AtCoder Beginner Contest 084 D - 2017-like Number【数论/素数/前缀和】

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

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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
【题意】：当N和（N + 1）/ 2都是素数时，奇数N与2017相似。 你有Q个查询。 在第i个查询中，给定两个奇数Li和Ri，找到与2017相似的奇数x的数目，使得li≤x≤ri。
【分析】：筛素数用埃筛，查询用前缀和。
【代码】：

#include<bits/stdc++.h>
using namespace std;
bool f [100001];
int c [100002];
int N,L,R ;
int main ()
{
    for (int i =2; i<=100000; i++)
        if (!f[i])
        for (int j = i+i ;j<=100000; j+=i )
                f[j]= true ;

    for (int i =3; i<=100000; i+=2)
        if (!f[i] && !f[(i+1)/2])
            c[i]++;

    for (int i =3; i <=100000; i ++)
            c[i]+=c[i-1];

    scanf ("%d",&N);
    while (N--)
    {
        scanf ("%d%d" ,&L ,&R );
        printf ("%d\n" , c[R] - c[L-1]);
    }
}