题目链接:http://codeforces.com/contest/381/problem/E  E. Sereja and Brackets time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja has a bracket sequence s1, s2, ..., sn, or, in other words, a…

C. Sereja and Brackets time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja has a bracket sequence s1, s2, ..., sn, or, in other words, a string s of length n, consisting of characters "(…

C. Drazil and Park 题目连接: http://codeforces.com/contest/516/problem/C Description Drazil is a monkey. He lives in a circular park. There are n trees around the park. The distance between the i-th tree and (i + 1)-st trees is di, the distance between t…

题目链接: http://codeforces.com/problemset/problem/444/C J. DZY Loves Colors time limit per test:2 secondsmemory limit per test:256 megabytes 问题描述 DZY loves colors, and he enjoys painting. On a colorful day, DZY gets a colorful ribbon, which consists of…

D. Vika and Segments 题目连接: http://www.codeforces.com/contest/610/problem/D Description Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet and drew n black hor…

题目链接:http://codeforces.com/contest/610/problem/D 就是给你宽度为1的n个线段,然你求总共有多少单位的长度. 相当于用线段树求面积并,只不过宽为1,注意y和x的最大都要+1,这样才相当于求面积. #include <iostream> #include <cstdio> #include <cstring> #include <map> #include <algorithm> using names…

题目链接:http://codeforces.com/problemset/problem/242/E 给你n个数,m个操作,操作1是查询l到r之间的和,操作2是将l到r之间的每个数xor与x. 这题是线段树成段更新,但是不能直接更新,不然只能一个数一个数更新.这样只能把每个数存到一个数组中,长度大概是20吧,然后模拟二进制的位操作.仔细一点就行了. #include <iostream> #include <cstdio> #include <cmath> #incl…

E. Kefa and Watch Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/580/problem/E Description One day Kefa the parrot was walking down the street as he was on the way home from the restaurant when he saw something glittering b…

题目地址:http://codeforces.com/contest/474/problem/E 第一次遇到这样的用线段树来维护DP的题目.ASC中也遇到过,当时也非常自然的想到了线段树维护DP,可是那题有简单方法,于是就没写.这次最终写出来了.. 这题的DP思想跟求最长上升子序列的思想是一样的.仅仅只是这里的找前面最大值时会超时,所以能够用线段树来维护这个最大值,然后因为还要输出路径,所以要用线段树再来维护一个每一个数在序列中所在的位置信息. 手残了好多地方,最终调试出来了... 代码例如以下…

题目:http://codeforces.com/problemset/problem/356/A 题意:首先给你n,m,代表有n个人还有m次描述,下面m行,每行l,r,x,代表l到r这个区间都被x所击败了(l<=x<=r),被击败的人立马退出游戏让你最后输出每个人是被谁击败的,最后那个胜利者没被 人击败就输出0 思路:他的每次修改的是一个区间的被击败的人,他而且只会记录第一次那个被击败的人,用线段树堕落标记的话他会记录最后一次的,所以我们倒着来修改, 然后因为那个区间里面还包含了自己,在线段…