codeforces961D. Pair Of Lines_geometry

最新推荐文章于 2021-02-06 13:09:36 发布

原创最新推荐文章于 2021-02-06 13:09:36 发布 · 134 阅读

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本文探讨了一个关于在笛卡尔平面上用两条直线覆盖给定点集的问题，并提供了一种算法解决方案。该问题要求所有整数坐标点至少位于所画的两条直线上之一。文章通过具体的示例说明了算法的有效性。

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You are given n points on Cartesian plane. Every point is a lattice point (i. e. both of its coordinates are integers), and all points are distinct.

You may draw two straight lines (not necessarily distinct). Is it possible to do this in such a way that every point lies on at least one of these lines?

Input

The first line contains one integer n (1 ≤ n ≤ 10⁵) — the number of points you are given.

Then n lines follow, each line containing two integers x_i and y_i (|x_i|, |y_i| ≤ 10⁹)— coordinates of i-th point. All n points are distinct.

Output

If it is possible to draw two straight lines in such a way that each of given points belongs to at least one of these lines, print YES. Otherwise, print NO.

Examples

Input

Copy

Output

Copy

YES

Input

Copy

Output

Copy

NO

Note

In the first example it is possible to draw two lines, the one containing the points 1, 3 and 5, and another one containing two remaining points.

$\text{[math]}$

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define all(x) begin(x),end(x)
ll rd(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

struct Point{
    ll x,y;
    void in(){
        x=rd();y=rd();
    }
    Point operator+(const Point &a)const{
        return Point({x+a.x,y+a.y});
    }
    Point operator-(const Point &a)const{
        return Point({x-a.x,y-a.y});
    }
    ll operator*(const Point &a)const{
        return x*a.y-y*a.x;
    }
};

const int N=1e5+10;
int n;
Point a[N];
Point b[N];

bool solve(Point P,Point p){
    int m=0;
    for(int i=0;i<n;i++)
        if((a[i]-P)*p!=0)
            b[m++]=a[i];
    if(m<=2)return true;
    P=b[0];
    p=b[1]-b[0];
    for(int i=0;i<m;i++)
        if((b[i]-P)*p!=0)
            return false;
    return true;
}

int main(){
    freopen("in.txt","r",stdin);
    n=rd();
    for(int i=0;i<n;i++)
    a[i].in();
    if(n<=4)return puts("YES"),0;
    if(solve(a[0],a[1]-a[0])||solve(a[0],a[2]-a[0])||solve(a[1],a[2]-a[1]))
        return puts("YES"),0;
    return puts("NO"),0;
}