模幂运算（递归）

给定三个数字 a、b 和 c，我们需要找到 (a ^b ) % c
现在为什么要在求幂后使用“% c”，因为即使 a、b 的值相对较小，a ^b也会非常大，这是一个问题，因为我们尝试编码该问题的语言的数据类型很可能不会让我们存储这么大的数字。
例子：

Input : a = 2312 b = 3434 c = 6789
Output : 6343

Input : a = -3 b = 5 c = 89 
Output : 24

这个想法基于以下属性。
属性 1：
(m * n) % p 有一个非常有趣的特性：
(m * n) % p =((m % p) * (n % p)) % p
属性 2：
如果 b 是偶数：
(a ^ b) % c = ((a ^ b/2) * (a ^ b/2))%c ?这表明分而治之
如果 b 是奇数：
(a ^ b) % c = (a * (a ^( b-1))%c
属性 3：
如果我们必须返回绝对值小于 y 的负数 x 的 mod：
然后 (x + y) % y 会起作用
笔记：
另外由于 (a ^ b/2) * (a ^ b/2) 和 a * (a ^( b-1) 的乘积可能会导致溢出，因此我们必须小心这些情况

C++

// Recursive C++ program to compute modular power
#include 
using namespace std;
 
int exponentMod(int A, int B, int C)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
 
    // If B is even
    long y;
    if (B % 2 == 0) {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
 
    // If B is odd
    else {
        y = A % C;
        y = (y * exponentMod(A, B - 1, C) % C) % C;
    }
 
    return (int)((y + C) % C);
}
 
// Driver code
int main()
{
    int A = 2, B = 5, C = 13;
    cout << "Power is " << exponentMod(A, B, C);
    return 0;
}
 
// This code is contributed by SHUBHAMSINGH10

C

// Recursive C program to compute modular power
#include 
 
int exponentMod(int A, int B, int C)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
 
    // If B is even
    long y;
    if (B % 2 == 0) {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
 
    // If B is odd
    else {
        y = A % C;
        y = (y * exponentMod(A, B - 1, C) % C) % C;
    }
 
    return (int)((y + C) % C);
}
 
// Driver program to test above functions
int main()
{
   int A = 2, B = 5, C = 13;
   printf("Power is %d", exponentMod(A, B, C));
   return 0;
}

Java

// Recursive Java program
// to compute modular power
import java.io.*;
 
class GFG
{
static int exponentMod(int A,
                       int B, int C)
{
         
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
     
    // If B is even
    long y;
    if (B % 2 == 0)
    {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
     
    // If B is odd
    else
    {
        y = A % C;
        y = (y * exponentMod(A, B - 1,
                             C) % C) % C;
    }
     
    return (int)((y + C) % C);
}
 
// Driver Code
public static void main(String args[])
{
    int A = 2, B = 5, C = 13;
    System.out.println("Power is " +
                        exponentMod(A, B, C));
}
}
 
// This code is contributed
// by Swetank Modi.

Python3

# Recursive Python program
# to compute modular power
def exponentMod(A, B, C):
     
    # Base Cases
    if (A == 0):
        return 0
    if (B == 0):
        return 1
     
    # If B is Even
    y = 0
    if (B % 2 == 0):
        y = exponentMod(A, B / 2, C)
        y = (y * y) % C
     
    # If B is Odd
    else:
        y = A % C
        y = (y * exponentMod(A, B - 1,
                             C) % C) % C
    return ((y + C) % C)
 
# Driver Code
A = 2
B = 5
C = 13
print("Power is", exponentMod(A, B, C))
     
# This code is contributed
# by Swetank Modi.

C#

// Recursive C# program
// to compute modular power
class GFG
{
static int exponentMod(int A, int B, int C)
{
         
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
     
    // If B is even
    long y;
    if (B % 2 == 0)
    {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
     
    // If B is odd
    else
    {
        y = A % C;
        y = (y * exponentMod(A, B - 1,
                             C) % C) % C;
    }
     
    return (int)((y + C) % C);
}
 
// Driver Code
public static void Main()
{
    int A = 2, B = 5, C = 13;
    System.Console.WriteLine("Power is " +
                    exponentMod(A, B, C));
}
}
 
// This code is contributed
// by mits

Javascript

 
// Recursive Javascript program
// to compute modular power
 
// Function to check if a given
// quadilateral is valid or not
function exponentMod(A, B, C)
{
     
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
     
    // If B is even
    var y;
    if (B % 2 == 0)
    {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
     
    // If B is odd
    else
    {
        y = A % C;
        y = (y * exponentMod(A, B - 1,
                             C) % C) % C;
    }
     
    return parseInt(((y + C) % C));
}
 
// Driver code
var A = 2, B = 5, C = 13;
document.write("Power is " +
               exponentMod(A, B, C));
     
// This code is contributed by Khushboogoyal499
 

输出：

Power is 6

迭代模幂运算。

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