D. CGCDSSQ time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output
Given a sequence of integers a1,?...,?an and q queries x1,?...,?xq on it. For each query xi you have to count the number of pairs (l,?r)such that 1?≤?l?≤?r?≤?n and gcd(al,?al?+?1,?...,?ar)?=?xi.
is a greatest common divisor of v1,?v2,?...,?vn, that is equal to a largest pZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"brush:java;">3 2 6 3 5 1 2 3 4 6 output
1
2
2
0
1
input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
output
14
0
2
2
2
0
2
2
1
1
题意：给出n个数，然后给q个询问，每次询问给出一个x，问有多少个区间的GCD是x
思路：比赛的时候yy的一个做法
首先预处理出所有值的区间个数，这个用map存一下就好了，设为ans
然后再开两个map 分别为mp mp2
mp存的是以Xi结尾的所有区间的GCD的数的个数
每次从Xi转移到Xi+1，只需要累加以Xi结尾的区间的所有mp值与Xi+1的GCD的个数就好了，可以临时赋给mp2，然后再赋给mp
按照这个方法从左往右for一遍就好了
然后查询的时候直接在ans里查就好了
复杂度大概是O(N*(每个数的因子个数))