• 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
    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;
    				}
    			}
    		}
    	}
    }
    


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  • 原文地址:https://www.cnblogs.com/suncoolcat/p/3313264.html
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