HDUOJ-----Difference Between Primes

Difference Between Primes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 832    Accepted Submission(s): 267

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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

Recommend

liuyiding

 

快速打素数表:

代码：

代码敲上去较为匆忙！,请自己优化......62ms

 1 #include<iostream>

 2 #include<cstdio>

 3 #define maxn 1000000

 4 using namespace std;

 5 int prime[78500];

 6 bool bo[maxn+5];

 7 int prime_table()

 8 {

 9     int i,j,flag=0;

10     memset(bo,0,sizeof bo);

11     bo[0]=bo[1]=1;

12     for(i=2; i*i<=maxn;i++)

13     {

14         if(!bo[i])

15         {

16           for(j=i*i;j<=maxn;j+=i)

17               bo[j]=1;

18         }

19     }

20      for(i=2;i<=maxn;i++)

21         if(!bo[i]) prime[flag++]=i;

22    return flag;

23 }

24 bool isprime(int a)

25 {

26     for(int i=0;prime[i]*prime[i]<=a;i++)

27     {

28         if(a%prime[i]==0)

29             return 0;

30     }

31     return 1;

32 }

33 

34 int main()

35 {

36   int i,t,b,num;

37   num=prime_table();

38   scanf("%d",&t);

39   while(t--)

40   {

41     scanf("%d",&b);

42     if(b>=0)

43     {

44      for(i=0;i<num;i++)

45       {

46       

47         if(isprime(b+prime[i]))

48        {

49            printf("%d %d\n",prime[i]+b,prime[i]);

50            break;

51        }

52      }

53     }

54     else

55     {

56       for(i=0;i<num;i++)

57       {

58       

59         if(isprime(prime[i]-b))

60        {

61            printf("%d %d\n",prime[i],prime[i]-b);

62            break;

63        }

64     }

65     }

66   }

67   return 0;

68 }

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