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

D - 2017-like Number

---

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.

---

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.

---

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.

---

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.

---

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 [];
int c [];
int N,L,R ;
int main ()
{
    for (int i =; i<=; i++)
        if (!f[i])
        for (int j = i+i ;j<=; j+=i )
                f[j]= true ;

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

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

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

前缀和