【CODEFORCES】 D. CGCDSSQ

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.

$\text{[math]}$ is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.

Input

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

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

Output

For each query print the result in a separate line.

Sample test(s)

input
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

题解：题目大意是说给你一个序列。然后有q个询问，每一个询问描写叙述成一个正整数c，问你在这个序列中，区间上的最大公约数等于c的区间共同拥有多少个。

由于我们仅仅关心个数而并不关心区间，所以能够递推来写。如果我们知道了第以i个数为结尾的全部区间上的最大公约数，我们就能够统计他的个数。而我们在讨论i+1的时候，我们是能够推出以第i+1个数结尾的全部最大公约数的（和第i个数的结果依次取GCD即可），然后统计个数（事实上就是直接加进去）。

最后记得每次把第i个数的结果加到ANS里面即可了。

代码并不长。时间也不长。果然是由于map太快了么= =||

#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>

using namespace std;

int a[100005],n,q,c;

int gcd(int a,int b)
{
    if (!b) return a;
    else return gcd(b,a%b);
}

map<int,long long> ans;
map<int,int> last;
map<int,int> now;
map<int,int>::iterator itr;

int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++) scanf("%d",&a[i]);
    ans.clear();
    for (int i=0;i<n;i++)
    {
        swap(last,now);
        now.clear();
        for (itr=last.begin();itr!=last.end();itr++)
            now[gcd(itr->first,a[i])]+=itr->second;
        now[a[i]]++;
        for (itr=now.begin();itr!=now.end();itr++)
            ans[itr->first]+=itr->second;
    }
    scanf("%d",&q);
    for (int i=0;i<q;i++)
    {
        scanf("%d",&c);
        printf("%I64d
",ans[c]);
    }
    return 0;
}