Rikka with Phi Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Problem Description Rikka and Yuta are interested in Phi function (which is known as Euler's totient function). Yuta gives Rikka an array A[1..n] of po…

// 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…

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,…

Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem. There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the bo…

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…

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…

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…

区间交 Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 849    Accepted Submission(s): 377 Problem Description 小A有一个含有n个非负整数的数列与m个区间.每个区间可以表示为li,ri. 它想选择其中k个区间, 使得这些区间的交的那些位置所对应的数的和最大. 例如样例中,选择[2,5]…

Snacks HDU 5692 dfs序列+线段树 题意 百度科技园内有n个零食机,零食机之间通过n−1条路相互连通.每个零食机都有一个值v,表示为小度熊提供零食的价值. 由于零食被频繁的消耗和补充,零食机的价值v会时常发生变化.小度熊只能从编号为0的零食机出发,并且每个零食机至多经过一次.另外,小度熊会对某个零食机的零食有所偏爱,要求路线上必须有那个零食机. 为小度熊规划一个路线,使得路线上的价值总和最大 输入输出: 输入数据第一行是一个整数T(T≤10),表示有T组测试数据. 对于每组数据,…

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…