获得给定总和所需的最少位数

给定一个整数N ，任务是找到生成一个数字总和等于N所需的最小位数。

例子：

Input: N = 18 
Output: 2 
Explanation: 
The number with smallest number of digits having sum of digits equal to 18 is 99.

Input: N = 28 
Output: 4 
Explanation: 
4-digit numbers like 8884, 6877, etc are the smallest in length having sum of digits equal to 28.

方法：该问题可以通过以下观察来解决：

1. 将计数增加9 。因此，现在计数等于最短数字中的 9 的数量。减少N到N % 9
2. 现在，如果N超过 0，则将count增加1 。
3. 最后，打印计数作为答案。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to return the
// minimum count of digits
void mindigits(int n)
{
    // IF N is divisible by 9
    if (n % 9 == 0) {
 
        // Count of 9's is the answer
        cout << n / 9 << endl;
    }
    else {
 
        // If remainder is non-zero
        cout << (n / 9) + 1 << endl;
    }
}
 
// Driver Code
int main()
{
    int n1 = 24;
    int n2 = 14;
    mindigits(n1);
    mindigits(n2);
}

Java

// Java program to implement
// the above approach
// required to make the given sum
import java.util.*;
 
class Main {
 
    // Function to print the minimum
    // count of digits
    static void mindigits(int n)
    {
 
        // IF N is divisible by 9
        if (n % 9 == 0) {
 
            // Count of 9's is the answer
            System.out.println(n / 9);
        }
        else {
 
            // If remainder is non-zero
            System.out.println((n / 9) + 1);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n1 = 24;
        int n2 = 18;
        mindigits(n1);
        mindigits(n2);
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function to print the minimum
# count of digits
def mindigits(n):
     
    # IF N is divisible by 9
    if (n % 9 == 0):
 
        # Count of 9's is the answer
        print(n // 9);
    else:
 
        # If remainder is non-zero
        print((n // 9) + 1);
 
# Driver Code
if __name__ == '__main__':
     
    n1 = 24;
    n2 = 18;
     
    mindigits(n1);
    mindigits(n2);
 
# This code is contributed by amal kumar choubey

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to print the minimum
// count of digits
static void mindigits(int n)
{
     
    // IF N is divisible by 9
    if (n % 9 == 0)
    {
         
        // Count of 9's is the answer
        Console.WriteLine(n / 9);
    }
    else
    {
         
        // If remainder is non-zero
        Console.WriteLine((n / 9) + 1);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int n1 = 24;
    int n2 = 18;
     
    mindigits(n1);
    mindigits(n2);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript program to implement
// the above approach
// required to make the given sum
 
    // Function to print the minimum
    // count of digits
    function mindigits(n)
    {
  
        // IF N is divisible by 9
        if (n % 9 == 0) {
  
            // Count of 9's is the answer
            document.write(Math.floor(n / 9) + "<br/>");
        }
        else {
  
            // If remainder is non-zero
            document.write(Math.floor(n / 9) + 1 + "<br/>");
        }
    }
 
// Driver Code
 
        let n1 = 24;
        let n2 = 18;
        mindigits(n1);
        mindigits(n2);
            

输出：

3
2

时间复杂度： O(1)
辅助空间： O(1)