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…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5877 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 nodes (u,v) is said to be weak if  (1) u…

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…

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…

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,…

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…

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…

//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  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…