A = { 2, 4 };
    int N = A.Length;
     
    cntWaysConsArray(A, N);
}
}
 
// This code is contributed by shikhasingrajput

输出：

5

时间复杂度： O(N)
辅助空间： O(1)