We have two rows. There are a dots on the top row and b dots on the bottom row. We draw line segmentsconnecting every dot on the top row with every dot on the bottom row. The dots are arranged in sucha way that the number of internal intersections am…

  How Many Points of Intersection?  We have two rows. There are a dots on the top row and b dots on the bottom row. We draw line segments connecting every dot on the top row with every dot on the bottom row. The dots are arranged in such a way that t…

 How Many Points of Intersection?  We have two rows. There are a dots on the toprow andb dots on the bottom row. We draw line segments connecting every dot on the top row with every dot on the bottom row. The dots are arranged in such a way that the…

uva10790:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1731 题意:两条水平线,分别有n,m个点,点之间两两连线,求有多少个交点. 题解:手动模拟一下,然后用不完全归纳法,就可以得到公式ans=(n-1)*n/2*(m-1)*m/2; #include<iostream> #include<cstdio>…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

113 - Power of Cryptography import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub BigInteger num1, num2; int n; int kc = 1000000000; int l, r, mid, t; Sc…

Robotics Perception Professor Kostas and Jianbo Shi week 1: camera model 凸透镜成像原理:凸透镜焦点与焦距是固定的,这是物理性质.物距u.像距v.焦距f的关系为1/f=1/u+1/v perspective drawing bi-perspectograph construction changing 1. change the distance from the objects: OS 2. change the foca…

A General Polygon Clipping Library Version 2.32    http://www.cs.man.ac.uk/~toby/alan/software/gpc.html Alan Murta Advanced Interfaces Group Department of Computer Science University of Manchester Manchester M13 9PL, UK Abstract: This document descri…

XTools Pro 是一套为ArcGIS平台设计的矢量空间分析. 形状转换和表管理扩展工具,大大增强了 ArcGIS 的功能,使用该工具能够提高 ArcGIS 用户的效率和性能. XTools Pro 包括矢量空间分析. 形状转换. 并表管理的 70 多个工具和超过 40 地理处理工具.在一个单一的工具栏为 ArcGIS 用户收集的最基本的功能.新的实用工具有助于节省您的时间和提高工作效率! 该工具相当于把ArcGIS Desktop平台软件更加的实用化,业务化,更贴合用户的应用,非常多相似的…

Circles and Pi Introduction id: intro-1 For as long as human beings exist, we have looked to the sky and tried to explain life on Earth using the motion of stars, planets and the moon. The Greeks were the first to discover that all celestial objects…

题目链接 https://www.luogu.org/problemnew/show/U56187 注:题面参考了网上的其他博客,并非原题题面,因此数据范围可能有误.数据为原创数据. 题解 其实就是许多板子码到一起. 首先对于边缘上的任意一点 \(u\),假设离它最远的顶点为 \(A\),那么我们称点 \(u\) 位于顶点 \(A\) 的控制范围之中.我们考虑在没有石雕的情况下怎么求出每个顶点的控制范围.对于除顶点 \(A\) 之外的任意一个顶点 \(B\),连接 \(AB\) 并作 \(AB\…

The information in this topic is useful if you're trying to programmatically find a built-in command, menu, or toolbar. The ICommandBars.Find and ICommandBar.Find methods can be used to get a reference to a specific toolbar, menu, or command. Both of…

Background\text{Background}Background Last night, lots of students from primary school came to our class to study OI.\text{Last night, lots of students from primary school came to our class to study OI.}Last night, lots of students from primary schoo…

Matlab geom3d函数解析 geom3d函数库 geom3d库的目的是处理和可视化三维几何原语,如点.线.平面.多面体等.它提供了操作三维几何原语的底层功能,使得开发更复杂的几何算法变得更加容易. | 序号 | 函数名 | 作用 | 输入 | 输出 | | --- | --- | --- | --- | --- | | 1 | area = triangleArea3d(pt1, pt2, pt3) | 获取三角形面积 | 三个点坐标 [10 10 10], [30 10 10], [1…

Cool Points We have a circle of radius R and several line segments situated within the circumference of this circle. Let’s define a cool point to be a point on the circumference of this circle so that the line segment that is formed by this point and…

UVA.10325 The Lottery (组合数学 容斥原理) 题意分析 首先给出一个数n,然后给出m个数字(m<=15),在[1-n]之间,依次删除给出m个数字的倍数,求最后在[1-n]之间还剩下多少个数字(包括1和n),已知m个数字中不会包含1(否则全部都被刷掉了). 前置技能 1. 给出数字s,在[1-n]之间,s的倍数有n/s个. 2. 给出数字s1,和s2,在[1-n]之间,既是s1的倍数,又是s2的倍数,有n/lcm(s1,s2)个. 3. 给出数字s1,s2--sk(共k个数字…

