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 1 | Sample Output 1 |
|---|---|
3 |
6 |
大致意思就是由前面的那个素数变到后面的那个素数，每次只能变一位数，变化后的数也应该是一个素数（无论是不是所求的数），求变化次数。
直接BFS即可。
//Asimple
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = ;
ll n, m, s, res, ans, len, T, k;
int x, y;
int pr[maxn]; int P(int n) {
for(int i=; i*i<=n; i++) {
if( n%i== ) return ;
}
return ;
}
//将k位化0
int change(int n, int k) {
char s[] = {};
sprintf(s, "%d", n);
s[k] = '';
sscanf(s, "%d", &n);
return n;
} int solve(int s, int e) {
queue<int> q;
int dis[maxn] = {};
q.push(s);
dis[s] = ;
while( q.size() ) {
s = q.front();
q.pop();
if( s == e ) return dis[s]-;
int t = ;
for(int i=; i<; i++) {
int k = change(s, i);
for(int j=; j<; j++) {
int a = k+j*t;
if( pr[a]== && dis[a]== ) {
q.push(a);
dis[a] = dis[s]+;
}
}
t /= ;
}
}
return -;
} void input() {
for(int i=; i<maxn; i++) pr[i] = P(i);
cin >> T;
while( T -- ) {
cin >> x >> y;
ans = solve(x, y);
if( ans==- ) puts("Impossible");
else cout << ans << endl;
}
} int main(){
input();
return ;
}