divisor_ranges.cpp

/*
 *
 *    J1K7_7
 * 
 */
//Ad Infinitum 11 - Math Programming Contest
//https://www.hackerrank.com/contests/infinitum11/challenges/divisor-ranges
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <cstring>
#include <list>
#include <map>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
using namespace std;
typedef long long          ll;
typedef unsigned long long ull;
typedef long double        ld;
typedef pair<int,int>      pii;
typedef pair<ll,ll>        pll;
typedef vector<int>        vi;
typedef vector<long long>  vll;
#define l(x) 		       (x << 1) + 1
#define r(x) 		       (x << 1) + 2
#define mid(l, r) 	       ((l + r) >> 1)
#define mp                 make_pair
#define pb                 push_back
#define all(a)             a.begin(),a.end()
#define pr(n)              printf("%d",n)
#define s(n)               scanf("%d",&n)
#define debug(x)	       {cerr <<#x<<" = " <<x<<"\n"; }
#define debug2(x, y)       {cerr <<#x<<" = " <<x<<", "<<#y <<" = " <<y <<"\n";}
#define ss                 second
#define ff                 first
#define m0(x) 		       memset(x,0,sizeof(x))
#define snuke(c,itr)       for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
const int mod=1e9+7;

const ll  mx_ll   = numeric_limits<ll> :: max();
const int mx_int  = numeric_limits<int> :: max();

const long double PI = (long double)(3.1415926535897932384626433832795);

inline bool ispow2(int x){return (x!=0 && (x&(x-1))==0);} //0 or 1

int msb(unsigned x){ union { double a; int b[2]; }; a = x; return (b[1] >> 20) - 1023; }

template<class T>
inline void cinarr(T a, int n){ for (int i=0;i<n;++i) cin >> a[i];}

inline ll powmod(ll a,ll b) {ll res = 1; while(b){if(b&1) res = (res*a)%mod;a = (a*a)%mod;b >>= 1;}return res;}

inline ll gcd(ll a,ll b){ll t;while(b){a=a%b;t=a;a=b;b=t;}return a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

inline int nextint(){ int x; scanf("%d",&x); return x; }
inline ll nextll(){ ll x; scanf("%lld",&x); return x; }

const int maxN = 1e5+7;
ll  dp[maxN];
int pos[maxN];
int n;

void precompute()
{
    for (int k = 1; k <= n; k++)
    {
        
        vi V;
        for(int j = k; j <= n; j += k )
        {
            V.pb(pos[j]);
        }
        sort(all(V));

        // find contigous sequences ( len x)  in this and add (x+1) * x / 2 to the answer

        ll cnt = 1;
        ll ans = 0;
        
        for (int i = 1; i < V.size(); i++ )
        {
            if ( V[i] == V[i-1] + 1)
            {
                cnt++;
            }
            else
            {
                ans +=  (cnt + 1) * cnt / 2;
                cnt = 1;
            }
        }

        ans +=  (cnt + 1) * cnt / 2;

        dp[k] = ans;

    }
}

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);

    n = nextint();
    for(int i = 1; i <= n; i++)
    {
        int x = nextint();
        pos[x] = i;
    }
    int q = nextint();

    precompute();

    while(q--)
    {
        int k = nextint();
        
        cout << dp[k] << "\n";
    }
    return 0;
}