将数字表示为以9结尾的正数之和

给定一个整数N ，任务是检查N是否可以表示为以9作为最后一位数字(9、19、29、39…)的整数之和。如果发现为真，则找到获得N所需的此类整数的最小计数。否则打印-1 。

例子：

Input: N = 156
Output: 4
Explanation:
156 = 9 + 9 + 9 + 129

Input: N = 60
Output: -1
Explanation:
No possible way to obtain sum 60 from numbers having 9 as the last digit.

天真的方法：这个问题可以看作是硬币找零问题的一种变体。对于这个问题，硬币可以用[9，19，29，39…。代替。直到最后一个小于N的数字(以9结尾)。

时间复杂度： O(N ² )
辅助空间： O(N)

高效的方法：可以基于以下观察条件来优化上述方法：如果N的最后一位是K ，则形成N至少需要(10-K)个最小数。

A sum N can be obtained by adding 10 – K numbers, where K is the last digit of N. 
Therefore, sum N can be obtained by adding 9, (9 – K) times and adding N – (9 * (9 – K)) finally.

请按照以下步骤解决问题：

1. 提取给定数字的最后一位， K = N％10
2. 根据上述观察，总共需要(10 – K)个数字。现在，计算9 *(9-K) ，因为获得N所需的前9-K个数是9 。
3. 现在，计算N – 9 *(9 – K)并存储在变量中，例如z 。如果z大于或等于9，并有9个为最后一个数字，打印10 – K的答案。否则，打印-1 。

下面是上述方法的实现：

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
// Function to find the minimum count
// of numbers ending with 9 to form N
int minCountOfNumbers(int N)
{
    // Extract last digit of N
    int k = N % 10;
  
    // Calculate the last digit
    int z = N - (9 * (9 - k));
  
    // If the last digit
    // satisfies the condition
    if (z >= 9 && z % 10 == 9) {
        return 10 - k;
    }
    else
        return -1;
}
  
// Driver Code
int main()
{
    int N = 156;
    cout << minCountOfNumbers(N);
  
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to find the minimum count
// of numbers ending with 9 to form N
static int minCountOfNumbers(int N)
{
      
    // Extract last digit of N
    int k = N % 10;
  
    // Calculate the last digit
    int z = N - (9 * (9 - k));
  
    // If the last digit
    // satisfies the condition
    if (z >= 9 && z % 10 == 9)
    {
        return 10 - k;
    }
    else
        return -1;
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 156;
    System.out.print(minCountOfNumbers(N));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
  
# Function to find the minimum count
# of numbers ending with 9 to form N
def minCountOfNumbers(N):
      
    # Extract last digit of N
    k = N % 10
  
    # Calculate the last digit
    z = N - (9 * (9 - k))
  
    # If the last digit
    # satisfies the condition
    if (z >= 9 and z % 10 == 9):
        return 10 - k
    else:
        return -1
  
# Driver Code
if __name__ == '__main__':
      
    N = 156
      
    print(minCountOfNumbers(N))
  
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the minimum count
// of numbers ending with 9 to form N
static int minCountOfNumbers(int N)
{
      
    // Extract last digit of N
    int k = N % 10;
  
    // Calculate the last digit
    int z = N - (9 * (9 - k));
  
    // If the last digit
    // satisfies the condition
    if (z >= 9 && z % 10 == 9)
    {
        return 10 - k;
    }
    else
        return -1;
}
  
// Driver Code
public static void Main(String[] args)
{
    int N = 156;
      
    Console.Write(minCountOfNumbers(N));
}
}
  
// This code is contributed by 29AjayKumar

输出：

4

时间复杂度： O(1)
辅助空间： O(1)