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, ..., 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 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.

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.

题意：

给你N个数a
，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
;///记录i之前的素数能整除的个数
int num
;///记录i出现的次数
bool is
;
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;
}