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.
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).
One line for each case, either with a number stating the minimal cost or containing the Word Impossible.
3 1033 8179 1373 8017 1033 1033
6 7 0
直接根據(jù)題意走一遍就行了
#include<stdio.h>#include<string.h>#include<queue>#define MAX_N 10000#define MIN_N 1000using namespace std;bool prim[MAX_N];int best[MAX_N];int M[]={1,10,100,1000,10000};void init(){ memset(prim,true,sizeof(prim)); for(int i=2;i*i<=MAX_N;i++) if(prim[i]) for(int j=i*i;j<=MAX_N;j+=i) prim[j]=false;}int bfs(int s,int e){ memset(best,-1,sizeof(best)); queue<int> que;//now que.push(s); best[s]=0; while(!que.empty()){ int t=que.front();que.pop(); int cnt=best[t]; if(t==e) return cnt; for(int i=0;i<4;i++){ int tmp=(t%M[i+1])/M[i]; for(int j=1;j<10-tmp;j++){ int k=j*M[i]+t; if(best[k]==-1&&prim[k]){ best[k]=cnt+1; que.push(k); } } for(int j=1;j<=tmp;j++){ int k=-j*M[i]+t; if(k>MIN_N&&best[k]==-1&&prim[k]){ best[k]=cnt+1; que.push(k); } } } } return -1;}int main(){ int T,s,e;init(); scanf("%d",&T); while(T--){ scanf("%d%d",&s,&e); int ans=bfs(s,e); if(ans>=0) printf("%d/n",ans); else puts("Impossible"); } return 0;}新聞熱點
疑難解答
