最小的N位数字，是5的倍数

给定整数N≥1 ，任务是找到最小的N位数字，该数字是5的倍数。
例子：

Input: N = 1 
Output: 5
Input: N = 2 
Output: 10 
Input: N = 3 
Output: 100 
 

方法：

• 如果N = 1，则答案为5 。
• 如果N> 1，那么答案将是(10 ^{(N – 1)} )，因为最小的5的倍数的序列将继续为10、100、1000、10000、100000等。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the smallest n digit
// number which is a multiple of 5
int smallestMultiple(int n)
{
    if (n == 1)
        return 5;
    return pow(10, n - 1);
}
 
// Driver code
int main()
{
    int n = 4;
    cout << smallestMultiple(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the smallest n digit
    // number which is a multiple of 5
    static int smallestMultiple(int n)
    {
        if (n == 1)
            return 5;
        return (int)(Math.pow(10, n - 1));
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 4;
        System.out.println(smallestMultiple(n));
    }
}

Python3

# Python3 implementation of the approach
 
# Function to return the smallest n digit
# number which is a multiple of 5
def smallestMultiple(n):
 
    if (n == 1):
        return 5
    return pow(10, n - 1)
 
# Driver code
n = 4
print(smallestMultiple(n))

C#

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the smallest n digit
    // number which is a multiple of 5
    static int smallestMultiple(int n)
    {
        if (n == 1)
            return 5;
        return (int)(Math.Pow(10, n - 1));
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4;
        Console.Write(smallestMultiple(n));
    }
}

Javascript

// Javascript implementation of the approach
 
// Function to return the smallest n digit
// number which is a multiple of 5
function smallestMultiple(n)
{
    if (n == 1)
        return 5;
    return Math.pow(10, n - 1);
}
 
// Driver code
var n = 4;
document.write(smallestMultiple(n));
 
// This code is contributed by noob2000.

输出：

1000

时间复杂度： O(1)