CodeForces - 593B Anton and Lines (数学方程&技巧) 判断直线是否相交

CodeForces - 593B

Anton and Lines

Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

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Description

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 = k_i·x + b_i. 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' = k_i * x' + b_i, that is, point (x', y') belongs to the line number i;
• y' = k_j * x' + b_j, 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 k_i, b_i ( - 1 000 000 ≤ k_i, b_i ≤ 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 k_i ≠ k_j, orb_i ≠ b_j.

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 Input

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

Hint

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

Source

Codeforces Round #329 (Div. 2)

//题意：直线方程为y=k*x+b.

先输入n，再输入x1,x2（左端点、右端点的横坐标，y1,y2可以根据下面给的k,b求出）,然后输入n组数，每组数都是k  b。问组成的这n条直线是否会相交。

//思路：

首先根据方程和给出的k  b求出y1,y2，将其存入结构体中，将其排序，在遍历这n条直线，如果不平行，就证明相交了。

#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3f3f#define ll long long#define N 100010#define M 1000000007using namespace std;struct zz{ll l;ll r;}p[N];int cmp(zz a,zz b){if(a.l==b.l)return a.r<b.r;return a.l<b.l;}int main(){int n;while(scanf("%d",&n)!=EOF){ll l,r,k,b;scanf("%lld%lld",&l,&r);for(int i=0;i<n;i++){scanf("%lld%lld",&k,&b);ll y1=l*k+b;ll y2=r*k+b;p[i].l=y1;p[i].r=y2;}int flag=0;sort(p,p+n,cmp);for(int i=1;i<n;i++){if(p[i].r<p[i-1].r)//因为左端点在排序后已经严格按照从小到大的顺序排好序了，只用判断右端点就行了。 {flag=1;break;}}printf(flag?"YES\n":"NO\n");}return 0;}