欧拉因式分解方法

给定数字N ，任务是找到N的因数。
例子：

Input: N = 1000009 
Output: 293 3413 
Explanation: 
293 * 3413 = 1000009
Input: N = 100000 
Output: 800 125 
Explanation: 
800 * 125 = 100000 
 

欧拉因式分解方法：欧拉因式分解方法的原理是，可以将以两种不同方式写为两个幂的和的所有数字N可以分解为两个数字，即N = A ² + B ² = C ² + D ² ，其中A！= C和A！= D，则对于N存在两个因素。
算法的工作：令N为我们需要查找因子的数字。

• 因此，最初，我们需要找到两种方式将N表示为两个数字的幂的和。

N = A2 + B2
N = C2 + D2
Therefore,
N = A2 + B2 = C2 + D2

• 现在，对上述方程式执行代数运算，以将方程式转换为：

N = A2 + B2 = C2 + D2
-> N = A2 - C2 = D2 - B2
-> N = (A - C)(A + C) = (D - B)(D + B)

• 令K为(A – C)和(D – B)的GCD。所以，

A - C = K * L
D - B = K * M
where GCD(L, M) is 1.

• 显然，L =(A – C)/ K，M =(D – B)/ K。将其代入初始方程式时：

N = K * L * (A + C) = K * M * (D + B)
-> L * (A + C) = M * (D + B)
->l (A + C)/(D + B) = M/L

• 所以：

(A + C) = M * H 
(D + B) = L * H
where,
H = gcd((A + C), (D + B))

• 令Q ＝(K ² + H ² )(L ² + M ² )。

-> ((KL)2 + (KM)2 + (HL)2 + (HM)2)
-> ((A - C)2 + (D - B)2 + (D + B)2 + (A + C)2)
-> ((2 * A)2 + (2 * B)2 + (2 * C)2 + (2 * D)2)
-> 4 * N

• 所以，

N = ((K/2)2 + (H/2)2)(L2 + M2)

• 这样就存在一对K和H，它们都是偶数。

让我们通过一个例子来形象化上述方法。令N = 221。

1. 221 = 11 ² + 10 ² = 5 ² + 14 ²
2. 根据以上公式：

A = 11 - 5 = 6
B = 11 + 5 = 15
C = 14 - 10 = 4
D = 14 + 10 = 24

• 因此，以上值可用于计算K，H，L和M的值。

K = GCD(6, 4) = 2
H = GCD(16, 24) = 8
L = GCD(6, 24) = 3
M = GCD(16, 4) = 2

• 所以：

221 = ((2/2)2 + (8/2)2) * (32 + 22)
221 = 17 * 13

方法：为了实现上述方法，需要执行以下步骤：

1. 通过迭代从1到sqrt(N)的循环来求平方和，因为[sqrt(N)，N]之间不存在除N之外的因子，并找到两对平方和等于N的对。
2. 将值存储在A，B，C，D中。
3. 使用上述方法中提到的公式找到K，H，L和M的值。
4. 使用K，H，L和M的值查找因素。检查两个数字均是偶数的对，并将它们对半除，然后找出因子。

下面是上述方法的实现：

C++

// C++ program to implement Eulers
// Factorization algorithm
 
#include 
using namespace std;
 
// Function to return N as the sum of
// two squares in two possible ways
void sumOfSquares(int n, vector >& vp)
{
    // Iterate a loop from 1 to sqrt(n)
    for (int i = 1; i <= sqrt(n); i++) {
 
        // If i*i is square check if there
        // exists another integer such that
        // h is a perfect square and i*i + h = n
        int h = n - i * i, h1 = sqrt(h);
 
        // If h is perfect square
        if (h1 * h1 == h) {
 
            // Store in the sorted way
            int a = max(h1, i), b = min(h1, i);
 
            // If there is already a pair
            // check if pairs are equal or not
            if (vp.size() == 1 && a != vp[0].first)
                vp.push_back(make_pair(a, b));
 
            // Insert the first pair
            if (vp.size() == 0)
                vp.push_back(make_pair(a, b));
 
            // If two pairs are found
            if (vp.size() == 2)
                return;
        }
    }
}
 
// Function to find the factors
void findFactors(int n)
{
 
    // Get pairs where a^2 + b^2 = n
    vector > vp;
    sumOfSquares(n, vp);
 
    // Number cannot be represented
    // as sum of squares in two ways
    if (vp.size() != 2)
        cout << "Factors Not Possible";
 
    // Assign a, b, c, d
    int a, b, c, d;
 
    a = vp[0].first;
    b = vp[0].second;
 
    c = vp[1].first;
    d = vp[1].second;
 
    // Swap if a < c because
    // if a - c < 0,
    // GCD cant be computed.
    if (a < c) {
        int t = a;
        a = c;
        c = t;
        t = b;
        b = d;
        d = t;
    }
 
