标签：|| oss imp continue ring any orm UI format
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 }
G - Number Transformation(BFS+素数)
标签：|| oss imp continue ring any orm UI format
原文地址：https://www.cnblogs.com/ruruozhenhao/p/8799057.html
