map+pair Bayan 2015 Contest Warm Up D题

最新推荐文章于 2018-05-21 13:05:47 发布

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最新推荐文章于 2018-05-21 13:05:47 发布

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分类专栏：数据结构 codeforces

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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;
}