HDU 4715 Difference Between Primes（我不管，水题我也要写博客）

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

题意：把一个偶数分解成两个素数之差，输出最小的两个素数

思路：用埃氏筛法打个素数表，然后从最小的素数枚举就好.
AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
const int pmax=1e5+10;
const int vmax=1e6+10;
int prime[pmax],vis[vmax],psize=0;
void getprime()
{
memset(vis,0,sizeof vis);
vis[0]=vis[1]=1;
for(int i=2;i<sqrt(vmax+0.5);i++) if(!vis[i])
for(int j=i*i;j<vmax;j+=i)
vis[j]=1;
for(int i=2;i<vmax;i++)
if(!vis[i])
prime[psize++]=i;
}

int main()
{
getprime();
int n;
scanf("%d",&n);
while(n--)
{
int x,a,b;
scanf("%d",&x);
for(int i=0;;i++)
{
if(!vis[(prime[i]+abs(x))])
{
a=prime[i];
b=abs(x)+prime[i];
break;
}
}
if(x>0)
swap(a,b);
cout<<a<<' '<<b<<endl;
}
return 0;
}