Create 数论（GCD和LCM关系）

数论（GCD和LCM关系）

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+#include <iostream>
+#include <vector>
+#include <algorithm>
+
+using namespace std;
+
+// 函数：计算两个数的最大公约数 (GCD)
+int gcd(int a, int b) {
+    while (b != 0) {
+        int temp = b;
+        b = a % b;
+        a = temp;
+    }
+    return a;
+}
+
+int main() {
+    const int g = 12;   // 已知的 gcd
+    const int l = 360;  // 已知的 lcm
+
+    // 检查输入是否合法：必须满足 g*l = a*b 的约束，但这里我们直接找 x*y = l/g
+    int product_xy = l / g;
+
+    if (l % g != 0) {
+        cout << "Invalid input: LCM must be a multiple of GCD." << endl;
+        return 0;
+    }
+
+    cout << "Finding all pairs (a, b) such that:" << endl;
+    cout << "  gcd(a, b) = " << g << ", lcm(a, b) = " << l << endl;
+    cout << "  a and b are positive integers, and a <= b" << endl << endl;
+
+    vector<pair<int, int>> solutions;
+
+    // 枚举所有 x*y = product_xy，且 gcd(x, y) == 1，x <= y
+    for (int x = 1; x * x <= product_xy; ++x) {
+        if (product_xy % x == 0) {
+            int y = product_xy / x;
+
+            if (gcd(x, y) == 1) {
+                int a = g * x;
+                int b = g * y;
+
+                if (a <= b) {
+                    solutions.emplace_back(a, b);
+                }
+            }
+        }
+    }
+
+    // 输出结果
+    cout << "All possible (a, b) pairs are:" << endl;
+    for (const auto& p : solutions) {
+        cout << "(" << p.first << ", " << p.second << ")" << endl;
+    }
+
+    cout << "\nTotal number of solutions: " << solutions.size() << endl;
+
+    return 0;
+}