Line.h #pragma once //Microsoft Visual Studio 2015 Enterprise //根据两点式方法求直线,并求两条直线的交点 #include"BoundaryPoint.h" #include"Coordinates.h" class Line { public: Line GetLine(BoundaryPoint sourcePoint, BoundaryPoint endPoint); Line GetLine(C
Area Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 225 Accepted Submission(s): 77 Problem Description 电子科大清水河校区是电子科大大力兴建的未来主校区,于07年秋正式迎接学生入住,目前有07.08级本科生及部分研究生在此校区学习.生活.清水河校区位于成都高新西区的中部地带,占
两次DFS求树直径方法见 这里. 这里的直径是指最长链包含的节点个数,而上一题是指最长链的路径权值之和,注意区分. K <= R: ans = K − 1; K > R: ans = R − 1 + ( K − R ) ∗ 2; #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; ; struct n
python中对两个 list 求交集,并集和差集: 1.首先是较为浅白的做法: >>> a=[1,2,3,4,5,6,7,8,9,10] >>> b=[1,2,3,4,5] >>> intersection=[v for v in a if v in b] >>> intersection [1, 2, 3, 4, 5] >>> union=b.extend([v for v in a]) >>>
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13549 Accepted Submission(s): 6645 Problem Description Many geometry(几何)problems were designed in the ACM/
def diff(listA,listB): #求交集的两种方式 retA = [i for i in listA if i in listB] retB = list(set(listA).intersection(set(listB))) print "retA is: ",retA print "retB is: ",retB #求并集 retC = list(set(listA).union(set(listB))) print "retC1 is
B. Our Tanya is Crying Out Loud time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Right now she actually isn't. But she will be, if you don't solve this problem. You are given integers n, k,
Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common. Input Input begins with a number T show
(1)获取两个时间相差天数(没有上午下午区分) var d1=ABS_DATESTRING(FStartTime,'yyyy/MM/dd'); var d2=ABS_DATESTRING(FEndTime,'yyyy/MM/dd'); var date1= new Date(d1); var date2=new Date(d2); var time=date2.getTime()-date1.getTime(); var day=time/(1000*60*60*24);day (2)求历时几小