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
Output
Sample Input
3 1033 8179 1373 8017 1033 1033
Sample Output
6 7 0
#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
bool prime[10000];
int state[10000];
int pre,obj;
queue<int> record;
void Init(){
for(int i=0;i<10000;i++) prime[i]=true;
for(int i=2;i<10000;i++){
if(prime[i]){
for(int j=2;i*j<10000;j++) prime[i*j]=false;
}
}
}
int BFS(){
if(pre==obj) return 0;
int multi[4]={1,10,100,1000};
int cur;
state[pre]=1;
record.push(pre);
while(!record.empty()){
cur=record.front();
record.pop();
int d[4];
d[0]=(cur/10)*10;
d[1]=(cur/100)*100+cur%10;
d[2]=(cur/1000)*1000+cur%100;
d[3]=cur%1000;
int digit;
for(int a=0;a<4;a++){
for(int b=0;b<=9;b++){
if(a==3&&b==0) continue;
digit=d[a]+multi[a]*b;
if(digit==obj) return state[cur];
if(prime[digit]&&!state[digit]){
record.push(digit);
state[digit]=state[cur]+1;
}
}
}
}
return -1;
}
int main(){
Init();
int ncase;
cin>>ncase;
while(ncase--){
cin>>pre>>obj;
while(!record.empty()) record.pop();
memset(state,0,sizeof(state));
int res;
res=BFS();
if(res==-1) cout<<"Impossible"<<endl;
else cout<<res<<endl;
}
}