BestCoder Round #60 GT and numbers （模拟）

GT and numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 133    Accepted Submission(s): 39

Problem Description

You are given two numbers N and M.

Every step you can get a new N in
the way that multiply N by
a factor of N.

Work out how many steps can N be
equal to M at
least.

If N can't be to M forever,print −1.

 

Input

In the first line there is a number T.T is
the test number.

In the next T lines
there are two numbers N and M.

T≤1000, 1≤N≤1000000,1≤M≤263.

Be careful to the range of M.

You'd better print the enter in the last line when you hack others.

You'd better not print space in the last of each line when you hack others.

 

Output

For each test case,output an answer.

 

Sample Input

3
1 1
1 2
2 4

 

Sample Output

0
-1
1

 

Source

BestCoder Round #60

 

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解析：n要变成m，即可表示为：n*(m/n)==m

           以n=2  m=32为例演示这个转换过程：

                 n=2  m/n=16

                1.  n=n*gcd(n,m/n)=4

                2.  n=n*gcd(n,m/n)=16

                3.  n=n*gcd(n,m/n)=32

          其实就是一直重复 n=n*gcd(n,m/n) 这个过程，直到n==m。        

代码：

#include<cstdio>
#include<iostream>
using namespace std;

typedef unsigned long long ull;
ull n,m;

ull gcd(ull x,ull y)
{
ull t;
while(y!=0)t=x%y,x=y,y=t;
return x;
}

void work()
{
if(n==m)
{
cout<<0<<endl;
return;
}
if(n==1 || m%n)
{
cout<<-1<<endl;
return;
}

ull ans=0,k; m=m/n;
while(m>1)
{
k=gcd(n,m);
if(k==1)
{
cout<<-1<<endl;
return;
}
n*=k,m/=k,ans++;
}
cout<<ans<<endl;
}

int main()
{
freopen("2.in","r",stdin);
int t;
cin>>t;
while(t--)
{
cin>>n>>m;
work();
}
return 0;
}