http://acm.hdu.edu.cn/showproblem.php?pid=4715
Difference Between Primes
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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 x at the next n lines. The absolute value of x is not greater than 10^6.
Output
For each number x tested, outputs two primes a and b at one line separated with 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
分析：
这道题就是求一个整数用两个素数（大于等于2）的差表示出来，要求两个素数在满足条件的情况下最小。
AC代码：
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <vector>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#pragma warning(disable:4786) using namespace std; const int INF = 0x3f3f3f3f;
const int MAX = + ;
const double eps = 1e-;
const double PI = acos(-1.0); int a[MAX]; int main()
{
int i , j ;
memset(a , , sizeof(a));
a[] = ; a[] = ;
for(i = ;i <= sqrt(MAX);i++)
{
if(a[i] == )
for(j = i * i;j <= MAX;j += i)
a[j] = ;
}
int n,m;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
int t = abs(m);
for(i = t;i < MAX;i++)
if(!a[i] && !a[i - t])
{
if(m > )
{
printf("%d %d\n",i,i - t);
break;
}
else
{
printf("%d %d\n",i - t,i);
break;
}
}
if(i == MAX)
printf("FAIL\n");
}
return ;
}