[codeforces 360]A. Levko and Array Recovery 试题描述 Levko loves array a1, a2, ... , an, consisting of integers, very much. That is why Levko is playing with array a, performing all sorts of operations with it. Each operation Levko performs is of one of…

原题: Description Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex coverproblem very interesting. Suppose the graph G is given. Subset A of its vertices is called a vertex cover of this graph, if for ea…

http://codeforces.com/problemset/problem/442/B (题目链接) 题意 n个人,每个人有p[i]的概率出一道题.问如何选择其中s个人使得这些人正好只出1道题的概率最大. Solution 很显然的概率dp,过了样例即可AC..话说我为什么要刷B题→_→ 代码 // codeforces442B #include<algorithm> #include<iostream> #include<cstdlib> #include<…

Another Rock-Paper-Scissors Problem 题目连接: http://codeforces.com/gym/100015/attachments Description Sonny uses a very peculiar pattern when it comes to playing rock-paper-scissors. He likes to vary his moves so that his opponent can't beat him with hi…

当在Codeforces上做题的时,有时会无意撇到右侧的Problem tags边栏,但是原本并不希望能够看到它. 能否把它屏蔽了呢?答案是显然的,我们只需要加一段很短的CSS即可. span.tag-box{ display: none; } 这样,当这段CSS生效时,标签将会变成如下所示的样子. 这样目的就达到了. 那么,如何将CSS应用到浏览器中使其生效呢?对于不同的浏览器有不同的方法.下面以主流浏览器为例. 对于Safari,将上述代码另存为.css文件,然后在偏好设置->高级->样式…

这道题是这种,给主人公一堆事件的成功概率,他仅仅想恰好成功一件. 于是,问题来了,他要选择哪些事件去做,才干使他的想法实现的概率最大. 我的第一个想法是枚举,枚举的话我想到用dfs,但是认为太麻烦. 于是想是不是有什么规律,于是推导了一下,推了一个出来,写成代码提交之后发现是错的. 最后就没办法了,剩下的时间不够写dfs,于是就放弃了. 今天看thnkndblv的代码,代码非常短,于是就想肯定是有什么数学规律,于是看了一下, 果然如此. 是这种,还是枚举,当然是有技巧的,看我娓娓道来. 枚举的话…

http://codeforces.com/contest/426/problem/B 对称标题的意思大概是.应当指出的,当线数为奇数时,答案是线路本身的数 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 1005 using namespace std; char a[maxn][maxn]; int ans; void solve…

题目链接:http://codeforces.com/problemset/problem/803/G 大致就是线段树动态开节点. 然后考虑到如果一个点还没有出现过,那么这个点显然未被修改,就将这个点所代表的区间定位到原序列中,利用ST表查一下区间最小值就可以了. 定位: llg minn(llg l,llg r) { >=n) return mt; l%=n;if(!l) l=n; r%=n;if(!r) r=n; ,r)); else return gw(l,r); } 其中${gw(l,r…

题目传送门 传送门I 传送门II 传送门III 题目大意 给定一个网络.网络分为$A$,$B$两个部分,每边各有$n$个点.对于$A_{i} \ (1\leqslant i < n)$会向$A_{i + 1}$连一条容量为$x_{i}$的有向边,对于$B_{i} \ (1\leqslant i < n)$会向$B_{i + 1}$连一条容量为$y_{i}$的有向边.$A$和$B$之间有$m$条边,起点为$A_{u_{i}}$,终点为$B_{v_{i}}$,容量为$w_{i}$的有向边.要求支持…

Adieu l'ami. Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around the abandoned Eikou Cram School building, Oshino's makeshift residence. The space is represented by a rectangular grid of n × m cells, arranged into n…