通过添加任何奇数或减去任何偶数来最小化将 A 转换为 B 的操作

给定两个正整数A和B 。任务是找到将数字A转换为B所需的最小操作数。在一次移动中，可以对数字A应用以下任一操作：

选择任何奇数x (x>0) 并将其添加到 A 即(A+x) ；
或者，选择任何偶数 y (y>0) 并将其从 A 中减去，即(Ay) 。

例子：

Input: A = 2, B = 3
Output: 1
Explanation: Add odd x = 1 to A to obtain B (1 + 2 = 3).

Input: A = 7, B = 4
Output: 2
Explanation: Two operations are required:
Subtract y = 4 from A (7 – 4 = 3).
Add x = 1 to get B (3 + 1 = 4).

编程需要懂一点英语

方法：这是一个基于实现的问题。请按照以下步骤解决给定的问题。

将AB的绝对差值存储在变量diff中。
检查A是否等于B 。由于两个整数相等，因此操作总数将为0 。
否则检查是否A < B 。
- 如果是，请检查它们的差异是奇数还是偶数。
  - 如果 diff 是奇数，则应用A+x操作一次（ x是等于diff的奇数）。操作总数为1 。
  - 否则 diff 是偶数，应用 A +x操作两次，首先是x是diff -1 （奇数），其次是 x 是1 。或者， A+x操作可以跟在Ay操作之后。在任何一种情况下，操作的总数都是2。
否则，如果A> B ，则应用相反的一组操作，即
- 如果 diff 为偶数，则应用一次Ay操作。
- 否则，可以应用Ay操作，然后执行A+x操作。或者，可以应用两次Ay操作。

因此，操作数将始终为0、1或 2。

以下是上述方法的实现。

C++

// C++ program for the given approach
#include 
using namespace std;
 
// Function to find
// minimum number of operations
int minOperations(int A, int B)
{
    // Variable to store
    // difference of A and B
    int diff;
 
    if (A == B)
        return 0;
    else if (A < B) {
 
        // A+x operation first
        diff = B - A;
        if (diff % 2 != 0)
            return 1;
        return 2;
    }
    else {
 
        // A-y operation first
        diff = A - B;
        if (diff % 2 == 0)
            return 1;
        return 2;
    }
}
 
// Driver code
int main()
{
    // Declaring integers A and B
    int A, B;
 
    // Initialising
    A = 7;
    B = 4;
 
    // Function call
    int ans = minOperations(A, B);
 
    // Displaying the result
    cout << ans;
    return 0;
}

Java

// Java program for the given approach
import java.util.*;
class GFG{
 
// Function to find
// minimum number of operations
static int minOperations(int A, int B)
{
   
    // Variable to store
    // difference of A and B
    int diff;
 
    if (A == B)
        return 0;
    else if (A < B) {
 
        // A+x operation first
        diff = B - A;
        if (diff % 2 != 0)
            return 1;
        return 2;
    }
    else {
 
        // A-y operation first
        diff = A - B;
        if (diff % 2 == 0)
            return 1;
        return 2;
    }
}
 
// Driver code
public static void main(String[] args)
{
   
    // Declaring integers A and B
    int A, B;
 
    // Initialising
    A = 7;
    B = 4;
 
    // Function call
    int ans = minOperations(A, B);
 
    // Displaying the result
    System.out.print(ans);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach
 
# Function to find
# minimum number of operations
def minOperations(A, B):
 
    # Variable to store
    # difference of A and B
    diff = None
 
    if (A == B):
        return 0;
    elif (A < B):
 
        # A+x operation first
        diff = B - A;
        if (diff % 2 != 0):
            return 1;
        return 2;
    else:
 
        # A-y operation first
        diff = A - B;
        if (diff % 2 == 0):
            return 1;
        return 2;
     
# Driver code
 
# Initialising A and B
A = 7;
B = 4;
 
# Function call
ans = minOperations(A, B);
 
# Displaying the result
print(ans);
 
# This code is contributed by gfgking

C#

// C# program for the given approach
using System;
class GFG{
 
// Function to find
// minimum number of operations
static int minOperations(int A, int B)
{
   
    // Variable to store
    // difference of A and B
    int diff;
 
    if (A == B)
        return 0;
    else if (A < B) {
 
        // A+x operation first
        diff = B - A;
        if (diff % 2 != 0)
            return 1;
        return 2;
    }
    else {
 
        // A-y operation first
        diff = A - B;
        if (diff % 2 == 0)
            return 1;
        return 2;
    }
}
 
// Driver code
public static void Main()
{
   
    // Declaring integers A and B
    int A, B;
 
    // Initialising
    A = 7;
    B = 4;
 
    // Function call
    int ans = minOperations(A, B);
 
    // Displaying the result
    Console.Write(ans);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

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