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
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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 }