G - Number Transformation(BFS+素数)

In this problem, you are given an integer number s. You can transform any integer number A to another integer number B by adding x to A. This x is an integer number which is a prime factor of A (please note that 1 and A are not being considered as a factor of A). Now, your task is to find the minimum number of transformations required to transform s to another integer number t.

Input

Input starts with an integer T (≤ 500), denoting the number of test cases.

Each case contains two integers: s (1 ≤ s ≤ 100) and t (1 ≤ t ≤ 1000).

Output

For each case, print the case number and the minimum number of transformations needed. If it's impossible, then print -1.

Sample Input

2

6 12

6 13

Sample Output

Case 1: 2

Case 2: -1

AC代码：

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<queue>
 5 using namespace std;
 6
 7 int c[1010], vis[1010];
 8 int a, b, f;
 9
10 struct note
11 {
12     int x, s;
13 };
14
15 int prime()
16 {
17     memset(c, -1, sizeof(c));
18     c[0] = 0, c[1] = 0;
19     for(int i = 2; i*i < 1005; i++)
20     {
21         if(c[i])
22             for(int j = i+i; j < 1005; j += i)
23                 c[j] = 0;
24     }
25 }
26
27 void bfs(int x)
28 {
29     queue<note>Q;
30     note p,q;
31     p.x = x;
32     p.s = 0;
33     vis[x] = 1;
34     Q.push(p);
35     while(!Q.empty())
36     {
37         p = Q.front();
38         Q.pop();
39         if(p.x == b)
40         {
41             f = p.s;
42             break;
43         }
44         for(int i = 2; i < p.x; i++)
45         {
46             if(c[i] == 0 || p.x + i > b || p.x % i != 0 || vis[p.x+i])
47                 continue;
48             q.x = p.x + i;
49             vis[q.x] = 1;
50             q.s = p.s + 1;
51             Q.push(q);
52         }
53     }
54     return ;
55 }
56
57 int main()
58 {
59     int k = 0, flag = 0;
60     prime();
61
62     int t;
63     cin >> t;
64     while(t--)
65     {
66         f = -1;
67         memset(vis, 0, sizeof(vis));
68         cin >> a >> b;
69         bfs(a);
70         cout << "Case " << ++flag << ": " << f << endl;
71     }
72
73     return 0;
74 }