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  • Codeforces Round #291 (Div. 2) B. Han Solo and Lazer Gun

    因为是x,y均为整数因此对于同一直线的点,其最简分数x/y是相同的(y可以为0,这里不做除法)于是将这些点不断求最简分数用pair在set中去重即可。

    #include <cmath>
    #include <cstdio>
    #include <cstdlib>
    #include <cassert>
    #include <cstring>
    #include <set>
    #include <map>
    #include <list>
    #include <queue>
    #include <string>
    #include <iostream>
    #include <algorithm>
    #include <functional>
    #include <stack>
    using namespace std;
    typedef long long ll;
    #define T int t_;Read(t_);while(t_--)
    #define dight(chr) (chr>='0'&&chr<='9')
    #define alpha(chr) (chr>='a'&&chr<='z')
    #define INF (0x3f3f3f3f)
    #define maxn (100005)
    #define hashmod 100000007
    #define ull unsigned long long
    #define repne(x,y,i) for(i=x;i<y;++i)
    #define repe(x,y,i) for(i=x;i<=y;++i)
    #define ri register int
    void Read(int &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if(chr=='-')sign=-1;
        for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;}
    void Read(ll &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if
        (chr=='-')sign=-1;
        for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;}
    int n,x,y,a,b;
    int gcd(int x,int y){return y==0?x:gcd(y,x%y);}
    set<pair<int,int> > se1,se2;
    int main()
    {
        freopen("a.in","r",stdin);
        freopen("b.out","w",stdout);
        ri i;
        Read(n),Read(x),Read(y);
        repe(1,n,i){
            Read(a),Read(b);
            int ex = abs(x - a),ey = abs(y - b),g = gcd(ex,ey);
            if((x - a >= 0 && y - b >= 0) || (x - a <= 0 && y - b <= 0)) se1.insert(make_pair(ex/g,ey/g));
            else se2.insert(make_pair(ex/g,ey/g));
        }
        printf("%d
    ",(int)se1.size() + (int)se2.size());
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhuiyicc/p/9588899.html
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