map+pair Bayan 2015 Contest Warm Up D题

本文主要是介绍map+pair Bayan 2015 Contest Warm Up D题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Given a sequence of integers a₁, ..., a_n andq queries x₁, ..., x_q on it. For each queryx_i you have to count the number of pairs(l, r) such that 1 ≤ l ≤ r ≤ n and gcd(a_l, a_l + 1, ..., a_r) = x_i.

$\text{[math]}$ is a greatest common divisor ofv₁, v₂, ..., v_n, that is equal to a largest positive integer that divides allv_i.

Input

The first line of the input contains integer n, (1 ≤ n ≤ 10⁵), denoting the length of the sequence. The next line containsn space separated integers a₁, ..., a_n, (1 ≤ a_i ≤ 10⁹).

The third line of the input contains integer q, (1 ≤ q ≤ 3 × 10⁵), denoting the number of queries. Then followsq lines, each contain an integer x_i, (1 ≤ x_i ≤ 10⁹).

Output

For each query print the result in a separate line.

Sample test(s)

Input

Output

Input

7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000

Output

题意给你一个序列和q个询问,输出有多少个区间[l,r]满足区间gcd为询问的值。

思路：处理出每个区间的gcd(要注意合并区间)，如果有询问则直接加上去。

#include <bits/stdc++.h>
using namespace std;
std::map<int, int> id;
vector<pair<int,int> >gcdList;
const int MAXN=100000+5;
int query[MAXN*3],A[MAXN];
long long ans[MAXN*3];
int _gcd(int a,int b){return b?_gcd(b,a%b):a;}
int ID(int x){if(!id[x])id[x]=id.size();return id[x];
}
int main(int argc, char const *argv[])
{int n;ios_base::sync_with_stdio(false);gcdList.clear();id.clear();cin>>n;for(int i=1;i<=n;i++)cin>>A[i];int q;cin>>q;for(int i=1;i<=q;i++){cin>>query[i];query[i]=ID(query[i]);}for(int i=1;i<=n;i++){for(int j=0;j<gcdList.size();j++)gcdList[j].first=_gcd(gcdList[j].first,A[i]);gcdList.push_back(make_pair(A[i],i));int cnt=1;//sort(gcdList.begin(), gcdList.end());for(int j=1;j<gcdList.size();j++){if(gcdList[j].first!=gcdList[cnt-1].first)gcdList[cnt++]=gcdList[j];}gcdList.resize(cnt);for(int j=0;j<gcdList.size();j++){if(id[gcdList[j].first]){int r=i+1;if(j+1<gcdList.size())r=gcdList[j+1].second;ans[id[gcdList[j].first]]+=r-gcdList[j].second;}}}for(int i=1;i<=q;i++)cout<<ans[query[i]]<<endl;return 0;
}

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