CF1152C Neko does Maths

思路：

假设a <= b，lcm(a + k, b + k) = (a + k) * (b + k) / gcd(a + k, b + k) = (a + k) * (b + k) / gcd(b - a, a + k)（根据gcd(a, b) = gcd(b - a, a)），所以枚举b - a的所有因子即可。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll INF = 0x3f3f3f3f3f3f3f3f;
 ll cal(ll x, ll y, ll i, ll & minn, ll & ans)
 {
     if (i > x)
     {
         if (y + i - x < minn)
         {
             minn = y + i - x;
             ans = i - x;
         }
     }
     else
     {
         ll k = ;
         if (x % i) k = (x / i + ) * i - x;
         ll t = (x + k) * (y + k) / __gcd(y - x, x + k);
         if (t < minn)
         {
             minn = t;
             ans = k;
         }
     }
 }
 int main()
 {
     ll a, b;
     while (cin >> a >> b)
     {
         ll x = min(a, b), y = max(a, b);
         if (y % x == ) { cout <<  << endl; continue; }
         ll minn = INF, ans = ;
         for (int i = ; i * i <= y - x; i++)
         {
             if ((y - x) % i == )
             {
                 cal(x, y, i, minn, ans);
                 if ((y - x) / i != i) cal(x, y, (y - x) / i, minn, ans);
             }
         }
         cout << ans << endl;
     }
     return ;
 }