Difference Between Primes
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 860 Accepted Submission(s): 278
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
思路 : 打表
总结: 不要忘了有负数的情况,
使用 Scanner sc = new Scanner(new BufferedInputStream(System.in)); 和
System.out.println(); 程序执行时间如下图
使用: BufferedReader bu=new BufferedReader(new InputStreamReader(System.in)); 和
PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out),true); 程序如下图
import java.io.*; import java.util.*; public class Main { int max=(int)Math.pow(10, 6)+10; boolean a[]=new boolean[max]; public static void main(String[] args) throws IOException{ new Main().work(); } void work() throws IOException{ BufferedReader bu=new BufferedReader(new InputStreamReader(System.in)); PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out),true); isPrime(); int n=Integer.parseInt(bu.readLine()); while(n--!=0){ int x=Integer.parseInt(bu.readLine()); int m=x>0?x:Math.abs(x); boolean boo=true; int i=2; for(;i<max;i++){ if(a[i+m]&&a[i]){ boo=false; break; } } if(!boo){ if(x>0) pw.println((i+m)+" "+i); else pw.println(i+" "+(i+m)); } else pw.println("FAIL"); } } //素数表 void isPrime(){ Arrays.fill(a,true); for(int i=2;i<max;i++){ if(a[i]){ for(int j=2*i;j<max;j+=i){ a[j]=false; } } } } }