Anton and Lines（模拟，有点贪心的思想）

B. Anton and Lines

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 are1 ≤ 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.

Input

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.

Output

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).

Sample test(s)

input

41 21 21 00 10 2

output

NO

input

21 31 0-1 3

output

YES

input

21 31 00 2

output

YES

input

21 31 00 3

output

NO

Note

In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.

                                                                         

思路：

     

For problem B, I think that there's no need to use EPS. The announced message during the contest maybe lead contestants the wrong ways.

My solution is all using int. In general, a Line Y = a * X + b, because [a, b] are Integers, [X1, X2] are also Integers, so we can sure that [Y1, Y2] are Integers either.

So now you have ([Yi1, Yi2]), sort that array.

Two lines (i, i + 1) are intersected strictly in range(X1, X2) must have a order that Yi1 <= Y(i+1)1 andYi2 >= Y(i+1)2. Because all lines are distinct, so these pairs are not identical.

Remember to use long long int.

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<vector>#include<cstdio>#include<cmath>using namespace std;#define CRL(a) memset(a,0,sizeof(a))typedef __int64 ll;#define T 200005pair<ll,ll> p[T];int main(){#ifdef zsc    freopen("input.txt","r",stdin);#endifll n,i,k,x1,x2,b;while(~scanf("%I64d",&n)){scanf("%I64d%I64d",&x1,&x2);for(i=0;i<n;++i){scanf("%I64d%I64d",&k,&b);p[i] = make_pair(k*x1+b,k*x2+b);}sort(p,p+n);bool flag = false;for(i=1;i<n;++i){if(p[i-1].second>p[i].second){flag = true;break;}}if(flag)printf("YES\n");elseprintf("NO\n");}return 0;}