CodeForces - 385C Bear and Prime Numbers (埃氏筛的美妙用法)

Recently, the bear started studying data structures and faced the following problem.

You are given a sequence of integers x₁, x₂, ..., x_n of length n and m queries, each of them is characterized by two integers l_i, r_i. Let's introduce f(p) to represent the number of such indexes k, that x_k is divisible by p. The answer to the query l_i, r_i is the sum: , where S(l_i, r_i) is a set of prime numbers from segment [l_i, r_i] (both borders are included in the segment).

Help the bear cope with the problem.

Input

The first line contains integer n (1 ≤ n ≤ 10⁶). The second line contains n integers x₁, x₂, ..., x_n (2 ≤ x_i ≤ 10⁷). The numbers are not necessarily distinct.

The third line contains integer m (1 ≤ m ≤ 50000). Each of the following m lines contains a pair of space-separated integers, l_i and r_i (2 ≤ l_i ≤ r_i ≤ 2·10⁹) — the numbers that characterize the current query.

Output

Print m integers — the answers to the queries on the order the queries appear in the input.

Example

Input

6
5 5 7 10 14 15
3
2 11
3 12
4 4

Output

9
7
0

Input

7
2 3 5 7 11 4 8
2
8 10
2 123

Output

0
7

Note

Consider the first sample. Overall, the first sample has 3 queries.

1. The first query l = 2, r = 11 comes. You need to count f(2) + f(3) + f(5) + f(7) + f(11) = 2 + 1 + 4 + 2 + 0 = 9.
2. The second query comes l = 3, r = 12. You need to count f(3) + f(5) + f(7) + f(11) = 1 + 4 + 2 + 0 = 7.
3. The third query comes l = 4, r = 4. As this interval has no prime numbers, then the sum equals 0.

题意:给n个数,给出m组l,r.设i为l-r之间的一个质数,这个质数能被n个数中的x个数整除,令f(i)=x.求l-r区间中所有f(i)的总和.

 

思路:首先考虑暴力做法,筛出1-10000000质数,然后用n个数除以l-r内的所有质数,加起来求出答案.复杂度很高,尤其是使用O(n√n)的滑稽筛法.这里可以用埃氏筛来筛质数,复杂度O(nloglogn).由于l-r有重复,所以可以事先记录,用前缀和进行优化,复杂度仍不过关,因为遍历n的复杂度仍太高.这时候可以用一些埃氏筛的美妙性质:

    for(i=2;i<N;i++)
    {
        if(!vis[i])
        {
            for(j=i;j<N;j+=i)
            {
                vis[j]=1;  
            }
        }
    }

上图是一个埃氏筛的模板,我们发现当i是质数时,该程序会遍历所有i的倍数,如果我们开一个数组cnt,cnt[i]记录n中第i个数出现的数量.若j扫到cnt[j]>0,sum[i]+=cnt[j](就是有新的cnt[j]个数能整除i)

    for(i=2;i<N;i++)
    {
        if(!vis[i])
        {
            for(j=i;j<N;j+=i)
            {
                if(cnt[j])
                {
                    sum[i]+=cnt[j];
                }
                vis[j]=1;
            }
        }
    }

就这么简单!

但这样竟然还是RE了,smg?!

仔细看题目,你会发现,尼玛l,r<=2*10^9....

数组开小了!

那么是滑稽的开2*10^9然后拿MLE回家还是再优化?

我们发现sum[r](r>10^7)的值和sum[10^7]有区别吗....

显然是没有的,因为不存在比10^7大的n.

所以就在优化一下了~

最后代码:

#include<cstdio>
#define N 10000010
using namespace std;

int k,cnt[N],vis[N],sum[N],n,i,j;

int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&k);
        cnt[k]++;
    }
    for(i=2;i<N;i++)
    {
        if(!vis[i])
        {
            for(j=i;j<N;j+=i)
            {
                if(cnt[j])
                {
                    sum[i]+=cnt[j];
                }
                vis[j]=1;
            }
        }
    }
    for(i=1;i<=N;i++)
    {
        sum[i]+=sum[i-1];
    }
    int t,l,r;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&l,&r);
        if (l>N)
        {
            l=N-1;
        }
        if (r>N)
        {
            r=N;
        }
        long long ans=sum[r]-sum[l-1];
        printf("%lld\n",ans);
    }
}

 

 

 

 

 

 

每天刷题,身体棒棒!