poj1244Slots of Fun

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几何的简单题，建立坐标，判断相等以及不共线

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 200
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 char s[N],sr[];
 struct Point
 {
     double x,y;
     Point(double x=,double y=):x(x),y(y) {}
 }p[N];
 vector<Point>ed[];
 typedef Point pointt;
 pointt operator + (Point a,Point b)
 {
     return Point(a.x+b.x,a.y+b.y);
 }
 pointt operator - (Point a,Point b)
 {
     return Point(a.x-b.x,a.y-b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     else return x<?-:;
 }
 double dis(Point a)
 {
     return sqrt(a.x*a.x+a.y*a.y);
 }
 double cross(Point a,Point b)
 {
     return a.x*b.y-a.y*b.x;
 }
 double mul(Point p0,Point p1,Point p2)
 {
     return cross(p1-p0,p2-p0);
 }
 int main()
 {
     int n,i,j,g,e;
     while(scanf("%d",&n)&&n)
     {
         getchar();
         for(i = ; i <  ; i++)
         ed[i].clear();
         for(i =  ; i <= n*(n+)/; i++)
         scanf("%c",&s[i]);
         int g = ;
         double tx=n,ty = sqrt(3.0)*(n-);
         for(i =  ;i <= n; i++)
         {
             double x0 = tx-i+;
             double y0 = ty-(i-)*sqrt(3.0);
             for(j = ; j <= i ;j++)
             {
                 Point pp;
                 pp.y = y0;
                 pp.x = x0;
                 x0+=;
                 g++;
                 ed[s[g]-'a'].push_back(pp);
             }
         }
         int cnt = ;
         for(i = ; i <  ; i++)
         {
             if(ed[i].size()<) continue;
             int k = ed[i].size();
             for(j =  ;j < k ; j++)
                 for(g = j+ ; g < k ; g++)
                     for(e = g+ ; e < k ; e++)
                     {
                         Point p1 = ed[i][j],p2 = ed[i][g],p3 = ed[i][e];
 //                        cout<<p1.x<<" "<<p1.y<<" "<<p2.x<<" "<<p2.y<<" "<<p3.x<<" "<<p3.y<<endl;
 //                        cout<<i<<" "<<dis(p1-p2)<<" "<<dis(p1-p3)<<" "<<dis(p2-p3)<<endl;
                         if(dcmp(dis(p1-p2)-dis(p1-p3))==&&dcmp(dis(p1-p2)-dis(p2-p3))==&&dcmp(mul(p1,p2,p3))!=)
                         {
                             sr[cnt++] = i+'a';
                         }
                     }
         }
         if(cnt==)
         {
             puts("LOOOOOOOOSER!");
             continue;
         }
         for(i =  ; i < cnt ; i++)
         printf("%c",sr[i]);
         puts("");
     }
     return ;
 }