Description After taking a modern art class, Farmer John has become interested in finding geometric patterns in everything around his farm. He carefully plots the locations of his N cows (2 <= N <= 1000), each one occupying a distinct point in the 2…

Symmetry Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description   The figure shown on the left is left-right symmetric as it is possible to fold the sheet of paper along a vertical line, drawn as a dashed line, and to cut…

题目链接:fzu 2035 Axial symmetry 题目大意:给出n个点,表示n边形的n个顶点,判断该n边形是否为轴对称图形.(给出点按照图形的顺时针或逆时针给出. 解题思路:将相邻两个点的中点有序的加入点集,即保证点是按照图形的顺时针或逆时针出现的,然后枚举i和i + n两点的直线作为对称轴.判断其他所有点是否对称即可. #include <stdio.h> #include <string.h> #include <stdlib.h> #include <…

题意:平面上给若干点,问它们是不是关于某垂直于x轴的直线对称. 代码:(Wrong Answer, –ms) //UVa1595 - Symmetry #include<iostream> #include<vector> #include<algorithm> using namespace std; struct point { int x, y; bool operator <(point& that)const { if (y == that.y)…

针对对称性PDB 3UKM,使用make_symmdef_file.pl脚本,可以执行产生对称单元及对称文件: $> $ROSETTA3/src/apps/public/symmetry/make_symmdef_file.pl -m NCS -a A -i B -p 3UKM.pdb > 3UKM.symm 参数意义: Here we are using symmetry to model an already-symmetric starting protein. This is call…

A. Clear Symmetry 题目连接: http://codeforces.com/contest/201/problem/A Description Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from l…

The main reasons why hydraulic motors have this symmetrical internal structure are as follows: The     Hydraulic Motor manufacturers    states: 1. Because this product is in use, it needs to be able to achieve forward and reverse. In order to achieve…

1595 - Symmetry The figure shown on the left is left-right symmetric as it is possible to fold the sheet of paper along a vertical line, drawn as a dashed line, and to cut the figure into two identical halves. The figure on the right is not left-righ…

「JSOI2015」symmetry 传送门 我们先考虑构造出原正方形经过 \(4\) 种轴对称变换以及 \(2\) 种旋转变换之后的正方形都构造出来,然后对所得的 \(7\) 个正方形都跑一遍二维哈希,这样我们就可以通过哈希,在 \(O(n ^ 2)\) 时间内判断原正方形中是否存在某一类型的某一大小的子正方形. 但是如果我们枚举边长,复杂度就会达到 \(O(n ^ 3)\) 级别,显然过不了. 考虑优化:我们发现对于任意一种类型的正方形,它把最外面一圈去掉之后还是满足原来的性质,所以我们可以…

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