BNU 7543 && HDU 3425 Wax (split room + analogue)

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"The main topic"

Give a rectangle, a width*hight rectangle, over the bottom edge of a point (dor,0) to do m-1 Ray, the area of the rectangle evenly divided into m parts, to find these rays and the rectangle (m-1) intersection of the coordinates

The idea of "problem solving" is given in the code:

Judging from left to right, each additional unit area state coordinates//can prove each area is a triangle # include <iostream> #include <algorithm> #include <bits/stdc ++.h>using namespace Std;const double eps=1e-6;struct node{double x, y;}    Map[105];int Main () {int i,j;    Double W,h,door,worker;          while (~SCANF ("%lf%lf%lf%lf", &w,&h,&door,&worker), (W+h+door+worker)) {double sum=w*h;     Total area double S=sum/worker;          The area of each room is double res=0.0;            for (i=0; i<worker-1; i++) {res+=s;       if (res<= (W-door) *h/2.0)///border to the right of the coordinates {MAP[I].X=W;            The area is derived high map[i].y=res*2.0/(W-door);                } else if (res<= (2*w-door) *h/2.0)///border to the left of the coordinates {map[i].x=w-(Res*2.0/h-w+door);            Map[i].y=h;                The else//coordinates have a boundary {double ress=sum-res;//the remaining area map[i].x=0;            Map[i].y=ress*2.0/door;     }   } for (i=0; i<worker-1; i++) {if (i) printf ("");        printf ("%.3lf%.3lf", map[i].x,map[i].y);    } printf ("\ n"); } return 0;}

BNU 7543 && HDU 3425 Wax (split room + analogue)