CF1142A The Beatles

思路：

令p表示步数，l表示步长。由于p是使(l * p) % (n * k) == 0的最小的p，所以p = (n * k) / gcd(n * k, l).

设l = k * x + r，则由题意可知r有四种可能的取值，分别是(a + b) % k, ((-a + b) % k + k) % k, ((a - b) % k + k) % k, ((-a - b) % k + k) % k，枚举各种情况计算即可。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll INF = 0x3f3f3f3f3f3f3f3f;
 ll n, k;
 void solve(ll r, ll & minn, ll & maxn)
 {
     for (int x = r ?  : ; k * x + r <= n * k; x++)
     {
         ll ans = n * k / __gcd(n * k, k * x + r);
         minn = min(minn, ans);
         maxn = max(maxn, ans);
     }
 }
 int main()
 {
     ll a, b;
     while (cin >> n >> k >> a >> b)
     {
         ll minn = INF, maxn = ;
         solve((a + b) % k, minn, maxn);
         solve(((-a + b) % k + k) % k, minn, maxn);
         solve(((a - b) % k + k) % k, minn, maxn);
         solve(((-a - b) % k + k) % k, minn, maxn);
         cout << minn << " " << maxn << endl;
     }
     return ;
 }