C - Number Transformation

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：献给一个T，表示有T组数据，然后又给两个数s，t；s经过加素因数变成c（素因数是指1~c内的因子（这些必须是素数）），问最少需要加几次。不能加到t时输出-1.

题解：这是一道简单的广搜题，但是首先需要素数打表（否则会时间超限）。

 

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int n,m,flag,ss[1100],book[1100];
void sushu()
{
    memset(ss,0,sizeof(ss));
    for(int i=2;i<1100;i++)
    {
        if(ss[i]==0)
        {
            for(int j=i*2;j<1100;j+=i)
                ss[j]=1;
        }
    }
}
struct node
{
    int x,f;
};
void bfs(int x1)
{
    queue<node>q;
    node a,b;
    a.x=x1;
    a.f=0;
    book[x1]=1;
    while(!q.empty())
        q.pop();
    q.push(a);
    while(!q.empty())
    {
        a=q.front();
        q.pop();
        for(int i=2;i<a.x;i++)
        {
            if(a.x%i==0&&ss[i]==0)
            {
                b.x=a.x+i;
                if(book[b.x]==1||b.x>m)
                    continue;
                book[b.x]=1;
                b.f=a.f+1;
                if(b.x==m)
                {
                    flag=b.f;
                    return ;
                }
                q.push(b);
            }
        }
    }
    return ;
}
int main()
{
    int T,k=1;
    scanf("%d",&T);
    sushu();
    while(T--)
    {
        flag=0;
        memset(book,0,sizeof(book));
        scanf("%d%d",&n,&m);
        if(n==m)
            printf("Case %d: 0\n",k++);
        else
        {
            bfs(n);
            if(flag==0)
                printf("Case %d: -1\n",k++);
            else
                printf("Case %d: %d\n",k++,flag);
        }
    }
    return 0;
}