HDU 4715 Difference Between Primes

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
   
   
   
   
   
   
   
   


   
   
   
   

模拟就能做，先保存后查询。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
map<int,int>a;

bool isprime(int num)
{
    if (num == 2 || num == 3)
    {
        return true;
    }
    if (num % 6 != 1 && num % 6 != 5)
    {
        return false;
    }
    for (int i = 5; i*i <= num; i += 6)
    {
        if (num % i == 0 || num % (i+2) == 0)
        {
            return false;
        }
    }
    return true;
}
int main()
{
    for(int i=2; i<=3110011; i++)
    {
        if(isprime(i))
            a[i]++;
    }
    int t,x;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&x);
        map<int,int>::iterator ai;
        int flag=0;
        if(x>0)
        {

            for(ai=a.begin(); ai!=a.end(); ai++)
            {

                if(isprime(ai->first+x))
                {
                    printf("%d %d\n",ai->first+x,ai->first);
                    flag=1;
                    break;

                }
            }
            if(!flag)
            {
                printf("FAIL\n");
            }
        }
        else
        {

            for(ai=a.begin(); ai!=a.end(); ai++)
            {

                if(isprime(ai->first-x))
                {
                    printf("%d %d\n",ai->first,ai->first-x);
                    flag=1;
                    break;

                }
            }
            if(!flag)
            {
                printf("FAIL\n");
            }
        }

    }
    return 0;
}