ICPC Asia Urumqi 2017 G.The Mountain

All as we know, a mountain is a large landform that stretches above the surrounding land in a limited area. If we as the tourists take a picture of a distant mountain and print it out, the image on the surface of paper will be in the shape of a particular polygon.

From mathematics angle we can describe the range of the mountain in the picture as a list of distinct points, denoted by (x1,y1) to (xn,yn). The first point is at the original point of the coordinate system and the last point is lying on the x-axis. All points else have positive y coordinates and incremental x coordinates. Specifically, all x coordinates satisfy 0=x1<x2<x3<...<xn. All y coordinates are positive except the first and the last points whose y coordinates are zeroes.

The range of the mountain is the polygon whose boundary passes through points (x1,y1) to (xn,yn) in turn and goes back to the first point. In this problem, your task is to calculate the area of the range of a mountain in the picture.

Input

The input has several test cases and the first line describes an integer t (1 ≤ t ≤ 20) which is the total number of cases.

In each case, the first line provides the integer n (1 ≤ n ≤100) which is the number of points used to describe the range of a mountain. Following n lines describe all points and the i-th line contains two integers xi and yi yi (0 ≤ xi,yi ≤1000) indicating the coordinate of the i-th point.

Output

For each test case, output the area in a line with the precision of 66 digits.

样例输入

3
3
0 0
1 1
2 0
4
0 0
5 10
10 15
15 0
5
0 0
3 7
7 2
9 10
13 0

样例输出

1.000000
125.000000
60.500000

题解

给出n的值和n个点的坐标，求这几个点组成的多边形的面积。其中所有的点横纵坐标都大于0，且最后一个点在y轴上，故这个多边形在第一象限上。

公式：area = |(x1y2 - y1x2) + (x2y3 - y2x3) + ... (xn-1yn - yn-1xn)/2|

代码：

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import java.util.Scanner;
public class Polygon {
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    int c = input.nextInt();
    while(c-- > 0) {
      int pn = input.nextInt();
      int[][] p = new int[pn][2];
      for(int i = 0; i < pn; i ++) {
        p[i][0] = input.nextInt();
        p[i][1] = input.nextInt();
      }
      double sum = 0;
      for(int i = 0; i < p.length - 1; i++) {
        sum += p[i][0] * p[i+1][1] - p[i][1] * p[i+1][0];
      }
      sum /= 2.00;
      System.out.printf("%.6f\n",Math.abs(sum));
    }
  }
}