4715 Difference Between Primes 打表

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

 

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ACode：

#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 1100000
#define maxx 500000
#define ll long
using namespace std;
ll p[maxn];
bool vis[maxn];
int main(){
    memset(p,0,sizeof(p));
    memset(vis,0,sizeof(vis));
    for(ll i=2;i<=maxn;++i){
        if(!p[i])p[++p[0]]=i;
        for(int j=1;j<=p[0]&&p[j]<maxn/i;++j){
            p[p[j]*i]=1;
            if(i%p[j]==0)break;
        }
    }
   // cout<<p[3252433];
    for(int i=1;i<maxn;++i){vis[p[i]]=true;}
    int loop;
    scanf("%d",&loop);
    while(loop--){
        ll a;
        cin>>a;
        if(a==0)printf("2 2\n");
        else if(a<0){
                a=-a;
             bool flag=true;
            for(int i=1;flag&&i<=maxn;++i){
                    //cout<<a*p[i]<<" "<<vis[a*p[i]]<<'\12';
            if(vis[a+p[i]]!=0){
                cout<<p[i]<<" "<<a+p[i]<<'\12';
                flag=false;
            }
        }
        }
        else{
            bool flag=true;
            for(int i=1;flag&&i<=maxn;++i){
                    //cout<<a*p[i]<<" "<<vis[a*p[i]]<<'\12';
            if(vis[a+p[i]]!=0){
                cout<<a+p[i]<<" "<<p[i]<<'\12';
                flag=false;
            }
        }
        }
    }
    return 0;
}