Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studyin…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5869 Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting.…

还是想不到,真的觉得难,思路太巧妙 题意:给你一串数和一些区间,对于每个区间求出区间内每段连续值的不同gcd个数(该区间任一点可做起点,此点及之后的点都可做终点) 首先我们可以知道每次添加一个值时gcd要么不变要么减小,并且减小的幅度很大,就是说固定右端点时最多只能有(log2 a)个不同的gcd,而且我们知道gcd(gcd(a,b),c)=gcd(a,gcd(b,c)),所以我们可以使用n*(log2 n)的时间预处理出每个固定右端点的不同gcd的值和位置.解法就是从左到右,每次只需要使用上一…

Function Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Problem Description The shorter, the simpler. With this problem, you should be convinced of this truth.    You are given an array A of N postive integers,…

Football Games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description A mysterious country will hold a football world championships---Abnormal Cup, attracting football teams and fans from all around th…

Sparse Graph Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Problem Description In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if…

Weak Pair Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Problem Description You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negative value ai is assigned.An ordered pair of no…

Weak Pair Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1468    Accepted Submission(s): 472 Problem Description You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node…

树状数组... Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1541    Accepted Submission(s): 599 Problem Description This is a simple problem. The teacher gives Bob a lis…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 828    Accepted Submission(s): 300 Problem Description  and aN. In other words, ai,ai+1,⋯,aj−1,aj is a subarray of a,…

题目链接  2016 青岛网络赛  Problem C 题意  给出一些敏感词,和一篇文章.现在要屏蔽这篇文章中所有出现过的敏感词,屏蔽掉的用$'*'$表示. 建立$AC$自动机,查询的时候沿着$fail$指针往下走,当匹配成功的时候更新$f[i]$ $f[i]$表示要屏蔽以第$i$个字母结尾的长度为$f[i]$的字符串. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i &l…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1328    Accepted Submission(s): 504 Problem Description This is a simple problem. The teacher gives Bob a list of pro…

Different GCD Subarray Query Problem Description   This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a…

//大连网络赛 1006 // 吐槽:数据比较水.下面代码可以AC // 但是正解好像是:排序后,前i项的和大于等于i*(i-1) #include <bits/stdc++.h> using namespace std; #define LL long long typedef pair<int,int> pii; const double inf = 123456789012345.0; const LL MOD =100000000LL; ; #define clc(a,b)…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5869 Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 This is a simple problem. The teacher gives Bob a list of problems about GCD (Great…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5869 问你l~r之间的连续序列的gcd种类. 首先固定右端点,预处理gcd不同尽量靠右的位置(此时gcd种类不超过loga[i]种). 预处理gcd如下代码,感觉真的有点巧妙... ; i <= n; ++i) { int x = a[i], y = i; ; j < ans[i - ].size(); ++j) { ][j].first); if(gcd != x) { ans[i].push_…

很简单的一个题的,结果后台数据有误,自己又太傻卡了3个小时... 题意:给你一串数a再给你一些区间(lef,rig),求出a[lef]%a[lef+1]...%a[rig] 题解:我们可以发现数字a对数字b取模时:如果a<b,则等于原数,否则a会变小至少一半.就是说a最多成功取模(log2 a)次,所以我们只需要每次在区间内找到最前面一个小于等于a的值,接着更新a与区间左端点,直到没有值比a小或者区间取模完成. 我们可以使用线段树求出区间内小于某个值的最前一个位置,具体方法就是:父节点记录区间最…

先讲1007,有m个人,n种石头,将n种石头分给m个人,每两个人之间要么是朋友关系,要么是敌人关系,朋友的话他们必须有一种相同颜色的石头,敌人的话他们必须所有石头的颜色都不相同.另外,一个人可以不拥有任何一种石头.求m个人的所有关系是不是都能用n种石头表示出来.比赛当时找的关系是n种石头可以表示n+1个人的关系.但是一直WA,因为考虑不周. 我们考虑这样的一种情况,我们把人分为左边和右边两部分,每边的人里面都互相为敌人,同时左边的任意一个人和右边的任意一个人都是朋友.举个例子,左边有3人,右边两…

