Anton and Lines(思维)

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 = 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, or b_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).

Examples

input

4 1 2 1 2 1 0 0 1 0 2

output

NO

input

2 1 3 1 0 -1 3

output

YES

input

2 1 3 1 0 0 2

output

YES

input

2 1 3 1 0 0 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.

题解：如果两个直线在x1--x2间有交点，必然对于直线A的y1,y2,包含直线B的y1,y2；只要知道这就可以了，求出每个点的y1,y2排序判断就好：

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
typedef __int64 LL;
const int MAXN=100100;
struct Node{
    LL l,r;
    bool operator < (const Node b) const{
        if(l!=b.l)return l<b.l;
        else return r<b.r;
    }
};
Node dt[MAXN];
int main(){
    int N,x1,x2;
    LL k,b;
    while(~scanf("%d",&N)){
        scanf("%d%d",&x1,&x2);
        for(int i=0;i<N;i++){
            scanf("%I64d%I64d",&k,&b);
            dt[i].l=k*x1+b;dt[i].r=k*x2+b;
        }
        sort(dt,dt+N);
        int flot=0,k=0;
        for(int i=0;i<N;i++){
            if(dt[i].l>dt[k].l&&dt[i].r<dt[k].r)flot=1;
            while(dt[i].r>dt[k].r)k++;
        }
        if(flot)puts("YES");
        else puts("NO");
    }
    return 0;
}

暴力必然超时：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
typedef __int64 LL;
double x1,x2;
const int MAXN=100010;
struct Node{
    int k,b;
};
Node dt[MAXN];
bool js(Node x,Node y){
    if(x.k==y.k)return false;
    double temp=1.0*(y.b-x.b)/(x.k-y.k);
//    printf("%lf\n",temp);
    if(temp>x1&&temp<x2)return true;
    else return false;
}
int main(){
    int n;
    while(~scanf("%d",&n)){
        scanf("%lf%lf",&x1,&x2);
        for(int i=0;i<n;i++){
            scanf("%d%d",&dt[i].k,&dt[i].b);
        }
        int flot=0;

        for(int i=0;i<n;i++){
            if(flot)break;
            for(int j=i+1;j<n;j++){
                if(js(dt[i],dt[j]))flot=1;
                if(flot)break;
            }
        }
        if(flot)puts("YES");
        else puts("NO");
    }
}