2016暑假多校联合---Rikka with Sequence (线段树) Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has an array A with n numbers. Then he make…

// 判断相同区间(lazy) 多校8 HDU 5828 Rikka with Sequence // 题意:三种操作,1增加值,2开根,3求和 // 思路:这题与HDU 4027 和HDU 5634 差不多 // 注意开根号的话,遇到极差等于1的,开根号以后有可能还是差1.如 // 2 3 2 3... // 8 9 8 9... // 2 3 2 3... // 8 9 8 9... // 剩下就是遇到区间相等的话,就直接开根号不往下传 #include <bits/stdc++.h> u…

Rikka with Sequence 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5828 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has an array A wi…

Rikka with Sequence 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5828 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has an array A wi…

Rikka with sequence Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5204 Description 众所周知,萌萌哒六花不擅长数学,所以勇太给了她一些数学问题做练习,其中有一道是这样的:现在有一个序列,因为这个序列很任性,开始时空的.接下来发生了n个事件,每一个事件是以下两种之一:1.勇太利用黑炎龙的力量在序列的开头.结尾以及每相邻两个元素之间都插入…

Rikka with Sequence Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2777    Accepted Submission(s): 503 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situati…

Problem DescriptionAs we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has an array A with n numbers. Then he makes m operations on it. There are three ty…

分析:这个题和bc round 73应该是差不多的题,当时是zimpha巨出的,那个是取phi,这个是开根 吐槽:赛场上写的时候直接维护数值相同的区间,然后1A,结果赛后糖教一组数据给hack了,仰慕 由于是开根,所以存在两个数刚开始差为1,加上某数再开根依旧是差1,这样维护相同数区间的就没用了 比如(2,3) +6-->(8,9)开根-->(2,3)如果全是这样的操作,即使维护相同的数,每次开根的复杂度都是O(N),不T才怪 这样只需要维护区间最大值最小值,当差1的时候,看看是否开根后还是差…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5828 给你n个数,三种操作.操作1是将l到r之间的数都加上x:操作2是将l到r之间的数都开方:操作3是求出l到r之间的和. 操作1和3就不说了,关键是开方操作. 一个一个开方,复杂度太高,无疑会T.所以我们来剪枝一下. 我们可以观察,这里一个数最多开方4,5次(loglogx次)就会到1,所以要是一段区间最大值为1的话,就不需要递归开方下去了.这是一个剪枝. 如果一段区间的数都是一样大小(最大值等于…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5828 [题目大意] 给出一个数列,要求支持区间加法,区间开方和区间和查询操作. [题解] 考虑开方后会出现成片相同的数字,因此我们在每个区间额外维护区间内数字是否相同的tag,如果区间内数字均相同,那么开方就可以转化为减法的标记,即t=sqrt(t)转化为tag=sqrt(t)-t.剩下就是简单的标记维护了. [代码] #include <cstdio> #include <algori…