题意:给你n个点m条边形成一个无向图,问你求出给定点在此图的补图上到每个点距离的最小值,每条边距离为1 补图:完全图减去原图 完全图:每两个点都相连的图 其实就是一个有技巧的bfs,我们可以看到虽然点很多但边很少,就使用vector存下每个点在原图中可以到达其他的哪些点,再使用bfs寻找此时起点可以到的其他点(每个距离都是1,所以越早到距离越短),接着更新起点继续查找:我们需要使用数组记录此时起点不能到的一些点(就是vector中原图起点可以到的点),但每次通过起点判断其他所有的点会超时,因此我…

正难则反的思想还是不能灵活应用啊 题意:给你n个点,每个点有一个权值,接着是n-1有向条边形成一颗有根树,问你有多少对点的权值乘积小于等于给定的值k,其中这对点必须是孩子节点与祖先的关系 我们反向思考,可以知道任一点都只对其每个祖先有贡献.所以我们可以转化为求每个点与其每个祖先的乘积小于等于给定的值k的对数. 我们dfs遍历这颗树使用树状数组维护,dfs遍历孩子就添点回溯就删点,接着对每个点计算树状数组里不大于(k/此点)的个数.注意值太大我们需要离散化,而且我们可以把每个点m与k/m都离散化出…

Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a new problem about GCD. Easy as…

Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a new problem about GCD. Easy as…

I Count Two Three 31.1% 1000ms 32768K   I will show you the most popular board game in the Shanghai Ingress Resistance Team. It all started several months ago. We found out the home address of the enlightened agent Icount2three and decided to draw hi…

http://acm.hdu.edu.cn/showproblem.php?pid=5869 题意:给定一个数组,然后给出若干个询问,询问[L, R]中,有多少个子数组的gcd是不同的. 就是[L, R]中不同区间的gcd值,有多少个是不同的. 给个样例 3 37 7 71 21 33 3 数学背景: 一个数字和若N个数字不断GCD,其结果只有loga[i]种,为什么呢?因为可以把a[i]质因数分解,其数目最多是loga[i]个数字相乘.(最小的数字是2,那么loga[i]个2相乘也爆了a[i]…

题目大意:求区间$[L,R]$中所有子区间产生的最大公因数的个数. ------------------------- 对于$gcd$,我们知道$gcd(a,b,c)=gcd(gcd(a,b),c)$.所以我们可以利用$gcd$的传递性来求区间的$gcd$.如果$gcd$相同,那么保留下来位置相对靠右的那一个,这与我们查询的方式有关.我们在查询时是$O(n)$正向遍历的,询问的区间按照右端点进行关键字排序,每次维护一个新的$gcd$最靠右的位置并让这个位置+1,让之前的位置-1即可. 因为每次$…

题意:将匹配的串用'*'代替 tips: 1 注意内存的使用,据说g++中指针占8字节,c++4字节,所以用g++交会MLE 2 注意这种例子, 12abcdbcabc 故失败指针要一直往下走,否则会丢弃一些串 3 当出现非英文字符时应先将指针指向根节点,否则出现 11cyc,,,,,,,y 时结果为c,,,,,,**(由不止一种Bug造成的),正确的应为c,,,,,,,y,然而很多题解这样也过了~ #include<cstdio> #include<iostream> #incl…

离线操作,树状数组,$RMQ$. 这个题的本质和$HDU$ $3333$是一样的,$HDU$ $3333$要求计算区间内不同的数字有几个. 这题稍微变了一下,相当于原来扫描到$i$的之后是更新$a[i]$的情况,现在是更新$log$级别个数的数字(因为以$i$为结尾的区间,最多只有$log$级别种不同的$gcd$). 求区间$gcd$可以用$RMQ$预处理一下,然后就可以$O(1)$查询了. #pragma comment(linker, "/STACK:1024000000,102400000…

odd-even number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even nu…

Barricade Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description The empire is under attack again. The general of empire is planning to defend his castle. The land can be seen as N towns and M roads, a…

QSC and Master Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Every school has some legends, Northeastern University is the same. Enter from the north gate of Northeastern University,You are…