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.

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

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.

15 20

3

14 8

-1

6 6

0

其實這就是一道數學題，題意可以看成“給兩個數a，b，讓它們除以2，3,5，最後相等，最少除幾次”。

　　那麽我們先求出他們的最大公約數c（因為要出的次數盡可能小），再看從原數到C分別用幾步，或者能否除到C。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio> 
#include<queue>
#include<math.h>
using namespace std;
int a,b,ans;
int maxyin;
void chu(int a,int b)
{
    int aa=a%b;
    if(aa==0)    maxyin=b;
    else chu(b,aa);
}
int main()
{
    cin>>a>>b;    
    int c;
    if(b>a)
    {
        c=a;
        a=b;
        b=c;
    }
    
    if(a==b)//特判    
    {
        cout<<0;
        return 0; 
    }
    if(a%b==0)//特潘 
    {
        c=a/b;        
        while(c%5==0)
            c/=5,ans++;
        while(c%3==0)
            c/=3,ans++;
        while(c%2==0)
            c/=2,ans++;
        if(c==1)
            cout<<ans;
        else cout<<-1;
        return 0;
    }
    chu(a,b);
    c=a/maxyin;int d =b/maxyin;
    if(((c%5)&&(c%3)&&(c%2))||((d%5)&&(d%3)&&(d%2)))    
    {
        cout<<-1;
        return 0;
    }
    ans=0;
    while(c%5==0)
        c/=5,ans++;
    while(c%3==0)
        c/=3,ans++;
    while(c%2==0)
        c/=2,ans++;
    while(d%5==0)
        d/=5,ans++;
    while(d%3==0)
        d/=3,ans++;
    while(d%2==0)
        d/=2,ans++;
    cout<<ans;
    return 0;
}

來個好看的代碼

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio> 
#include<queue>
#include<math.h>
using namespace std;
int a,b;
int a2,a3,a5,b2,b3,b5;
int x;
void gcd(int a,int b)
{
    int aa=a%b;
    if(aa==0)    {x=b;return ;} 
    else gcd(b,aa); 
}
int main()
{
    cin>>a>>b;
    if(a==b)
    {
        cout<<0;
        return 0;
    }
    gcd(a,b);
    int m=a/x,n=b/x;
    while(m%3==0)    m/=3,a3++;
    while(m%2==0)    m/=2,a2++;
    while(m%5==0)    m/=5,a5++;
    while(n%3==0)    n/=3,b3++;
    while(n%2==0)    n/=2,b2++;
    while(n%5==0)    n/=5,b5++;
    if(m*n==1)
    {
        cout<<(a3+a2+a5+b2+b3+b5);
        return 0;
    }
    cout<<-1;
    return 0;
}

Codeforces 371BB. Fox Dividing Cheese