View Code of Problem 15

#include <iostream>
#include <iomanip>
#include <string>
#include <stdio.h>
#include <math.h>
 
using namespace std;
 
int main()
{
  int n;
  int index = 1;
  string x;
  while (cin >> n)
  {
      if (n == 0)
              break;
      double s = 0;//总面积
      for(int i = 0;i<n;i++)
      {
              double x1 = 0, y1= 0, x2=0, y2=0;//默认每个图形都是正多边形，所以只需记录一份边长即可
              int m;//表示本平面点的个数
              double a = 0;//边长
 
              cin >> m >> x1 >> y1 >> x2 >> y2;
              //这里需要对输入缓存做一次清空,不然本次输入多余的部分将会对下一次读取影响
              getline(cin, x);
 
              a = sqrt((y1 - y2)*(y1 - y2) + (x1 - x2)*(x1 - x2));
              
              switch (m)
              {
                      case 3:
                              s += 1.73205 / 4 * a * a;
                              break;
                      case 4:
                              s += a * a;
                              break;
                      case 5:
                              s += 1.72048 * a * a;
                              break;
                      case 6:
                              s += 1.73205 / 4 * a*a * 6;
                              break;
                      case 7:
                              s += 3.63391 * a * a;
                              break;
                      case 8:
                              s += 4.82843 * a*a;
                              break;
                      default:
                              break;
              }
      }
      printf("Case #%d: %.4f\n", index++, s);
  }
}