/*思路完全搞乱，开始就没想清楚就写。我晕，各种WA。 思路： 　　找到所有点中最下边一层点里边最靠左的点d。然后求d到其他每个点连线与x轴的 夹角Θ（0 <= Θ <= π 因为d的纵坐标最小）。然后从小到大排序，找到存角度的数组 里n/2号点就是要找的另一个点。*/ My Code: #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           #define pi 3.1415926535 using namespace std; const int N = 10010; struct point{
   double x; double y; }p[N]; struct angle{
   double ang; int i; }a[N]; point tmp; bool cmp(angle a, angle b){
   return a.ang < b.ang; } int main(){
   //freopen("data.in", "r", stdin); int n, i, d, j; double b, c;     scanf("%d", &n);     scanf("%lf%lf", &p[1].x, &p[1].y);     b = p[1].x; c = p[1].y; d = 1; for(i = 2; i <= n; i++){
           scanf("%lf%lf", &p[i].x, &p[i].y); if(c > p[i].y){
               b = p[i].x; c = p[i].y; d = i;         } if(c == p[i].y && b > p[i].x){
               b = p[i].x; c = p[i].y; d = i;         }     } for(j = 1, i = 1; i <= n; i++){
   if(i != d){
               a[j].i = i;             b = p[i].x - p[d].x;             c = p[i].y - p[d].y; if(b == 0)    {a[j++].ang = pi/2; continue;} if(c == 0)    {a[j++].ang = 0; continue;} if(p[i].x < p[d].x)                 a[j++].ang = pi - atan(c/b); else                 a[j++].ang = atan(c/b);         }     }     sort(a+1, a+n, cmp); /*for(i = 1; i < n; i++){
            printf("%lf %d\n", a[i].ang, a[i].i);     }*/     printf("%d %d\n", d, a[n/2].i); return 0; }