/* poj 2187 Beauty Contest 凸包:寻找每两点之间距离的最大值 这个最大值一定是在凸包的边缘上的! 求凸包的算法: Andrew算法! */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Point{ Point(){} Point(int x, int y){ this->…

题目链接: Beauty of Sequence Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 813    Accepted Submission(s): 379 Problem Description Sequence is beautiful and the beauty of an integer sequence is def…

Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill bet…

题目传送门 /* DP:dp 表示当前输入的x前的包含x的子序列的和, 求和方法是找到之前出现x的位置(a[x])的区间内的子序列: sum 表示当前输入x前的所有和: a[x] 表示id: 详细解释:http://blog.csdn.net/u013050857/article/details/45285515 */ #include <cstdio> #include <algorithm> #include <cmath> #include <iostrea…

http://poj.org/problem?id=2187 显然直径在凸包上(黑书上有证明).(然后这题让我发现我之前好几次凸包的排序都错了QAQ只排序了x轴.....没有排序y轴.. 然后本题数据水,暴力也能过... (之前一直以为距离是单增的,其实并不是,应该是三角形面积单增...) 考虑旋转卡壳 一篇好的文章:http://www.cnblogs.com/Booble/archive/2011/04/03/2004865.html 首先对踵点就是两条平行线夹紧凸包的两个点(或者3个点或4…

One more question, rational beauty should come from a dedicated brain and mind. Should there be more awards. The other problem, the world is full of eclipse, the collector should be able to make some of them a whole, that`s where all envy comes from.…

http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5496 The 12th Zhejiang Provincial Collegiate Programming Contest - D Beauty of Array Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward has an array A with N integers. He defines…

Beauty of Array Time Limit: 2 Seconds Memory Limit: 65536 KB Edward has an array A with N integers. He defines the beauty of an array as the summation of all distinct integers in the array. Now Edward wants to know the summation of the beauty of all…

链接:http://poj.org/problem?id=2187 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the wor…

Beauty Contest 题意:给你一个数据范围在2~5e4范围内的横纵坐标在-1e4~1e4的点,问你任意两点之间的距离的最大值的平方等于多少? 一道卡壳凸包的模板题,也是第一次写计算几何的题,就看了些模板,关于预备知识;我是直接找到左下角的点,排好序之后,就直接形成凸包,之后调用rotating_calipers()求解:里面注意在凸包构造好之后,因为是++top的,所以在后面卡壳里面%top会出现问题,所以循环之后再一次++top;开始看graham()里面的while循环看了很久,其…

/** Author: Oliver ProblemId: ZOJ 3872 Beauty of Array */ /* 需求: 求beauty sum,所谓的beauty要求如下: 1·给你一个集合,然后把所有子集合的美丽和求出来: 2·上例子,2.3.3.->2. 3. 3. 2.3. 3. 2.3. 思路: 1·既然要连续的,暴力也会爆,那么自然而然的优化,朝着递推想一下: 2·哥们把之前的算好了,那我们是不是可以用算好的值来求现在的的需要呢: 3·想好了,但是过程我不造该怎么说; 步骤:…

Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25256   Accepted: 7756 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 31214   Accepted: 9681 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty con…

Beauty of Array Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward has an array A with N integers. He defines the beauty of an array as the summation of all distinct integers in the array. Now Edward wants to know the summation of the beauty of…

Problem地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5520 根据题目的要求,需要算出所有连续子数组的the beauty的总和. 那么要求这个这个总和,刚开始最容易想到的就是这样: for( int i=1; i<=N; i++ ) { for( int j = 1; j<=i; j++ ) { ... //排除重复的数计算总和 } } 这样子的结果实际上是 Time Limit Exceeded 因此采取…

也是一道比赛时候没有写出来的题目,队友想到了解法不过最后匆匆忙忙没有 A 掉 What a pity... 题意:定义Beauty数是一个序列里所有不相同的数的和,求一个序列所有字序列的Beauty和 1 <= N <= 100000 解题思路:由于数据比较大,常规方法求字序列和肯定是行不通的,我们不妨这样想:因为要区别于不同的数 ,可以看成序列里的数是一个一个加进去的,每次加入一个数,统计前面序列里第一次出现新加入的这个数的位置,表达的不好, 举个例子: 1 2 3 定义dp(当前元素前面(…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27276   Accepted: 8432 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

Description Edward has an array A with N integers. He defines the beauty of an array as the summation of all distinct integers in the array. Now Edward wants to know the summation of the beauty of all contiguous subarray of the array A. Input There a…

id=2187">Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27218   Accepted: 8410 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27507   Accepted: 8493 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 36113   Accepted: 11204 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bes…

题面 Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farm…

You are given an array a consisting of n integers. A subarray (l, r) from array a is defined as non-empty sequence of consecutive elements al, al + 1, ..., ar. The beauty of a subarray (l, r) is calculated as the bitwise AND for all elements in the s…

题目链接:http://poj.org/problem?id=2187 Time Limit: 3000MS Memory Limit: 65536K Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of…

转化为beauty模式 var jsonText = $('#json').val(); $('#json').val(JSON.stringify(JSON.parse(jsonText), null, 4)); 转为ugly模式 var jsonText = $('#json').val(); $('#json').val(JSON.stringify(JSON.parse(jsonText))); jsonmate的网址 json编辑可视化http://jsonmate.com/…

P1452 Beauty Contest 题意 求平面\(n(\le 50000)\)个点的最远点对 收获了一堆计算几何的卡点.. 凸包如果不保留共线的点,在加入上凸壳时搞一个相对栈顶,以免把\(n\)号点删走了 旋转卡壳和凸包都想一下更新的时候带不带等于号阿 Code: #include <cstdio> #include <algorithm> using std::max; const int N=5e4+10; int s[N],tot,n,p; struct Vector…

Pretty Smart? Why We Equate Beauty With Truth With some regularity we hear about the latest beauty-pageant contestant who has responded to a softball of a question with an epic fail of a mistake, a bizarre opinion or an incoherent ramble. There's the…

HYMN TO INTELLECTUAL BEAUTY III No voice from some sublimer world hath ever ⁠To sage or poet these responses given- ⁠Therefore the names of Demon, Ghost, and Heaven, Remain the records of their vain endeavour, Frail spells-whose uttered charm might n…

计算几何/旋转卡壳 学习旋转卡壳请戳这里~感觉讲的最好的就是这个了…… 其实就是找面积最大的三角形?...并且满足单调…… 嗯反正就是这样…… 这是一道模板题 好像必须写成循环访问?我在原数组后面复制了一遍点,结果挂了……改成cur=cur%n+1就过了QAQ //其实是不是数组没开够所以复制的方法就爆了? UPD:(2015年5月13日 20:40:45) 其实是我点保存在1~n里面,所以复制的时候不能写p[i+n-1]=p[i]; 而应该是p[i+n]=p[i];……QAQ我是傻逼 Sour…

25482: Beauty 时间限制: 1 Sec  内存限制: 128 MB献花: 7  解决: 3[献花][花圈][TK题库] 题目描述 一年一度的星哥选美又拉开了帷幕 N个人报名参加选拔,每个人都有着各自的相貌参数和身材参数(不大于10000的正整数).你的任务是尽可能让更多人被星哥选中,而唯一要求就是,在这只队伍里面的每个人,都需满足以下不等式: A (H−h)+B(W−w)≤C 其中H和W为这个人的相貌和身材,h和w为选中者中的最小相貌参数和最小身材参数,而A.B.C为三个不大于100…