Codeforces 385C Bear and Prime Numbers(数论）

C. Bear and Prime Numbers

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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

You are given a sequence of integers x1, x2, ..., xn of length n and m queries, each of them is characterized by two integers li, ri. Let's introduce f(p) to represent the number of such indexes k, that xk is divisible by p. The answer to the query li, ri is the sum: , where S(li, ri) is a set of prime numbers from segment [li, ri] (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, ..., xn (2 ≤ xi ≤ 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, li and ri(2 ≤ li ≤ ri ≤ 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.

Examples

input

65 5 7 10 14 1532 113 124 4

output

970

input

72 3 5 7 11 4 828 102 123

output

07

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：

给出一个长度为n的序列，然后有m次询问。每次询问给出a，b。求【a，b】中的所有质数F（x） 的和；F（x）为序列中有几个数为x的倍数；

思路：

因为内存空间限制比较大，所以可以开的下10^7的数组，然后用筛选法求素数的同时计算个数。

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int MAX_N = 1e7+9;int vec[MAX_N];int res[MAX_N];bool prime[MAX_N];int main(){    int N,M,T;    memset(prime,true,sizeof(prime));    prime[0] = prime[1] =  false;    for(int i=2;i<MAX_N;i++)    {        if(prime[i])        {            for(int j=2*i;j<MAX_N;j+=i)            {                prime[j] = false;            }        }    }    while(cin>>N)    {        memset(vec,0,sizeof(vec));        memset(res,0,sizeof(res));        int temp;        for(int i=0;i<N;i++)        {            scanf("%d",&temp);            res[temp]++;        }        for(int i=2;i<MAX_N;i++)        {            if(prime[i])            {                vec[i] += res[i];                for(int j=2*i;j<MAX_N;j+=i)                {                    vec[i] += res[j];                }            }        }        for(int i=2;i<MAX_N;i++)        {            vec[i] += vec[i-1];        }        int l,r;        cin>>M;        for(int i=0;i<M;i++)        {            scanf("%d%d",&l,&r);            if(r>=MAX_N) r= MAX_N-1;            if(l>=MAX_N) l= MAX_N-1;            printf("%d\n",vec[r]-vec[l-1]);        }    }    return 0;}