HDUOJ-----Difference Between Primes

Gxjun

发布于 2018-03-21 04:50:58

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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

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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 }