Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7281 | Accepted: 4128 |
Description
— 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
Source
#include <iostream> #include <string.h> #include <math.h> #include <queue> using namespace std; int isprime( int n){ int i,t; t=sqrt(n+0.0); for (i=2;i<=t;i++){ if (n%i==0) return -2; } return -1; }//prime:-1; not prime:-2; arrived:>=0; const int S=10050; int prime[S],f[S],w[4]; int main(){ int i,T,x,y,tmp,a,b,c,d,t; for (i=2;i<S;i++) prime[i]=isprime(i); cin>>T; while (T--){ cin>>x>>y; memcpy(f,prime,S*sizeof(int)); f[x]=0; queue<int> q; q.push(x); while (!q.empty()&&(f[y]==-1)) { t=q.front(); tmp=t; q.pop(); for (i=0;i<=3;i++) {w[i]=tmp%10;tmp/=10;} //ge_wei for (i=0;i<=9;i++) { tmp=i+10*w[1]+100*w[2]+1000*w[3]; if (f[tmp]==-1) { f[tmp]=f[t]+1; q.push(tmp); } } //shi_wei for (i=0;i<=9;i++) { tmp=w[0]+10*i+100*w[2]+1000*w[3]; if (f[tmp]==-1) { f[tmp]=f[t]+1; q.push(tmp); } } for (i=0;i<=9;i++) { tmp=w[0]+10*w[1]+100*i+1000*w[3]; if (f[tmp]==-1) { f[tmp]=f[t]+1; q.push(tmp); } } for (i=1;i<=9;i++) { tmp=w[0]+10*w[1]+100*w[2]+1000*i; if (f[tmp]==-1) { f[tmp]=f[t]+1; q.push(tmp); } } } if (f[y]!=-1) cout<<f[y]<<endl;else cout<<"Impossible\n"; } return 0; }