2017-like Number

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D - 2017-like Number

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

We say that a odd number N is similar to 2017 when both N and (N+1)⁄2 are prime.

You are given Q queries.

In the i-th query, given two odd numbers li and ri, find the number of odd numbers x similar to 2017 such that li≤x≤ri.

Constraints

• 1≤Q≤105
• 1≤li≤ri≤105
• li and ri are odd.
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

Q
l1 r1
:
lQ rQ

Output

Print Q lines. The i-th line (1≤i≤Q) should contain the response to the i-th query.

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Sample Input 1

Copy

1
3 7

Sample Output 1

Copy

2

• 3 is similar to 2017, since both 3 and (3+1)⁄2=2 are prime.
• 5 is similar to 2017, since both 5 and (5+1)⁄2=3 are prime.
• 7 is not similar to 2017, since (7+1)⁄2=4 is not prime, although 7 is prime.

Thus, the response to the first query should be 2.

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Sample Input 2

Copy

4
13 13
7 11
7 11
2017 2017

Sample Output 2

Copy

1
0
0
1

Note that 2017 is also similar to 2017.

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Sample Input 3

Copy

6
1 53
13 91
37 55
19 51
73 91
13 49

Sample Output 3

Copy

4
4
1
1
1
2

这个题的题意很简单就不多说了，就问L~R之间素数的个数。

下边的这个是常规写法，遍历L~R，每个数都判断一遍。  这个肯定会超时

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <list>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x3f3f3f3f;
int fun(int x)
{
    int w=sqrt(x);
    for(int i=2;i<=w;i++)
        if(x%i==0)
            return 0;
    return 1;
}
int main() 
{
    int t;
    cin>>t;
    while(t--)
    {
        int l,r;
        cin>>l>>r;
        int ans=0;
        if(l%2==0)
            l++;
        for(int i=l;i<=r;i+=2)
        {
            if(fun(i)&&fun((i+1)/2))
            {
                ans++;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

但是这是多组数据，时间2s，看似很长，其实很短。

由于种种原因，我需要预处理，把素数不是素数区分做标记，这样先处理一次就行了，就避免了重复计算了。

这个就是——素数筛选法

#include <bits/stdc++.h>
using namespace std;
const int n=1e5+1;
int c[n],s[n]; 
int main()
{
    for(int i=2;i<n;i++)//素数筛选法，不是素数的标记为1
    {
        if(!c[i])
        {
            if(i!=2&&!c[(i+1)/2])
                s[i]=1;
            for(int j=i+i;j<n;j+=i)
                c[j]=1;
        }
    }
    for(int i=2;i<n;i++)//记录此坐标与之前的值素数个数
    {
        s[i]+=s[i-1];
    }
    int q;
    cin>>q;
    while(q--)//q次查询
    {
        int l,r;
        cin>>l>>r;
        cout<<s[r]-s[l-1]<<endl;//l与r之间的素数个数
    }
    return 0;
}