找到gcd(a ^ n，c)，其中a，n和c的范围可以从1到10 ^ 9

问题问题表明找到两个数字的gcd()，其中一个数字可能与(10 ^ 9)^(10 ^ 9)一样大，而该数字不能存储在C++中的long long int之类的数据类型中

例子：

Input : 1 1 1
Output : 1

Input : 10248585 1000000 12564
Output : 9

从Euclid算法中可以知道，gcd(a，b)= gcd(a％b，b)。现在问题仍然是找到一个n mod c。这可以使用具有O(logn)复杂度的模幂运算来完成。

C++

// CPP program to find GCD of a^n and b.
#include 
#define ll long long int
using namespace std;
  
/* Iterative Function to calculate (x^y)%p in O(log y) */
ll modPower(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;
}
  
// Finds GCD of a and b
ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
  
// Finds GCD of a^n and c
ll gcdPow(ll a, ll n, ll c)
{
    // check if c is a divisor of a
    if (a % c == 0)
        return c;
  
    // First compute (a^n) % c
    ll modexpo = modPower(a, n, c);
  
    // Now simply return GCD of modulo 
    // power and c.
    return gcd(modexpo, c);
}
  
// Driver code
int main()
{
    ll a = 10248585, n = 1000000, c = 12564;
    cout << gcdPow(a, n, c);
    return 0;
}

Java

// Java program to find 
// GCD of a^n and b.
class GFG
{
/* Iterative Function to calculate 
(x^y)%p in O(log y) */
static long modPower(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;
}
  
// Finds GCD of a and b
static long gcd(long a, long b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
  
// Finds GCD of a^n and c
static long gcdPow(long a, 
                   long n, long c)
{
    // check if c is a divisor of a
    if (a % c == 0)
        return c;
  
    // First compute (a^n) % c
    long modexpo = modPower(a, n, c);
  
    // Now simply return GCD 
    // of modulo power and c.
    return gcd(modexpo, c);
}
  
// Driver code
public static void main(String[] args)
{
    long a = 10248585, 
         n = 1000000, c = 12564;
    System.out.println(gcdPow(a, n, c));
}
}
  
// This code is contributed by mits

Python 3

# Python3 program to find 
# GCD of a^n and b.
  
# Iterative Function to
# calculate (x^y)%p in O(log y) 
def modPower(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
  
# Finds GCD of a and b
def gcd(a, b):
  
    if (b == 0):
        return a
    return gcd(b, a % b)
  
# Finds GCD of a^n and c
def gcdPow(a, n, c):
  
    # check if c is a divisor of a
    if (a % c == 0):
        return c
  
    # First compute (a^n) % c
    modexpo = modPower(a, n, c)
  
    # Now simply return GCD of 
    # modulo power and c.
    return gcd(modexpo, c)
  
# Driver code
if __name__ == "__main__":
    a = 10248585
    n = 1000000
    c = 12564
    print(gcdPow(a, n, c))
  
# This code is contributed 
# by ChitraNayal

C#

// C# program to find 
// GCD of a^n and b.
using System;
  
class GFG
{
/* Iterative Function to calculate 
(x^y)%p in O(log y) */
static long modPower(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;
}
  
// Finds GCD of a and b
static long gcd(long a, long b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
  
// Finds GCD of a^n and c
static long gcdPow(long a, 
                   long n, long c)
{
    // check if c is a divisor of a
    if (a % c == 0)
        return c;
  
    // First compute (a^n) % c
    long modexpo = modPower(a, n, c);
  
    // Now simply return GCD 
    // of modulo power and c.
    return gcd(modexpo, c);
}
  
// Driver code
public static void Main()
{
    long a = 10248585, 
         n = 1000000, c = 12564;
    Console.Write(gcdPow(a, n, c));
}
}
  
// This code is contributed
// by ChitraNayal

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;
}
  
// Finds GCD of a and b
function gcd($a, $b)
{
    if ($b == 0)
        return $a;
    return gcd($b, $a % $b);
}
  
// Finds GCD of a^n and c
function gcdPow($a, $n, $c)
{
    // check if c is a divisor of a
    if ($a % $c == 0)
        return $c;
  
    // First compute (a^n) % c
    $modexpo = modPower($a, $n, $c);
  
    // Now simply return GCD 
    // of modulo power and c.
    return gcd($modexpo, $c);
}
  
// Driver code
$a = 10248585; 
$n = 1000000; 
$c = 12564;
echo gcdPow($a, $n, $c);
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)
?>

输出：

9