LightOJ 1141 Program E

Description

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

1. #include <iostream>
2. #include <cstdio>
3. #include <cstring>
4. #include <algorithm>
5. #include <queue>
6. #include <cmath>
7. using namespace std;
8. #define MAX 0x3f3f3f3f
9. typedef struct{
10. int step;
11. int now;
12. }Node;
13. int s, t;
14. bool prime[1010];
15. int marks[1001];
16. int BFS(){
17. queue<Node> q;
18. Node start;
19. start.now = s;
20. start.step = 0;
21. q.push( start );
22. memset( marks, 0x3f, sizeof( marks ) );
23. int ans = MAX;
24. while( !q.empty() ){
25. Node n = q.front();
26. q.pop();
27. if( n.now == t ){
28. ans = min( ans, n.step );
29. continue;
30. }
31. for( int i = 2; i < n.now; i++ ){
32. if( n.now % i != 0 || !prime[i] ){
33. continue;
34. }
35. if( n.now + i > t ){
36. continue;
37. }
38. if( marks[n.now+i] > n.step + 1 ){
39. Node temp;
40. temp.now = n.now + i;
41. temp.step = n.step + 1;
42. marks[n.now+i] = temp.step;
43. q.push( temp );
44. }
45. }
46. }
47. if( ans < MAX ){
48. return ans;
49. }
50. return -1;
51. }
52. int main(){
53. int T, Case = 1;
54. memset( prime, true, sizeof( prime ) );
55. for( int i = 2; i <= 1000; i++ ){
56. for( int j = 2; i * j <= 1000; j++ ){
57. prime[i*j] = false;
58. }
59. }
60. cin >> T;
61. while( T-- ){
62. cin >> s >> t;
63. cout << "Case " << Case++ << ": " << BFS() << endl;
64. }
65. return 0;
66. }