Codeforce Round #226 Div2 C

Codeforces Round #226 (Div. 2)
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Submission	Time	Verdict
5814433	01/26/2014 10:30PM	Accepted
5796818	01/24/2014 09:24PM	Time limit exceeded on pretest 6
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bitmasks

brute force

data structures

implementation

math

number theory

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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: $\text{[math]}$ , 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.

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.

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <queue>
 6 #include <cstring>
 7 #include <iostream>
 8 #include <algorithm>
 9 using namespace std;
10 #define INF 0x7fffffff 
11 #define mod 1000000007
12 #define ll long long
13 #define maxn 10000025
14 #define pi acos(-1.0)
15 int n, m;
16 int a[1000025], pri[1000025], prv[maxn],b[maxn];
17 ll c[maxn];
18 void add(int x, ll d){
19     while (x < maxn){
20         c[x] += d;
21         x += x&-x;
22     }
23 }
24 ll sum(ll x){
25     ll res = 0;
26     while (x > 0){
27         res += c[x];
28         x -= x&-x;
29     }
30     return res;
31 }
32 int main(){
33     for (int i = 2; i*i < maxn;i++)
34     if (!prv[i]){
35         for (int j = i*i; j < maxn;j+=i)
36         if (!prv[j])prv[j] = 1;
37     }
38     int k = 0;
39     for (int i = 2; i < maxn;i++)
40     if (!prv[i])pri[k++] = i;
41     scanf("%d", &n);
42     for (int i = 0; i < n; i++)scanf("%d", &a[i]);
43     for (int i = 0; i < n; i++)b[a[i]]++;
44     for (int i=0;i<k;i++)
45         for (int j = pri[i]; j < maxn; j += pri[i])
46             if(b[j])add(pri[i], b[j]);
47     scanf("%d", &m);
48     while (m--){
49         int l, r;
50         scanf("%d%d", &l, &r);
51         if (l >= maxn){printf("0\n"); continue;}
52         if (r >= maxn)r = maxn - 1;
53         printf("%I64d\n", sum(r) - sum(l - 1));
54     }
55     return 0;
56 }

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posted @ 2014-01-27 02:35 HaibaraAi 阅读(184) 评论(0) 收藏举报

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