A. Vladik and Courtesy time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. Afte…

A. Vladik and Courtesy 题面 At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same…

http://codeforces.com/contest/811/problem/C 题意: 给出一行序列,现在要选出一些区间来(不必全部选完),但是相同的数必须出现在同一个区间中,也就是说该数要么不选,选了就必须出现在同一个区间,最后累加区间不同的数的异或值. 思路: 先预处理,求出每个数的左位置和右位置. d[i]表示分析到第 i 位时的最大值. #include<iostream> #include<algorithm> #include<cstring> #i…

地址:http://codeforces.com/contest/811/problem/D 题目: D. Vladik and Favorite Game time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output This is an interactive problem. Vladik has favorite game, in…

A. Vladik and Courtesy 2 seconds 256 megabytes   At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less gen…

CodeForces - 811A A. Vladik and Courtesy time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his can…

B. Vladik and Complicated Book time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vladik had started reading a complicated book about algorithms containing n pages. To improve understanding o…

A. Vladik and Courtesy time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. Afte…

那个人第一步肯定要么能向下走,要么能向右走.于是一定可以判断出上下是否对调,或者左右是否对调. 然后他往这个方向再走一走就能发现一定可以再往旁边走,此时就可以判断出另一个方向是否对调. 都判断出来以后,跑个spfa或者bfs就行了. 细节较多……有一些边界情况需要处理.比如终点在第一行或者第一列的情况. #include<cstdio> #include<queue> #include<cstring> #include<algorithm> using n…

划分那个序列,没必要完全覆盖原序列.对于划分出来的每个序列,对于某个值v,要么全都在该序列,要么全都不在该序列.  一个序列的价值是所有不同的值的异或和.整个的价值是所有划分出来的序列的价值之和.    求整个的价值的最大值   f(i)表示最后一个划分序列的右端点为i时,1~i的答案. f(i)=max{max{f(j)}(1<=j<i)+xorsum(j+1,i)(j+1到i的区间合法)}(1<=i<=n) 需要在转移的时候,顺便处理f(i)的前缀max. 最终的答案就是所有f…