Difference Between Primes（素数打表，素数表与数表通用）

Difference Between Primes

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2735 Accepted Submission(s): 768



Problem Description
All you know Goldbach conjecture.That is to say, Every even integer greater than 2 can be expressed as the sum of two primes. Today, skywind present a new conjecture: every even integer can be expressed as the difference
of two primes. To validate this conjecture, you are asked to write a program.


Input
The first line of input is a number nidentified the count of test cases(n<10^5). There is a even number xat the next nlines. The absolute value of xis not greater than 10^6.



Output
For each number xtested, outputstwo primes aand bat one line separatedwith one space where a-b=x. If more than one group can meet it, output the minimum group. If no primes can satisfy it, output 'FAIL'.



Sample Input

3
6
10
20

Sample Output

11 5
13 3
23 3

Source
2013 ACM/ICPC
Asia Regional Online —— Warmup

感觉这个题特别好！注意数据大！

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;
int prime[1000050],num[1000050];
int main()
{
    int t,n,len=0,k,l,f,f1,m;
    long long i,j;
    memset(num,0,sizeof(num));
    memset(prime,0,sizeof(prime));
    for(i=2;i<=sqrt(1000050);i++)
    {
        for(j=i*i;j<=1000050;j+=i)
        {
            num[j]=1;
        }
    }
    for(i=2;i<=1000050;i++)
    {
        if(!num[i])
        {
            prime[len++]=i;
        }
    }
   /* for(i=0;i<5;i++)
    {
        cout<<prime[i]<<" ";
    }*/
    cin>>t;
    while(t--)
    {
        f=f1=0;
        cin>>n;
        m=n;
        if(n<0)
        {
            n=-n;
        }//注意负数！
        if(n==0)
        {
            cout<<"2 2"<<endl;
            continue;
        }
        for(k=0;k<len;k++)
        {
           if(!num[prime[k]+n])//这样比较快哦！
           {
               if(m<0)
               {
                   cout<<prime[k]<<" "<<prime[k]+n<<endl;;
               }
               else
               {
                   cout<<prime[k]+n<<" "<<prime[k]<<endl;
               }
               f=1;
               break;
           }
        }
        if(!f)
        {
            cout<<"FAIL"<<endl;
        }
    }
}