AtCoder Beginner Contest 084 D - 2017-like Number (区间问题 a[r]-a[l-1])

最新推荐文章于 2019-07-30 17:36:00 发布

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最新推荐文章于 2019-07-30 17:36:00 发布

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分类专栏： Atcoder 算法欠缺弱项—区间问题

本文链接：https://blog.youkuaiyun.com/deepseazbw/article/details/78940985

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本文介绍了一个算法挑战，任务是在指定范围内找出所有与2017类似的素数。所谓2017类似的素数是指该数及其一半加一均为素数。通过预先计算并存储这些特殊素数，可以高效地处理多次查询。

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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

1
3 7
Sample Output 1

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

4
13 13
7 11
7 11
2017 2017
Sample Output 2

1
0
0
1
Note that 2017 is also similar to 2017.

Sample Input 3
6
1 53
13 91
37 55
19 51
73 91
13 49
Sample Output 3

4
4
1
1
1

题意：判断[l,r]内与2017类似的数字，2017的性质：2017为素数，（2017+1）/2仍为素数。

思路：打表算出来1-10^5内各自的数字，然后结果为a[r]-a[l-1].（区间问题一直是我的一个弱项，类似于[l,r]找是3和7的倍数是多少个的问题，我在HNUST上WA了20发！！！！）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#define N 100005
using namespace std;
typedef long long ll;
int vis[N],a[N];
void creat()
{
    for(int i=2; i<=400; i++)
    {
        if(!vis[i])
        {
            for(int j=i*2; j<=N; j+=i)
                vis[j]=1;
        }
    }
}
int main()
{

    memset(vis,0,sizeof(vis));
    creat();
    vis[1]=1;
    int sum=0,n;
    for(int i=1; i<=N; i++)
    {
        if(i%2==1)
        {
            if(vis[i]==0&&vis[(i+1)/2]==0)
                sum++;
        }

        a[i]=sum;
    }
    while(scanf("%d",&n)==1)
    {
        while(n--)
        {
            int c,b;
            scanf("%d%d",&c,&b);

            printf("%d\n",a[b]-a[c-1]);
        }
    }
}