在两个给定的整数的和加到N的幂之间以及它们的差之间找到GCD

给定三个正整数P ， Q和N ，任务是求模10 ⁹ + 7下的(P ^N + Q ^N )和(P – Q)的GCD。

例子：

Input: p = 10, q = 6, n = 5
Output: 4
Explanation: pⁿ + qⁿ = 10⁵ + 6⁵ = 107776  and p – q = 4. GCD of 107776 and 4 is 4.

Input: p = 7, q = 2 and n = 5
Output: 1
Explanation: pⁿ + qⁿ = 7⁵ + 2⁵  = 16839 and p – q = 5. GCD of 16839 and 5 is 1.

方法：由于数量(p ⁿ + q ⁿ )可能非常大，因此无法以任何数据类型存储如此大的数量，因此无法使用欧几里得算法来计算GCD。因此，可以使用模块化算术来找到答案。

请按照以下步骤解决此问题：

• 要找到大数和小数的GCD，请在O(√pq)中找到小数的除数。
• 这些数字是潜在的GCD候选者。
• 现在，检查这些潜在的GCD中是否有任何一个将较大的数相除。将两个数字相除的最大数字是最终答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
#define mod 1000000007
using namespace std;
 
// Function to find the value of (a ^ n) % d
long long int power(long long a, long long n,
                    long long int d)
{
 
    // Stores the value
    // of (a ^ n) % d
    long long int res = 1;
 
    // Calculate the value
    // of (a ^ n) % d
    while (n) {
 
        // If n is odd
        if (n % 2) {
 
            // Update res
            res = ((res % d) * (a % d)) % d;
        }
 
        // Update a
        a = ((a % d) * (a % d)) % d;
 
        // Update n
        n /= 2;
    }
 
    return res;
}
 
// Function to find the GCD of (p ^ n + q ^ n)
// and p - q mod d
long long int gcd(long long p, long long q,
                  long long n)
{
 
    // If p and q are equal
    if (p == q) {
        return (power(p, n, mod)
                + power(q, n, mod))
               % mod;
    }
 
    // Stores GCD of (p ^ n + q ^ n)
    // and (p - q) mod d
    long long int candidate = 1;
 
    // Stores the value of (p-q)
    long long int num = p - q;
 
    // Stores square root of num
    long long int sq = sqrt(num);
 
    // Find the divisors of num.
    for (long long i = 1; i <= sq; ++i) {
 
        // If i divides num
        if (num % i == 0) {
 
            // Stores power of (p ^ n) mod i
            long long int X = power(p, n, i);
 
            // Stores power of (q ^ n) mod i
            long long int Y = power(q, n, i);
 
            // Stores power of (p ^ n + q ^ n) mod i
            long long int temp = (X + Y) % i;
 
            // If (p^n + q^n) is divisible by i
            if (temp == 0) {
 
                // Calculate the largest divisor.
                candidate = max(candidate, i);
            }
 
            // If i divides num, (num/i) also divides
            // num. Hence, calculate temp.
            temp = (power(p, n, num / i)
                    + power(q, n, num / i))
                   % (num / i);
 
            // If (p^n + q^n) is divisible
            // by (num/i)
            if (temp == 0) {
 
                // Calculate the largest divisor.
                candidate = max(candidate, num / i);
            }
        }
    }
 
    return candidate % mod;
}
 
// Driver Code
int main()
{
 
    // Given p, q and n
    long long int p, q, n;
    p = 10;
    q = 6;
    n = 5;
 
    // Function Call
    cout << gcd(p, q, n);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
static int mod = 1000000007;
       
// Function to find the value of (a ^ n) % d
static int power(int a, int n, int d)
{
 
    // Stores the value
    // of (a ^ n) % d
    int res = 1;
 
    // Calculate the value
    // of (a ^ n) % d
    while (n != 0) {
 
        // If n is odd
        if ((n % 2) !=0) {
 
            // Update res
            res = ((res % d) * (a % d)) % d;
        }
 
        // Update a
        a = ((a % d) * (a % d)) % d;
 
        // Update n
        n /= 2;
    }
 
    return res;
}
 
// Function to find the GCD of (p ^ n + q ^ n)
// and p - q mod d
static int gcd(int p, int q, int n)
{
 
    // If p and q are equal
    if (p == q) {
        return (power(p, n, mod)
                + power(q, n, mod))
               % mod;
    }
 
