codeforces 691F F. Couple Cover(组合计数)

题目链接：

F. Couple Cover

time limit per test

3 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Couple Cover, a wildly popular luck-based game, is about to begin! Two players must work together to construct a rectangle. A bag withn balls, each with an integer written on it, is placed on the table. The first player reaches in and grabs a ball randomly (all balls have equal probability of being chosen) — the number written on this ball is the rectangle's width in meters. This ball is not returned to the bag, and the second player reaches into the bag and grabs another ball — the number written on this ball is the rectangle's height in meters. If the area of the rectangle is greater than or equal some threshold p square meters, the players win. Otherwise, they lose.

The organizers of the game are trying to select an appropriate value for p so that the probability of a couple winning is not too high and not too low, but they are slow at counting, so they have hired you to answer some questions for them. You are given a list of the numbers written on the balls, the organizers would like to know how many winning pairs of balls exist for different values of p. Note that two pairs are different if either the first or the second ball is different between the two in pair, and two different balls with the same number are considered different.

Input

The input begins with a single positive integer n in its own line (1 ≤ n ≤ 10^6).

The second line contains n positive integers — the i-th number in this line is equal to ai (1 ≤ ai ≤ 3·106), the number written on the i-th ball.

The next line contains an integer m (1 ≤ m ≤ 106), the number of questions you are being asked.

Then, the following line contains m positive integers — the j-th number in this line is equal to the value of p (1 ≤ p ≤ 3·10^6) in the j-th question you are being asked.

Output

For each question, print the number of winning pairs of balls that exist for the given value of p in the separate line.

Examples

input

5
4 2 6 1 3
4
1 3 5 8

output

20
18
14
10

input

2
5 6
2
30 31

output

2
0

题意：

给一个数列,问这里面有多少对的积大于等于p;


思路：

由于询问太多,所以要降低复杂度,可以反方向考虑,可以用总的方案数减去小于p的方案数,小于p的方案数;
可以先把相同的数压缩,最的的复杂度是O（nlogn+m）

AC代码：

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=3e6+10;
const int maxn=3e6;
const double eps=1e-10;

int m,a[N],b[N];
LL num[N],sum[N],n;
void Init()
{
    sort(a+1,a+n+1);
    int cnt=0;
    For(i,1,n)
    {
        if(a[i]==a[i-1])num[cnt]++;
        else 
        {
            ++cnt;
            b[cnt]=a[i];
            num[cnt]=1;
        }
    }
    For(i,1,cnt)
    {
        For(j,1,i)
        {
            if((LL)b[i]*b[j]>=maxn)break;
            if(i==j)sum[b[i]*b[j]]+=num[i]*(num[i]-1);
            else sum[b[i]*b[j]]+=2*num[i]*num[j];
        }
    }
    //for(int i=1;i<30;i++)print(sum[i]);
    For(i,1,maxn)sum[i]+=sum[i-1];
}
int main()
{
    read(n);
    For(i,1,n)read(a[i]);
    Init();
    read(m);
    int p;
    For(i,1,m)
    {
        read(p);
        print((LL)n*(n-1)-sum[p-1]);
    }
        return 0;
}