为 X 选择的值的计数，以便在给定操作后将 N 减少到 0

给定一个正整数N ，任务是找到整数X的数量，以便在执行以下操作序列后将N的值减少到0 。

• 从m N 中减去 X的值。
• 如果X是个位数的整数，然后停止施加动作。
• 将 X的值更新为X 的数字总和。

例子：

Input: N = 9940
Output: 1
Explanation:
Considering the value of  X  as 9910, the value of N can be reduced to 0 as:

1. Operation 1: Update N as N – X, N = 9939 – 9910 = 30, Update X as sum of digit of X, X = 9 + 9 + 1 + 0 = 19.
2. Operation 2:  Update N as N – X, N = 30 – 19 = 11, Update X as sum of digit of X, X = 9 + 1 = 10.
3. Operation 3: Update N as N – X, N = 11 – 10 = 1, Update X as sum of digit of X, X = 1 + 0 = 1.
4. Operation 4:  Update N as N – X, N = 1 – 1 = 0, Stop applying operations.

Therefore, there exists only one value of X as 1.

Input: N = 9939
Output: 3

方法：可以使用基于以下观察的贪心方法来解决给定的问题：

• 可以观察到，选择大于N的 X值不会将N减少到0 。
• 可以肯定地说，如果N 中的位数是K ，那么 X 的任何值都不会小于(N – 10 * K * (K + 1) / 2)并且可以将N转换为0 。

从上面的观察来看，想法是在范围[N – 10 * K * (K + 1) / 2, N]上进行迭代，并计算该范围内在执行给定操作后可以将 N转换为0 的那些值。

下面是上述方法的实现：

Python3

# Python program for the above approach
  
# Function to check if the value of X
# reduces N to 0 or not
def check(x, N):
  
    while True:
  
        # Update the value of N as N-x
        N -= x
  
        # Check if x is a single digit integer
        if len(str(x)) == 1:
            break
  
        # Update x as sum digit of x
        x = sum(list(map(int, str(x))))
  
    if len(str(x)) == 1 and N == 0:
  
        return 1
  
    return 0
  
# Fuction to find the number of values
# X such that N can be reduced to 0
# after performing the given operations
def countNoOfsuchX(N):
  
    # Number of digits in N
    k = len(str(N))
  
    # Stores the count of value of X
    count = 0
  
    # Iterate over all possible value
    # of X
    for x in range(N - k*(k + 1)*5, N + 1):
  
        # Check if x follow the conditions
        if check(x, N):
            count += 1
  
    # Return the total count
    return count
  
  
# Driver Code
N = 9939
print(countNoOfsuchX(N))

输出：

3

时间复杂度： O(log N)
辅助空间： O(log N)