UVA.11806 Cheerleaders (组合数学 容斥原理 二进制枚举) 题意分析 给出n*m的矩形格子,给出k个点,每个格子里面可以放一个点.现在要求格子的最外围一圈的每行每列,至少要放一个点,并且放在角上的点,同时算那个角所在的行和所在的列.不允许剩下点,求总共的方案数量,结果对1000007取模. 数据范围2 ≤ M,N ≤ 20,K ≤ 500. 考虑到要求组合数目,首先就需要预处理500以内的组合数.正向求解可能有些困难,这样考虑: 不管三七二十一,先求解出所有情况的总和,即C…

option=com_onlinejudge&Itemid=8&category=471&page=show_problem&problem=4224" style="">题目链接:uva 1478 - Delta Wave 题目大意:对于每一个位置来说,能够向上,水平,向下.坐标不能位负.每次上下移动最多为1. 给定n问说有多少种不同的图.结果对10100取模. 解题思路:由于最后都要落回y=0的位置,所以上升的次数和下降的次数是同样的…

UVA 10869 - Brownie Points II 题目链接 题意:平面上n个点,两个人,第一个人先选一条经过点的垂直x轴的线.然后还有一个人在这条线上穿过的点选一点作垂直该直线的线,然后划分出4个象限,第一个人得到分数为1.3象限,第二个人为二四象限.问第一个个人按最优取法,能得到最小分数的最大值,和这个值下还有一个人的得分可能情况 思路:树状数组,能够枚举一点,假设能求出右上和左下点的个数就好办了,其有用一个树状数组,把y坐标离散化掉,然后记录进来,然后把点按x从左往右,每次删掉点后…

We have a circle of radius R and several line segments situated within the circumference of this circle. Let’s define a cool point to be a point on the circumference of this circle so that the line segment that is formed by this point and the centre…

A pair of numbers has a unique LCM but a single number can be the LCM of more than one possiblepairs. For example 12 is the LCM of (1, 12), (2, 12), (3,4) etc. For a given positive integer N, thenumber of different integer pairs with LCM is equal to N…

Problem A Chess Queen Input: Standard Input Output: Standard Output You probably know how the game of chess is played and how chess queen operates. Two chess queens are in attacking position when they are on same row, column or diagonal of a chess bo…

UVA: https://vjudge.net/problem/UVA-11806 AC代码 #include <bits/stdc++.h> #define pb push_back #define mp make_pair #define fi first #define se second #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define huan pr…

从左往右处理,左半部分记为left, 右半部分记为right,若i,i -1均为1, 贡献为ans += (left + 1) + right * (1ll << (i - 1)); 否则贡献为ans += right * (1ll << (i - 1)); import java.math.BigInteger; import java.util.Scanner; public class Main{ public static void main(String[] args){…

题意:长度为n的序列,前m位恰好k位正确排序,求方法数 前m位选k个数正确排,为cm[m][k],剩余m - k个空位,要错排,这m - k个数可能是前m个数中剩下的,也可能来自后面的n - m个数 考虑这样一个问题,共n个数,前i位错排的方法数,显然dp[i][0] = i! 递推考虑:处理到第i个数时,等价于前i - 1个数错排的方法数减去在前i - 1个数错排的情况下第i位恰好为i的方法数,后者相当于n - 1个数前i - 1位错排 所以 dp[n][i] = dp[n][i - 1] -…

根据组合数公式C(m,n),由于m可能达到20万,因此转换为ln,之后可以表达为ln(m!)-ln(n!)-ln((m-n)!); 求每一个c[n]时,也要根据杨辉三角求组合数进行转化. 注意long double输出一般要用cout, printf不好使. #include <cstdio> #include <iostream> #include <cstring> #include <cmath> #define repu(i,a,b) for(int…

In a galaxy far far away there is an ancient game played among the planets. The specialty of the gameis that there is no limitation on the number of players in each team, as long as there is a captain inthe team. (The game is totally strategic, so so…

/* * UVA_10098.cpp * * Created on: 2013年10月8日 * Author: Administrator */ #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; char s[11]; int l; bool get() { int i = l - 1; int j; while…

题意:有n个人,选不少于一个人参加比赛,其中一人当队长,有多少种选择方案. 思路:我们首先C(n,1)选出一人当队长,然后剩下的 n-1 人组合的总数为2^(n-1),这里用快速幂解决 代码: #include <iostream> #define ll long long using namespace std; ; ll qmod(ll a, ll b) { ll ans=; while(b) { ) { ans=(ans*a)%mod; } b=b/; a=(a*a)%mod; } re…

题意:给你N个数,求把他们的全排列加和为多少 思路:对于这道题,假设数字k1在第一位,然后求出剩下N-1位的排列数num1,我们就可以知道k1在第一位时 排列有多少种为kind1, 同理,假设数字k2在第一位然后求出剩下N-1位的排列数num2, 我们就可以知道k2在第一位时的排列有多少种为kind2, k1*num1+k1*num2.....+kn*numn 就是我们要求的这些数对第一位的所有贡献, 我们知道第一位的贡献=对第二位的贡献=第三位的贡献..... 把所有贡献加和,就能求出结果 知…