题目：

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

题意：

      问从a最少需要改变几次才能变成b，每次只能改变一位数字，且改变后的数字必须为素数。

 

题解：

     用广搜来解决，每次都从千位到各位依次改变一位判断是否满足题目要求，满足即加入队列。

 

代码：

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;int a,b;int turn[4];          //将每次从队列中取出的数转化为数组的形式，便于改变每一位的数字bool isprime[10003];   //打表，将素数记录下来int visited[10000];   //判断某一数字是否出现过int step[10000];      //判断每一位数字从初始状态到此翻转了几次queue
           
            q;void prime(){ int i,j; for(i=1000;i<=10000;i++){ for(j=2;j
            
             =0;i--) { turn[i]=u%10; u/=10; }}int bfs(){ int u,i,j; q.push(a); visited[a]=1; while(!q.empty()) { u=q.front(); q.pop(); if(u==b) return step[u]; turned(u); //将数字转换成数组形式 for(i=0;i<4;i++) //依次遍历千百十个位 { int x=turn[i]; //x用来还原数组的数 for(j=0;j<=9;j++) { if(i==0&&j==0) continue; if(j==x) continue; turn[i]=j; int v=turn[0]*1000+turn[1]*100+turn[2]*10+turn[3]; //v表示转换后的数 if(isprime[v]&&!visited[v]) { step[v]=step[u]+1; visited[v]=1; q.push(v); if(v==b) return step[v]; } turn[i]=x; //还原 } } } return -1;}int main(){ int n; prime(); cin>>n; while(n--) { cin>>a>>b; memset(turn,0,sizeof(turn)); memset(visited,0,sizeof(visited)); memset(step,0,sizeof(step)); while(!q.empty()) q.pop(); int ans=bfs(); if(ans==-1) printf("Impossible\n"); else cout<
             
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