托一个学弟的福,学了一下他的最简便三分写法,然后找了一道三分的题验证了下,AC了一题,写法确实方便,还是我太弱了,漫漫AC路!各路大神,以后你们有啥好的简便写法可以在博客下方留个言或私信我,谢谢了! Turn the corner Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3196    Accepted Submission(s…

传送门 Turn the corner Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2081    Accepted Submission(s): 787 Problem Description Mr. West bought a new car! So he is travelling around the city. One da…

Problem Description Mr. West bought a new car! So he is travelling around the city. One day he comes to a vertical corner. The street he is currently in has a width x, the street he wants to turn to has a width y. The car has a length l and a width d…

Turn the corner Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2229    Accepted Submission(s): 856 Problem Description Mr. West bought a new car! So he is travelling around the city. One day h…

传送门 分析 根据这张图,我们只要使得h<=y即可,可以发现h是一个凸函数,故使用三分,具体见代码 代码 #include<cstdio> #include<cstring> #include<vector> #include<algorithm> using namespace std; const double eps = 1e-7; const double pi = acos(-1.0); #define ll long long #defin…

                                                               B. The Meeting Place Cannot Be Changed The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost build…

Turn the corner Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1930    Accepted Submission(s): 736 Problem Description Mr. West bought a new car! So he is travelling around the city. One day h…

Turn the corner Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 151 Accepted Submission(s): 61   Problem Description Mr. West bought a new car! So he is travelling around the city. One day he come…

Problem Description Mr. West bought a new car! So he is travelling around the city. One day he comes to a vertical corner. The street he is currently in has a width x, the street he wants to turn to has a width y. The car has a length l and a width d…

HDU 4869 Turn the pokers 题目链接 题意:给定n个翻转扑克方式,每次方式相应能够选择当中xi张进行翻转.一共同拥有m张牌.问最后翻转之后的情况数 思路:对于每一些翻转,假设能确定终于正面向上张数的情况,那么全部的情况就是全部情况的C(m, 张数)之和.那么这个张数进行推理会发现,事实上会有一个上下界,每隔2个位置的数字就是能够的方案,由于在翻牌的时候,相应的肯定会有牌被翻转,而假设向上牌少翻一张,向下牌就要多翻一张.奇偶性是不变的,因此仅仅要每次输入张数,维护上下界,最后…

Turn the corner Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 654 Accepted Submission(s): 291   Problem Description Mr. West bought a new car! So he is travelling around the city. One day he com…

这题是道三分的题,首先要分析满足条件的情况,这个就是平面几何的功夫了.要想车子能够转弯成功,最上面那个点到水平线的距离要小于等于y.这里h和s的公式就是利用平面几何的知识求出来的:s=l*cos(a)+w*sin(a)-x;s=l*cos(a)+w*sin(a)-x;其中s为最右边的那个点到拐角处的水平距离.因为角度和高度h满足凸函数的关系,因此想到利用角度采用三分的方法进行求解. #include"iostream" #include"stdio.h" #incl…

Problem Description Mr. West bought a new car! So he is travelling around the city. One day he comes to a vertical corner. The street he is currently in has a width x, the street he wants to turn to has a width y. The car has a length l and a width d…

Problem Description Mr. West bought a new car! So he is travelling around the city.One day he comes to a vertical corner. The street he is currently in has a width x, the street he wants to turn to has a width y. The car has a length l and a width d.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4717 说明下为啥满足三分: 设y=f(x) (x>0)表示任意两个点的距离随时间x的增长,距离y的变化.则f(x)函数单调性有两种:1.先单减,后单增.2.一直单增. 设y=m(x) (x>0)表示随时间x的增长,所有点的最大距离y的变化.即m(x)是所有点对构成的f(x)图像取最上面的部分.则m(x)的单调性也只有两种可能:1.先单减,后单增.2.一直单增. 这个地方的证明可以这样:假如时刻t1…

Turn the pokers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1265    Accepted Submission(s): 465 Problem Description During summer vacation,Alice stay at home for a long time, with nothing t…