    // Compute the values of k, h, l, m
    // using the formula mentioned
    // in the approach
    int k, h, l, m;
    k = __gcd(a - c, d - b);
    h = __gcd(a + c, d + b);
    l = (a - c) / k;
    m = (d - b) / k;
 
    // Print the values of a, b, c, d
    // and k, l, m, h
    cout << "a = " << a
         << "\t\t(A) a - c = " << (a - c)
         << "\t\tk = gcd[A, C] = "
         << k << endl;
 
    cout << "b = " << b
         << "\t\t(B) a + c = " << (a + c)
         << "\t\th = gcd[B, D] = "
         << h << endl;
 
    cout << "c = " << c
         << "\t\t(C) d - b = " << (d - b)
         << "\t\tl = A/k = "
         << l << endl;
 
    cout << "d = " << d
         << "\t\t(D) d + b = " << (d + b)
         << "\t\tm = c/k = "
         << m << endl;
 
    // Printing the factors
    if (k % 2 == 0 && h % 2 == 0) {
        k = k / 2;
        h = h / 2;
 
        cout << "Factors are: "
             << ((k) * (k) + (h) * (h))
             << " " << (l * l + m * m)
             << endl;
    }
    else {
        l = l / 2;
        m = m / 2;
 
        cout << "Factors are: "
             << ((l) * (l) + (m) * (m))
             << " " << (k * k + h * h)
             << endl;
    }
}
 
// Driver code
int main()
{
    int n = 100000;
 
    findFactors(n);
 
    return 0;
}

Java

// Java program to implement Eulers
// Factorization algorithm
import java.util.*;
class GFG{
 
static class pair
{
  int first, second;
  public pair(int first,
              int second) 
  {
    this.first = first;
    this.second = second;
  }   
}
     
// Recursive function to
// return gcd of a and b 
static int __gcd(int a, int b) 
{ 
  return b == 0 ? a :
         __gcd(b, a % b);    
}
 
// Function to return N as the sum of
// two squares in two possible ways
static void sumOfSquares(int n,
                         Vector vp)
{
  // Iterate a loop from 1 to Math.sqrt(n)
  for (int i = 1; i <= Math.sqrt(n); i++)
  {
    // If i*i is square check if there
    // exists another integer such that
    // h is a perfect square and i*i + h = n
    int h = n - i * i, h1 = (int)Math.sqrt(h);
 
    // If h is perfect square
    if (h1 * h1 == h)
    {
      // Store in the sorted way
      int a = Math.max(h1, i),
          b = Math.min(h1, i);
 
      // If there is already a pair
      // check if pairs are equal or not
      if (vp.size() == 1 &&
          a != vp.get(0).first)
        vp.add(new pair(a, b));
 
      // Insert the first pair
      if (vp.size() == 0)
        vp.add(new pair(a, b));
 
      // If two pairs are found
      if (vp.size() == 2)
        return;
    }
  }
}
 
// Function to find the factors
static void findFactors(int n)
{
  // Get pairs where a^2 + b^2 = n
  Vector vp = new Vector<>();
  sumOfSquares(n, vp);
 
  // Number cannot be represented
  // as sum of squares in two ways
  if (vp.size() != 2)
    System.out.print("Factors Not Possible");
 
  // Assign a, b, c, d
  int a, b, c, d;
 
  a = vp.get(0).first;
  b = vp.get(0).second;
 
  c = vp.get(1).first;
  d = vp.get(1).second;
 
  // Swap if a < c because
  // if a - c < 0,
  // GCD cant be computed.
  if (a < c)
  {
    int t = a;
    a = c;
    c = t;
    t = b;
    b = d;
    d = t;
  }
 
  // Compute the values of k, h, l, m
  // using the formula mentioned
  // in the approach
  int k, h, l, m;
  k = __gcd(a - c, d - b);
  h = __gcd(a + c, d + b);
  l = (a - c) / k;
  m = (d - b) / k;
 
  // Print the values of a, b, c, d
  // and k, l, m, h
  System.out.print("a = " + a +
                   "\t\t(A) a - c = " + 
                   (a - c) +
                   "\t\tk = gcd[A, C] = " +
                   k + "\n");
 
  System.out.print("b = " + b +
                   "\t\t(B) a + c = " + 
                   (a + c) +
                   "\t\th = gcd[B, D] = " +
                   h + "\n");
 
  System.out.print("c = " +  c +
                   "\t\t(C) d - b = " + 
                   (d - b) +
                   "\t\tl = A/k = " +
                   l + "\n");
 
  System.out.print("d = " +  d +
                   "\t\t(D) d + b = " +
                   (d + b) +
                   "\t\tm = c/k = " +
                   m + "\n");
 
  // Printing the factors
  if (k % 2 == 0 && h % 2 == 0)
  {
    k = k / 2;
    h = h / 2;
    System.out.print("Factors are: " +
                     ((k) * (k) + (h) * (h)) +
                     " " + (l * l + m * m) + "\n");
  }
  else
  {
    l = l / 2;
    m = m / 2;
    System.out.print("Factors are: " +
                     ((l) * (l) + (m) * (m)) +
                     " " + (k * k + h * h) + "\n");
  }
}
 
// Driver code
public static void main(String[] args)
{
  int n = 100000;
  findFactors(n);
}
}
 
// This code is contributed by gauravrajput1

C#

// C# program to implement Eulers
// Factorization algorithm
using System;
using System.Collections.Generic;
class GFG{
 
public class pair
{
  public int first, second;
  public pair(int first,
              int second) 
  {
    this.first = first;
    this.second = second;
  }   
}
     
// Recursive function to
// return gcd of a and b 
static int __gcd(int a, int b) 
{ 
  return b == 0 ? a :
         __gcd(b, a % b);    
}
 
// Function to return N as the sum of
// two squares in two possible ways
static void sumOfSquares(int n,
                         List vp)
{
  // Iterate a loop from 1 to Math.Sqrt(n)
  for (int i = 1; i <= Math.Sqrt(n); i++)
  {
    // If i*i is square check if there
    // exists another integer such that
    // h is a perfect square and i*i + h = n
    int h = n - i * i, h1 = (int)Math.Sqrt(h);
 
    // If h is perfect square
    if (h1 * h1 == h)
    {
      // Store in the sorted way
      int a = Math.Max(h1, i),
          b = Math.Min(h1, i);
 
      // If there is already a pair
      // check if pairs are equal or not
      if (vp.Count == 1 &&
          a != vp[0].first)
        vp.Add(new pair(a, b));
 
      // Insert the first pair
      if (vp.Count == 0)
        vp.Add(new pair(a, b));
 
      // If two pairs are found
      if (vp.Count == 2)
        return;
    }
  }
}
 
// Function to find the factors
static void findFactors(int n)
{
  // Get pairs where a^2 + b^2 = n
  List vp = new List();
  sumOfSquares(n, vp);
 
  // Number cannot be represented
  // as sum of squares in two ways
  if (vp.Count != 2)
    Console.Write("Factors Not Possible");
 
  // Assign a, b, c, d
  int a, b, c, d;
 
  a = vp[0].first;
  b = vp[0].second;
 
  c = vp[1].first;
  d = vp[1].second;
 
  // Swap if a < c because
  // if a - c < 0,
  // GCD cant be computed.
  if (a < c)
  {
    int t = a;
    a = c;
    c = t;
    t = b;
    b = d;
    d = t;
  }
 
  // Compute the values of k, h, l, m
  // using the formula mentioned
  // in the approach
  int k, h, l, m;
  k = __gcd(a - c, d - b);
  h = __gcd(a + c, d + b);
  l = (a - c) / k;
  m = (d - b) / k;
 
  // Print the values of a, b, c, d
  // and k, l, m, h
  Console.Write("a = " + a +
                "\t\t(A) a - c = " + 
                (a - c) +
                "\t\tk = gcd[A, C] = " +                   
                k + "\n");
 
  Console.Write("b = " + b +
                "\t\t(B) a + c = " + 
                (a + c) +
                "\t\th = gcd[B, D] = " +
                h + "\n");
 
  Console.Write("c = " +  c +
                "\t\t(C) d - b = " + 
                (d - b) +
                "\t\tl = A/k = " +
                l + "\n");
 
  Console.Write("d = " +  d +
                "\t\t(D) d + b = " +
                (d + b) +
                "\t\tm = c/k = " +
                m + "\n");
 
  // Printing the factors
  if (k % 2 == 0 && h % 2 == 0)
  {
    k = k / 2;
    h = h / 2;
    Console.Write("Factors are: " +
                  ((k) * (k) + (h) * (h)) +
                  " " + (l * l + m * m) + "\n");
  }
  else
  {
    l = l / 2;
    m = m / 2;
    Console.Write("Factors are: " +
                  ((l) * (l) + (m) * (m)) +
                  " " + (k * k + h * h) + "\n");
  }
}
 
// Driver code
public static void Main(String[] args)
{
  int n = 100000;
  findFactors(n);
}
}
 
// This code is contributed by gauravrajput1

输出：

a = 316        (A) a - c = 16        k = gcd[A, C] = 8
b = 12        (B) a + c = 616        h = gcd[B, D] = 56
c = 300        (C) d - b = 88        l = A/k = 2
d = 100        (D) d + b = 112        m = c/k = 11
Factors are: 800 125

复杂度分析：
时间复杂度： O(sqrt(N)) ，其中N是给定的数字
空间复杂度： O(1)