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;
}
}
}
}
}