D. The Beatles
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n⋅kn⋅k cities. The cities are numerated from 11to n⋅kn⋅k, the distance between the neighboring cities is exactly 11 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are nn fast food restaurants on the circle, they are located in the 11-st, the (k+1)(k+1)-st, the (2k+1)(2k+1)-st, and so on, the ((n−1)k+1)((n−1)k+1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is kk km.
Sergey began his journey at some city ss and traveled along the circle, making stops at cities each ll km (l>0l>0), until he stopped in ss once again. Sergey then forgot numbers ss and ll, but he remembers that the distance from the city ss to the nearest fast food restaurant was aakm, and the distance from the city he stopped at after traveling the first ll km from ss to the nearest fast food restaurant was bb km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer xx is the minimum number of stops (excluding the first) Sergey could have done before returning to ss. The second integer yy is the maximum number of stops (excluding the first) Sergey could have done before returning to ss.
Input
The first line contains two integers nn and kk (1≤n,k≤1000001≤n,k≤100000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers aa and bb (0≤a,b≤k20≤a,b≤k2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers xx and yy.
Examples
input
Copy
2 3 1 1
output
Copy
1 6
input
Copy
3 2 0 0
output
Copy
1 3
input
Copy
1 10 5 3
output
Copy
5 5
Note
In the first example the restaurants are located in the cities 11 and 44, the initial city ss could be 22, 33, 55, or 66. The next city Sergey stopped at could also be at cities 2,3,5,62,3,5,6. Let's loop through all possible combinations of these cities. If both ss and the city of the first stop are at the city 22 (for example, l=6l=6), then Sergey is at ss after the first stop already, so x=1x=1. In other pairs Sergey needs 1,2,31,2,3, or 66 stops to return to ss, so y=6y=6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so ll is 22, 44, or 66. Thus x=1x=1, y=3y=3.
In the third example there is only one restaurant, so the possible locations of ss and the first stop are: (6,8)(6,8) and (6,4)(6,4). For the first option l=2l=2, for the second l=8l=8. In both cases Sergey needs x=y=5x=y=5 stops to go to ss.
题意：给一个有n*k个点的环，环上从1开始，每k个点有一个餐厅，你每次单向移动L个点，现在不知道你的初始位置和L的值，但是知道你初始位置离最近的餐厅的距离a和你第一次移动之后离最近的餐厅的距离b。你移动若干次后能回到起点，求这个值的最小值和最大值。
分析：sb暴力题。
考虑L有两种情况
（1）L=mk+b-a
（2）L=mk-b-a
m取1到n
原因画个图就知道了，那么移动次数=lmc(n*k,l)/l=n*k/gcd(nk,l)，n<=1e5，暴力枚举即可。
#include "bits/stdc++.h"
using namespace std;
int main()
{
long long n,k;
cin>>n>>k;
long long a,b;
cin>>a>>b;
long long ansmi=8e18,ansma=0;
long long l1=b-a,l2=k-b-a;
if(l1<=0)l1+=k;
if(l2<=0)l2+=k;
for (int i = 0; i < n; ++i) {
ansmi=min(ansmi,n*k/(__gcd(n*k,l1)));
ansmi=min(ansmi,n*k/(__gcd(n*k,l2)));
ansma=max(ansma,n*k/(__gcd(n*k,l1)));
ansma=max(ansma,n*k/(__gcd(n*k,l2)));
l1+=k,l2+=k;
}
printf("%lld %lld\n",ansmi,ansma);
}