CF 371B Fox Dividing Cheese[数论]

B. Fox Dividing Cheese

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

15 20

output

3

input

14 8

output

-1

input

6 6

output

0

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分蛋糕，可以吃掉剩下1/2,1/3,1/5，求相同的最小次数

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官方题解
It is easy to see that the fox can do three type of operations: divide by 2, divide by 3 and divide by 5. Let’s write both given numbers in form a = x·2a2·3a3·5a5, b = y·2b2·3b3·5b5, where x and y are not dibisible by 2, 3 and 5. If x ≠ y the fox can’t make numbers equal and program should print -1. If x = y then soluion exists. The answer equals to |a2 - b2| + |a3 - b3| + |a5 - b5|, because |a2 - b2| is the minimal number of operations to have 2 in the same power in both numbers, |a3 - b3| is the minimal number of operations to have 3 in the same power in both numbers, and |a5 - b5| is the same for 5.

稍微不同的思路

如果有解，a b除了2,3,5之外的因子一定在最后的相同里，并且2,3,5因子相同也可以不分，那么最后分成的一定是最大公约数

//
//  main.cpp
//  cf371b
//
//  Created by Candy on 9/15/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
using namespace std;
int a,b;
inline int fac(int &x,int p){
    int cnt=0;
    while(x%p==0) {x/=p;cnt++;}
    return cnt;
}
int main(int argc, const char * argv[]) {
    scanf("%d%d",&a,&b);
    if(a==b){printf("0");return 0;}
    
    int a2=fac(a,2),a3=fac(a,3),a5=fac(a,5);
    int b2=fac(b,2),b3=fac(b,3),b5=fac(b,5);
    if(a!=b){printf("-1");return 0;}
    
    printf("%d",abs(a2-b2)+abs(a3-b3)+abs(a5-b5));
    return 0;
}