题意:对数列有三种操作: Print operation l, r. Picks should write down the value of . Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). Set operation k, x. Picks should set the value of a[k] to x (in other words…

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks. Fortunately, Picks remembers how to repair the sequence. Initi…

传送门 题目大意: 给你一个序列,要求在序列上维护三个操作: 1)区间求和 2)区间取模 3)单点修改 这里的操作二很讨厌,取模必须模到叶子节点上,否则跑出来肯定是错的.没有操作二就是线段树水题了. 既然必须模到叶子节点,那我们就模咯. 显然,若$b<c$,则$b%c=b$. 因此我们同时维护一个区间最大值,若某区间内最大值小于模数,就把该分支剪掉. 若$a=b%c$,那么肯定有$a \leq \frac{b}{2}$成立. 也就是说,一个数最多被模$\log_2 x$次.总的时间复杂度为$O(…

D. The Child and Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/problem/D Description At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important…

D. The Child and Sequence   At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks. Fortunately, Picks remembers how t…

题意 题目链接 单点修改,区间mod,区间和 Sol 如果x > mod ,那么 x % mod < x / 2 证明: 即得易见平凡, 仿照上例显然, 留作习题答案略, 读者自证不难. 反之亦然同理, 推论自然成立, 略去过程Q.E.D., 由上可知证毕. 然后维护个最大值就做完了.. 复杂度不知道是一个log还是两个log,大概是两个吧(线段树一个+最多改log次.) #include<bits/stdc++.h> #define Pair pair<int, int&g…

题目链接:http://codeforces.com/problemset/problem/438/D 给你n个数,m个操作,1操作是查询l到r之间的和,2操作是将l到r之间大于等于x的数xor于x,3操作是将下标为k的数变为x. 注意成段更新的时候,遇到一个区间的最大值还小于x的话就停止更新. #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef __…

给定数列,区间查询和,区间取模,单点修改. n,m小于10^5 ...当区间最值小于模数时,就直接返回就好啦~ #include<cstdio> #include<iostream> #define R register int #define ls (tr<<1) #define rs (tr<<1|1) using namespace std; inline long long g() { register ,fix=; register :fix; +…

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…

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…