CF C. Bear and Prime Numbers 筛选素数并计算

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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.

Sample test(s)

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.

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.

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.

The third query comes l = 4, r = 4. As this interval
has no prime numbers, then the sum equals 0.

素数快速打表，打表的同时求出每个数满足条件的总数。

//951 ms	88000 KB
#include<stdio.h>
#include<string.h>
#define M 10000010
int b[M],count[M];
bool visit[M];
int num=0;
void init_prim()
{
memset(visit,true,sizeof(visit));
for(int i=2; i<M;++i)
{
if(visit[i]==true)
{
if(b[i])count[i]+=b[i];
for(int j=i*2; j<M; j+=i)
{
count[i]+=b[j];//同时加上能够整除i的个数
visit[j]=false;
}
}
}
for(int i=2; i<M; i++)
count[i]+=count[i-1];
}
int main()
{
int n,q,l,r,s;
init_prim();
while(scanf("%d",&n)!=EOF)
{
memset(b,0,sizeof(b));
for(int i=1; i<=n; i++)
{
scanf("%d",&s);
b[s]++;
}
init_prim();
scanf("%d",&q);
for(int cas=0; cas<q; cas++)
{
scanf("%d%d",&l,&r);
if(l>M)l=M-1;
if(r>M)r=M-1;
printf("%d\n",count[r]-count[l-1]);
}
}
return 0;
}