一个数字的GCD升为一个幂而另一个数字

给定三个数字a，b，n。找出GCD(a ⁿ ，b)。
例子：

Input : a = 2, b = 3, n = 3
Output : 1
2^3 = 8. GCD of 8 and 3 is 1. 

Input : a = 2, b = 4, n = 5
Output : 4

第一种方法：蛮力方法是先计算a ^ n，然后计算a ^ n和b的GCD。

C++

// CPP program to find GCD of a^n and b.
#include 
using namespace std;
 
typedef long long int ll;
 
ll gcd(ll a, ll b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Returns GCD of a^n and b.
ll powGCD(ll a, ll n, ll b)
{
    for (int i = 0; i < n; i++)
        a = a * a;
 
    return gcd(a, b);
}
 
// Driver code
int main()
{
    ll a = 10, b = 5, n = 2;
    cout << powGCD(a, n, b);
    return 0;
}

Java

// Java program to find GCD of a^n and b.
 
import java.io.*;
 
class GFG {
 
 
static long gcd(long a, long b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Returns GCD of a^n and b.
static long powGCD(long a, long n, long b)
{
    for (int i = 0; i < n; i++)
        a = a * a;
 
    return gcd(a, b);
}
 
// Driver code
    public static void main (String[] args) {
    long a = 10, b = 5, n = 2;
    System.out.println(powGCD(a, n, b));
    }
}
// This code is contributed by anuj_67..

Python3

# Python 3 program to find
# GCD of a^n and b.
def gcd(a, b):
    if (a == 0):
        return b
    return gcd(b % a, a)
 
# Returns GCD of a^n and b.
def powGCD(a, n, b):
    for i in range(0, n + 1, 1):
        a = a * a
 
    return gcd(a, b)
 
# Driver code
if __name__ == '__main__':
    a = 10
    b = 5
    n = 2
    print(powGCD(a, n, b))
     
# This code is contributed
# by Surendra_Gangwar

C#

// C# program to find GCD of a^n and b.
using System;
 
class GFG
{
public static long gcd(long a, long b)
{
    if (a == 0)
    {
        return b;
    }
    return gcd(b % a, a);
}
 
// Returns GCD of a^n and b.
public static long powGCD(long a,
                          long n, long b)
{
    for (int i = 0; i < n; i++)
    {
        a = a * a;
    }
 
    return gcd(a, b);
}
 
// Driver code
public static void Main(string[] args)
{
    long a = 10, b = 5, n = 2;
    Console.WriteLine(powGCD(a, n, b));
}
}
 
// This code is contributed
// by Shrikant13

Javascript

// javascript program to find GCD of a^n and b.   
function gcd(a , b)
{
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    // Returns GCD of a^n and b.
    function powGCD(a , n , b)
    {
        for (i = 0; i < n; i++)
            a = a * a;
 
        return gcd(a, b);
    }
 
    // Driver code
        var a = 10, b = 5, n = 2;
        document.write(powGCD(a, n, b));
 
// This code is contributed by gauravrajput1

输出：

5

但是，如果n非常大(例如> 10 ^ 9)，该怎么办。模幂是方法。我们知道(a * b)％m =((a％m)*(b％m))％m)。我们也知道gcd(a，b)= gcd(b％a，a) 。因此，我们使用模幂运算来代替计算pow(a，n)。

C++

// C++ program of the above approach
#include 
using namespace std;
 
typedef long long int ll;
 
/* Calculates modular exponentiation, i.e.,
   (x^y)%p in O(log y) */
ll power(ll x, ll y, ll p)
{
    ll res = 1; // Initialize result
 
    x = x % p; // Update x if it is more than or
    // equal to p
 
    while (y > 0) {
 
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
 
ll gcd(ll a, ll b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Returns GCD of a^n and b
ll powerGCD(ll a, ll b, ll n)
{
    ll e = power(a, n, b);
    return gcd(e, b);
}
 
// Driver code
int main()
{
    ll a = 5, b = 4, n = 2;
    cout << powerGCD(a, b, n);
    return 0;
}

