题意：给你两个四位数的素数，让你从第一个素数变成第二个，每次操作只能改动一位数字，且需要保证改动后的数字任为为素数；输出需要操作多少次；
题解：先素数打表，BFS处理时将满足条件的数字放入队列中（未被访问过且为素数，与队列front（）的元数只有一位数字不同）；
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#define INF 0x7ffffff
using namespace std;
const int maxn = 1e4+5;
bool prime[maxn];
int T, s, e;
void is_prime()
{memset(prime, 1, sizeof(prime));int m = sqrt(maxn);prime[0] = prime[1] = false;for (int i = 2; i < m; i++){if(prime[i]){for (int j = i * i; j < maxn; j += i){prime[j] = false;}}}
}
bool Judge(int a, int b)
{int cnt = 0;if(a%10 != b%10) cnt++;if(a%100/10 != b%100/10) cnt++;if(a%1000/100 != b%1000/100) cnt++;if(a/1000 != b/1000) cnt++;if(cnt == 1) return true;return false;
}
int bfs()
{bool vis[10000];memset(vis, 0, sizeof(vis));pair<int, int> Now;Now.first = s, Now.second = 0;queue<pair<int, int> > q;vis[s] = true;q.push(Now);while(!q.empty()){Now = q.front();if(Now.first == e){return Now.second;}q.pop();for (int i = 1000; i < 10000; i++){if(prime[i] && !vis[i] && Judge(i, Now.first)){vis[i] = true;q.push(make_pair(i, Now.second + 1));}}}return -1;
}
int main()
{is_prime();scanf("%d", &T);while(T--){scanf("%d%d", &s, &e);int ans = bfs();if(ans != -1)printf("%d\n",ans);elseprintf("Impossible\n");}
}