很明显 kkk 是 A,BA,BA,B 的公约数，也就是 gcd⁡(A,B)\gcd(A,B)gcd(A,B) 的约数，直接枚举即可。要注意约数成对出现，所以复杂度为 O(gcd⁡(A,B)O(\sqrt{\gcd(A,B})O(gcd(A,B ). Code:

递归，辗转相除法

#include <bits/stdc++.h> using namespace std; int gcd(int a, int b) { while (b != 0) { int temp = b; b = a % b; a = temp; } return a; } int main() { int A, B; cin >> A >> B; int g = gcd(A, B); vector<int> divisors; for (int i = 1; i * i <= g; ++i) { if (g % i == 0) { divisors.push_back(i); if (i != g / i) { divisors.push_back(g / i); } } } sort(divisors.begin(), divisors.end()); for (int k : divisors) { cout << k << " "; } return 0; }