摘要:
int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy) //testx,testy 是要判断的点{ int i, j, c = 0; for (i = 0, j = n... 阅读全文
摘要:
转自https://www.cnblogs.com/jbelial/archive/2011/08/05/2128625.htmldouble Multiply(POINT p1 , POINT p2 , POINT p3) // 叉积 { return ... 阅读全文
摘要:
挑战程序设计P258代码https://blog.csdn.net/sepnine/article/details/46804299将圆的x左边左端和右端存储起来 for(int i=0;i >::iterator it=outers.lower_bound(make... 阅读全文