莫队算法,预处理出每个数字往后的gcd情况,每个数字的gcd只可能是他的因子,因此后面最多只可能有logn种,可以先预处理出,然后套莫队算法,复杂度O(n*sqrt(n)*log(n)). 代码 #include<cstdio> #include<cmath> #include<vector> #include<algorithm> #define N 100000 using namespace std; ],i,j,tmp,l,r; long long…

题意:http://acm.hdu.edu.cn/showproblem.php?pid=5381 思路:这个题属于没有修改的区间查询问题,可以用莫队算法来做.首先预处理出每个点以它为起点向左和向右连续一段的gcd发生变化的每个位置,不难发现对每个点A[i],这样的位置最多logA[i]个,这可以利用ST表用nlognlogA[i]的时间预处理,然后用二分+RMQ在nlogn的时间内得到.然后就是区间变化为1时的转移了,不难发现区间变化为1时,变化的答案仅仅是以变化的那一个点作为左端点或右端点的…

[题目] The sum of gcd Problem Description You have an array A,the length of A is nLet f(l,r)=∑ri=l∑rj=igcd(ai,ai+1....aj) Input There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each…

The sum of gcd Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description You have an array A,the length of A is nLet f(l,r)=∑ri=l∑rj=igcd(ai,ai+1....aj)   Input There are multiple test cases. The first li…

Sum Of Gcd 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=4676 Description Given you a sequence of number a1, a2, ..., an, which is a permutation of 1...n. You need to answer some queries, each with the following format: Give you two numbers L, R, y…

Given you a sequence of number a 1, a 2, ..., a n, which is a permutation of 1...n. You need to answer some queries, each with the following format: Give you two numbers L, R, you should calculate sum of gcd(a[i], a[j]) for every L <= i < j <= R.…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=4676 Sum Of Gcd Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 908    Accepted Submission(s): 438 Problem Description Given you a sequence of numb…

The sum of gcd Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 23    Accepted Submission(s): 4 Problem Description You have an array A,the length of A is n Let f(l,r)=∑ri=l∑rj=igcd(ai,ai+1....a…

题目链接:hdu 5381 The sum of gcd 将查询离线处理,依照r排序,然后从左向右处理每一个A[i],碰到查询时处理.用线段树维护.每一个节点表示从[l,i]中以l为起始的区间gcd总和.所以每次改动时须要处理[1,i-1]与i的gcd值.可是由于gcd值是递减的,成log级,对于每一个gcd值记录其区间就可以.然后用线段树段改动,可是是改动一个等差数列. #include <cstdio> #include <cstring> #include <vecto…

大神题解: http://blog.csdn.net/u014800748/article/details/47680899 The sum of gcd Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 526    Accepted Submission(s): 226 Problem Description You have an…