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  • 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.

    Sample test(s)
    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:


    题意:给出武器位置和敌人位置,要求计算武器小灭所有敌人所需的最少直线方向,其实就是求坐标系中所有敌人组成的不同斜率有多少;

    #include<cstring>
    #include<cstdio>
    #include<iostream>
    #include<set>
    
    using namespace std;
    
    int main()
    {
        int n;
        double xx,yy;
        while(cin>>n>>xx>>yy)
        {
            int p=0;
            set<double>s;
            for(int i=0;i<n;i++)
            {
                double a,b;
                cin>>a>>b;
                if(fabs(a-xx)<=1e-10) p=1;
                else s.insert((b-yy)/(a-xx));
            }
            cout<<s.size()+p<<endl;
        }
    }
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  • 原文地址:https://www.cnblogs.com/wsaaaaa/p/4292896.html
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