1040: Alex and Asd fight for two pieces of cake
Time Limit: 1 Sec
Memory Limit: 128 MB
Submit: 27
Solved: 12
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Description
Alex and Asd have found two pieces of cake on table of weight a and b grams.They are so greedy that they all want the larger piece. A fight may happenes.
Now the smart person-Radical comes in and starts the dialog: "Stupid people, wait a little, I will make your pieces equal"
"Wow,you are so amazing, how are you going to do that?", Alex and Asd ask. Radical says"Ok,listen to me,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".
Alougth they are not som smart , they got it.So they agrees to his proposal, but on one condition: Radical should make the pieces equal as quickly as possible.
Find the minimum number of operations Radical needs to make pieces equal.
Input
Output
If it is impossible to make the pieces equal, print -1. Otherwise, print the required minimum number of operations. If the pieces of the cake are initially equal, the required number is 0.
Sample Input
36 30
7 8
11 11
Sample Output
3
-1
0
题意：两个人的一定要分到相等的蛋糕，否则输出-1，若初始值就相等， 输出0。跟狐狸给两只熊分饼一个道理，每次吃掉1/2或2/3或4/5。
那么此题就可以理解为每次将初始值乘以1/2或1/3或1/5，Alex和Asd乘以这几个数的次数可以不一样，每次乘的值也可以不一样，求最少的次数让这两个人相等。
首先感觉是贪心，但是后来感觉2、3、5都是质数，2^a和3^b和5^c次的公因数都是1，应该不是贪心。
比如例一、36与30，gcd为6，6/36=1/6，6/30=1/5。
再进一步，题意就成了用1/2，1/3，1/5来凑gcd(Alex,Asd)/Alex（或Asd，可行状态下这两个假分数肯定相等且最简式分子为1）且项数最少。
再进一步，就是求上述分母分解为2、3、5的个数（感觉由于三个数互质，只有唯一解，不存在最大最小的问题。）
代码：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
LL list[3]={5,3,2};//为了循环方便用个数组
LL gcd(LL a,LL b)
{
return b?gcd(b,a%b):a;
}
int main (void)
{
LL a,b;
while (cin>>a>>b)
{
LL g=gcd(a,b);
LL ca,cb,ag,bg,fenzia,fenzib,fenmua,fenmub;
if(a==b)
{
cout<<0<<endl;
continue;
}
else
{
map<LL,LL>lista;//记录Alex分母的分解情况
map<LL,LL>listb;//记录Asd分母的分解情况
ag=gcd(a,g);
bg=gcd(b,g);
fenmua=a/ag;//得到Alex最简分式的分母
fenmub=b/bg;//得到Asd最简分式的分母
for (int i=0; i<3; i++)//Alex分解
{
while (fenmua>=list[i])
{
if(fenmua%list[i]==0)
{
fenmua/=list[i];
lista[list[i]]++;
}
else
break;
}
}
for (int i=0; i<3; i++)//Asd分解
{
while (fenmub>=list[i])
{
if(fenmub%list[i]==0)
{
fenmub/=list[i];
listb[list[i]]++;
}
else
break;
}
}
if(fenmua==1&&fenmub==1)
cout<<lista[2]+lista[3]+lista[5]+listb[2]+listb[3]+listb[5]<<endl;//输出操作次数(Alex+Asd)
else
cout<<-1<<endl;
}
}
return 0;
}
