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：

 

 

 1 #include <stdio.h>

 2 #include <algorithm>

 3 #include <iostream>

 4 #include <string.h>

 5 #include <string>

 6 #include <math.h>

 7 #include <stdlib.h>

 8 #include <queue>

 9 #include <stack>

10 #include <set>

11 #include <map>

12 #include <list>

13 #include <iomanip>

14 #include <vector>

15 #pragma comment(linker, "/STACK:1024000000,1024000000")

16 #pragma warning(disable:4786)

17 

18 using namespace std;

19 

20 const int INF = 0x3f3f3f3f;

21 const int MAX = 1000000 + 10;

22 const double eps = 1e-8;

23 const double PI = acos(-1.0);

24 

25 int a[MAX];

26 

27 int main()

28 {

29     int i , j ;

30     memset(a , 0 , sizeof(a));

31     a[1] = 1; a[0] = 1;

32     for(i = 2;i <= sqrt(MAX);i++)

33     {

34         if(a[i] == 0)

35             for(j = i * i;j <= MAX;j += i)

36                 a[j] = 1;

37     }

38     int n,m;

39     scanf("%d",&n);

40     while(n--)

41     {

42         scanf("%d",&m);

43         int t = abs(m);

44         for(i = t;i < MAX;i++)

45             if(!a[i] && !a[i - t])

46             {

47                 if(m > 0)

48                 {

49                     printf("%d %d\n",i,i - t);

50                     break;

51                 }

52                 else

53                 {

54                     printf("%d %d\n",i - t,i);

55                     break;

56                 }

57             }

58         if(i == MAX)

59             printf("FAIL\n");

60     }

61     return 0;

62 }

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