POJ 3126 - Prime Path

F - Prime Path
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status

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.

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

#define N 10005

bool a
;
bool vis
;

void prime(){
memset(a,true,sizeof(a));
for (int i=2;i<N;i++)
if (a[i]){
for (int j=i+i;j<N;j+=i)
a[j]=false;
}
for (int i=0;i<1000;i++)
a[i]=false;
}

struct node{
int val,step;
node(int v,int s=0):val(v),step(s){}
};

int b[]={1,10,100,1000,10000};

void bfs(){
int x,y;
scanf("%d%d",&x,&y);
queue<node> q;
memset(vis,false,sizeof(vis));
q.push(node(x,0));
vis[x]=true;

while (!q.empty()){
int val=q.front().val,step=q.front().step;
q.pop();

//cout<<val<<' '<<step<<endl;

if (val==y){
printf("%d\n",step);
return;
}

for (int i=0;i<4;i++){
int temp=val/b[i+1]*b[i+1]+val%b[i];

for (int j=0;j<10;j++){
if (j)
temp+=b[i];
if (!vis[temp] && a[temp]){
vis[temp]=true;
q.push(node(temp,step+1));
}
}
}
}

printf("Impossible\n");
}

int main(){
prime();
int n;
scanf("%d",&n);
while(n--)
bfs();
return 0;
}