Twin Primes

Twin primes are pairs of primes of the form (p; p + 2). The term \twin prime" was coined by Paul
Stckel (1892-1919). The rst few twin primes are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43).
In this problem you are asked to nd out the S-th twin prime pair where S is an integer that will be
given in the input.
Input
The input will contain less than 10001 lines of input. Each line contains an integers S (1 S 100000),
which is the serial number of a twin prime pair. Input le is terminated by end of le.
Output
For each line of input you will have to produce one line of output which contains the S-th twin prime
pair. The pair is printed in the form (p1,<space>p2). Here <space> means the space character (ASCII
32) . You can safely assume that the primes in the 100000-th twin prime pair are less than 20000000.
Sample Input
1
2
3
4
Sample Output
(3, 5)
(5, 7)
(11, 13)
(17, 19)Twin primes are pairs of primes of the form (p; p + 2). The term \twin prime" was coined by Paul
Stckel (1892-1919). The rst few twin primes are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43).
In this problem you are asked to nd out the S-th twin prime pair where S is an integer that will be
given in the input.
Input
The input will contain less than 10001 lines of input. Each line contains an integers S (1 S 100000),
which is the serial number of a twin prime pair. Input le is terminated by end of le.
Output
For each line of input you will have to produce one line of output which contains the S-th twin prime
pair. The pair is printed in the form (p1,<space>p2). Here <space> means the space character (ASCII
32) . You can safely assume that the primes in the 100000-th twin prime pair are less than 20000000.
Sample Input
1
2
3
4
Sample Output
(3, 5)
(5, 7)
(11, 13)
(17, 19)

题意：

定义双素数(p,p+2)，p,p+2都为素数。

先输入若干个n，输出第n对双素数。

分析：

时间限制为3000ms，数据量为2e7，可以直接暴力求解。

首先用欧拉筛筛出所有素数，然后暴力枚举所有素数，判断是否是双素数即可。

AC code:

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
P p[20000005];
bool u[20000005];
int su[20000005];
int num;
void olas()
{
    num=1;
    memset(u,true,sizeof(u));
    for(int i=2;i<=20000000;i++)
    {
        if(u[i])    su[num++]=i;
        for(int j=1;j<num;j++)
        {
            if(i*su[j]>20000000)    break;
            u[i*su[j]]=false;
            if(i%su[j]==0)    break;
        }
    }
}
int main()
{
    //freopen("input.txt","r",stdin);
    int n;
    olas();
    int num=0;
    for(int i=2;i<=20000000-2;i++)
    {
        if(u[i]&&u[i+2])
        {
            p[num++]=P(i,i+2);
        }
    }
    while(~scanf("%d",&n))
    {
        printf("(%d, %d)\n",p[n-1].first,p[n-1].second);
    }
    return 0;
}

View Code