temp

g = gcd(a+k,b+k),设 a+k = g*x,b + k = g*y,则g(x-y) = a-b, 所以 g|(a-b),枚举a-b的所有约数，按照题意取出最小答案

#include "bits/stdc++.h"

using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 100;
ll a, b;
vector<ll> yu;
ll maxx = 0x3f3f3f3f3f3f3f3f;
ll ans = 0x3f3f3f3f3f3f3f3f;

int main() {
    //freopen("in.txt", "r", stdin);
    cin >> a >> b;
    if (a == b) {
        cout << 0 << endl;
        return 0;
    }
    if (__gcd(a, b) == a || __gcd(a, b) == b) {
        cout << 0 << endl;
        return 0;
    }
    if (a < b) swap(a, b);

    ll n = a - b;
    for (ll i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            yu.push_back(i);
            if (i != n / i) yu.push_back(n / i);
        }
    }
    ll x, y, cha;
    for (auto g:yu) {
        cha = n / g;
        y = (b + g - 1) / g;
        if ((y + cha) * g >= a) {
            x = y + cha;
            if (g * x * y < maxx) {
                maxx = g * x * y;
                ans = g * x - a;
            } else if (g * x * y == maxx && g * x - a < ans) {
                ans = g * x - a;
            }
        } else {
            x = (a + g - 1) / g;
            y = x - cha;
            if (g * x * y < maxx) {
                maxx = g * x * y;
                ans = g * x - a;
            } else if (g * x * y == maxx && g * x - a < ans) {
                ans = g * x - a;
            }
        }
    }
    cout << ans << endl;
    return 0;
}