传送门 这道题维护区间加,区间开根,区间求和. 线段树常规操作. 首先回忆两道简单得多的线段树. 第一个:区间覆盖,区间加,区间求和. 第二个:区间开根,区间求和. 这两个是名副其实的常规操作. 但这道题如果学习没有区间加的做法维护最大值很容易卡掉. 所以怎么做呢? 区间加和区间求和就略了. 考虑到开根的性质,显然一段区间的数多开几次根差就不大了. 这样的话,我们维护区间最大值和区间最小值,如果当前区间的最大值与最小值的差不大于1的话就直接进行开根操作,否则继续递归. 开根时要分类讨论. 我们令…

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…

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…

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…

Color it Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others) Problem Description Do you like painting? Little D doesn't like painting, especially messy color paintings. Now Little B is painting. To prevent him from dr…

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…

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…

Cows Time Limit: 3000MS Memory Limit: 65536K Description Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good. Farmer John ha…

// HDU5634 Rikka with Phi 线段树 // 思路:操作1的时候,判断一下当前区间是不是每个数都相等,在每个数相等的区间上操作.相当于lazy,不必更新到底. #include <bits/stdc++.h> using namespace std; #define clc(a,b) memset(a,b,sizeof(a)) #define inf 0x3f3f3f3f #define lson l,mid,rt<<1 #define rson mid+1,r…

传送门 线段树维护区间取模,单点修改,区间求和. 这题老套路了,对一个数来说,每次取模至少让它减少一半,这样每次单点修改对时间复杂度的贡献就是一个log" role="presentation" style="position: relative;">loglog,所以维护区间最大值剪枝,然后每次单点暴力取模,这样的话时间复杂度为O（nlogn）" role="presentation" style="posi…