hdu 5505 GT and numbers

GT and numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1520    Accepted Submission(s): 381

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

 题意：给出n和m，n每次和自身的因子相乘得到一个新的n，问最少多少次可使n==m

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define MAX 10100
#define INF 0x3f3f3f
#define LL unsigned long long
using namespace std;
LL gcd(LL a,LL b)
{
	LL c;
	while(b)
	{
		c=a%b;
		a=b;
		b=c;
	}
	return a;
}
int main()
{
	LL n,m,j,i,t,k;
	scanf("%lld",&t);
	while(t--)
	{
		scanf("%lld%lld",&n,&m);
		if(n>m||n==0||m%n)
		{
			printf("-1
");
			continue;
		}
		if(n==m)
		{
			printf("0
");
			continue;
		}
		int flag=1;
		k=0;
		while(n!=m)
		{
			k++;
		    if(gcd(m/n,n)==1)
		    {
		    	flag=0;
		    	break;
		    }
			n*=gcd(m/n,n);
		}
		if(flag)
		printf("%lld
",k);
		else
		printf("-1
");
	}
	return 0;
}