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 是否可以表示为最后一位为9的整数之和(9, 19, 29, 39 …)，或者不能。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.

朴素的方法：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 以 9 结尾的数字]。

时间复杂度： O(N ² )
辅助空间： O(N)

有效的方法：上述方法可以根据观察进行优化，如果N的最后一位数字是 K ，那么正好需要(10 – K) 个最小数字来形成N 。

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. 使用上述观察，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 个数字是9 。
3. 现在，计算N – 9 * (9 – K)并存储在一个变量中，比如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

Javascript

// java script  program for the above approach
 
// Function to find the minimum count
// of numbers ending with 9 to form N
function minCountOfNumbers(N){
     
    // Extract last digit of N
    let k = N % 10;
 
    // Calculate the last digit
    let 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   
    let N = 156;
    document.write(minCountOfNumbers(N));
 
// This code is contributed by sravan kumar

输出：

4

时间复杂度： O(1)
辅助空间： O(1)