计算 N 位数字，使得每个位置都可以被该位置的数字整除

给定一个正整数N ，任务是计算N位数字的数量，使得数字中的每个索引(基于 1 的索引)都可以被出现在该索引处的数字整除。由于法院可以非常大，打印它模10 ⁹ + 7 。

例子：

Input: N = 2
Output: 2
Explanation: The numbers 11 and 12 are the only 2-digit numbers satisfying the condition.

Input: N = 5
Output: 24

朴素方法：解决问题的最简单方法是生成所有可能的N位数字，并对每个这样的数字，检查其所有数字是否满足要求的条件。

时间复杂度： O(10 ^N *N)
辅助空间： O(1)

高效的方法：优化上述方法的思路是观察这样一个事实，即对于每个位置，从 1 到 9 有 9 个可能的选项。检查每个可能的选项并找到结果。
请按照以下步骤解决此问题：

• 将变量ans初始化为1以存储答案。
• 使用变量索引迭代范围[1, N]并执行以下任务：
  • 初始化变量，比如说选择为0，存储的该特定指数期权的数量。
  • 使用变量数字迭代范围[1, 9]并执行以下任务：
    • 如果index%digit等于0，则将选项的值增加1。
  • 将ans的值设置为(ans*1LL*choices)%mod 。
• 执行完上述步骤后，打印ans的值作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
const int mod = 1e9 + 7;
 
// Function to count N-digit numbers
// such that each position is divisible
// by the digit occurring at that position
void countOfNumbers(int N)
{
    // Stores the answer.
    int ans = 1;
 
    // Iterate from indices 1 to N
    for (int index = 1; index <= N; ++index) {
 
        // Stores count of digits that can
        // be placed at the current index
        int choices = 0;
 
        // Iterate from digit 1 to 9
        for (int digit = 1; digit <= 9; ++digit) {
 
            // If index is divisible by digit
            if (index % digit == 0) {
                ++choices;
            }
        }
 
        // Multiply answer with possible choices
        ans = (ans * 1LL * choices) % mod;
    }
 
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 5;
 
    // Function call
    countOfNumbers(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.lang.*;
import java.util.*;
  
class GFG{
 
static int mod = 1000000007;
   
// Function to count N-digit numbers
// such that each position is divisible
// by the digit occurring at that position
static void countOfNumbers(int N)
{
     
    // Stores the answer.
    int ans = 1;
 
    // Iterate from indices 1 to N
    for(int index = 1; index <= N; ++index)
    {
         
        // Stores count of digits that can
        // be placed at the current index
        int choices = 0;
 
        // Iterate from digit 1 to 9
        for(int digit = 1; digit <= 9; ++digit)
        {
             
            // If index is divisible by digit
            if (index % digit == 0)
            {
                ++choices;
            }
        }
 
        // Multiply answer with possible choices
        ans = (ans * choices) % mod;
    }
    System.out.println(ans);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int N = 5;
 
    // Function call
    countOfNumbers(N);
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# python program for the above approach
mod = 1000000000 + 7
 
# Function to count N-digit numbers
# such that each position is divisible
# by the digit occurring at that position
def countOfNumbers(N):
   
    # Stores the answer.
    ans = 1
     
    # Iterate from indices 1 to N
    for index in range(1, N + 1):
       
        # Stores count of digits that can
        # be placed at the current index
        choices = 0
         
        # Iterate from digit 1 to 9
        for digit in range(1, 10):
           
            # If index is divisible by digit
            if (index % digit == 0):
                choices += 1
 
        # Multiply answer with possible choices
        ans = (ans * choices) % mod
    print(ans)
 
# Driver Code
 
# Given Input
N = 5
 
# Function call
countOfNumbers(N)
 
# This code is contributed by amreshkumar3.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
  static int mod = 1000000007;
   
// Function to count N-digit numbers
// such that each position is divisible
// by the digit occurring at that position
static void countOfNumbers(int N)
{
    // Stores the answer.
    int ans = 1;
 
    // Iterate from indices 1 to N
    for (int index = 1; index <= N; ++index) {
 
        // Stores count of digits that can
        // be placed at the current index
        int choices = 0;
 
        // Iterate from digit 1 to 9
        for (int digit = 1; digit <= 9; ++digit) {
 
            // If index is divisible by digit
            if (index % digit == 0) {
                ++choices;
            }
        }
 
        // Multiply answer with possible choices
        ans = (ans * choices) % mod;
    }
 
    Console.Write(ans);
}
 
// Driver Code
public static void Main()
{
    // Given Input
    int N = 5;
 
    // Function call
    countOfNumbers(N);
}
}
 
// This code is contributed by bgangwar59.

Javascript

 
        // JavaScript program for the above approach
        const mod = 1e9 + 7;
 
        // Function to count N-digit numbers
        // such that each position is divisible
        // by the digit occurring at that position
        function countOfNumbers(N)
        {
         
            // Stores the answer.
            let ans = 1;
 
            // Iterate from indices 1 to N
            for (let index = 1; index <= N; ++index) {
 
                // Stores count of digits that can
                // be placed at the current index
                let choices = 0;
 
                // Iterate from digit 1 to 9
                for (let digit = 1; digit <= 9; ++digit) {
 
                    // If index is divisible by digit
                    if (index % digit == 0) {
                        ++choices;
                    }
                }
 
                // Multiply answer with possible choices
                ans = (ans * 1 * choices) % mod;
            }
 
            document.write(ans);
        }
 
        // Driver Code
 
        // Given Input
        let N = 5;
 
        // Function call
        countOfNumbers(N);
 
    // This code is contributed by Potta Lokesh
    

输出：

24

时间复杂度： O(10 * N)
辅助空间： O(1)