题意: 给你正n边形上的三个点,问n最少为多少 思路: 三个点在多边形上,所以三个点的外接圆就是这个正多边形的外接圆,余弦定理求出每个角的弧度值,即该角所对边的圆周角,该边对应的圆心角为圆心角的二倍.同时这个圆心角应为正多边形的每条边对应圆心角的整数倍,即2*pi/i的整数倍,遍历for i 1 to 1000 ,判断A*i/pi是否均为整数即可 坑点: 注意判断double是否为整数:return a-(int)(a+eps); 代码: #include<iostream> #include…

思路:三角形的圆心角可以整除(2*pi)/n #include<cstdio> #include<cstring> #include<iostream> #include<queue> #include<stack> #include<algorithm> using namespace std; #define clc(a,b) memset(a,b,sizeof(a)) #define inf 0x3f3f3f3f ; #defi…

Given a list of points that form a polygon when joined sequentially, find if this polygon is convex (Convex polygon definition). Note: There are at least 3 and at most 10,000 points. Coordinates are in the range -10,000 to 10,000. You may assume the…

Given a list of points that form a polygon when joined sequentially, find if this polygon is convex (Convex polygon definition). Note: There are at least 3 and at most 10,000 points. Coordinates are in the range -10,000 to 10,000. You may assume the…

The Triangle Division of the Convex Polygon 题意:求 n 凸多边形可以有多少种方法分解成不相交的三角形,最后值模 m. 思路:卡特兰数的例子,只是模 m 让人头疼,因为 m 不一定是素数,所以不一定存在逆元. 解法:式子为f(n) =  ( C( 2*(n-2),  (n-2) ) / (n-1))   % m :令 p = n-2, 式子可化为:f(p) = ((2*p)! / ( p! * (p+1)! ) ) % m; 对 s!分解质因素,统计个…

65536K   Teemo is very interested in convex polygon. There is a convex n-sides polygon, and Teemo connect every two points as diagonal lines, and he want to kown how many segments which be divided into intersections. Teemo ensure that any three diago…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 计算向量夹角 日期 题目地址:https://leetcode-cn.com/problems/convex-polygon/ 题目描述 Given a list of points that form a polygon when joined sequentially, find if this polygon is convex (Convex…

from:http://blog.csdn.net/qb371/article/details/8102109 Locate the following tool - ArcToolbox > Conversion Tools > To Shapefile > Feature Class To Shapefile (multiple) Launch the tool and add your Polygon ZM shapefiles to the input features. Sel…

http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11562&courseid=0 求n边形分解成三角形的方案数. 就是求n-2个卡特兰数,从大神那盗取了一份模板,效率极高.同时也很复杂. #include <cstdio> #include <cmath> #include <stdlib.h> #include <memory.h> typedef int ty…

题意: 求第n-2个Catalan数 模上 m. 思路: Catalan数公式: Catalan[n] = C(n, 2n)/(n+1) = (2n)!/[(n+1)!n!] 因为m是在输入中给的,所以我们不能用求阶乘和其逆元的方法来求.因为当m不是素数的时候,可能不存在逆元. 这里,我们把阶乘做质因数分解,然后上下两边约分,即可求出解. 怎么来把这个n!因式分解呢? 我们知道 n!中某个因子x的数量可以用log(n)的方法来求. 详见:这里 代码: #include <iostream> #…