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…

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…

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…

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 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…

题目链接:hdu 4983 Wow! Such Sequence! 题目大意:就是三种操作 1 k d, 改动k的为值添加d 2 l r, 查询l到r的区间和 3 l r. 间l到r区间上的所以数变成近期的斐波那契数,相等的话取向下取. 解题思路:线段树.对于每一个节点新增一个bool表示该节点下面的位置是否都是斐波那契数. #include <cstdio> #include <cstring> #include <cstdlib> #include <algo…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=6089 题解 这波强行维护搞得我很懵逼... 扫描线,只考虑每个点能走到左上方(不包括正上方,但包括正左方)的哪些点,然后旋转四次坐标系处理 所有询问和操作点按照先\(x\)后\(y\)坐标的顺序排序,然后枚举每一行,按\(y\)从小到大的顺序枚举这一行每个点 对于一个询问点找出前面最后一个操作点,那么要求的就是一个矩形减去一个区间内所有后缀最大值的和 然后这个东西可以用线段树直接维护,记录个区间最大…

题意:bc round 73 div1 D 中文题面 分析:注意到10^7之内的数最多phi O(log(n))次就会变成1, 因此可以考虑把一段相同的不为1的数缩成一个点,用平衡树来维护. 每次求phi的时候就在平衡树上取出这个区间然后暴力求phi,如果一段数变成了1, 就在平衡树里面删掉它,最后统计答案的时候只要把区间中被删去的1加回答案即可, 时间复杂度O((n + m)logn) 注:平衡树,写起来麻烦(然后其实我也不会写) 但是题解当中说把一段相同的数缩成一个点,就很好 所以用线段树,…