【CODEFORCES】 D. CGCDSSQ

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 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，每个询问描述成一个正整数c，问你在这个序列中，区间上的最大公约数等于c的区间共有多少个。

因为我们只关心个数而并不关心区间，所以可以递推来写。假设我们知道了第以i个数为结尾的所有区间上的最大公约数，我们就可以统计他的个数。而我们在讨论i+1的时候，我们是可以推出以第i+1个数结尾的所有最大公约数的（和第i个数的结果依次取GCD即可），然后统计个数（其实就是直接加进去）。最后记得每次把第i个数的结果加到ANS里面就行了。

代码并不长，时间也不长。果然是因为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\n",ans[c]);
}
return 0;
}