LightOJ 1141 Program E

Description

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;

#define MAX 0x3f3f3f3f

typedef struct{
    int step;
    int now;
}Node;

int s, t;
bool prime[1010];
int marks[1001];

int BFS(){
    queue<Node> q;
    Node start;
    start.now = s;
    start.step = 0;
    q.push( start );

    memset( marks, 0x3f, sizeof( marks ) );

    int ans = MAX;

    while( !q.empty() ){
        Node n = q.front();
        q.pop();

        if( n.now == t ){
            ans = min( ans, n.step );
            continue;
        }

        for( int i = 2; i < n.now; i++ ){
            if( n.now % i != 0 || !prime[i] ){
                continue;
            }
            if( n.now + i > t ){
                continue;
            }
            if( marks[n.now+i] > n.step + 1 ){
                Node temp;
                temp.now = n.now + i;
                temp.step = n.step + 1;
                marks[n.now+i] = temp.step;
                q.push( temp );
            }
        }
    }

    if( ans < MAX ){
        return ans;
    }

    return -1;
}

int main(){
    int T, Case = 1;

    memset( prime, true, sizeof( prime ) );

    for( int i = 2; i <= 1000; i++ ){
        for( int j = 2; i * j <= 1000; j++ ){
            prime[i*j] = false;
        }
    }

    cin >> T;
    while( T-- ){
        cin >> s >> t;
        cout << "Case " << Case++ << ": " << BFS() << endl;
    }
    return 0;
}