Codeforces Round #218 (Div. 2) B. Fox Dividing Cheese

最新推荐文章于 2022-12-16 16:19:21 发布

转载最新推荐文章于 2022-12-16 16:19:21 发布 · 100 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/liyangtianmen/p/3464223.html

本文介绍了一个有趣的算法问题B.FoxDividingCheese，并提供了详细的解题思路和实现代码。两只熊找到两块重量不同的奶酪，狐狸通过特定的操作使两块奶酪重量相等，挑战在于用最少的步骤完成这一过程。

Two little greedy bears have found two pieces of cheese in the forest of weight a and b grams, correspondingly. The bears are so greedy that they are ready to fight for the larger piece. That's where the fox comes in and starts the dialog: "Little bears, wait a little, I want to make your pieces equal" "Come off it fox, how are you going to do that?", the curious bears asked. "It's easy", said the fox. "If the mass of a certain piece is divisible by two, then I can eat exactly a half of the piece. If the mass of a certain piece is divisible by three, then I can eat exactly two-thirds, and if the mass is divisible by five, then I can eat four-fifths. I'll eat a little here and there and make the pieces equal".

The little bears realize that the fox's proposal contains a catch. But at the same time they realize that they can not make the two pieces equal themselves. So they agreed to her proposal, but on one condition: the fox should make the pieces equal as quickly as possible. Find the minimum number of operations the fox needs to make pieces equal.

Input

The first line contains two space-separated integers a and b (1 ≤ a, b ≤ 109).

Output

If the fox is lying to the little bears and it is impossible to make the pieces equal, print -1. Otherwise, print the required minimum number of operations. If the pieces of the cheese are initially equal, the required number is 0.

Sample test(s)

input

15 20

output

input

14 8

output

-1

input

6 6

output

挺好的一个题目，思想简单。

#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <set>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
const int inf = 1000000000 ;
map<int ,int> meA ,meB ;
map<int,int>::iterator it ;
set<int>st ;
void dfs(int x,int step,map<int,int> &me){
    if(me.find(x) != me.end()){
       if(me[x] > step)
          me[x] = Min(me[x],step) ;
       else
          return  ;
    }
    else
       me[x] = step ;
    for(int i = 2 ; i <= 5 ; i++){
        if(i == 4)
          continue ;
        if(x%i == 0)
           dfs(x/i,step+1,me) ;
    }
}

int main(){
   int ans ,a ,b ;
   while(cin>>a>>b){
        meA.clear() ;
        meB.clear() ;
        st.clear() ;
        dfs(a,0,meA) ;
        dfs(b,0,meB) ;
        ans = inf ;
        for(it = meA.begin() ; it != meA.end() ; it++)
            st.insert(it->first) ;
        for(it = meB.begin() ; it != meB.end() ; it++){
            int x = it->first ;
            if(st.find(x) != st.end()){
                ans = Min(ans , meA[x] + meB[x]) ;
            }
        }
        if(ans == inf)
          cout<<"-1"<<endl ;
        else
          cout<<ans<<endl ;
   }
   return 0 ;
}

转载于:https://www.cnblogs.com/liyangtianmen/p/3464223.html