查找N的最大N位数倍数

给定数字N ，任务是找到N的最大N位数倍数。
例子：

Input: N = 2 
Output: 98 
Explanation: 
98 is the largest multiple of 2 and is of 2 digits.
Input: N = 3 
Output: 999 
Explanation: 
999 is the largest multiple of 3 and is of 3 digits. 
 

方法：想法是观察。

• 如果我们仔细观察，将会形成一个序列，分别是9、98、999、9996、99995，…
• 在以上系列中，第N个项可以计算为：

$N*\lfloor \frac{10^N-1}{N} \rfloor$
 

• 因此，将数字N作为输入并实现以上公式。

下面是上述方法的实现：

C++

// C++ program to find largest multiple
// of N containing N digits
#include 
#include 
using namespace std;
 
// Function to find the largest
// N digit multiple of N
 
void smallestNumber(int N)
{
    cout << N * floor((pow(10, N) - 1) / N);
}
 
// Driver code
int main()
{
    int N = 2;
    smallestNumber(N);
 
    return 0;
}

Java

// Java program to find largest multiple
// of N containing N digits
import java.util.*;
class GFG{
 
// Function to find the largest
// N digit multiple of N
static void smallestNumber(int N)
{
    System.out.print(N * Math.floor((
                         Math.pow(10, N) - 1) / N));
}
 
// Driver code
public static void main(String args[])
{
    int N = 2;
    smallestNumber(N);
}
}
 
// This code is contributed by Nidhi_biet

Python3

# Python3 program to find largest multiple
# of N containing N digits
from math import floor
 
# Function to find the largest
# N digit multiple of N
def smallestNumber(N):
    print(N * floor((pow(10, N) - 1) / N))
 
# Driver code
if __name__ == '__main__':
    N = 2
    smallestNumber(N)
 
# This code is contributed by Mohit Kumar

C#

// C# program to find largest multiple
// of N containing N digits
using System;
class GFG{
 
// Function to find the largest
// N digit multiple of N
static void smallestNumber(int N)
{
    Console.Write(N * Math.Floor((
                      Math.Pow(10, N) - 1) / N));
}
 
// Driver code
public static void Main()
{
    int N = 2;
    smallestNumber(N);
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// javascript program to find largest multiple
// of N containing N digits
 
 
// Function to find the largest
// N digit multiple of N
 
function smallestNumber( N)
{
    document.write( N * Math.floor((Math.pow(10, N) - 1) / N));
}
 
// Driver code
let N = 2;
    smallestNumber(N);
 
// This code is contributed by todaysgaurav
 

输出：

98