475D: CGCDSSQ (Time limit exceeded on test 9)

##Problem Description:

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

Solution:

The solution based on the following assumption:

gcd(a1, a2, ..., an) = gcd(gcd(a1, a2, ..., aj), gcd(aj+1, aj+2, ..., an))

so all xi can be got by:

|a1                     |a2             |a3          |a4|  ... |an-2               |an-1         |an|
|gcd(a1, a2)            |gcd(a2, a3)    |gcd(a3, a4) |     ... |gcd(an-2, an-1)    |gcd(an-1, an)|
|gcd(a1, a2, a3)        |gcd(a2, a3, a4)|                  ... |gcd(an-2, an-1, an)|
...
|gcd(a1, a2, a3, ... an)|

Souce code of the solution:

#include <map>
#include <vector>
#include <iostream>

using namespace std;

static int gcd(int a, int b)
{
    if (a == 1 || b == 1) {
        return 1;
    }

    while (b != 0) {
        int r = b;
        b = a % b;
        a = r;
    }
    return a;
}

int main(int argc, char const *argv[])
{
    int n, t;

    while (cin>>n) {
        vector v(n);
        map<int, int> db;

        for (int i = 0; i < n; ++i) {
            cin >> v[i];
            db[v[i]]++;
        }

        for (int i = n-1; i >=0; --i) {
            for (int j = 0; j < i; ++j) {
                v[j] = gcd (v[j], v[j+1]);
                db[v[j]]++;
            }
        }
        cin>> n;
        for (int i = 0; i < n; ++i) {
            cin >> t;
            cout << db[t] << endl;
        }
   }
   return 0;
}
Written on October 12, 2014