hduoj 4715 Difference Between Primes 2013 ACM/ICPC Asia Regional Online —— Warmup

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

2013 ACM/ICPC Asia Regional Online —— Warmup

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 ;
 }