Prime Path

The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. 
— It is a matter of security to change such things every now and then, to keep the enemy in the dark. 
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know! 
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door. 
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime! 
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds. 
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime. 

Now, the minister of finance, who had been eavesdropping, intervened. 
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound. 
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you? 
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above. 

1033 
1733 
3733 
3739 
3779 
8779 
8179

The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.

Input

One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).

Output

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

Sample Input

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

题目意思是：给你两个素数，从一个素数变为另一个素数需要几步。条件是：1、个位数一定为奇数。2、千位不能为0.

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxnn=100005;
int v[maxnn];
int n,m;
struct node
{
    int x;
    int step;
};
queue<node>Q;
bool prime(int y)//判断素数
{
    if(y==0||y==1)
        return false;
    else   if(y==2||y==3)
        return true;
    else
    {
        double s=sqrt((double)y);
        for(int i=2; i<=s; i++)
            if(y%i==0)
                return false;
        return true;
    }
}
void bfs()
{
    int X,STEP;
    while(!Q.empty())
    {
        node a;
        a = Q.front();
        Q.pop();
        X=a.x;
        STEP=a.step;
        if(X==m)
        {
            printf("%d\n",STEP);
            return ;
        }
        for(int i=1; i<=9; i+=2) //个位
        {
            int y=X/10*10+i;
            if(y!=X&&!v[y]&&prime(y))
            {
                v[y]=1;
                node a;
                a.x=y;
                a.step=STEP+1;
                Q.push(a);

            }
        }
        for(int i=0; i<=9; i++) //十位
        {
            int y=X/100*100+i*10+X%10;
            if(y!=X&&!v[y]&&prime(y))
            {
                v[y]=1;
                node a;
                a.x=y;
                a.step=STEP+1;
                Q.push(a);
            }
        }
        for(int i=0; i<=9; i++) //百位
        {
            int y=X/1000*1000+i*100+X%100;
            if(y!=X&&!v[y]&&prime(y))
            {
                v[y]=1;
                node a;
                a.x=y;
                a.step=STEP+1;
                Q.push(a);

            }
        }
        for(int i=1; i<=9; i++) //千位
        {
            int y=i*1000+X%1000;
            if(y!=X&&!v[y]&&prime(y))
            {
                v[y]=1;
                node a;
                a.x=y;
                a.step=STEP+1;
                Q.push(a);

            }
        }

    }
    printf("Impossible\n");
    return ;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        while(!Q.empty())
            Q.pop();
        scanf("%d%d",&n,&m);
        memset(v,0,sizeof(v));
        v[n]=1;
        node a;
        a.x=n;
        a.step=0;
        Q.push(a);
        bfs();
    }
    return 0;
}