Java

// Java program of the above approach
import java.util.*;
class Solution{
   
   
/* Calculates modular exponentiation, i.e.,
   (x^y)%p in O(log y) */
static long power(long x, long y, long p)
{
    long res = 1; // Initialize result
   
    x = x % p; // Update x if it is more than or
    // equal to p
   
    while (y > 0) {
   
        // If y is odd, multiply x with result
        if ((y & 1)!=0)
            res = (res * x) % p;
   
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
   
   
static long gcd(long a, long b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
   
// Returns GCD of a^n and b
static long powerGCD(long a, long b, long n)
{
    long e = power(a, n, b);
    return gcd(e, b);
}
   
// Driver code
public static void main(String args[])
{
    long a = 5, b = 4, n = 2;
    System.out.print( powerGCD(a, b, n));
 
}
}
//contributed by Arnab Kundu

Python3

# Python3 program of the above approach
  
# Calculates modular exponentiation, i.e.,
 # (x^y)%p in O(log y)
def power( x,  y,  p):
 
    res = 1  # Initialize result
  
    x = x % p # Update x if it is more than or
    # equal to p
  
    while (y > 0) :
  
        # If y is odd, multiply x with result
        if (y & 1):
            res = (res * x) % p
  
        # y must be even now
        y = y >> 1   # y = y/2
        x = (x * x) % p
     
    return res
  
  
def gcd(a,  b):
 
    if (a == 0):
        return b
    return gcd(b % a, a)
  
# Returns GCD of a^n and b
def powerGCD( a,  b,  n):
 
    e = power(a, n, b)
    return gcd(e, b)
  
# Driver code
if __name__ == "__main__":
 
    a = 5
    b = 4
    n = 2
    print (powerGCD(a, b, n))

C＃

// C# program of the above approach
using System;
class GFG
{
 
/* Calculates modular exponentiation,
i.e.,  (x^y)%p in O(log y) */
static long power(long x, long y, long p)
{
    long res = 1; // Initialize result
 
    x = x % p; // Update x if it is more
               // than or equal to p
 
    while (y > 0)
    {
 
        // If y is odd, multiply x
        // with result
        if ((y & 1) != 0)
            res = (res * x) % p;
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
static long gcd(long a, long b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Returns GCD of a^n and b
static long powerGCD(long a, long b,
                             long n)
{
    long e = power(a, n, b);
    return gcd(e, b);
}
 
// Driver code
public static void Main()
{
    long a = 5, b = 4, n = 2;
    Console.Write( powerGCD(a, b, n));
}
}
 
// This code is contributed
// by Akanksha Rai

的PHP

 0)
    {
 
        // If y is odd, multiply x
        // with result
        if ($y & 1)
            $res = ($res * $x) % $p;
 
        // y must be even now
        $y = $y >> 1; // y = y/2
        $x = ($x * $x) % $p;
    }
    return $res;
}
 
function gcd ($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
 
// Returns GCD of a^n and b
function powerGCD($a, $b, $n)
{
    $e = power($a, $n, $b);
    return gcd($e, $b);
}
 
// Driver code
$a = 5;
$b = 4;
$n = 2;
echo powerGCD($a, $b, $n);
 
// This code is contributed by Sachin.
?>

Java脚本

 
// Javascript program of the above approach
 
    /*
      Calculates modular exponentiation,
      i.e., (x^y)%p in O(log y)
     */
    function power(x , y , p) {
        var res = 1; // Initialize result
 
        x = x % p; // Update x if it is more than or
        // equal to p
 
        while (y > 0) {
 
            // If y is odd, multiply x with result
            if ((y & 1) != 0)
                res = (res * x) % p;
 
            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }
        return res;
    }
 
    function gcd(a , b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    // Returns GCD of a^n and b
    function powerGCD(a , b , n) {
        var e = power(a, n, b);
        return gcd(e, b);
    }
 
    // Driver code
     
        var a = 5, b = 4, n = 2;
        document.write(powerGCD(a, b, n));
 
 
// This code contributed by Rajput-Ji
 

输出：

1