F. Heroes of Making Magic III time limit per test:3 seconds memory limit per test:256 megabytes input:standard input output:standard output I’m strolling on sunshine, yeah-ah! And doesn’t it feel good! Well, it certainly feels good for our Heroes of…

这个题目一开始感觉还是有点难的,这个模数这么大,根本就不知道怎么写,然后去搜了题解,知道了怎么去求当x很大的时候x的平方对一个数取模怎么样不会爆掉. 然后还顺便发现了一个规律就是当一个数更新一定次数之后就不会变化了. 然后这个题目就很好写了,就是一个区间求和和一个区间修改.现在还不确定如果不加一个找到的规律是不是会超时. 现在写完了,写的过程你会发现,这个每次必须更新到叶节点才可以,不然这个是有问题的,因为我们要求和, 所以如果不更新到叶节点,那就无法求和,然后我们再计算一下复杂度,如果直接是m…

F. Heroes of Making Magic III time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output I’m strolling on sunshine, yeah-ah! And doesn’t it feel good! Well, it certainly feels good for our Heroes of…

magic balls Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 323    Accepted Submission(s): 90 Problem Description The town of W has N people. Each person takes two magic balls A and B every day.…

题目地址:HDU 5266 这题用转RMQ求LCA的方法来做的很easy,仅仅须要找到l-r区间内的dfs序最大的和最小的就能够.那么用线段树或者RMQ维护一下区间最值就能够了.然后就是找dfs序最大的点和dfs序最小的点的近期公共祖先了. 代码例如以下: #include <iostream> #include <string.h> #include <math.h> #include <queue> #include <algorithm>…

[题目分析] GSS1的基础上增加修改操作. 同理线段树即可,多写一个函数就好了. [代码] #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <map> #include <set> #include <queue> #include <string> #include <iostream&…

Pog and Szh are playing games. Firstly Pog draw a tree on the paper. Here we define 1 as the root of the tree.Then Szh choose some nodes from the tree. He wants Pog helps to find the least common ancestor (LCA) of these node.The question is too diffi…

A - Magic Number Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3622 Appoint description:   Description A positive number y is called magic number if for every positive integer x it satisfies tha…

SPOJ - GSS3 Can you answer these queries III Description You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations: modify the i-th element in the sequence or for giv…

题目链接 给出n个数, 2种操作, 一种是将第x个数改为y, 第二种是询问区间[x,y]内的最大连续子区间. 开4个数组, 一个是区间和, 一个是区间最大值, 一个是后缀的最大值, 一个是前缀的最大值. 合并起来好麻烦...... #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <…