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2015

极角排序的几种常见的方式

·378 words·1 min
20190211update:工作的时候看同事ocr的代码,发现有一段就是极角排序orz…所以说算法还是有用的… 先介绍几种极角排序: 1.利用叉积的正负来作cmp.(即是按逆时针排序).此题就是用这种方法

hdu 1086 A - You can Solve a Geometry Problem too (线段的规范相交&&非规范相交)

·583 words·2 mins
A - You can Solve a Geometry Problem too **Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u Submit Status Description Many geometry(几何)problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :) Give you N (1<=N<=100) segments(线段), please output the number of all intersections(交点). You should count repeatedly if M (M>2) segments intersect at the same point.

hdoj 2036 (计算几何,叉积求面积)

·778 words·2 mins
改革春风吹满地 # **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24179 Accepted Submission(s): 12504 ** Problem Description " 改革春风吹满地, 不会AC没关系; 实在不行回老家, 还有一亩三分地。 谢谢!(乐队奏乐)"

codeforces #329 div 2 B. Anton and Lines(几何)

·961 words·2 mins
B. Anton and Lines time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The teacher gave Anton a large geometry homework, but he didn’t do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = k__i*x + b__i. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between _x_1 < _x_2. In other words, is it true that there are1 ≤ i < j ≤ n and x’, y’, such that:

codeforces #329 div 2 A. 2Char (暴力)

·700 words·2 mins
A. 2Char time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn’t written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.

hdu 1754 I Hate It (线段树)

·746 words·2 mins
I Hate It # **Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 53991 Accepted Submission(s): 21180 ** Problem Description 很多学校流行一种比较的习惯。老师们很喜欢询问,从某某到某某当中,分数最高的是多少。 这让很多学生很反感。

codeforces 589 B - Layer Cake

·698 words·2 mins
B - Layer Cake **Time Limit:**6000MS **Memory Limit:**524288KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 589B Description Dasha decided to bake a big and tasty layer cake. In order to do that she went shopping and bought n rectangular cake layers. The length and the width of the i-th cake layer were a__i and b__i respectively, while the height of each cake layer was equal to one. From a cooking book Dasha learned that a cake must have a form of a rectangular parallelepiped constructed from cake layers of the same sizes.

zoj 3634 Bounty hunter(dp,已经想明白)

·846 words·2 mins
M - Bounty hunter **Time Limit:**5000MS **Memory Limit:**65536KB 64bit IO Format:%lld & %llu Submit Status Description Bounty hunter is a hero who always moves along cities to earn money by his power. One day he decides to N cities one by one At the beginning ,Bounty hunter has X money and Y points of Attack force. At day 1, he will goes to city 1, then city 2 at day 2, city 3 at day 3, … At last ,he goes to city N at day N and leaves it at day N+1. In each city, he can increase his attack force by money and earn some money by accepting a task. In the city i, it costs him ai money to increase one point of attack force. And he can gets bi*yi money after finishing the task in city i while yi is his attack force after his increasing at city i.