长度为 N 的数的计数，在奇数索引处具有素数，在偶数索引处具有奇数

给定一个数字N ，任务是计算长度为 N的数字的计数，这些数字在奇数索引处具有素数，在偶数索引处具有奇数。

示例：

Input : N = 1
Output: 5
Explanation : All valid numbers length 1 are 1, 3, 5, 7, 9, here we have only 1 odd index, therefore we have 5 valid numbers.

Input: N = 2
Output: 20
Explanation: There are 20 valid numbers of length 2.

编程需要懂一点英语

方法：这个问题可以在组合学的帮助下解决。奇数索引的数字有 4 个选择，偶数索引的数字有 5 个选择。
请按照以下步骤解决问题：

偶数索引有 5 种选择（1、3、5、7、9），奇数索引有 4 种选择（2、3、5、7）。
对于长度为N 的数字，将有N/2 个奇数索引和 ( N/2 + N%2 ) 个偶数索引。
因此，填充 N/2 奇数索引的方法数为4 ^N/2 。
填充偶数索引的方法数量为5 ^{(N/2 + N%2)} 。
因此，所有有效数字的总数将为4 ^N/2 * 5 ^{(N/2 + N%2)} 。

以下是上述方法的实现：

C++

// c++ program to Count of numbers of length
// N having prime numbers at odd indices and
// odd numbers at even indices
#include
using namespace std;
// function to find total number of ways
int find_Numb_ways(int n)
{
    // No of odd indices in n-digit number
    int odd_indices = n/2;
   
    // No of even indices in n-digit number
    int even_indices = (n / 2) + (n % 2);
   
    //  No of ways of arranging prime number
    //  digits in odd indices
    int arr_odd = pow(4, odd_indices);
   
    //   No of ways of arranging odd number
    //  digits in even indices
    int arr_even = pow(5, even_indices);
   
    // returning the total number of ways
    return arr_odd * arr_even;
}
 
// drive code
int main()
{
    int n = 4;
    cout << find_Numb_ways(n) << endl;
    return 0;
}
 
// This code is contributed by kondamrohan02.

Java

// Java program to Count of numbers of length
// N having prime numbers at odd indices and
// odd numbers at even indices
import java.util.*;
 
class GFG
{
 
// function to find total number of ways
static int find_Numb_ways(int n)
{
   
    // No of odd indices in n-digit number
    int odd_indices = n/2;
   
    // No of even indices in n-digit number
    int even_indices = (n / 2) + (n % 2);
   
    //  No of ways of arranging prime number
    //  digits in odd indices
    int arr_odd = (int)Math.pow(4, odd_indices);
   
    //   No of ways of arranging odd number
    //  digits in even indices
    int arr_even = (int)Math.pow(5, even_indices);
   
    // returning the total number of ways
    return arr_odd * arr_even;
}
 
    // Driver Code
    public static void main(String[] args) {
        int n = 4;
     System.out.print(find_Numb_ways(n));
 
    }
}
 
// This code is contributed by code_hunt.

Python3

# python program for above approach
def count(N):
 
    # No of odd indices in N-digit number
    odd_indices = N//2
 
    # No of even indices in N-digit number
    even_indices = N//2 + N % 2
 
    # No of ways of arranging prime number
    # digits in odd indices
    arrange_odd = 4 ** odd_indices
 
    # No of ways of arranging odd number
    # digits in even indices
    arrange_even = 5 ** even_indices
 
    # returning the total number of ways
    return arrange_odd * arrange_even
 
 
# Driver code
if __name__ == "__main__":
 
    N = 4
    # calling the function
    print(count(N))

C#

// C# program to Count of numbers of length
// N having prime numbers at odd indices and
// odd numbers at even indices
using System;
using System.Collections.Generic;
 
class GFG{
 
// function to find total number of ways
static int find_Numb_ways(int n)
{
    // No of odd indices in n-digit number
    int odd_indices = n/2;
   
    // No of even indices in n-digit number
    int even_indices = (n / 2) + (n % 2);
   
    //  No of ways of arranging prime number
    //  digits in odd indices
    int arr_odd = (int)Math.Pow(4, odd_indices);
   
    //   No of ways of arranging odd number
    //  digits in even indices
    int arr_even = (int)Math.Pow(5, even_indices);
   
    // returning the total number of ways
    return arr_odd * arr_even;
}
 
// drive code
public static void Main()
{
    int n = 4;
    Console.Write(find_Numb_ways(n));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

输出

时间复杂度：O(1)，（恒定时间操作）
辅助空间：O(1)，（不需要额外空间）