Prime Path
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.
— 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
Source
1 //2017-02-23 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <queue> 6 7 using namespace std; 8 9 struct node 10 { 11 int num, step; 12 }; 13 14 bool isPrime(int n) 15 { 16 for(int i = 2; i*i <= n; i++) 17 if(n%i==0)return false; 18 return true; 19 } 20 21 int pow(int a, int n) 22 { 23 int ans = 1; 24 for(int i = 0; i < n; i++) 25 ans *= a; 26 return ans; 27 } 28 29 int main() 30 { 31 int n, x, y; 32 bool book[10005]; 33 cin>>n; 34 while(n--) 35 { 36 cin>>x>>y; 37 queue<node> q; 38 node tmp; 39 tmp.num = x; 40 tmp.step = 0; 41 q.push(tmp); 42 int num, step; 43 bool ok = false; 44 memset(book, 0, sizeof(book)); 45 book[x] = 1; 46 while(!q.empty()){ 47 num = q.front().num; 48 step = q.front().step; 49 q.pop(); 50 if(num == y){ 51 ok = true; 52 cout<<step<<endl; 53 break; 54 } 55 for(int i = 0; i < 4; i++){ 56 for(int p = 0; p < 10; p++){ 57 if(i==3 && p==0)continue; 58 int nowNum = num-((num/pow(10, i))%10)*pow(10, i)+p*pow(10, i); 59 if(!book[nowNum] && isPrime(nowNum)){ 60 tmp.num = nowNum; 61 tmp.step = step+1; 62 q.push(tmp); 63 book[nowNum] = 1; 64 } 65 } 66 } 67 if(ok)break; 68 } 69 } 70 71 return 0; 72 }