分别在奇数和偶数位置计算由偶数和质数组成的N位数字

给定一个正整数N ，任务是找到在奇数索引处具有偶数位且在偶数索引处具有质数位的N 位整数的数量。

例子：

Input: N = 2
Output: 20
Explanation:
Following are the possible number of 2-digits satisfying the given criteria {20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 50, 52, 54, 56, 58, 70, 72, 74, 76, 78}. Therefore, the count of such number is 20.

Input: N = 5
Output: 1600

方法：可以使用排列和组合的概念来解决给定的问题，通过观察以下事实：偶数位置只有4 个选择[2, 3, 5, 7] ，奇数位置只有5 个选择[0, 2, 4, 6, 8] 。因此，满足给定标准的 N 位数字的计数由下式给出：

total count = 4^P5^Q, where P and Q is the number of even and odd positions respectively.

下面是上述方法的实现：

C++

// C++ program for the above approache
#include
using namespace std;
 
int m = 1000000007;
 
// Function to find the value of x ^ y
int power(int x, int y)
{
     
    // Stores the value of x ^ y
    int res = 1;
 
    // Iterate until y is positive
    while (y > 0)
    {
         
        // If y is odd
        if ((y & 1) != 0)
            res = (res * x) % m;
 
        // Divide y by 2
        y = y >> 1;
 
        x = (x * x) % m;
    }
     
    // Return the value of x ^ y
    return res;
}
 
// Function to find the number of N-digit
// integers satisfying the given criteria
int countNDigitNumber(int N)
{
     
    // Count of even positions
    int ne = N / 2 + N % 2;
 
    // Count of odd positions
    int no = floor(N / 2);
 
    // Return the resultant count
    return power(4, ne) * power(5, no);
}
 
// Driver Code
int main()
{
    int N = 5;
    cout << countNDigitNumber(N) % m << endl;
}
 
// This code is contributed by SURENDRA_GANGWAR

Java

// Java program for the above approach
import java.io.*;
class GFG {
 
static int m = 1000000007;
 
// Function to find the value of x ^ y
static int power(int x, int y)
{
     
    // Stores the value of x ^ y
    int res = 1;
 
    // Iterate until y is positive
    while (y > 0)
    {
         
        // If y is odd
        if ((y & 1) != 0)
            res = (res * x) % m;
 
        // Divide y by 2
        y = y >> 1;
 
        x = (x * x) % m;
    }
     
    // Return the value of x ^ y
    return res;
}
 
// Function to find the number of N-digit
// integers satisfying the given criteria
static int countNDigitNumber(int N)
{
     
    // Count of even positions
    int ne = N / 2 + N % 2;
 
    // Count of odd positions
    int no = (int)Math.floor(N / 2);
 
    // Return the resultant count
    return power(4, ne) * power(5, no);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 5;
    System.out.println(countNDigitNumber(N) % m);
}
}
 
// This code is contributed by sanjoy_62.

Python3

# Python program for the above approach
 
import math
m = 10**9 + 7
 
# Function to find the value of x ^ y
def power(x, y):
   
    # Stores the value of x ^ y
    res = 1
     
    # Iterate until y is positive
    while y > 0:
       
        # If y is odd
        if (y & 1) != 0:
            res = (res * x) % m
             
        # Divide y by 2
        y = y >> 1
         
        x = (x * x) % m
         
    # Return the value of x ^ y
    return res
 
# Function to find the number of N-digit
# integers satisfying the given criteria
def countNDigitNumber(n: int) -> None:
     
    # Count of even positions
    ne = N // 2 + N % 2
     
    # Count of odd positions
    no = N // 2
     
    # Return the resultant count
    return power(4, ne) * power(5, no)
 
# Driver Code
if __name__ == '__main__':
       
    N = 5
    print(countNDigitNumber(N) % m)

C#

// C# program for the above approach
using System;
 
class GFG{
 
static int m = 1000000007;
 
// Function to find the value of x ^ y
static int power(int x, int y)
{
     
    // Stores the value of x ^ y
    int res = 1;
 
    // Iterate until y is positive
    while (y > 0)
    {
         
        // If y is odd
        if ((y & 1) != 0)
            res = (res * x) % m;
 
        // Divide y by 2
        y = y >> 1;
 
        x = (x * x) % m;
    }
     
    // Return the value of x ^ y
    return res;
}
 
// Function to find the number of N-digit
// integers satisfying the given criteria
static int countNDigitNumber(int N)
{
     
    // Count of even positions
    int ne = N / 2 + N % 2;
 
    // Count of odd positions
    int no = (int)Math.Floor((double)N / 2);
 
    // Return the resultant count
    return power(4, ne) * power(5, no);
}
 
 
// Driver Code
public static void Main()
{
    int N = 5;
    Console.Write(countNDigitNumber(N) % m);
}
}
 
// This code is contributed by splevel62.

Javascript

 
        // JavaScript program for the above approache
 
 
        var m = 10 ** 9 + 7
 
        // Function to find the value of x ^ y
        function power(x, y) {
 
            // Stores the value of x ^ y
            var res = 1
 
            // Iterate until y is positive
            while (y > 0) {
 
                // If y is odd
                if ((y & 1) != 0)
                    res = (res * x) % m
 
                // Divide y by 2
                y = y >> 1
 
                x = (x * x) % m
            }
            // Return the value of x ^ y
            return res
        }
        // Function to find the number of N-digit
        // integers satisfying the given criteria
        function countNDigitNumber(N) {
 
            // Count of even positions
            var ne = Math.floor(N / 2) + N % 2
 
            // Count of odd positions
            var no = Math.floor(N / 2)
 
            // Return the resultant count
            return power(4, ne) * power(5, no)
        }
        // Driver Code
 
 
        let N = 5
        document.write(countNDigitNumber(N) % m);
 
 
// This code is contributed by Potta Lokesh
    

输出：

1600

时间复杂度： O(log N)
辅助空间： O(1)