Codeforces Round #226 (Div. 2):Problem 385C - Bear and Prime Numbers (素数刷法+前缀和)

Time Limit: 2000ms

Memory Limit: 524288KB

This problem will be judged on CodeForces. Original ID:
385C

64-bit integer IO format: %I64d      Java class name:
(Any)

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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:
, 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 Input

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

Hint

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：

给你N个数a[N]，m个操作，每个操作给你一个区间[l,r],求a[i]能被[l,r]内素数整除的总数。

题解：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 10000010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;

const int INF = 1000010;

using namespace std;

int sum[N];///记录i之前的素数能整除的个数
int num[N];///记录i出现的次数
bool is[N];
int n;

void build()
{
    memset(is,0,sizeof is);
    memset(sum,0,sizeof sum);
    is[0]=is[1]=1;
    for(int i=2; i<=n; i++)
    {
        if(!is[i])
        {
            for(int j=i; j<=n; j+=i)
            {
                sum[i]+=num[j];
                if(j!=i)
                    is[j]=1;
            }
        }
    }
    for(int i=1; i<=n; i++)
    {
        sum[i]=sum[i-1]+sum[i];
    }
}
int main()
{
    int m;
    while(~scanf("%d",&m))
    {
        memset(num,0,sizeof num);
        int x;
        n=0;
        for(int i=0; i<m; i++)
        {
            scanf("%d",&x);
            if(x>n)
                n=x;
            num[x]++;
        }
        build();
        int w;
        scanf("%d",&w);
        int l, r;
        while(w--)
        {
            scanf("%d%d",&l,&r);
            if(r>n)
                r=n;
            if(l>n)
            {
                printf("0\n");
                continue;
            }
            printf("%d\n",sum[r]-sum[l-1]);
        }
    }
    return 0;
}