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 3 1033 8179 1373 8017 1033 1033 Sample Output 6 7 0题意是 一个四位的素数，每次更改一位最终更改到另一个素数，并且更改的过程中的数也得是素数。
bfs 修改一位对每个符合条件的数枚举记录步数就好了
编译器竟然不识别万能头文件，itoa函数也不识别，搞得编译错误了好几次。。。。
#include<math.h> #include<stdio.h> #include<queue> #include<string.h> #include<stdlib.h> using namespace std; int vis[10005]; struct node { int v,step; }; int s,e; bool isp(int x) { for(int i=2;i<=(int)sqrt(double(x));i++) { if(x%i==0) return 0; } return 1; } int bfs(int s) { node a,next; a.v=s; a.step=0; queue<node> q; q.push(a); vis[a.v]=1; while(!q.empty()) { a=q.front(); q.pop(); if(a.v==e&&a.step!=0) return a.step; char num[10]; for(int i=0;i<4;i++) { sprintf(num,"%d",a.v); // itoa(a.v,num,10); for(char ch='0';ch<='9';ch++) { if((i==0||i==3)&&ch=='0') continue; num[i]=ch; sscanf(num,"%d",&next.v); if(!vis[next.v]&&isp(next.v)) { next.step=a.step+1; q.push(next); } vis[next.v]=1; } } } return 0; } int main() { int t; scanf("%d",&t); while(t--) { memset(vis,0,sizeof(vis)); scanf("%d%d",&s,&e); int ans=bfs(s); printf("%d\n",ans); } }
