Prime Friend

Prime Friend

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4638    Accepted Submission(s): 947

 

Problem Description

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.

 

 

Input

The 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 <= 150

 

 

Output

For 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

AC代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstring>
using namespace std;
int su[20000005];
bool a[20000010];
bool b[1000];




int main()
{
	int T;int t=0;
	cin>>T;
	 memset(a,true,sizeof(a));
	a[0]=a[1]=false;
	for(int i=2;i<=20000000;i++)
	{
		if(a[i])
		{
			su[t++]=i;
			for(int j=i+i;j<=20000000;j+=i)
			a[j]=false;
		}
	}	memset(b,false,sizeof(b));
	for(int i=0;i<t-1;i++)
        {
	        if(su[i+1]-su[i]<=150)
         	b[su[i+1]-su[i]]=true;
        }
	for(int k=1;k<=T;k++)
	{
		int x,y,ans=-1;
		cin>>x>>y;
		cout<<"Case "<<k<<": ";
		if(x<y) swap(x,y);
		
		if(!b[x-y]||x==y)
		{
			cout<<"-1"<<endl;
		}
		else
		{
			for(int i=0;i<t-1;i++)
		{
			if(su[i+1]-su[i]==x-y&&y<=su[i])
			{
				ans=su[i]-y;break;
			}
			
		}cout<<ans<<endl;
		}
		
		
		
	}
}