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…

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…

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> #…

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…

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…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5914 Problem Description Mr. Frog has n sticks, whose lengths are 1,2, 3⋯n respectively. Wallice is a bad man, so he does not want Mr. Frog to form a triangle with three of the sticks here. He decides t…

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…

题意是逆时针方向给你两个多边形,问你这两个多边形通过旋转和平移能否拼成一个凸包. 首先可以想到的便是枚举边,肯定是有一对长度相同的边贴合,那么我们就可以n2枚举所有边对,接下来就是旋转点对,那么假设多边型在这条向量的左侧,那么我们可以根据叉积正负判断旋转的方向. 然后就是如何判断了,显然有一种情况是凸包里扣除整个多边形,那么这种情况我们需要对重合的边进行删除,可以发现,如果有重合的边,他一定是成对出现,有一条进去的边也有一条出来的边,那么我们可以直接通过vector不断插入的过程和之前那条边比较…