题意： 给你一个整数A和B，A可以加上自己本身的质因数（不包括1和本身）【这点很重要】，从而等于B，问A通过加的操作的最少次数。我是用优先队列实现的，代码如下：
/* 题目要去一个数能否通过加除了自身和零的质因数==另一个数*/
#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
int a,b,out,res;
int vis[1020],num[1020];
struct node{
int x;
int temp;
bool friend operator < (node a,node b){
return a.temp>b.temp;
}
};
void fenjie(int n){
res=0;
int m=n;
for(int i=2;i*i<=n;i++)
{
if(n%i==0)
{
num[res++]=i;
while(n%i==0)
n/=i;
}
}
if(n!=1&&n!=m) //这题的坑在n!=m这一片，一定不能忘记！
num[res++]=n;
}
void BFS(){
memset(vis,0,sizeof(vis));
priority_queue<node>que;
node e1,e2;
e1.x=a,e1.temp=0;
vis[e1.x]=1;
que.push(e1);
int ans=-1;
while(!que.empty())
{
memset(num,0,sizeof(num));
e2=que.top();
que.pop();
if(e2.x==b)
{
ans=e2.temp;
break;
}
fenjie(e2.x);
for(int i=0;i<res;i++)
{
e1.x=e2.x+num[i];
e1.temp=e2.temp+1;
if(e1.x<=b && !vis[e1.x])
{
vis[e1.x]=1;
que.push(e1);
}
}
}
if(ans==-1)
cout<<-1<<endl;
else
cout<<ans<<endl;
}
void solve()
{
int n;
cin>>n;
out=1;
while(n--){
cin>>a>>b;
cout<<"Case "<<out++<<": ";
if(a>b)
{
cout<<-1<<endl;;
continue;
}
if(a==b)
{
cout<<0<<endl;
continue;
}
BFS();
}
}
int main()
{
solve();
return 0;
}ps:想带你去洛阳！
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