JZOJ 6290. 倾斜的线

题目

Description

 

Input

Output

 

Sample Input

6 15698 17433
112412868 636515040
122123982 526131695
58758943 343718480
447544052 640491230
162809501 315494932
870543506 895723090 
 

Sample Output

193409386/235911335
 

 

Data Constraint

分析

 

• 我们可以对于每个点都做一条斜率为p/q的线
• 然后截距排序 然后答案的直线肯定是相邻两个点
• 求最小就好了
• 那么怎么求截距呢
• 截距=a[i].y-p/q*a[i].x

 

代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
const int N=2e5;
const double INF=1e9+7;
struct sb
{
    double x,y;
    double b;
}a[N+10];
bool cmp(sb a,sb b){return a.b<b.b;};
double P,Q,ans,nowx,nowy,now=INF;
int n;
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
int main ()
{
    freopen("slope.in","r",stdin);
    freopen("slope.out","w",stdout);
    cin>>n>>P>>Q;
    ans=P/Q;
    for (int i=1;i<=n;i++) cin>>a[i].x>>a[i].y,a[i].b=a[i].y-P/Q*a[i].x;
    sort(a+1,a+1+n,cmp);
    for (int i=1;i<n;i++)
    {
        double k=(a[i].y-a[i+1].y)/(a[i].x-a[i+1].x);
        if (abs(k-ans)<now) now=abs(k-ans),nowx=abs((a[i].x-a[i+1].x)),nowy=(a[i].y-a[i+1].y);
    }
    long long ansx=(long long)nowx,ansy=(long long)nowy;
    long long g=gcd(ansx,ansy);
    ansx/=g,ansy/=g;
    cout<<abs(ansy)<<"/"<<abs(ansx);
}