1179. Extrusion

                                   1179. Extrusion

Description 

The Acme Extrusion Company specializes in the production of steel bars with custom-designed cross-sections. The manufacturing process involves cutting a hole in a thick metal plate, the shape of the hole being determined by the customer's specifications.

Molten metal is then forced through the hole to form a long bar. The shape of the hole determines the shape of the cross-section of the resulting bar.

Given a description of a polygonal hole and the volume of molten metal available, determine how long a bar can be formed by this process.

 

Input

Input consists of one or more data sets consisting of the following information:

• An integer, N , indicating the number of vertices making up the polygon. End of input is signaled by any N less than 3.
• Next are N lines, each containing a pair of floating-point numbers, (x_i, y_i) , each denoting one vertex of the polygon. Vertices will be presented in clockwise order (relative to the closest interior point) proceeding around the perimeter of the polygon. The x_iand y_i values are in units of meters.
• The data set is terminated by a floating point value indicating the amount of molten metal available (in cubic meters).

Output

For each data set, the program should produce a single line of output of the form: BAR LENGTH: x where ``x " is the maximum bar length, a floating point number expressed with two digits precision.

Sample Input

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4 0.0 0.0 0.0 0.1 0.1 0.1 0.1 0.0 1.0 7 0.5 1.25 0.9 1.6 0.9  1.1 0.85 1.0 0.9 0.85 0.9 0.5 0.5 0.75 100.0 0

Sample Output

BAR LENGTH: 100.00 BAR LENGTH: 318.73

题目描述：

给出n个点，按顺时针顺序连接组成一个多边形，求它的面积，然后根据给出的体积，求高。

主要是求多边形面积
公式：|(x2*y1-y2*x1+x3*y2-y3*x2+…+x1*yn-y1*xn)|/2

#include<iostream>#include<algorithm>#include<iomanip>using namespace std;int main(){    double x,y,x0,y0,x1,y1;    int n;    double tiji;    while(cin>>n)    {     if(n==0) return 0;     cin>>x>>y;     x0=x;     y0=y;     double  note=0;     for(int i=1;i<n;i++)     {      cin>>x1>>y1;      note+=x1*y-y1*x;      x=x1;      y=y1;      }      note+=x0*y-y0*x;            cin>>tiji;      double re=tiji*2/note;      cout<<"BAR LENGTH: ";      cout << fixed << setprecision(2) << re << endl;    }    system("pause");    return 0;}