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：

1 #include 
     
       2 using namespace std; 3 typedef long long ll; 4 const ll INF = 0x3f3f3f3f3f3f3f3f; 5 ll n, k; 6 void solve(ll r, ll & minn, ll & maxn) 7 { 8     for (int x = r ? 0 : 1; k * x + r <= n * k; x++) 9     {10         ll ans = n * k / __gcd(n * k, k * x + r);11         minn = min(minn, ans);12         maxn = max(maxn, ans);13     }14 }15 int main()16 {17     ll a, b;18     while (cin >> n >> k >> a >> b)19     {20         ll minn = INF, maxn = 0;21         solve((a + b) % k, minn, maxn);22         solve(((-a + b) % k + k) % k, minn, maxn);23         solve(((a - b) % k + k) % k, minn, maxn);24         solve(((-a - b) % k + k) % k, minn, maxn);25         cout << minn << " " << maxn << endl;26     }27     return 0;28 }