HDU 4715：Difference Between Primes

Difference Between Primes

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

Total Submission(s): 3339    Accepted Submission(s): 953

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

 

迷失在幽谷中的鸟儿，独自飞翔在这偌大的天地间，却不知自己该飞往何方……

#include<iostream>
using namespace std;
const long N = 1000005;
long prime[N],num_prime;
int isNotPrime[N]= {1, 1};
void getprime()
{
    for(long i = 2 ; i < N ; i ++)
    {
        if(!isNotPrime[i])prime[num_prime ++]=i;
        for(long j = 0 ; j < num_prime && i * prime[j] <  N ; j ++)
        {
            isNotPrime[i * prime[j]] = 1;
            if(!(i%prime[j]))break;
        }
    }
}
int main()
{
    getprime();
    int T,n,s,a;
    cin>>T;
    while(T--)
    {
        cin>>n;
        a=n>0?n:-n;
        s=0;
        for(int i=0; i<100000; i++)
            if(!isNotPrime[prime[i]+a])
            {
                if(n>0) cout<<prime[i]+a<<" "<<prime[i]<<endl;
                else cout<<prime[i]<<" "<<prime[i]+a<<endl;
                s=1;
                break;
            }
        if(s==0)cout<<"FAIL"<<endl;
    }
    return 0;
}