題目：

B. Han Solo and Lazer Gun time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

There are?n?Imperial stormtroopers on the field. The battle field is a plane with Cartesian coordinate system. Each stormtrooper is associated with his coordinates?(x,?y)?on this plane.

Han Solo has the newest duplex lazer gun to fight these stormtroopers. It is situated at the point?(x0,?y0). In one shot it can can destroy all the stormtroopers, situated on some line that crosses point?(x0,?y0).

Your task is to determine what minimum number of shots Han Solo needs to defeat all the stormtroopers.

The gun is the newest invention, it shoots very quickly and even after a very large number of shots the stormtroopers don't have enough time to realize what's happening and change their location.

Input

The first line contains three integers?n,?x0?и?y0?(1?≤?n?≤?1000,??-?104?≤?x0,?y0?≤?104) — the number of stormtroopers on the battle field and the coordinates of your gun.

Next?n?lines contain two integers each?xi,?yi?(?-?104?≤?xi,?yi?≤?104) — the coordinates of the stormtroopers on the battlefield. It is guaranteed that no stormtrooper stands at the same point with the gun. Multiple stormtroopers can stand at the same point.

Output

Print a single integer — the minimum number of shots Han Solo needs to destroy all the stormtroopers.

Examples input 4 0 0 1 1 2 2 2 0 -1 -1 output 2 input 2 1 2 1 1 1 0 output 1 Note

Explanation to the first and second samples from the statement, respectively:

分析：

數據結構里面找到的題...然而并沒想到數據結構解法？...

判有多少種不同的斜率 考慮到浮點數帶來的誤差，可用最簡分數來描述斜率

代碼：

#include<bits/stdc++.h> using namespace std; typedef pair<int,int> P;set<P> s;int gcd(int a,int b){ // if(b>a) swap(a,b);//交換的話負數會出問題 if(b==0) return a;else return gcd(b,a%b); }int main(){ios_base::sync_with_stdio(false);int n,x,y;int a,b,ans=0;s.clear();cin>>n>>x>>y;while(n--){cin>>a>>b;a-=x,b-=y;int mod=gcd(a,b);a/=mod;b/=mod;P p=P(a,b); // cout<<p.first<<" "<<p.second<<endl;if(!s.count(p)) ans++,s.insert(p);}cout<<ans<<endl;return 0; }

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