    // Stores GCD of (p ^ n + q ^ n)
    // and (p - q) mod d
    int candidate = 1;
 
    // Stores the value of (p-q)
    int num = p - q;
 
    // Stores square root of num
    int sq = (int)Math.sqrt(num);
 
    // Find the divisors of num.
    for (int  i = 1; i <= sq; ++i) {
 
        // If i divides num
        if (num % i == 0) {
 
            // Stores power of (p ^ n) mod i
            int X = power(p, n, i);
 
            // Stores power of (q ^ n) mod i
            int Y = power(q, n, i);
 
            // Stores power of (p ^ n + q ^ n) mod i
            int temp = (X + Y) % i;
 
            // If (p^n + q^n) is divisible by i
            if (temp == 0) {
 
                // Calculate the largest divisor.
                candidate = Math.max(candidate, i);
            }
 
            // If i divides num, (num/i) also divides
            // num. Hence, calculate temp.
            temp = (power(p, n, num / i)
                    + power(q, n, num / i))
                   % (num / i);
 
            // If (p^n + q^n) is divisible
            // by (num/i)
            if (temp == 0) {
 
                // Calculate the largest divisor.
                candidate = Math.max(candidate, num / i);
            }
        }
    }
    return candidate % mod;
}
   
// Driver code
public static void main(String[] args)
{
    // Given p, q and n
    int p, q, n;
    p = 10;
    q = 6;
    n = 5;
 
    // Function Call
    System.out.println(gcd(p, q, n));
}
}
 
// This code is contributed by code_hunt.

Python3

# Python program for the above approach
import math
mod = 1000000007;
 
# Function to find the value of (a ^ n) % d
def power(a, n, d):
 
    # Stores the value
    # of (a ^ n) % d
    res = 1;
 
    # Calculate the value
    # of (a ^ n) % d
    while (n != 0):
 
        # If n is odd
        if ((n % 2) != 0):
 
            # Update res
            res = ((res % d) * (a % d)) % d;
         
        # Update a
        a = ((a % d) * (a % d)) % d;
 
        # Update n
        n /= 2;
    return res;
 
# Function to find the GCD of (p ^ n + q ^ n)
# and p - q mod d
def gcd(p, q, n):
 
    # If p and q are equal
    if (p == q):
        return (power(p, n, mod) + power(q, n, mod)) % mod;
     
    # Stores GCD of (p ^ n + q ^ n)
    # and (p - q) mod d
    candidate = 1;
 
    # Stores the value of (p-q)
    num = p - q;
 
    # Stores square root of num
    sq = (int)(math.sqrt(num));
 
    # Find the divisors of num.
    for i in range(1, sq):
 
        # If i divides num
        if (num % i == 0):
 
            # Stores power of (p ^ n) mod i
            X = power(p, n, i);
 
            # Stores power of (q ^ n) mod i
            Y = power(q, n, i);
 
            # Stores power of (p ^ n + q ^ n) mod i
            temp = (X + Y) % i;
 
            # If (p^n + q^n) is divisible by i
            if (temp == 0):
 
                # Calculate the largest divisor.
                candidate = max(candidate, i);
             
 
            # If i divides num, (num/i) also divides
            # num. Hence, calculate temp.
            temp = (power(p, n, num / i) + power(q, n, num / i)) % (num / i);
 
            # If (p^n + q^n) is divisible
            # by (num/i)
            if (temp == 0):
 
                # Calculate the largest divisor.
                candidate = max(candidate, num / i);           
    return candidate % mod;
 
# Driver code
if __name__ == '__main__':
 
  # Given p, q and n
    p = 10;
    q = 6;
    n = 5;
 
    # Function Call
    print((int)(gcd(p, q, n)));
 
# This code contributed by aashish1995

C#

// C# program for the above approach
using System;
 
class GFG
{
    static int mod = 1000000007;
 
    // Function to find the value of (a ^ n) % d
    static int power(int a, int n, int d)
    {
 
        // Stores the value
        // of (a ^ n) % d
        int res = 1;
 
        // Calculate the value
        // of (a ^ n) % d
        while (n != 0)
        {
 
            // If n is odd
            if ((n % 2) != 0)
            {
 
                // Update res
                res = ((res % d) * (a % d)) % d;
            }
 
            // Update a
            a = ((a % d) * (a % d)) % d;
 
