在对 A 和 B 进行交替 AND-OR 操作后找到 A 的最终值

给定 3 个整数A 、 B和N ，任务是对A和B执行交替的 AND-OR运算，然后将每个运算的结果分配给A 。找到A的最终值

例子：

Input: A = 4, B = 5, N = 1
Output: 4
Explanation: Perform 1 operation i.e A = A & B, therefore A = 4 & 5 = 4
Input: A = 4, B = 5, N = 1000
Output: 5

天真的方法：可以使用观察来解决任务。可以观察到，如果数字N是Odd ，答案将是A AND B 。如果数字 N 是偶数，答案将是A OR B。这是因为 AND-OR 操作交替运行。

下面是上述方法的实现：

C++

/*package whatever //do not write package name here */
#include 
using namespace std;
 
int main()
{
    int A = 4;
    int B = 5;
    int n = 1000;
    int N = 1;
    for (N = 1; N <= n; N++) {
        if ((N % 2) != 0) {
            A = A & B;
        }
        else {
            A = A | B;
        }
    }
 
    cout << "Output is:" << A;
    return 0;
}
 
// This code is contributed by rakeshsahni

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    public static void main (String[] args) {
        int A=4;
                int B=5;
                int n =1000;
                int N = 1;
                for(N=1;N<=n;N++)
        {
            if ((N % 2) != 0) {
                A = A & B;
            } else {
                A = A | B;
            }}
 
            System.out.println("Output is:" + A);
    }
}

Python3

# Python program for above approach
A = 4
B = 5
n = 1000
N = 1
 
for N in range(1, n+1):
    if ((N % 2) != 0):
        A = A & B
 
    else:
        A = A | B
 
print("Output is:", end="")
print(A)
 
# This code is contributed by ninja_hattori.

C#

// C# program to implement the approach
using System;
class GFG {
 
  public static void Main()
  {
        int A=4;
                int B=5;
                int n =1000;
                int N = 1;
                for(N=1;N<=n;N++)
        {
            if ((N % 2) != 0) {
                A = A & B;
            } else {
                A = A | B;
            }}
 
        Console.Write("Output is:" + A);
  }
}
 
// This code is contributed by sanjoy_62.

Javascript

     // JavaScript program for above approach
        var A = 4;
        var B = 5;
        var n = 1000;
        var N = 1;
        for (N = 1; N <= n; N++) {
            if ((N % 2) != 0) {
                A = A & B;
            } else {
                A = A | B;
            }
        }
 
        document.write("Output is:" + A);
 
// This code is contributed by Rajput-Ji

输出

Output is:5

方法：可以使用观察来解决任务。可以观察到答案将是A ，仅当N为1时，对于其余N值，答案将为B 。这是因为，在第一次操作之后，两个数字都将等于B
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 resultant value of A
int find(int A, int B, int N)
{
    if (N == 1)
        return A & B;
    else
        return B;
}
 
// Driver Code
int main()
{
    cout << find(4, 5, 1000);
    return 0;
}

Java

// JAVA program for the above approach
import java.util.*;
class GFG
{
   
  // Function to find the resultant value of A
  public static int find(int A, int B, int N)
  {
    if (N == 1)
      return A & B;
    else
      return B;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    System.out.print(find(4, 5, 1000));
  }
}
 
// This code is contributed by Taranpreet

Python3

# Python code for the above approach
 
# Function to find the resultant value of A
def find(A, B, N):
    if (N == 1):
        return A & B;
    else:
        return B;
 
# Driver Code
print(find(4, 5, 1000));
 
# This code is contributed by Saurabh Jaiswal

C＃

// C# program for the above approach
using System;
class GFG
{
   
  // Function to find the resultant value of A
  static int find(int A, int B, int N)
  {
    if (N == 1)
      return A & B;
    else
      return B;
  }
 
  // Driver Code
  public static void Main()
  {
    Console.Write(find(4, 5, 1000));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

        // JavaScript code for the above approach
 
        // Function to find the resultant value of A
        function find(A, B, N) {
            if (N == 1)
                return A & B;
            else
                return B;
        }
 
        // Driver Code
        document.write(find(4, 5, 1000));
 
       // This code is contributed by Potta Lokesh
    

输出

5

输出

5

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