HDU 4715 Difference Between Primes (打表)

   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

361020

Sample Output

11 513 323 3

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

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);elsepw.println(i+" "+(i+m));}elsepw.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;}}}}}