            // Update n
            n /= 2;
        }
 
        return res;
    }
 
    // Function to find the GCD of (p ^ n + q ^ n)
    // and p - q mod d
    static int gcd(int p, int q, int n)
    {
 
        // If p and q are equal
        if (p == q)
        {
            return (power(p, n, mod) + power(q, n, mod))
                % mod;
        }
 
        // Stores GCD of (p ^ n + q ^ n)
        // and (p - q) mod d
        int candidate = 1;
 
        // Stores the value of (p-q)
        int num = p - q;
 
        // Stores square root of num
        int sq = (int)Math.Sqrt(num);
 
        // Find the divisors of num.
        for (int i = 1; i <= sq; ++i) {
 
            // If i divides num
            if (num % i == 0) {
 
                // Stores power of (p ^ n) mod i
                int X = power(p, n, i);
 
                // Stores power of (q ^ n) mod i
                int Y = power(q, n, i);
 
                // Stores power of (p ^ n + q ^ n) mod i
                int temp = (X + Y) % i;
 
                // If (p^n + q^n) is divisible by i
                if (temp == 0) {
 
                    // Calculate the largest divisor.
                    candidate = Math.Max(candidate, i);
                }
 
                // If i divides num, (num/i) also divides
                // num. Hence, calculate temp.
                temp = (power(p, n, num / i)
                        + power(q, n, num / i))
                       % (num / i);
 
                // If (p^n + q^n) is divisible
                // by (num/i)
                if (temp == 0) {
 
                    // Calculate the largest divisor.
                    candidate
                        = Math.Max(candidate, num / i);
                }
            }
        }
        return candidate % mod;
    }
 
    // Driver code
    static public void Main()
    {
 
        // Given p, q and n
        int p, q, n;
        p = 10;
        q = 6;
        n = 5;
 
        // Function Call
        Console.WriteLine(gcd(p, q, n));
    }
}
 
// This is contributed by Dharanendra L V

Javascript

// javascript program for the above approach
    const mod = 1000000007;
 
    // Function to find the value of (a ^ n) % d
    function power(a , n , d) {
 
        // Stores the value
        // of (a ^ n) % d
        var res = 1;
 
        // Calculate the value
        // of (a ^ n) % d
        while (n != 0) {
 
            // If n is odd
            if ((n % 2) != 0) {
 
                // Update res
                res = ((res % d) * (a % d)) % d;
            }
 
            // Update a
            a = ((a % d) * (a % d)) % d;
 
            // Update n
            n /= 2;
        }
 
        return res;
    }
 
    // Function to find the GCD of (p ^ n + q ^ n)
    // and p - q mod d
    function gcd(p , q , n)
    {
 
        // If p and q are equal
        if (p == q)
        {
            return (power(p, n, mod) + power(q, n, mod)) % mod;
        }
 
        // Stores GCD of (p ^ n + q ^ n)
        // and (p - q) mod d
        var candidate = 1;
 
        // Stores the value of (p-q)
        var num = p - q;
 
        // Stores square root of num
        var sq = parseInt( Math.sqrt(num));
 
        // Find the divisors of num.
        for (i = 1; i <= sq; ++i) {
 
            // If i divides num
            if (num % i == 0) {
 
                // Stores power of (p ^ n) mod i
                var X = power(p, n, i);
 
                // Stores power of (q ^ n) mod i
                var Y = power(q, n, i);
 
                // Stores power of (p ^ n + q ^ n) mod i
                var temp = (X + Y) % i;
 
                // If (p^n + q^n) is divisible by i
                if (temp == 0) {
 
                    // Calculate the largest divisor.
                    candidate = Math.max(candidate, i);
                }
 
                // If i divides num, (num/i) also divides
                // num. Hence, calculate temp.
                temp = (power(p, n, num / i) + power(q, n, num / i)) % (num / i);
 
                // If (p^n + q^n) is divisible
                // by (num/i)
                if (temp == 0) {
 
                    // Calculate the largest divisor.
                    candidate = Math.max(candidate, num / i);
                }
            }
        }
        return candidate % mod;
    }
 
    // Driver code
     
        // Given p, q and n
        var p, q, n;
        p = 10;
        q = 6;
        n = 5;
 
        // Function Call
        document.write(gcd(p, q, n));
 
// This code is contributed by todaysgaurav

输出：

4

时间复杂度： O(sqrt(p – q))
辅助空间： O(1)