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;
}