文件名称:POJ2500
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圆内四边形最大面积
//描述：圆上平均分配的n个点，选择其中m个，在这m个里取出4个，保证4个点组成的面积最大，并输出面积值。
//超时思路：先对m个数排序，再删除相同元素，对所有两两组合的点分别找出最大面积的点，再计算面积
//修改方案：因为对称性，保证0点必取，因此只需与0进行组合照最大面积。
-Circle quadrilateral area
Descr iption:// N point on circle average distribution, choose the m, out of 4 in the m, 4 points with the largest area, and the output of the area value.
//timeout ideas: first on the M number of the same sort, delete elements, the combination of all two two points respectively to find the largest area, then calculate the area
//modifications: because of symmetry, which will take 0 points, so just as the largest area and the 0 combination.
//描述：圆上平均分配的n个点，选择其中m个，在这m个里取出4个，保证4个点组成的面积最大，并输出面积值。
//超时思路：先对m个数排序，再删除相同元素，对所有两两组合的点分别找出最大面积的点，再计算面积
//修改方案：因为对称性，保证0点必取，因此只需与0进行组合照最大面积。
-Circle quadrilateral area
Descr iption:// N point on circle average distribution, choose the m, out of 4 in the m, 4 points with the largest area, and the output of the area value.
//timeout ideas: first on the M number of the same sort, delete elements, the combination of all two two points respectively to find the largest area, then calculate the area
//modifications: because of symmetry, which will take 0 points, so just as the largest area and the 0 combination.
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POJ2500.txt