Prime Friend

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

题解：这道题素数范围筛选很重要，一百万会 WA，二百万超内存，一百九十万就能AC，也是很玄学了；

这题的素数筛法选区也很重要，否者会超时；

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <map>
#include <string.h>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
using namespace std;


#define maxn 19000000
bool isprime[maxn];
long long prime[maxn],nprime;
void getprime(void)
{
    memset(isprime,true,sizeof(isprime));
    nprime=0;
    long long i,j;
    for(i=2;i<maxn;i++)
    {
        if(isprime[i])
        {
            prime[nprime++]=i;
            for(j=i*i;j<maxn;j+=i)
            {
                isprime[j]=false;
            }
        }
    }
}//666的素数表；

bool isd[150];
int main()
{
    getprime();
    int T,a,b;
    for(int i=1;i<nprime;i++)
    {
        if(prime[i]-prime[i-1]<150)
        {
            isd[prime[i]-prime[i-1]]=1;
        }
    }
    cin>>T;
    int num=1;
    while(1)
    {
        if(num>T) break;
        else { printf("Case %d: ",num); num++;}
        cin>>a>>b;
        if(a>b)
        {
            int t;
            t=a,a=b,b=t;
        }
        long long ans=-1,d=b-a;
        if(!isd[d] || a==b)
        {
            cout<<"-1"<<endl;
            continue;
        }
        for(int i=1;i<nprime;i++)
        {
            if(prime[i]-prime[i-1]==d && a<=prime[i-1])
            {
                ans=prime[i-1]-a;
                break;
            }
        }
        cout<<ans<<endl;
    }
}