求2 ^(2 ^ A)％B

给定两个整数A和B ，任务是计算2 ^2A ％B 。

例子：

Input: A = 3, B = 5 
Output: 1 
2²³ % 5 = 2⁸ % 5 = 256 % 5 = 1.

Input: A = 10, B = 13 
Output: 3 
 

方法：通过使用递归将问题分解为子问题而不会出现整数溢出，可以有效地解决该问题。

Let F(A, B) = 2^2A % B. 
Now, F(A, B) = 2^2A % B 
= 2^{2 * 2A – 1} % B 
= (2^{2A – 1 + 2A – 1}) % B 
= (2^{2A – 1} * 2^{2A – 1}) % B 
= (F(A – 1, B) * F(A – 1, B)) % B 
Therefore, F(A, B) = (F(A – 1, B) * F(A – 1, B)) % B. 
The base case is F(1, B) = 2²¹ % B = 4 % B. 
 

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define ll long long
 
// Function to return 2^(2^A) % B
ll F(ll A, ll B)
{
 
    // Base case, 2^(2^1) % B = 4 % B
    if (A == 1)
        return (4 % B);
    else
    {
        ll temp =  F(A - 1, B);
        return (temp * temp) % B;
    }
}
 
// Driver code
int main()
{
    ll A = 25, B = 50;
 
    // Print 2^(2^A) % B
    cout << F(A, B);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
    // Function to return 2^(2^A) % B
    static long F(long A, long B)
    {
     
        // Base case, 2^(2^1) % B = 4 % B
        if (A == 1)
            return (4 % B);
        else
        {
            long temp = F(A - 1, B);
            return (temp * temp) % B;
        }
    }
     
    // Driver code
    public static void main(String args[])
    {
        long A = 25, B = 50;
     
        // Print 2^(2^A) % B
        System.out.println(F(A, B));
    }
}
 
// This code is contributed by Ryuga

Python3

# Python3 implementation of the approach
 
# Function to return 2^(2^A) % B
def F(A, B):
 
    # Base case, 2^(2^1) % B = 4 % B
    if (A == 1):
        return (4 % B);
    else:
        temp = F(A - 1, B);
        return (temp * temp) % B;
 
# Driver code
A = 25;
B = 50;
 
# Print 2^(2^A) % B
print(F(A, B));
 
# This code is contributed by mits

C#

// C# implementation of the approach
class GFG
{
     
// Function to return 2^(2^A) % B
static long F(long A, long B)
{
 
    // Base case, 2^(2^1) % B = 4 % B
    if (A == 1)
        return (4 % B);
    else
    {
        long temp = F(A - 1, B);
        return (temp * temp) % B;
    }
}
 
// Driver code
static void Main()
{
    long A = 25, B = 50;
 
    // Print 2^(2^A) % B
    System.Console.WriteLine(F(A, B));
}
}
 
// This code is contributed by mits

---

Javascript

 
// Javascript implementation of the approach
 
// Function to return 2^(2^A) % B
function F(A, B)
{
 
    // Base case, 2^(2^1) % B = 4 % B
    if (A == 1)
        return (4 % B);
    else
    {
        var temp =  F(A - 1, B);
        return (temp * temp) % B;
    }
}
 
// Driver code
var A = 25, B = 50;
 
// Print 2^(2^A) % B
document.write( F(A, B));
 
// This code is contributed by noob2000
 

输出：

46

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