HDU 3823 Prime Friend（线性欧拉筛+打表）

Besides the ordinary Boy Friend and Girl Friend, here we define a more academic kind of friend: Prime Friend. We call a nonnegative integer A is the integer B’s Prime Friend when the sum of A and B is a prime. 
So an integer has many prime friends, for example, 1 has infinite prime friends: 1, 2, 4, 6, 10 and so on. This problem is very simple, given two integers A and B, find the minimum common prime friend which will make them not only become primes but also prime neighbor. We say C and D is prime neighbor only when both of them are primes and integer(s) between them is/are not.

InputThe first line contains a single integer T, indicating the number of test cases. 
Each test case only contains two integers A and B.

Technical Specification

1. 1 <= T <= 1000 
2. 1 <= A, B <= 150OutputFor each test case, output the case number first, then the minimum common prime friend of A and B, if not such number exists, output -1.Sample Input

2
2 4
3 6

Sample Output

Case 1: 1
Case 2: -1

题意：给出两个整数ab 使a+x b+x均为素数 且ab之间没有素数 求最小的符合条件的x

思路：java数组开大了就爆空间，注意空间；一个潜在条件a+x-（b+x）=a-b<150，即两个素数之差小于150；

代码：

import java.util.Scanner;
public class Main {
        static final int max=(int)16000000;
        static int prime[]=new int[1031131];
        static boolean is_prime[]=new boolean[max];
        static int k=0;
        public static void Prime(){
              is_prime[0]=is_prime[1]=true;
              for(int i=2;i<max;i++){
                    if(!is_prime[i]) prime[k++]=i;
                    for(int j=0;j<k&&prime[j]*i<max;j++){
                          is_prime[i*prime[j]]=true;
                          if(i%prime[j]==0) break;
                    }
              }
        }
        public static void main(String[] args) {
               Prime();
//             System.out.println(k);
               Scanner scan=new Scanner(System.in);
               int t=scan.nextInt();
               for(int i=1;i<=t;i++){
                     int a=scan.nextInt();
                     int b=scan.nextInt();
                     if(a>b) {
                           int tmp=a;
                           a=b;
                           b=tmp;
                     }
                     System.out.print("Case "+i+": ");
                     boolean flag=false;
                     int num=0;
                     for(int j=0;j<k-1;j++){
                          if(prime[j]>=a && prime[j+1]>=b &&prime[j]-a==prime[j+1]-b &&prime[j+1]-prime[j]<150){
                                 num=prime[j]-a;
                                 flag=true;
                                 break;
                          }
                     }
                     if(flag) System.out.println(num);
                     else System.out.println("-1");
               }
        }
}

原文地址：https://www.cnblogs.com/qdu-lkc/p/12195537.html