这题的俩种方法都是看别人的代码,方法可以学习学习,要多看看.. 几何题用到向量.. Points on Cycle Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1294    Accepted Submission(s): 455 Problem Description There is a cycle with its cente…

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=1700 题目: Points on Cycle Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2523    Accepted Submission(s): 972 Problem Description There is a cyc…

Problem Description There is a cycle with its center on the origin. Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other you may assume that the radius of the cycle…

                                            Points on Cycle Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description There is a cycle with its center on the origin. Now give you a point on the cycle, you are to find ou…

Points on Cycle Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1303    Accepted Submission(s): 459 Problem Description There is a cycle with its center on the origin. Now give you a point on t…

http://acm.hdu.edu.cn/showproblem.php?pid=1700 Points on Cycle Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1567    Accepted Submission(s): 570 Problem Description There is a cycle with its c…

L - Points on Cycle Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description There is a cycle with its center on the origin. Now give you a point on the cycle, you are to find out the other two points on it, to maximiz…

Problem Description There is a cycle with its center on the origin. Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other you may assume that the radius of the cycle…

Description There is a cycle with its center on the origin. Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other you may assume that the radius of the cycle will not…

题目链接:HDU 1700 Problem Description There is a cycle with its center on the origin. Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other you may assume that the radius…

题目链接 水题,卡了下下精度. #include <cstdio> #include <iostream> #include <cmath> using namespace std ; #define PI acos(-1.0) #define eps 1e-8 int judge(double x,double y) { double a; a = x-y; ) a = -a; if(a < eps) ; else ; } int main() { double…

题目链接:http://acm.hust.edu.cn/vjudge/contest/121192#problem/L 这是一道很有意思的题,就是给定一个以原点为圆心的圆,然后给定 一个点  求最大三角的 其他的坐标 ,很容易知道这个三角形一定是等边三角形,所以圆心就是三角形的重心,刚开始用直线的方程来写,还要解一个复杂的一元二次方程,十分复杂,后来又想到一种简单的方法 这样就可以得到一个点的坐标  又重心为圆点  则三点的横坐标之和为0 总坐标之和为0 ac代码: import java.io…

一个简单的几何题,自己在纸上列出方程解出结果的表达式,再用程序表达出来就行了. 不过老司机(老司机的woodcoding)说用旋转向量法比较简单,有时间要去看一看. 大致题意:一个圆心在原点的圆,半径未知,现在给你圆上的一点,让你在这个圆上找到另外两点,使得这三点构成的三角形的周长最长. 样例输入:(第一行为一个整数N,表示后面有N组案例,每个案例给出一组圆上点的坐标) 2 1.500        2.000 563.585    1.251 样例输出:(其他两个点的坐标) 0.982 -2.…

已知圆心(0,0)圆周上的一点,求圆周上另外两点使得三点构成等边三角形. 懒得推公式,直接用模板2圆(r1=dist,r2=sqrt(3)*dist)相交水过 #include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<iterator> using namespace std; #define eps 1e-6 typedef long lon…

http://acm.hdu.edu.cn/showproblem.php?pid=1700 题目大意: 二维平面,一个圆的圆心在原点上.给定圆上的一点A,求另外两点B,C,B.C在圆上,并且三角形ABC的周长是最长的. 解题思路: 我记得小学的时候给出个一个定理,在园里面正多边形的的周长是最长的,这个定理我不会证明. 所以这里是三角形,当三角形为正三角形的时候,周长是最长的. 因为圆心在原点,所以我就向量(x,y)绕原点逆时针旋转120度和顺时针旋转120度.给定的点A可以看成(x,y)向量.…

However, because the reference is weak, the object that self points to could be deallocated while theblock is executing.You can eliminate this risk by creating a strong local reference to self inside the block: __weak BNREmployee *weakSelf = self; //…

2.6 Given a circular linked list, implement an algorithm which returns the node at the beginning of the loop.DEFINITIONCircular linked list: A (corrupt) linked list in which a node's next pointer points to an earlier node, so as to make a loop in the…

  英文稿: The “Hype Cycle for Emerging Technologies” report is the longest-running annual Hype Cycle, providing a cross-industry perspective on the technologies and trends that IT managers should consider in developing emerging-technology portfolios (se…

1.Given a linked list, determine if it has a cycle in it. 2.Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 3.Given a linked list, return the length of the cycle, if there is no cycle, return 0; 题目要求: 1…

Spring Beans are the most important part of any Spring application. Spring ApplicationContext is responsible to initialize the Spring Beans defined in spring bean configuration file. Spring Context is also responsible for injection dependencies in th…

在使用JSONObject.fromObject的时候,出现“There is a cycle in the hierarchy”异常.   意思是出现了死循环,也就是Model之间有循环包含关系:   解决办法:   使用setCycleDetectionStrategy防止自包含   代码: JsonConfig jsonConfig=new JsonConfig();  jsonConfig.setIgnoreDefaultExcludes(false);    jsonConfig.se…

元素轮播效果是页面中经常会使用的一种效果.这个例子实现了通过元素的隐藏和显示来表现轮播效果.效果比较简单. 效果图如下: 源代码如下: <!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" Content="text/html; charset=utf-8;"> <title> cycle demo </title> <…

每一个实数都能用有理数去逼近到任意精确的程度,这就是有理数的稠密性.The rational points are dense on the number axis.…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 这道题给了我们一堆二维点,然后让我们求最大的共线点的个数,根据初中数学我们知道,两点确定一条直线,而且可以写成y = ax + b的形式,所有共线的点都满足这个公式.所以这些给定点两两之间都可以算一个斜率,每个斜率代表一条直线,对每一条直线,带入所有的点看是否共线并计算个数,这是整体的思路.但是还有…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 这个求单链表中的环的起始点是之前那个判断单链表中是否有环的延伸,可参见我之前的一篇文章 (http://www.cnblogs.com/grandyang/p/4137187.html). 还是要设…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 这道题是快慢指针的经典应用.只需要设两个指针,一个每次走一步的慢指针和一个每次走两步的快指针,如果链表里有环的话,两个指针最终肯定会相遇.实在是太巧妙了,要是我肯定想不出来.代码如下: C++ 解法: class Solution { public: bool hasCycle…

Given a linked list, determine if it has a cycle in it. ExampleGiven -21->10->4->5, tail connects to node index 1, return true Challenge Follow up:Can you solve it without using extra space? LeetCode上的原题,请参见我之前的博客Linked List Cycle. class Solution…

https://vjudge.net/problem/UVA-11090 平均权值最小的回路 为后面的做个铺垫 二分最小值,每条边权减去他,有负环说明有的回路平均权值小于他 spfa求负环的时候可以先把所有点加到队列里,d[i]=0 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; ,INF=1e9; inl…

题目链接 Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 分析:首先要注意的是,输入数组中可能有重复的点.由于两点确定一条直线,一个很直观的解法是计算每两个点形成的直线,然后把相同的直线合并,最后包含点最多的直线上点的个数就是本题的解.我们知道表示一条直线可以用斜率和y截距两个浮点数(垂直于x轴的直线斜率为无穷大,截距用x截距),同时还需要保存每…

题目要求 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 如何判断一个单链表中有环? Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle…