如果最后一位为 0，则通过将 N 减 1 来查找在 K 步中形成的数字，否则除以 10

给定两个整数N和K 。对N执行以下类型的操作：

• 如果N的最后一位非零，则将数字减一。
• 如果N的最后一位为零，则将该数字除以10 （即删除最后一位）。

任务是在K 个这样的操作之后打印结果。

例子：

Input: N = 512, K = 4
Output: 50
Explanation: Following are the operations performed K times to get the desired result.
Operation 1: Last digit of N i.e. 2 != 0. N is reduced by 1. ( N = 512  – 1 i.e.  511).
Operation 2: Last digit of N i.e. 1 != 0. N is reduced by 1. (N = 511 – 1 i.e.  510).
Operation 3: Last digit of N is 0. N is divided by 10. ( N = 510/10 i.e. 51).
Operation 4: Last digit of N i.e. 2 != 0. N is reduced by 1. (N = 51 – 1 i.e. 50).
Therefore, after 4 operations N = 50.

Input: N = 100, K = 2
Output: 1
Explanation: N is divided by 10 two times.

方法：这个问题是基于实现的，类似于数字的最后一位。请按照以下步骤解决给定的问题。

• 反复检查整数N的最后一位。
• 如果最后一位是0 ，则将N除以10 。
• 如果 最后一位不是 0 ，从N中减去1 。
• 重复以上步骤K次。

以下是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to perform operations K times
int decreaseNum(int N, int K)
{
    while (K--) {
        // Last digit is 0
        if (N % 10 == 0)
            N /= 10;
 
        // Last digit is not 0
        else
            N--;
    }
    return N;
}
 
// Driver Code
int main()
{
    // Declaration and initialisation
    int N, K;
    N = 512;
    K = 4;
 
    // Function call
    cout << decreaseNum(N, K);
 
    return 0;
}

Java

// Java program of the above approach
import java.util.*;
class GFG {
 
  // Function to perform operations K times
  public static int decreaseNum(int N, int K)
  {
    while (true) {
      K -= 1;
       
      // Last digit is 0
      if (N % 10 == 0)
        N /= 10;
 
      // Last digit is not 0
      else
        N--;
 
      if (K == 0)
        break;
    }
    return N;
  }
 
  // Driver Code
  public static void main(String args[])
  {
     
    // Declaration and initialisation
    int N, K;
    N = 512;
    K = 4;
 
    // Function call
    System.out.println(decreaseNum(N, K));
 
  }
}
 
    // This code is contributed by rakeshsahni

Python3

# python3 for above approach
 
# def Function to perform operations K times
def decreaseNum(N, K):
 
    while True:
        K -= 1
        # Last digit is 0
        if (N % 10 == 0):
            N //= 10
 
            # Last digit is not 0
        else:
            N -= 1
 
        if K == 0:
            break
 
    return N
 
# Driver Code
if __name__ == "__main__":
 
        # Declaration and initialisation
    N = 512
    K = 4
 
    # Function call
    print(decreaseNum(N, K))
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections;
 
class GFG
{
 
  // Function to perform operations K times
  public static int decreaseNum(int N, int K)
  {
    while (true) {
      K -= 1;
 
      // Last digit is 0
      if (N % 10 == 0)
        N /= 10;
 
      // Last digit is not 0
      else
        N--;
 
      if (K == 0)
        break;
    }
    return N;
  }
 
  // Driver Code
  public static void Main()
  {
    // Declaration and initialisation
    int N = 512;
    int K = 4;
 
    // Function call
    Console.Write(decreaseNum(N, K));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

      // JavaScript code for the above approach
 
      // Function to perform operations K times
      function decreaseNum(N, K)
      {
          while (K--)
          {
           
              // Last digit is 0
              if (N % 10 == 0)
                  N /= 10;
 
              // Last digit is not 0
              else
                  N--;
          }
          return N;
      }
 
      // Driver Code
 
      // Declaration and initialisation
      let N, K;
      N = 512;
      K = 4;
 
      // Function call
      document.write(decreaseNum(N, K));
 
// This code is contributed by Potta Lokesh
  

输出

50

时间复杂度： O(K)
辅助空间： O(1)