CF1152C Neko does Maths

思路：

假設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)（根據gcd(a, b) = gcd(b - a, a)），因此枚舉b - a的全部因子便可。

實現：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const ll INF = 0x3f3f3f3f3f3f3f3f;
 5 ll cal(ll x, ll y, ll i, ll & minn, ll & ans)
 6 {
 7     if (i > x)
 8     {
 9         if (y + i - x < minn)
10         {
11             minn = y + i - x;
12             ans = i - x;
13         }
14     }
15     else
16     {
17         ll k = 0;
18         if (x % i) k = (x / i + 1) * i - x;
19         ll t = (x + k) * (y + k) / __gcd(y - x, x + k);
20         if (t < minn)
21         {
22             minn = t;
23             ans = k;
24         }
25     }
26 }
27 int main()
28 {
29     ll a, b;
30     while (cin >> a >> b)
31     {
32         ll x = min(a, b), y = max(a, b);
33         if (y % x == 0) { cout << 0 << endl; continue; }
34         ll minn = INF, ans = 0;
35         for (int i = 1; i * i <= y - x; i++)
36         {
37             if ((y - x) % i == 0)
38             {
39                 cal(x, y, i, minn, ans);
40                 if ((y - x) / i != i) cal(x, y, (y - x) / i, minn, ans);
41             }
42         }
43         cout << ans << endl;
44     }
45     return 0;
46 }