通过从 B 中减少 A 或从 A 中减少 B 来最小化将 A 或 B 减少到 0 的操作

给定两个数字A和B ，任务是找到将 A 或 B 减少到 0 所需的最小操作数，其中如果A大于等于B，则每个操作A可以减少B ，反之亦然。

例子：

Input: A = 5, B = 4
Output: 5
Explanation:
Reduce B from A: A=1, B=4
Reduce A from B: A=1, B=3
Reduce A from B: A=1, B=2
Reduce A from B: A=1, B=1
Reduce B from A: A=0, B=1

Input: A=1, B=1
Output: 1
Explanation:
Reduce A from B: A=0, B=0

方法：解决这个问题的方法是简单地检查较大的数字并从中减少小数字。

• 重复以下操作，直到两个数字中的至少一个变为 0
  • 如果 A 大于等于 B，则从 A 减少 B
  • 如果 A 小于 A，则从 B 减少 A
  • 对于每个循环迭代，存储计数。
• 最后返回循环迭代的计数。

下面是上述方法的实现：

C++

// C++ program to find Minimum number
// Of operations required to reduce
// Either A or B to Zero
 
#include 
using namespace std;
 
int countOperations(int num1, int num2)
{
    int cnt = 0;
    while (num1 > 0 && num2 > 0) {
        if (num1 >= num2)
            num1 -= num2;
        else
            num2 -= num1;
        cnt++;
    }
    return cnt;
}
 
// Driver Code
int main()
{
 
    int A = 5, B = 4;
    cout << countOperations(A, B);
    return 0;
}

Java

// Java program to find Minimum number
import java.io.*;
 
class GFG {
 
  // Of operations required to reduce
  // Either A or B to Zero
 
  static int countOperations(int num1, int num2)
  {
    int cnt = 0;
    while (num1 > 0 && num2 > 0) {
      if (num1 >= num2)
        num1 -= num2;
      else
        num2 -= num1;
      cnt++;
    }
    return cnt;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int A = 5, B = 4;
    System.out.println(countOperations(A, B));
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach
def countOperations(num1, num2):
    cnt = 0
    while (num1 > 0 and num2 > 0):
        if (num1 >= num2):
            num1 -= num2
        else:
            num2 -= num1
        cnt += 1
 
    return cnt
 
# Driver Code
A,B = 5,4
print(countOperations(A, B))
 
# This code is contributed by shinjanpatra

C#

// C# program to find Minimum number
// Of operations required to reduce
// Either A or B to Zero
using System;
class GFG {
 
  static int countOperations(int num1, int num2)
  {
    int cnt = 0;
    while (num1 > 0 && num2 > 0) {
      if (num1 >= num2)
        num1 -= num2;
      else
        num2 -= num1;
      cnt++;
    }
    return cnt;
  }
 
  // Driver Code
  public static void Main()
  {
 
    int A = 5, B = 4;
    Console.Write(countOperations(A, B));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

       // JavaScript code for the above approach
       function countOperations(num1, num2) {
           let cnt = 0;
           while (num1 > 0 && num2 > 0) {
               if (num1 >= num2)
                   num1 -= num2;
               else
                   num2 -= num1;
               cnt++;
           }
           return cnt;
       }
 
       // Driver Code
       let A = 5, B = 4;
       document.write(countOperations(A, B));
 
      // This code is contributed by Potta Lokesh
   

输出

5

时间复杂度： 0(N)
辅助空间： 0(1)