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(…

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…

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…

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…

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…

题面 D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot…

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 __…

题意:给定一个n个数的序列,完成以下3个操作: 1.给定区间求和 2.给定区间对x取模 3.单点修改 对一个数取模,这个数至少折半.于是我们记一个最大值max,如果x>max则不做处理. #include<stdio.h> #include<algorithm> using namespace std; #define MAXN 1000000+10 typedef long long LL; ]; int n,m; LL a[MAXN]; void build(int k,…