HDUOJ-----Difference Between Primes

Difference Between Primes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 832    Accepted Submission(s): 267

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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liuyiding

 

快速打素数表:

代码：

代码敲上去较为匆忙！,请自己优化......62ms

 #include<iostream>
 #include<cstdio>
 #define maxn 1000000
 using namespace std;
 int prime[];
 bool bo[maxn+];
 int prime_table()
 {
     int i,j,flag=;
     memset(bo,,sizeof bo);
     bo[]=bo[]=;
     for(i=; i*i<=maxn;i++)
     {
         if(!bo[i])
         {
           for(j=i*i;j<=maxn;j+=i)
               bo[j]=;
         }
     }
      for(i=;i<=maxn;i++)
         if(!bo[i]) prime[flag++]=i;
    return flag;
 }
 bool isprime(int a)
 {
     for(int i=;prime[i]*prime[i]<=a;i++)
     {
         if(a%prime[i]==)
             return ;
     }
     return ;
 }

 int main()
 {
   int i,t,b,num;
   num=prime_table();
   scanf("%d",&t);
   while(t--)
   {
     scanf("%d",&b);
     if(b>=)
     {
      for(i=;i<num;i++)
       {

         if(isprime(b+prime[i]))
        {
            printf("%d %d\n",prime[i]+b,prime[i]);
            break;
        }
      }
     }
     else
     {
       for(i=;i<num;i++)
       {

         if(isprime(prime[i]-b))
        {
            printf("%d %d\n",prime[i],prime[i]-b);
            break;
        }
     }
     }
   }
   return ;
 }