The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
- y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
- y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
- x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
4
1 2
1 2
1 0
0 1
0 2
NO
2
1 3
1 0
-1 3
YES
2
1 3
1 0
0 2
YES
2
1 3
1 0
0 3
NO
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
Solution
两直线y1, y2交点横坐标在开区间(x1, x2)的充要条件：(y1(x1)-y2(x1))*(y1(x2)-y2(x2)) < 0
直线y对应二元组(y(x1), y(x2))，即直线y被x=x1, x=x2所截得的线段，将所有二元组按字典序排序只要判断是否有相邻两线段相交的情况即可。
Implementation
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod(1e9+); const int N(1e5+); int sign(LL x){
return x?x>?:-:;
} vector<pair<LL,LL>> seg; int main(){
ios::sync_with_stdio(false);
cin.tie(), cout.tie(); int n, x1, x2;
cin>>n>>x1>>x2;
LL k, b;
for(int i=; i<n; i++){
cin>>k>>b;
LL y1=k*x1+b, y2=k*x2+b;
seg.push_back({y1, y2});
} sort(seg.begin(), seg.end());
for(int i=; i<seg.size(); i++){
if(sign(seg[i].first-seg[i-].first)*sign(seg[i].second-seg[i-].second)<){
puts("YES");
return ;
}
}
puts("NO");
return ;
}