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[78500];
bool bo[maxn+5];
int prime_table()
{
    int i,j,flag=0;
    memset(bo,0,sizeof bo);
    bo[0]=bo[1]=1;
    for(i=2; i*i<=maxn;i++)
    {
        if(!bo[i])
        {
          for(j=i*i;j<=maxn;j+=i)
              bo[j]=1;
        }
    }
     for(i=2;i<=maxn;i++)
        if(!bo[i]) prime[flag++]=i;
   return flag;
}
bool isprime(int a)
{
    for(int i=0;prime[i]*prime[i]<=a;i++)
    {
        if(a%prime[i]==0)
            return 0;
    }
    return 1;
}

int main()
{
  int i,t,b,num;
  num=prime_table();
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d",&b);
    if(b>=0)
    {
     for(i=0;i<num;i++)
      {
      
        if(isprime(b+prime[i]))
       {
           printf("%d %d
",prime[i]+b,prime[i]);
           break;
       }
     }
    }
    else
    {
      for(i=0;i<num;i++)
      {
      
        if(isprime(prime[i]-b))
       {
           printf("%d %d
",prime[i],prime[i]-b);
           break;
       }
    }
    }
  }
  return 0;
}