Number Transformation

Number Transformation

In this problem, you are given an integer number s. You can transform any integer number A to another integer numberB 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 sto 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

题意：一个数与它的质因数相加等于t需要几步

#include<cstdio>#include<queue>#include<vector>#include<cstring>#include<algorithm>using namespace std;const int MAX=1e4;const int INF= 0x3f3f3f3f / 2;int a[MAX]={1,1};int vis[MAX];int s,t,ok,cut;vector<int > v[MAX];void search(){for(int i = 2; i < MAX ; i++)   {          if(a[i])          continue;          for(int j= i*i ; j< MAX ;j+=i)            {             a[j]=1;}   }   for(int i=2;i<MAX;i++)     for(int j=2 ; j< MAX ;j++)       {           if(i%j==0&&a[j]==0)           v[i].push_back(j);   }}struct node{ int x,y;} st,pt;void bfs(){     memset(vis,0,sizeof(vis));      queue<node> q;  vis[s]=1;      st.x=s,st.y=0;      q.push(st);      while(!q.empty())      {           st=q.front();           q.pop();           if(st.x==t)           {            ok=1;            cut=st.y;   }         for(int i=0;i<v[st.x].size();i++)        {         pt.x=st.x+v[st.x][i],    pt.y=st.y+1;          if(pt.x <= t&&!vis[pt.x])            {                  vis[pt.x]=1;                 q.push(pt);}}  }}int main(){           search();     int n;    scanf("%d",&n);    int w=n;   while(n--) {   scanf("%d %d",&s,&t);  if(s == t)   printf("Case %d: %d\n",w-n,0);  else if (!a[s] || !a[t])    printf("Case %d: %d\n",w-n,-1);    else     {     ok=0;     cut=INF;     bfs();     if(ok)  printf("Case %d: %d\n",w-n,cut);     else  printf("Case %d: %d\n",w-n,-1); } }return 0;}