NPY and shot Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description NPY is going to have a PE test.One of the test subjects is throwing the shot.The height of NPY is H meters.He can throw the shot at t…

http://acm.hdu.edu.cn/showproblem.php?pid=4717 大致题意:给出每一个点的坐标以及每一个点移动的速度和方向. 问在那一时刻点集中最远的距离在全部时刻的最远距离中最小. 比赛时一直以为是计算几何,和线段相交什么的有关.赛后队友说这是道三分,细致想了想确实是三分.试着画绘图发现它是一个凸性函数,存在一个最短距离. 然后三分时间就能够了. #include <stdio.h> #include <iostream> #include <m…

Turn the pokers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1064    Accepted Submission(s): 398 Problem Description During summer vacation,Alice stay at home for a long time, with nothing to…

[题意] 在一个星球(是一个球体)表面有一个飞机(坐标(x1,y1,z1),原点是星球中心),在空中有一个空间站(坐标(x2,y2,z2)),所有值均小于100,现在要使飞机与空间站相遇,飞机的速度是1,空间站速度是v,v是小于等于100的整数.飞机只能沿星球表面飞,而空间站可以任意飞,当然不能进入星球内部. [解答] 首先可以把三维图形转化为二维的,也就是飞机,空间站,原点确定的那一个平面, 为什么是含原点的平面?我们不妨把球体旋转一下,把飞机的初始位置固定在球体的最上边,空间站在球的侧边的上…

http://acm.hdu.edu.cn/showproblem.php?pid=3912 这个题我用递归深搜模拟,直接爆栈了.哭啊!为什么! 这个题最主要是能走重复格子,但是方向不一样. 我用的剪枝条件:         1.判断有没有走出迷宫,走出迷宫了直接退出.         2.由于题目要求只要判断有没有把所有格子走完,则可以用一个计数器统计走过格子的数量(重复走的不算).走完了所有格子则直接退出搜索.         3.在这个格子,当前走的方向以前走过,则重复了(判重) 别人的代…

Description After a day trip with his friend Dick, Harry noticed a strange pattern of tiny holes in the door of his SUV. The local American Tire store sells fiberglass patching material only in square sheets. What is the smallest patch that Harry nee…

我比较快速的想到了三分,但是我是从0到2*pi区间进行三分,并且漏了一种点到边距离的情况,一直WA了好几次 后来画了下图才发现,0到2*pi区间内是有两个极值的,每个半圆存在一个极值 以下是代码 #include <cstdio> #include <cmath> #include <algorithm> #define pi acos(-1.0) using namespace std; typedef struct { double x; double y; }po…

Turn the pokers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 196    Accepted Submission(s): 51 Problem Description During summer vacation,Alice stay at home for a long time, with nothing to…

http://acm.hdu.edu.cn/showproblem.php?pid=1432 题目大意: 2维平面上给定n个点,求一条直线能够穿过点数最多是多少. 解题思路: 因为题目给定的n(1~700),所以枚举,时间复杂度是O(n^3),不会超时. 枚举两个点,然后判断剩下的点是否在这条直线. AC代码: #include<cstdio> struct Point{ int x, y; Point(, ): x(x), y(y){} void scan(){ scanf("%d…

The Moving Points Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 72    Accepted Submission(s): 18 Problem Description There are N points in total. Every point moves in certain direction and cer…

http://www.lydsy.com/JudgeOnline/problem.php?id=3203 wa无数次QAQ,犯sb错....一是数组没有引用...二是输出转成了int(越界了sad)..三是叉积的顺序忘记了(cross(v, w)>0的话说明v在下边啊....)...sad..然后提交无数次.wa了无数次..QAQ 还有这题的思想真巧妙.. 首先看题我们得得到一个公式. 对于每一关,yi=max{(sum[i]-sum[j-1])/(x[i]+(i-j)*d)} 我一开始很好奇这…

Turn the pokers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 108    Accepted Submission(s): 21 Problem Description During summer vacation,Alice stay at home for a long time, with nothing to…

pid=4869" target="_blank">Turn the pokers 大意:给出n次操作,给出m个扑克.然后给出n个操作的个数a[i],每一个a[i]代表能够翻的扑克的个数,求最后可能出现的扑克的组合情况. Hint Sample Input: 3 3 3 2 3 For the this example: 0 express face down,1 express face up Initial state 000 The first result:00…

Geometry Problem HDU - 6242 Alice is interesting in computation geometry problem recently. She found a interesting problem and solved it easily. Now she will give this problem to you : You are given NN distinct points (Xi,Yi)(Xi,Yi) on the two-dimens…