Description
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.
1033The 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.
1733
3733
3739
3779
8779
8179
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#include <iostream> #include <cstring> #include <queue> using namespace std; bool num[10001],used[10001]; int s,e; void prim()//判断素数 { memset(num,true,sizeof(num)); for(int i=2;i*i<10001;i++) { if(num[i]) { for(int j=i+i;j<10000;j=j+i) num[j]=false; } } memset(num,false,1000); } struct node { int x,step; }; int d[4]={1,10,100,1000};//位数 int bfs(int x) { queue<node> que; node tp,num; int tx; num.x=x,num.step=0; used[x]=false,que.push(num); while(!que.empty()) { tp=que.front(); que.pop(); for(int i=0;i<4;i++) for(int j=0;j<=9;j++) if(i==3&&j==0) continue; else { if(j!=(tp.x/d[i]%10)) { tx=((tp.x/d[i]/10)*10+j)*d[i]+tp.x%d[i]; if(used[tx]) { used[tx]=false,num.x=tx,num.step=tp.step+1,que.push(num); if(num.x==e) return num.step; } } } } } int main() { int t; cin>>t; prim(); while(t--) { cin>>s>>e; memcpy(used,num,10000); if(s==e) cout<<"0"<<endl; else cout<<bfs(s)<<endl; } return 0; }