LightOJ 1141 Program E

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 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).

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

2

6 12

6 13

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.   
9. #define MAX 0x3f3f3f3f  
10.   
11. typedef struct{  
12.     int step;  
13.     int now;  
14. }Node;  
15.   
16. int s, t;  
17. bool prime[1010];  
18. int marks[1001];  
19.   
20. int BFS(){  
21.     queue<Node> q;  
22.     Node start;  
23.     start.now = s;  
24.     start.step = 0;  
25.     q.push( start );  
26.   
27.     memset( marks, 0x3f, sizeof( marks ) );  
28.   
29.     int ans = MAX;  
30.   
31.     while( !q.empty() ){  
32.         Node n = q.front();  
33.         q.pop();  
34.   
35.         if( n.now == t ){  
36.             ans = min( ans, n.step );  
37.             continue;  
38.         }  
39.   
40.         for( int i = 2; i < n.now; i++ ){  
41.             if( n.now % i != 0 || !prime[i] ){  
42.                 continue;  
43.             }  
44.             if( n.now + i > t ){  
45.                 continue;  
46.             }  
47.             if( marks[n.now+i] > n.step + 1 ){  
48.                 Node temp;  
49.                 temp.now = n.now + i;  
50.                 temp.step = n.step + 1;  
51.                 marks[n.now+i] = temp.step;  
52.                 q.push( temp );  
53.             }  
54.         }  
55.     }  
56.   
57.     if( ans < MAX ){  
58.         return ans;  
59.     }  
60.   
61.     return -1;  
62. }  
63.   
64. int main(){  
65.     int T, Case = 1;  
66.   
67.     memset( prime, true, sizeof( prime ) );  
68.   
69.     for( int i = 2; i <= 1000; i++ ){  
70.         for( int j = 2; i * j <= 1000; j++ ){  
71.             prime[i*j] = false;  
72.         }  
73.     }  
74.   
75.     cin >> T;  
76.     while( T-- ){  
77.         cin >> s >> t;  
78.         cout << "Case " << Case++ << ": " << BFS() << endl;  
79.     }  
80.     return 0;  
81. }