LightOJ

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
题意：
在这个问题中，给出一个整数s。 您可以将任何整数A转换为另一个整数B，将x添加到A.这个x是一个整数，它是A的素因子（请注意，1和A不被视为A的因子）。 现在，您的任务是找到将s转换为另一整数t所需的最小转换次数。（谷歌版）
思路：
普通队列就可以，没必要优先，主要是每次转换后的B就是下一个A，题意明确点，接下来就是搜了；

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#define INF 0x3f3f3f3fusing namespace std;int vis[1010],p[1010];int s,t,cnt,ans;struct node{    int x;    int step;    bool friend operator<(node a,node b)    {        return a.step>b.step;    }};void getp(int n){    cnt=0;    int m=n;    for(int i=2;i*i<=n;i++)    {         if(n%i==0)        {            p[cnt++]=i;            while(n%i==0)                n/=i;        }    }    if(n>1 && n!=m) p[cnt++]=n;}void bfs(){    bool flag=false;    priority_queue<node> q;    node now,next;    now.x=s;    vis[s]=1;    now.step=0;    q.push(now);    while(!q.empty())    {        next=q.top();        q.pop();        if(next.x==t)        {            ans=min(ans,next.step);            flag=true;            continue;        }        memset(p,0,sizeof(p));        getp(next.x);         for(int i=0;i<cnt;i++)        {            now.x=next.x+p[i];            now.step=next.step+1;            if(now.x<=t && !vis[now.x])            {                vis[now.x]=1;                q.push(now);            }        }    }    if(flag) printf("%d\n",ans);    else printf("-1\n");}int main(){    int T;    scanf("%d",&T);    while(T--)    {        static int res=1;        ans=INF;        memset(vis,0,sizeof(vis));        scanf("%d %d",&s,&t);        printf("Case %d: ",res++);        if(s==t)        {            printf("0\n");            continue;        }        if(s>t)        {            printf("-1\n");            continue;        }        bfs();    }    return 0;}